chem241 transcripts

Lecture 038: Exam Review 2

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:10:00 -0700

Lecture 038: Exam Review 2

Jean-Claude Bradley: OK, so for the review session, I didn't receive any emails so I'm just going to do the problems we're getting in class here today. The other thing I want to point out is that for chapter nine, the exam will cover chapter one through eight. I was looking at the material that we covered, and the quiz that I gave you, and I didn't quite cover enough ground to be able to do that. But if you're taking 242, anything you did learn in chapter nine, that's where you are going to start, so one to eight for the exam.

[Student1 suggests a problem.]

Jean-Claude Bradley: This one right here? Ok, the SO. All right, so I'm starting off with HN3.

What is the charge of the central atom in HN3?

So available electrons? We're going to have nitrogen to contribute five. Five times three plus one. So, the hydrogen will have 16 and we're going to need eight for each nitrogen plus two for the hydrogen. So that's 26. We're going to share 26 minus 16, that's 10. So we have five bonds.

We have three nitrogen's and one hydrogen's position. Again we have different choices for what we can do. We can consider making triangles or other structures that are cyclic, but we want to avoid that. So if I have three nitrogens and I can't make a triangle out of them, we're going to make a straight line. That's the first thing we're deciding to do.

Now, we have another three bonds to put in a hydrogen so if I notice where else I can put the next two bonds, I pretty much have to do that. Therefore the only place I could logically put the hydrogen would be on the ends, because I already have four bonds on the central nitrogen. So that's the only place that the H could go.

Now I put my five bonds. I have to complete the octets. I'm missing four electrons on the left nitrogen. I'm missing no electrons on the central and one lone pair on the last.

Finally, the charges. So the nitrogen on the left has one, two, three, four, five, six electrons for charges. Nitrogen only has five so I have an excess, minus 1. The nitrogen in the middle: one, two, three, four electrons. Nitrogen has five so I'm going to have a positive charge; it's missing an electron. And the one on the right, one, two, three, four, five is neutral, so the question is what is the charge on the central atom? The central atom is nitrogen with the plus, so the answer here would be plus one. Does that make sense? In fact, this problem is extremely similar to azide, N3-. This is hydrazoic acid. It is just azide with a proton on one end, basically.

I have another question here on charges.

The Lewis Structure of SO, the sulfur has how many lone pairs? So for the available electrons we'll have six plus six equals 12. We'll need 16. It tells us that we have two bonds. With only two atoms, there's not much choice. We have to put the two bonds in the middle. Then we complete the octets. If we look for charges, six electrons with charges on each. Both sulfur and oxygen have six so there's no charges. So the question is, how many lone pairs? It would be two. So the answer you drew on the paper is two. Did you get that wrong?

Student1: Yeah.

Jean-Claude Bradley: Yeah, that's definitely the answer; two lone pairs.

And the Lewis Structure of SO4 two minus, the sulfur atom has what charge?

Ok, so available it would have six for the sulfur, six for each oxygen, so that would be 30. But we also have to add the two electrons from the two minus. We're going to need eight times five. So again, like all sulfur-oxygen compounds that we looked at, we're going to put the sulfur in the middle, surrounded by four oxygen. That uses up our four bonds. I can see here that's where you made your mistake. Is it because you didn't count the two negative?

Student1: Yeah.

Jean-Claude Bradley: So, basically if you count the two negative, you'd end up with only four bonds and then complete the octets. Those will all be negative one. The central sulfur then would be plus two.

Student1 comments on the procedure outlined.

Jean-Claude Bradley: Yes, if you have a plus, you remove that, because you are counting electrons. So the more negative, the higher the number. All right, so this is the charge on the central sulfur so the answer was plus 2.

SO2 is polar, has an average charge of negative one on each oxygen, has a plus two charge on the sulfur, has Van Der Waals as a dominant molecular force or none of the above?

So let's work out the Lewis Structure for SO2. So available we're going to have six times three equals 18. We're going to need 24. Share six. We've got three bonds. It's almost tempting to make a triangle; we know not to do that! We're going to put the sulfur in the middle with the oxygens around it. That uses up only two bonds. We have to use up a third one so we have to create a double bond on one of those oxygen. Complete the octets.

We look at charges. We have a negative on the right oxygen, a plus on the sulfur. Now, that's not the whole picture because that's only one of the resonance forms, so we have to draw the other resonance hybrid.

Ok, now is this molecule polar? SO2? It comes down to the hybridization of the sulfur. The hybridization of the sulfur is sp2. It's got three groups of electrons around it so that means I have 120 degree angles between the two oxygens. So it's bent like this. When we work out the dipoles, remember when you have charges, the charges will be more important than anything else, and you'll go from plus to minus. So I'll have one dipole moment in that direction for that hybrid. I'll have one dipole moment in that direction. Those two arrows can only cancel each other out if they are at 180 degree to each other. They are not, they're at 120, so the dipole moments don't cancel out and therefore, yes, the molecule is polar.

The average charge on each oxygen would be what? How do you figure that out? Negative one half, right. You look at all the resonance hybrids and you average them out. The charge on this oxygen is zero on the left, minus one on the right. Minus one plus zero dided by two is negative one-half, so central sulfur is plus one on the left, plus one on the right. It's average charge will be plus one.

We already know 'A' is the answer, but let's work these out to shed some light here. Is Van Der Waals the dominant intramolecular force? Well, no. It has a dipole moment so therefore will be dipole-dipole so 'A' will be the correct answer. The other questions are on chapter nine, and like I said, chapter nine will not be on the exam. So anything else in chapters one through eight? You guys exhausted everything on Wednesday?

Student2: The last question?

Jean-Claude Bradley: The last question? Yeah, I'm going to ask you to draw out how far you got on it and then come and see me after. OK, anything else, any general questions? Are we all good? So, for some reason I set the exam to start at two o'clock this afternoon, but there's no reason for that. It will start at 10 o'clock if you wanted to take it early. All right. Well, good luck and hopefully I'll see a few of you next term.

Transcription by CastingWords

Lecture 037: exam review 1

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:09:00 -0700

Lecture 037: exam review 1

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Jean-Claude Bradley: Let's do this review session for the test. The average was 79 percent, so if you want to situate yourself relative to that. I received a number of questions both online and in class, the thing is you also have to be proactive in looking at the problems we did in class, because many of the problems are already done. So, if we see that we have done an identical problem on Monday, I am going to skip those. In the future I am going to require you, if you want me to go over a question, to show me the work you have done on that question. So save your papers after you come out of the test and bring it to me, because it really does not do anybody any good to repeat what has already been recorded. However if I can help you work out where you were blocking, I would certainly be happy to do that.

Jean-Claude Bradley: Let me do what I can here, with the problems we have. How many different molecules for 1, 3-dibromocyclohexane? This one looks awfully familiar, I think we've already done it, but let me go through this. So, 1, 3-dibromocyclohexane, worked out we got two chiral centers, we draw out all the molecules that might be possible, and then we need to determine if any two are the same. So, if I have an axis of rotation, I can rotate 180 degrees about that axis, I see that the first two structures are in fact the same. They can be superimposed. So, total, we would have 1, 2, 3, so we got three molecules for that one.

Jean-Claude Bradley: Reaction of two butene with water in the presence of acids, so this says it's cis or trans, it does really matter in this case. So, here we're going to do what, Markovnikov addition of water. Now, one of the carbons has one hydrogen, the other side has one hydrogen, so what happens in that case? You can't follow the Markovnikov rule, so what actually happens? I can think about the mechanism as the wide Markovnikov rules exists, because you generate a more stable carbocation, if you're going to generate a secondary carbocation, secondary in the right, you're just going to get random. So you apply the Markovnikov rule when you know you have a difference between the left and the right, but if you don't have a difference in the hydrocarbons between left and right, then you would get random distribution. Now in this particular case, the left and the right are the same, so it's not going to make any difference; but if you come across a problem where you have different groups and they're both going to generate secondary carbocations, then you would end up with both. So in this case we would just put the OH on the two positions.

Jean-Claude Bradley: Two butene, or propene, react with MCPBA. MCPBA, meta-Chloroperbenzoic acid, so that is an epoxidizing agent, it will put on epoxide. Reaction of two butene with ozone, then dimethylsulfide reacts, results in what? Again, it doesn't matter if it is cis or trans. First ozone then methylsulfide is standard conditions for ozonalysis, and we don't get further oxidation with ozone, so we just end up with ethaldehyde. Okay, one butene reacts with methyliodide and zinc, so that's the Simmons-Smith reaction, we make a cyclopropane from that reaction. We'd have ethyl-cyclopropane in this case. We are doing SN1 with hydroxide. So first step with SN1 is always the same, generate a secondary carbocation, and then we see we are going to do a 1-2 shift, and we generate a tertiary carbocation. The final step is the attack of the nucleophile.

Student: Does it matter if the molecules are chiral? From the perspective of the carbocation?

Jean-Claude Bradley: Is it chiral?

Student: It looks the same on the left and the right, does it matter?

Jean-Claude Bradley: Where do you have four different groups?

Student: Does it matter that the SN1 has to be chiral?

Jean-Claude Bradley: Well, SN1, if the chiral center is on the carbon that has a bromine, then when you make the carbocation you lose the chiral information, because you go from four to three groups; if the chiral center was somewhere else, then you retain it. It really depends where the chirality is. And I think we did a couple of problems with a few chiral molecules.

Jean-Claude Bradley: Again we have an SN1 reaction. I have secondary carbocation, I see there's one 1-2 shift that I can do to generate a tertiary carbocation, so I'm going to do that one. That's the only product of that. CH2 CH2 CHBr. Hydroxide E2. We look for the various possibilities, 1, 2, three anywhere I try to count three on the right, I can't find a proton to remove, so I have to go on the left. 1, 2, 3, so there's only one proton in this case that is suitable for an E2. We draw this one that we see that they are cis/trans isomers, so we don't need to worry about that. Because both of these have the same branching, we have a branching with two, a branching with two, they would be the major product, it would be the only product, there would be no minor products in this case.

Jean-Claude Bradley: With the synthesis of 1-butene, I could put the Widding up on the left, and it'll end up like this. Or I can put the Widding on the right. So far as the possible starting materials you could have bromomethene, propanal, 1-bromopropane, formaldehyde.

Jean-Claude Bradley: E2-butene, the following two groups are trans. So here's E2-butene, and we see that the methyl and the methyl group are trans; also the hydrogen are trans. 2-butene reacts with warm, concentrated permanganate. So we create the carbon-carbon double bond and then oxidize each carbon, as far as it will go, which is a carboxylic acid, so acetic acid will be the product. Are there any other questions that were not brought to me on paper or email?

Jean-Claude Bradley: That one we did on Monday, come see me after class. Anything else? Okay, so what I'd like you to do on Friday if you have any questions for me, definitely work out the problems as best as you can, and then it will be a lot better for you to understand how to solve these things. All right, I will hang around until the end of the class. If any of you want to take the make up or have extra questions I will be happy to answer them.

Transcription by CastingWords

Lecture 035: alkynes

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:08:00 -0700

Lecture 035: alkynes

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Jean-Claude Bradley: Let's get started with chapter nine, which are alkynes, carbon-carbon triple bonds. So, this chapter is going to be pretty similar to the alkene chapter. Alkynes and alkenes are pretty similar, but there are some notable differences, but you'll notice that a lot of the reactions are very, very similar. You're likely going to redo chapter nine if you take 242, so this will be kind of an introduction. We're not going to go in as much depth as you will when you redo this chapter later.

Let's start off with the basics, nomenclature. So the simplest alkyne would be just two carbons and hydrogens. So, this molecule is acetylene, but you can also name it by UPAC standards, where you would take the 'eth', for two carbons, and put a 'yne' at the end to show that it's an alkyne, so I can call that ethyne. It also has a common name ethylene, no actually that's fine ethyne or acetylene.

Okay, so you don't have a numbering system here because there's no other place a double bond could be, same thing with three carbons. There's only one place that I can have a triple bond, so there I would have propyne. I wouldn't have to use a numbering system.

Now, when I have four carbons, just like with alkenes, I have to specify where the triple bond is. You do it the same way as you do for alkenes. You start at the position where the triple bond starts. So, this would be position two, so this one would be 2-butyne. It's also simpler than alkenes because there's no scissor trans. We have two sp hybridizeds centers, so the acetylenes are always going to be straight. The nomenclature is simpler for that reason as well.

We spent some time in the first week looking at the orbitals that are involved here, so you have your sp hybridized carbon and you have two leftover p orbitals on each carbon. You have one sigma bond and two pi bonds that constitute that.

It turns out that having that triple bond there, having a high percentage s character on the carbon, actually can stabilize the negative charge on the carbon. It turns out if I treat a terminal acetylene - let me get this nomenclature out of the way here - this will be a terminal alkyne, meaning that it has at least one end with a hydrogen. If I had two alkyl groups, I would have an internal alkyne. That's our definitions for internal and terminal alkyne.

Now, back to the acidity, internal alkynes don't have hydrogen, so they're not going to be acidic, but if I have a terminal alkyne, then it is possible to remove that proton and put a negative charge. When I say that it's acidic, that's relative to an alkene or relative to an alkane, it's certainly not acidic in the sense of water or any carboxylic acid even. These are way beyond all these, so we need to use an extremely strong base. The traditional base that we use here is called sodamide, NaNH2. That's basically ammonia that acted as an acid. We removed the proton from ammonia, so you end up with NH2-. That's extremely basic and it is basic enough to be able to abstract the proton on terminal alkynes. The way this works is like this, you just abstract that.

This is advantageous because by putting a negative charge on the acetylide, we can make it nucleophyllic. Now, this can act just like any other nucleophile, like Cl- or OH-, yet we can do SN2 reactions in the same way. An example of that would be, let's say I react this with methyl bromide. That would give you a classic SN2 reaction. So that's a good way to synthesize alkynes. That would fall under the first preparation of alkynes.

Another way we can make alkynes, if we don't have one to start with, is to dehydrohalogenate, much in the same way that we made alkenes. But, instead of losing one HBr molecule, we're going to have to lose two HBr molecules to make the alkyne. Losing one HBr molecule isn't very difficult to create an alkene, however if we want to put that second double bond in, it turns out to be a little bit more difficult. We need to use a strong base, and in this case, we will use sodamide and furthermore, we're going to do it at high temperature.

So we're going to do it at 150 degrees. Previously, when we used sodamide, we were generally doing that in liquid ammonia, so about -33, this is different, this is much more intense. What that will do is remove two HBr molecules and we can create a triple bond.

There's a lot more to this reaction, and like I said, if you follow along to 242, you're going to start with this chapter. You'll go into a lot more detail about how this whole process happens.

Let's look at a few reactions of alkynes. The first is hydrogenation. If we use the same catalytic system as we did for alkenes, like palladium and hydrogen. Here 'xs' means literally that you have an excess of hydrogen, so you use up as much hydrogen as you need in the reaction. What will happen is that you will use up two equivalents of hydrogen to reduce the triple bond to a single bond, so you'll form a alkane if you use a standard catalyst like palladium.

If you want to do this reaction and you want to stop at the double bond and you don't want to go further, one tactic that you could use, which you'll see used again and again in chemistry, is to take a really good catalytic system and to make it less efficient. What happens is that it becomes less and less able to do all the competing reactions, and if you're lucky, you'll get the energies just right so it will do the reaction that you want, but not the ones you don't want.

Luckily, it's easier to reduce the triple bond to a double bond, than it is to reduce the double bond to a single bond. That means if we poison the catalyst, add a poison like quinoline, you can actually make the catalyst unable to reduce a double bond, but still reduce be able to reduce a triple bond.

Let's take a look at that. Same alkyne, palladium on barium sulfate and quinoline. This mixture is also called Lindlar's catalyst. What will happen is it will reduce to a double bond and it will stay there. That will be an example of partial hydrogenation.

Another thing we can do is add halogens across triple bonds. Adding an excess of bromine, we'll add the first bromine molecule across the triple bond to the double bond, and then all the way to having four bromine atoms added across. The mechanism is a little bit different than the addition onto an alkene. If you remember, when you added on an alkene, you would get trans addition, here you end up getting mixtures of cis and trans initially, but it doesn't matter because if you have excess bromine, you're only going to get one product.

Let me show you what I mean by that. If we get an addition like this - I'm going to put these in square brackets to indicate that those are not isolated; they're just transiently produced in the reaction mixture - both of these alkenes will act like normal alkenes and we will add bromine across the double bond and they will give the same product.

Another reaction we will revisit is Markovnikov addition of HBr. It will also work with alkynes. Let's say that we do Markovnikov addition of HBr, we just add HBr to this. If we have an excess, because it is hard to stop it after the first stage, so if you get just one product, we'll have an excess. We will first give it a Markovnikov addition, which means that the H will go where there are more hydrogens. In this case there is one hydrogen on the left, zero hydrogens on the right, so the H will go on the left and the Br will go in the middle. That is a transient product. That will also undergo an Markovnikov addition. The H, again, will go where there are more hydrogens on the left, so we will get 2Br bromopropane from that.

If we do Markovnikov addition of water, it's a little different. We're going to use an acidic aqueous solution again, the traditional mixture used is mercuric sulfate, HgSO4, and sulfuric acid, H2SO4, with water. Basically, you need to translate this into a Markovnikov addition with water. It's not exactly the same as the previous case because something interesting happens. The first step will be exactly as you would expect, Markovnikov addition of water, where we get H on the left and the OH in the middle. The thing is, is that this is not a normal alcohol; it is an alcohol that is on an alkene. That's a special kind of relationship called an enol. The enol doesn't survive very long, it basically will tautomerize, which is a word that refers to rearrange, in a sense of going from an enol to a ketone. In the mixture, what will happen is the hydrogen that is on the oxygen will leave from there and go on the double bond, on the carbons on the double bond. That's basically the movement of a proton.

All we did here is we moved the double bond to the oxygen and moved the proton from the OH onto the carbon, down here. The end result of the hydration of an alkyne is you end up with a ketone. If you are looking for the final product, what you need to remember is that you are going to get a methyl ketone in this case. If you have a terminal alkyne, you always get a methyl ketone like this.

If you want to do anti Markovnikov addition of water, you do something similar as what we did for alkenes. For alkenes, we use borane, BH3, and then we use peroxide to oxidize away the boron. Here we are going to do something similar, except that if we were to use borane, there would be a chance that we would add twice across the triple bond, because borane is a pretty small molecule, so I would end up with a alkene and then I would end up with two borons on that structure. To avoid that, what we use is a huge boron reagents that has two pentol groups on it, and the specific pentol groups are called siamyl. The traditional reagent will be this, sia 2BH, siamylborane. Again, the siamyl groups are just pentol groups; they are just really, really big, so because of this spheric hindrance, I can only put one.

If you remember, after you add the boron, you got oxidized, so we would do exactly the same thing in this case, H2O2 basic. Anti Markovnikov addition of water, the OH will go on the left, but again, we have an enol, so that is not the product we isolate. We end up with an aldehyde in this case. You can see that the reactions are very analogous, but there are slight twists to each and every one of these reactions. Any questions on this addition of water?

Next we have the permanganate oxidations. Cold end dilute will be similar to alkenes in that there is not any breakage of the carbon-carbon bond, and it is also similar in that we deliver oxygens, but the difference is instead of getting alcohols, we get ketones, or, I should say, carbon enol. We just replace the triple bond with two carbon enol groups. That's called end dilute. Warm and concentrated would be very similar to alkenes, in that we break the carbon-carbon triple bond and then oxidize each carbon as far as it can go. If we break this triple bond, then on the left we have one carbon. Whenever you have one carbon with concentrated permanganate, it oxidizes all the way to CO2. On the right, that carbon will be oxidized to carboxylic acid. That's very similar to the alkenes.

That's basically the theory that I want to cover for chapter nine. Did anyone bring their books?

Transcription by CastingWords

Lecture 034: review for test2 make-up

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:07:00 -0700

Lecture 034: review for test2 make-up

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Jean-Claude Bradley: I'm going to first go through the email questions. Many of these will be redundant, so we can get through these.

The first question is how many different molecules can be drawn for 1, 2 dibromopropane? Lets draw 1, 2 dibromopropane; when I draw it like this I'm not showing any chiral information, I'm just trying to see where the atoms are connected. The key thing is that we are going to have to count the chiral centers. How many chiral centers do we have in this molecule?

Student: One.

Jean-Claude Bradley: Just one? The carbon on the left has two identical groups, so that is not chiral; the right is a methyl group, thats not chiral. Only the one in the middle has four different groups on it. Now that we've determined how many chiral centers we have, with one it's easy; you're going to have an R and you're going to have an S. If we were to draw Fischer projection for that, we would have H and Br and we would have a set of enantiomers like this, so, the answer is two. Basically that's how you do these "how many molecules are there."

Next we have (R) 2-bromobutane reacts with hydroxide by an sn2 reaction. There's 2-bromobutane, we have to draw an (R)-2-bromobutane and use a Fischer projection for that. That's clockwise, and clockwise is normally R, but the lowest part of it is in the front. I actually drew the S isomer, so the one that I want to work with then would be the mirror image. This is the R isomer.

The question is we're doing sn2 with hydroxide. sn2 is one of the easier mechanisms, there is only one step. The nucleophile will attack the center and will kick out the R-. One thing you need to remember to do, when you have an sn2 reaction, is the Walden inversion. We will invert the chirality at the carbon where the attack is taking place. The OH will then go in the Fischer projection on the right side. If we determine the product, we have three clockwise, normally R, the lowest part of the group is in the, front so that's S. We have (S)2-butanol, which in the particular example I'm looking at was not listed, so it would be none of the above. We did have (R)-2-butanol listed, but (R)-2-butanol, we don't get any of that because of the inversion.

We have an E2 mechanism, CH2-CH, CH-Br, CH3. We are doing an E2 with hydroxide. Again, when you see that number 2, you should be happy because it means there's going to be a very short mechanism, it's going to be one step. We're doing E2, so we are going to lose HBr and generate a set of alkenes. The key thing to do here is to figure out how many different ways can we lose HBr. A convenient way to do it to start at the bromine and count to three until you hit a hydrogen. I can count one, two, three, and that's the candidate for the E2 elimination. I can also go on the left side; one, two, three, and there is no other way I can count to three to generate another another alkene, so there are two pathways that we will follow. I'll do them on the same molecule here in different colors.

By the red mechanism, we would generate an alkene just at the end of the molecule. That's one possiblity and there's no cis-trans products here, so that's all we get. If you go by the blue arrow, then we would have the double bond in a different position. On the left side I have a methyl and an ethyl group, and on the right side I have a hydrogen and a methyl. In this case, we do have cis-trans isomers, so we have to draw the cis isomer.

We look at the branching on the left structure and we have a branching of three; on the right structure we have a branching of one. The major products will be on the left hand side. This particular question asks for the major product, so we're going to have CH3-CH2-C, so it looks like it's B in this case. I'm not sure what quiz question this is, but this would be B.

We have the same molecule reacting by an sn1. Sn1 would always start off the same way; we would lose the R- and generate a carbocation in that position. We generate a carbocation, now we want to do a flip and see if we can do a 1-2 shift. We start a carbocation and count to two. I can move this hydride over and transform the secondary carbocation to a tertiary. Anything else I move, the methyl or the ethyl group, would just give me secondary carbocation. If I went to the other side, and moved one of these hydrides, I would I would go to primary, which definitely won't happen; there is only possible 1-2 shift, this one here. In sn1, the hydroxide acts as the nucleophile and finishes this. We end up with this tertiary alcohol. It looks like it's C in the answer set on the quiz.

Which of the following are impossible structures? We have cis cyclopentine. There's nothing stopping you from having a cis double bond in a ring of any size. That's fine; that certainly is a possible structure. In B, we have trans-cyclomelamine. It's harder to put a trans double bond in a cyclic structure; you need at least eight carbons to do that. We have cyclomelamine in this cause, so nine is more than eight, so that is also a possible structure.

Cis-cyclopropane. We know that three membered rings are going to be strained, but that doesn't mean that they are impossible, so cyclopropane would not necessarily be the most favorable molecule, but it would certainly be a possible structure because I have cis double bond in the ring.

Trans-cyclodecene. We have ten carbons, so it's very easy to put a double bond in the ring. In this particular question, the answer is none of the above; I dont have any impossible structures.

Propene reacts with reacts with HBr in the presence of peroxide. We need to determine what kind of reaction this is, HBr peroxide is anti Markovnikov addition of HBr. We went through the entire mechanism; we don't have to go through the mechanism to get the final product in this case, though you are responsible for the mechanism. Here, all we need to know is that HBr will add where the hydrogen will go - where there are fewer hydrogens. That means that the Br will go on the end and the H will go in the middle, so one bromopropane. That would be A.

Same question but now with H-Cl and peroxide, what happens in this case?

Student: Just a Markovnikov addition.

Jean-Claude Bradley: A normal Markovnikov addition. Just because there's peroxide there, it doesn't mean you will get anti; you'll only get anti Markovnikov with HBr, the energetics don't work out for H-Cl. Whether you have a perioxide or not will not change anything you'll just get a normal Markovnikov addition, so the answer is 2-chloropropane, which is D.

1-butene reacts with methylene, iodide, and zinc. This is the Simmons-Smith reaction. What kinds of products comes from that? This is how you add aCH2 group to a double bond to make a cycloprpane; wherever the double bond is, you just add a CH2. I would end up with this cyclopropane, so this is ethyl cyclopropane, and that's actually answer C in the quiz.

I think that's all I have for email questions; let me go on to the paper. I have a question here from the chapter seven quiz.

Which one of the followings can be used in a Wittig synthesis of 2-methylpropene? When your doing a Wittig synthesis, you're working backwards from the alkene. If it's unsymmetrical, like this one, there will be two different routes that you will have to work out. One of the things I could do is put the Wittig reagent on the left, and have an aldehyde on the right. You normally don't have a Wittig reagent laying around the lab, so you normally have to make it; you're going to make it from an alkyl halide. To do that, you would replace the C double bond P with an H and a Br. It's going to be two steps to that; first we have to treat it with triphenylphosphine, and then with n-butyl lithium. We went through that whole mechanism here; we don't need to do that, we just need to figure out the starting materials.

That would be one route; the other way would be to put the Wittig reagent on the right. That means that I would have acetone and ketone on the left, and again, we make the Wittig reagent the same way. We have four potential candidates as starting material for this Wittig synthesis. We have bromomethane, acetone, formaldehyde or 2-bromopropane. I see here in this particular example that B is formaldehyde, HCsO, so formaldehyde would be a canadite. But, any of these four products would be a correct answer to this question, so you really do have to work out both ways everytime you do these problems.

We're doing an E2 on this molecule. What could I do to abstarct an E2 in this molecule? I have g ot to count to three from the halide to a hydrogen. One, two; this one is too close. One, two, three; anywhere I count to three, I hit a methyl group. You will never lose a methyl group as the result of an acid base reaction. In E2 elimination, you can only lose a proton, so basically there's no way you can do an E2 on this molecule; it's just not possible, so there would be no products. E1 might be possible becaue I make a carbocation, then I get an rearrangement; then I might be able to an E1 but I can't do an E2 on this.

Student: [unintelligible]

Jean-Claude Bradley: Yes. This is my first step, do I do a 1-2shift on this? On the secondary, if I count to two from the carbocation, there's only one possible 1-2 shift; I have six identical methyl groups that can move. The question is, if they move do I end up with something more stable? I end up with a tertiary carbocation. So yes, you would defintely get a 1-2 shift. Now, you see what the 1-2 shift does for you, because without the 1-2 shift, you wouldn't do an E1 as well. There would be no way to do an elimination in this molecule. The shift enables a reaction to happen, so lets do that 1-2 shift.

Now I have tertiary carbocation, and now we will lose a proton. How many different ways can I lose a proton in this carbocation? Two. I can lose one of these hydrogens. These two methyl groups are the same; that would give the same product so we don't count them twice. On the right, I could lose this proton. Let's do the color thing here. In red, I'm going to lose a proton like this, so what do I have left? I have a methyl group and I have CH, CH3 and then a t-buty, so that would be one product. I don't have cis-trans isomers here, so that's just one. Alternatively, I can lose a proton in this direction. Two methyls on the left, a methyl and a t-butyl on the right. Again, this is trans isomer, so there're only two products from this reaction, and the right product has a branching of four, so that would be the major product.

How many different molecules can be drawn for one 3-dibromocyclohexane? I figure out that I have two chiral centers, I draw four possibilities. It doesn't mean that we have four molecules here. Any of these two the same and how can you tell? To be able to superimpose them, you have to be able to rotate them. When you have a ring, typically the kind of rotation you'll do is a flip. If you can flip the ring over according to an axis and generate the other structure, then that's what will happen. If you look at the top two structures, if I put an axis right here, and if I flip it 180 degrees, two bromines that are in the front go in the back. The first top two structures is really the same molecule. If I try to the same thing on the bottom molecule, when I flip them I end up with the same thing. I can't interconvert; the two bottom structures are different. This is just another example of having a meso compound, and then a set of enantiomers; so you would have three different molecules possible.

R-2-bromobutane reacts with hydroxide. Let's see, that's three, so that's R; and it reacts by an sn1 reaction with hydroxide. The first thing is we will lose the Br-; generate a secondary carbocation. If you check around there's no 1-2 shifts possible, so that will stay like that, but when we generated the carbocation, the carbocation is flat, it's not a chiral structure, we've lost the chiral information in that first step. Now this cation will react by hydroxide, hydroxide will come either from the top or the bottom; there's no reason to expect one to be more likely than the other, so we'll get a 50-50 mixture of R and S.

2-methyl-propane reacts with HBr. How does 2-methyl-propane react with HBr? It'll react nicely with an alkene, but there's no reason to expect a reaction between an alkane and HBr, so there's no reaction.

Reaction of bromine with cyclohexane in the presence of light? How many alkyl halide products are expected, not counting enantiomers and diastereomers? We need to determine how many different hydrogens we have. In cyclohexane how many different hydrogens? How many different ways can I put a single bromine? There's only one way. There's only one bromocyclohexane. There're no chiral centers in this molecule; well, we're not even counting chiral centers, but even if we were, there's just one way to put it, so that would be one product.

I'm seeing some repetition here, E-2-chloro-2-butene has the following two groups; which of the following two groups are trans? If I have E-2-chloro-2-butene, I compare chlorine versus carbon, chlorine has higher priority than carbon. On the right hand side, carbon is higher priority than hydrogen. My two highest priority groups are in opposite directions, so this is definitely the E isomer. Once you've drawn it correctly, let's take a look at our options. Which of the following two groups are trans? Methyl and ethyl; well, we don't even have an ethyl group here, so that's not it. Chlorine and butyl; we don't have a butyl group. H and Cl; the H and the Cl are cis. H and methyl; I see the H and the methyl group are trans, so that would be A.

(2R, 3R) 2, 3-dibromopentane, we actually could have done this problem really without drawing them, but let's take a look at them anyway. If you take a mirror image of the (2R, 3R) 2, 3-dibromopentane, you will definitely get (2S, 3S). So they're going to be either enantiomers or the same molecule. They clearly can't be the same molecule; the only time that kind of thing can happen is if you have meso compounds, in which case I would have to have a symmetrical molecule, which I don't. I drew them down here, you can see that there's no way I can rotate one into the other. These are definitely going to be enantiomers. 2, 3-dibromopentane; how many molecules? We're going to determine we have two chiral centers. Any of these the same molecule? That's what you have to do after you draw them. First of all, it's asymmetrical, so any of them that I attempt to rotate, the methyl group will end up on the bottom and the ethyl group would end up on top. The answer here is that we have four different molecules.

1-butene reacts with palladium and hydrogen. Pd H2, those are standard hydrogenation conditions; you will add hydrogen across the double bonds in a molecule. We would end up with butane, which is D in this particular example.

How many different molecules can be drawn for 1, 2 dibromoethane? First we determine the number of chiral centers, which is what for this one? No chiral centers, so there's only one way you can draw it. One molecule.

How many molecules can be drawn for 1, two dibromobutane? So that would be the same as the dibromopropane. Go back to that example.

1-bromo-2-methylbutane reacts by chloride by an SN1 reaction. Let's draw out the connection here first. 1-bromo-2-methyl. The R will refer to the chiral center that is the second carbon. I put CH2Br. That's the R isomer; so we do an SN1 reaction with hydroxide. We generate a carbocation that's next to a chiral center, we look for 1-2 shift; is there a 1-2 shift possible in this case? There is only one way that I can make a tertiary carbocation out of this, so I will. Now that I've made a carbocation on my carbon, that carbon is no longer chiral, so I can't use a Fischer projection to display it anymore. So, what do I have? I have two methyl groups and an ethyl group and I'm going to attack it with hydroxide at the end. There is an equal probability of attacking it from the top or the bottom, however even that doesn't matter in this case because I don't end up with a chiral product. I have two methyl groups that are the same. I would have CH3, COH, another methyl; so the answer in this case would be 2-chloro-2-methyl-butane, D, just replace the OH with a CO.

SN2 reaction, hydroxide. Again, you want to be happy that you have the two mechanism; only one step. In this particular example I did not give the chirality, so you don't have to worry about that. Let's see what we have; CH3, CO, H, H, C, so it looks like the answer is A, in this example.

1-butene can best be made from dehydrohalogenation of which alkyl halide? What we want to do in these kinds of problems is do an E-2 reaction, and see which one of the reactions will only give us one product. There are two ways I can start; I can start with the bromine on the first position or on the second position. If I have 1-bromobutane, that would certainly give me the product, and there would be no other side products from that. I would just get the alkene that I want, so that would be a good solution. Let's see if the other one is just as good. Unfortunately, the other possible starting material, 2-bromobutane, would generate the product that we want by losing HBr from the left, but we would also get some contamination with this elimination product. I get the top and the bottom, and because I didn't generate exclusively the product that I wanted, the right one would not be suitable. The only suitable one would be this, which is answer A.

Which one of the following compounds shows cis-trans isomerism? A methane, no. 1, 2-dibromocyclobutane? With 1, 2-dibromocyclobutane, I can either have the bromines on the same side or on the opposite. We're not looking at enantiomers on this particular question; this is from a previous chapter. We only want to look at cis-trans isomerism, so B would be the answer.

1-butene reacts with mercuric acetate; water followed by sodium borohydride. Those are standard conditions for Markovnikov addition of water without the possibility of rearrangement. That's all you do; you do Markovnikov addition of water. 2-butenol would be the product, which is B, in this question.

I think I covered all of the questions; is there anything else? That's it? Good luck on the test; we'll see you next week.

Transcription by CastingWords

Lecture 033: test 2 review

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:06:00 -0700

Lecture 033: test 2 review

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Jean-Claude Bradley: I'll start off with some questions that were requested for the test. I think these are all quiz questions.

2s-3s and 2-3 dibromopentane and 2s-3r. The question has to do with the relationship. It's either enantiomers, diastereomers, geometric isomers, the same molecule or none of the above. Pretty classic based on some of the problems we did from that chapter.

The first thing we do is we will write out the molecule an official projection. We have three clockwise. Lowest priority group is in the front, so the top one here is s. Going clockwise, clockwise is normally r, but the lowest priority group is in the front so that's s. Going counter clockwise, counterclockwise is normally s, but the lowest priority group is in the front so that's r. What I drew in this isomer is the 2s3r, basically, this one here. To draw the other one, both of the positions are inverted so I just need to switch the two bromine positions. This one would be r3s. Because each s becomes an r and each r becomes an s, we know that basically it's either going to be the same molecule or they will be enantiomers, so which is it in this case?

Student: [unintelligible]

Jean-Claude Bradley: Same molecule?

Student: [unintelligible]

Jean-Claude Bradley: How can we tell? If it's the same molecule I have to be able to rotate 180 degrees and can I do that and superimpose these two? I can't because I have an ethyl group on the bottom and an methyl group on top. If I had methyl groups on top and bottom then yes. I could rotate and superimpose it, so that's not the case. The answer in this case is that they are enantiomers. They are mirror images.

Another question has to do with the number of molecules that can be drawn for 2-3 dibromobutane. The first thing here is we need to determine is we've got two chiral centers in this molecule. If I have two chiral centers, that means if I do all the possible combinations there will be four structures that I can draw. A good way to draw them is as mirror images, it will make it easy to do the final count that the end. These two are mirror images, and then I can have these two possibilities. How many molecules can be drawn for 2-3 dibromobutane?

Student: Three.

Jean-Claude Bradley: Three, right. Part of the process you have to go through is after you draw everything, you need to determine if any two of these are identical. You'll notice the first two are the same molecule, because I can rotate 180 degrees and superimpose it, so the answer is three molecules.

Student: [unintelligible]

Jean-Claude Bradley: The first one, you mean this? Actually we can do it more easily in the second problem. A diastereomer would be any two molecules that would have the same framework. Like there would be two three dibromobutane but would not be an enantiomer. Any pair, for example, if I have A B C, A and C would be diastereomers. Now we have an sn2 reaction question. One bromo two 2-dimethylbutane reacts with chloride by an sn2 reaction.

We're going to draw out one bromo two 2-dimethylbutane. Sn2 mechanism is a short one, it's only one step. One of the lone pairs from the nucleophile will attack the electrolytic carbon connected to the bromine and simultaniously kick out the O-. When you do an sn2 reaction, you need to be wary about the Walden inversion. If I had a chiral center there, I would get an inversion, but I don't, so it's just going to replace the Br with a Cl.

The various answers we have are one chloro two 2-dimethylbutane, one chloro two 2-methylpentane, two chloro two methylpentane, three chloro three methylpentane. This is obviously one chloro two 2-dimethylbutane, so it would be A in this case. Be wary when you do these problems, there are many problems that look very similar, but they're not, so make sure that you carefully look at each answer and make sure it's really the molecule you've drawn.

We have an e1 example. Reacts with hydroxide via e1. If we're doing e1 there's going to be multiple steps. First step is always the generation of the carbocation. The second step is once you make the carbocation, you need to investigate whether or not you can get a 1-2 shift. Can we in this case?

Student: [unintelligible]

Jean-Claude Bradley: Right, so we had a secondary carbocation, there's only one shift that we can do that will give us a tertiary, and that is the one that we'll do. The 1-2 shift, I would have to move this hydrogen over. If I moved the methyl or the ethyl group, I would still have a secondary carbo cadine which you don't do, you always go more stable in these problems. We start drawing the arrow in the middle of the CH bond and end it on the carbon. I'm drawing out the carbon hydrogen bonds that I need to worry about at this point because we're doing an e1. At this point, the only thing left to do after the shift is to lose a proton and to generate all the various alkenes that are possible. This particular question asks you for the major product, but you could just as easily get the minor project, so let's to do the whole problem and see what we end up with.

In order to figure out which hydrogen we can lose we're going to count to two starting from the carbocation. When the H is on the CH2 on either the left ethyl group or the right ethyl group will give you the same product. The two ethyl groups are identical. The other possibility is one of the hydrogens from the methyl group. We have two sets of alkenes that we're going to make for that. I'll use different colors to show the two different mechanisms. In red, we would lose that proton. Actually, the mechanish states hydroxide, but of course, any base would do here.

We are going to make an alkene. The easiest way to do that is to first draw the two sps hybridized centers, then put the groups that are left. I'm losing this hydrogen, so what's left on the left hand side would be a hydrogen and a methyl group. On the right hand side I have an methyl group and an ethyl group.

Now, whenever you make an alkene you need to worry about cis-trans isomers. In this case we actually do have two products we're going to get from this. You need to remember that, especially if you're counting products.

Student: [unintelligible] one of the groups are cis and trans, you would consider it two products?

Jean-Claude Bradley: If you can have cis-trans isomers, you're going to get cis and trans. Or I should say e and z. It's not specifically cis and trans, it's just if you have geometric isomers.

Let's pick a different color to do the other side, blue. We want to grab the methyl group. Going the blue route, we would have another alkene. Lets say that we put the left side of the of the alkene from the bottom, what we have left are two hydrogens on one side. On the other side we have an ethyl group and an ethyl group. No cis-trans isomers for this product, so that means in total we would have three products from the e1 reaction. This specific question asks for the major products, so we need to look at the branching. Branching on the left structures is one, two, three. On the right structure the branching is only two. What we have are two major products and one minor product.

When you see this kind of question it will typically say which one of the following is a major product, so there may be five different answers to that question. It's really important you step through the whole problem to make sure you don't miss any. Any other questions on that? E1 is the longest, probably the most difficult problems. If you can do that you should be fine with all these problems.

Last one here, 1-butene reacts with mercury. In this problem we notice that we have a mercury reagent. Do we know what this product is going to be? When do we normally use the mercury reagent? I think we only used it one time in this class. Remember?

Student: The guaranteeing of the Markovnikov addition of water?

Jean-Claude Bradley: Of water, right. To guarantee the Markovnikov addition of water we use mercury. Now the only problem with this particular question is that the reagents are wrong in the second step. First, we use mercuric acetate, but after that we need to use sodium borohydride. If this question was asked in this way, it would be none of the above because we did look at the mechanism. This second one should be NaBH, the Markovnikov addition of water. The time we use the peroxide and sodium hydroxide is when we do anti Markovnikov addition of water, where we first we use borane and then we use peroxide.

That is what I have from the review problems, any other questions from chapters one through eight? Let me wrap up the ones that we...

Transcription by CastingWords

Lecture 030: Alkenes 3

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:05:00 -0700

Lecture 030: Alkenes 3

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Jean-Claude Bradley: So we left off with the Anti-Markovnikov addition of HBr to an alkene last time, so this time we'll talk about the mechanisms, so let's do that.

The way that works is if we've got an alkene that's unsymmetrical, in this sense, where we've got a methyl group on one side, we have two hydrogens on the other" there are a different number of hydrogens on either side. So anti-Markovnikov addition, the hydrogen would go on the side there are fewer hydrogens; it's the opposite of the Markovnikov rule. And to do that, we mention that we need peroxide" any peroxide will do.

"R" could be an alkyl group, "R" could be a hydrogen, it's all going to work.

If the H goes on the side were there are fewer hydrogens, the H will go in the middle. And the Br will go at the end.

The mechanism for this, as I said, is similar to one we've seen before, which involves free radicals. So let's take a look at that.

As in all free radical mechanisms, we're going to have initiation step, propagation, and termination. This time, there's more than one initiation step. There's no light in this reaction. So it's not like bromine separates into two bromine radicals. So what's happening in this case is that the peroxide is first going to break into radicals, and that doesn't need light. That just happens with heat. If you take a solution of hydrogen peroxide and heat it, you'll see oxygen coming out of it very quickly. So that reaction is pretty easy. And I'm going to use hydrogen peroxide as an example, but it would be the same with dimethyl peroxide or any other peroxide.

This is just a homolytic cleavage of the oxygen-oxygen single bond, this is actually really weak. Okay, so that's the first step, it's going to generate the OH radicals. Now, this OH radical is pretty reactive in itself, and it's going to react with the HBr.

When you're breaking a bond with free radicals, you're always going to have the same three arrow pattern, that we've seen before. So a good way to remember this is that you're making something really stable, in the second step, you're making water. The only way to make water from the reagents is by reacting the peroxy radical with the HBr. So it's the Br radical that is actually going to add onto the alkene. So those two steps are what we consider to be the initiation.

Next, the bromine radical is going to add onto the alkene. Now, that the alkene is unsymmetrical, we're going to end up with a radical, or two different possibilities for a radical. One of these radicals is going to be more stable than the other, and that's what going to decide the product. So this step right here.

Here's my alkene. The bromine could add onto the left or the right. So let's add it to the left, and see what happens.

Again, I'm using the same three arrow pattern, but now I'm actually breaking a double bond. So I haven't completely cleaved the carbon-carbon bond. I still have a single bond left there. But I have created a carbon-bromine bond on the left. So that generates that radical.

Now, let's consider the alternative, which I'll call three-prime (3'). The alternative is that the bromine would add on the right side. Okay, so there are our two alternatives. So you can see from this, the first reaction, we generate a secondary radical. And in the 3" we generate a primary radical.

Going back to the stability order, from the previous chapter, the secondary radical is always going to be more stable than the primary.

Because of this, step 3' simply doesn't happen. So it turns out you're going to get anti-Markovnikov products whenever you're faced with the difference in different kinds of radicals. So when you do anti-Markovnikov addition, you count the hydrogens on each side of the double bond, but this is ultimately the reason why it happens.

Basically, the product is decided. We just have to finish it. So in step four I'm going to take the radical that I generated in step 3, and now we're going to react it with HBr.

Actually we've generated the anti-Markovnikov product in this step. And we've also generated a bromine radical. So steps three and four then are the propagation steps, because the radical generated in step four will then be used in another step 3. So if you didn't run out of material or you didn't have any termination steps, then you wouldn't need any more initiation once you had that stuff gone through, right? That's how propagation works.

Of course, you do have termination steps, and that's going to work the same way we did free radicals. Each free radical coming together would be a termination step. So what do we have available here?

We have a bromine radical, and we have a carbon radical that has a bromine on it. That's pretty much all we have to work with in terms of terminations. We have two possibilities, two possible radicals I should say. The easiest one is the two bromines coming together.

Another example would be the bromine-carbon radical combining with a bromine radical. And finally we can have two carbon radicals coming together. Yes?

Student 1: [inaudible question]

Jean-Claude Bradley: Yeah, basically the termination involves terminating something involved in the propagations steps. So as you get it started, if you capture a hydroxyl radical, yeah if a hydroxyl radical reacted with a bromine, yes, that would be a termination. But that would create something unstable, so. Just focus on the one's that we did, that are analogous. Yes, any two radicals involved in the propagation.

Again with the termination, you get extremely small amounts of these compounds, you will get just trace amounts because so many cycles of propagation relative to termination, but that's what they would be.

Any questions on anti-Markovnikov addition of HBr?

Now, though you can add HCl Markovnikov, you can't add HCl anti-Markovnikov just by dumping in some peroxide. So this particular reaction is for HBr. And the reason for that has to do with step number 2. It's that the energies are right for the hydroxyl radical to make a bromine radical. It won't be right for HCl or HF.

Remember that when you're doing problems. If you add peroxide, you're not going to do much. Just Markovnikov addition for HCl. And we're going to do some of that in the problem set.

Let's talk about Markovnikov addition of water: Instead of adding like HBr, we want to add an H and an OH, we can still do that on an alkene. For adding water, all I have to do is add a reasonably strong acid.

If I have water, and I'll just put H+ here, but that can be sulfuric acid, that can be any strong acid you add to this, you just need to catalyze this process. So acid and water will give you Markovnikov addition of H2O. This means that the H goes where there are already more hydrogens, In this molecule on the left, it means the H would go in the middle.

That's all we mean by Markovnikov.

The mechanism for this is going to be really similar for what we did for HBr: We have the proton that can go either on the left or right, we already did that, so we're just going to show the one that happens. The proton is going to go on the carbon here to generate the secondary carbon cation. You form the secondary carbon cation, you look for the one that shifts, we don't have any in this example, and then we add the water to this.

You can't add OH- at this point because you're under acidic conditions. So, there will be very very little OH- floating around. Okay, so the nucleophile in this case will be water. Generally water is not that great of a nucleophile, but a carbyl cation is very tantalizing for a poor nucleophile like water, so that's what will happen.

So that means we need an additional step to complete the reaction, we now have to lose the proton that's on the alcohol.

Now, we have the neutral alcohol. So that's how Markovnikov addition with water works, similar to what we had with HBr.

One of the problems with this Markovnikov addition, using acid and water, is that if you do have the possibility of a 1, two shift, then you've got a problem, because you're not going to get the addition that you want. So it turns out there's a way around that, that you will get Markovnikov addition no matter, even if there's a possibility of a 1, two shift or not.

And that's using an oxymercuration approach, so let's do that one next.

We're not going to look at the full mechanism of this, we're just going to look at the reagents.

So the first step will be mercuric acetates and water, and the second step will be sodium borohydrate. So the OAC here stands for acetate, which is just that. But that's how we're typically going to abbreviate that when we do reactions. So those two steps, again when you have one and two on the same arrow, that means you have to do this sequentially, those two steps will always give you Markovnikov addition of water, no matter what. So that means I will end up with OH in the middle and not have to worry about 1, two shifts.

Any questions on that?

Student 2: [inaudible question]

Jean-Claude Bradley: Yes, the mechanism's in the chapter, that's not one I'm going to expect you to know. But if you look, the mercury adds, there's not a possibility, you never get a full carbyl cation. Some, but not enough to promote a 1, two shift.

We're going to look at reactions where we go through every step of the mechanism, like the one we just did this morning, and then you're increasingly going to see reactions where we don't look at every step of the mechanism, but you are expected to know what it does. I think a lot of these, if you're really curious, will be in your book. If you want to get down and see what it's like.

The only thing we haven't covered is anti-Markonikov addition of water.

It would be tempting to just think we could add peroxide to the acidic water, but that doesn't work -- again -- because the energies don't make any sense. The peroxide trick will work only for HBr.

So, in this case, we're going to be using again two sets of reactions that we're not going to look at the mechanism of, but that will always give you anti-Markovnikov addition of water.

The first of these is to add borane, BH3, and then we'll add a peroxide. The peroxide here is acting in a completely different way from the peroxide we had with HBr. Actually here it's a base, so, typically, that's written H2O2, and hydroxide, you have to be in a basic environment.

Those two steps will give you anti-Markovnikov addition of water, so that means the H will go where there are fewer Hs. So the H will go in the middle and the OH on the end.

Again, with the anti-Markovnikov additions, there is no need to worry about 1, two shifts. Because you know that radicals don't shift. Even though a secondary radical may be more stable than a primary radical, it's just not possible to do 1, two shifts as with carbyl cations. So again we don't have to worry about any kind of rearrangement.

That covers the reactions with alkenes, but there are additional reactions we are going to look at. The next one is catalytic hydrogenation.

Basically involves adding hydrogens across an alkene. For that you actually use molecular hydrogen, H2, but you need a catalyst. Here, we'll think of a heavy metal like platinum or palladium or iridium or something like that. That's pretty simple to do: It will happen for any normal, unconjugated alkene.

If I just have a carbon-carbon double bond, that will go fine, but if I have a hyper-stabilized double bond, like in an aromatic, that won't be enough. You'll have to use partial reduction.

For example, if I had styrene, and I used a normal catalyst, it's not going to do this, not the catalyst we're going to look at. But if we just took hydrogen and palladium, you would selectively reduce the carbon-carbon double bond that's outside the ring.

I'm writing one atmosphere here to show normal conditions, we're not trying to force it, then you get selectivity.

The Simmons-Smith reaction is the way of making a cyclopropane. So you would use methylene iodide, CH2I2, in the presence of zinc. And that CH2 unit would then become attached to the double bond, and make a three-member ring. So that's one way to make a cyclo propane.

Next is halogenation. Yes?

Student 3: [inaudible question]

Jean-Claude Bradley: You mean in the product? Let's make it clear by putting in the hydrogens. So you've got a double bond and the CH2 getting stuck on the top or bottom of the ring for that matter.

So, halogenation: Hydro-halogenation means to add Hx, halogenation just means to add x2. So, for example, adding Br2. I can add bromine across a double bond, just simply by adding it. You have to be careful with these reactions because this is in the dark: We know that bromine can react with basically alkyl groups, if you have light, so you have a complication there.

If you have a double bond, and you have other positions that would create stable radicals, then if you really want to do this, you have to do this in the dark. Without any light.

In this case, it wouldn't make any difference, because there are no stable radicals. If it just says Br2, and it doesn't say if it's light or dark, you can assume that it's in the dark. When we want to add light, we'll add it on purpose.

We are going to look at the mechanism of this reaction. This is kind of interesting: What happens here instead of opening the double bond, to the left or to the right, the way that we did a proton, we're actually going to go directly from the middle. We'll do something like this.

What happens here, we form this, and end up with bromine in a three-membered ring, with two other carbons: That's what's known as a bromonium ion. This is reactive as an electrophile, and what's going to attack it is going to be the Br- I generated at this step. So I kick out the Br- and the Br- comes back and reacts with the bromonium ion.

What is interesting about this mechanism is that, because it involves an Sn2 reaction, Sn2 reactions are always a backside attack.

That means that the two Brs are always going to end up on opposite sides, they'll always end up trans to each other, if there's a cis and trans possibility.

In ethylene, of course, there is no cis and trans in the final product, so regardless of what the mechanism would be I would end up with 1, two dibromyl ethane. So let's look at the possibility where you do have cis and trans, like in a ring.

Reacting cyclohexene to bromine in the dark, I would end up with two bromines on the opposite sides. So trans addition because of the mechanism.

With this mechanism, also, there are some competitive reactions that will happen. If I mix bromine with another nucleophile, like let's say water, the first step will happen, I'll still get the bromonium ion, but then I'll have a competition between the Br- and the water. So if we do the same reaction in the presence of water, we will end up with competitive reactions at the same time.

Let's take a look at that: We have considerably more water than bromine in this example, because we want competition to be one-sided in favor of water.

Let's take a look at the mechanism we would end up with: First, formation of the bromonium ion. Then, the water would come in and attack the electrophile from the opposite side of the ring. That's the final result: You get Br and OH added. Again, trans.

Lots of other things we could do with alkenes. We can also put an oxygen on it, the same way we made a bromonium ion.

In order to do that, we need what is called a peracid, it has CO3H. It's kind of a hybrid between a carboxylic acid and a peroxide. So whenever you see an oxygen-oxygen single bond, you know that's weak, so it's going to be easy to break that bond. And what the peroxy acid is going to do, it's going to deliver one of the oxygens on top of the carbon-carbon double bond.

That's called an epoxide. Epoxide is just a three-membered ring.

The thing about epoxides is that why don't we just call them ethers? Call a three-membered ring an ether? Because the nomenclature in organic chemistry is based upon behavior.

That means it has a different name because it doesn't act like a normal ether.

The fact that I have a three-membered ring means that it's strange. I have three SP3-hybridized atoms there, the angle wants to be 109.5 degrees, but it's forced into a triangle. So it's forced into being 60 degrees. For that reason, it's unstable. So epoxides can open up with a variety of nucleophiles. I think in the problem set we do a couple examples of that.

The "R" group here can really be anything, so as long as you have the CO3H, you can expect you're going to get epoxidation. In practice, typically you would use a compound called McPBA, which is metachloroperoxybenzoic acid. That would basically be an example of a peracid. It's common enough I expect you to recognize it, McPBA, in a problem. For epoxidation.

The next thing we'll look at a reaction with permanganate. There are two conditions we'll look at. The first one is cold and dilute. So that means you have potassium permanganate, but the conditions are very mild. And that will do something very different from when you heat it up. So in the cold, you're simply going to add OH groups on the double bond.

Now if you do the same thing under hot conditions, you will actually break the carbon-carbon double bond. "Conc" stands for concentrated, so if you have hot, concentrated potassium permanganate, you will break the carbon-carbon double bond, and furthermore you will oxidize either end as far as it can be oxidized.

In this example, when we break the double bond, first we replace C-double bond-C [C=C] with C-double bond-O [C=O]. So the initial products we get would be this. I'm going to put these in square brackets, this means that they're going to be formed transiently. In the reaction, you never actually isolate them, because soon as they're formed, they get oxidized. Aldehydes are easily oxidized through carboxylic acids.

So if you form an aldehyde at this step, you need to carry it through all the way to the carboxylic acid.

Permanganate is a pretty strong oxidizer, and it's going to do that. Okay, there are various permutations here, if you produce ketones, ketones will not be oxidized by permanganate. We'll do some of that when we do the problems.

While we're on this topic here, this top reaction, this hydroxylation reaction, can either be done with potassium permanganate under cold conditions, or it can be done with osmium salt and hydrogen peroxide.

Finally, if we treat the alkene with first ozone O3, and then dimethyl sulfide, you'll do pretty much what you did with the hot potassium permanganate, but you will not oxidize the carboxylic acids. So that means any double bond, you will break it here, and put oxygens on either side.

That's basically it for chapter 8, so we're at the problems which we'll do the next time we have class.

Transcription by CastingWords

Lecture 029: Alkenes 2

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:04:00 -0700

Lecture 029: Alkenes 2

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Jean-Claude Bradley: We looked at SN1, SN2 reactions. Those were called nucleophilic substitution reactions, because the reagents had a pair of electrons that was attacking the compound. In our case it was the alkyl halide. Those were called nucleophilic reactions. In the case of an alkene, it is the opposite. The double bond is actually electron rich. So instead of having nucleophilic reactions, we'll have electrophilic reactions. [writing]

Jean-Claude Bradley: And one of those would be electrophilic addition. So if I want to add HBr to an alkene, the way that happens is by first taking the electron out of the double bond and putting a proton on there. Let's take a look at an example of that. [drawing] If I react ethylene with HBr, let's think of the HBr as being dissociated, so the first thing, just like the proton goes on an oxygen in an alcohol, the proton will go on the alkene like this. [drawing]

Jean-Claude Bradley: I end up with a carbonation, and, again, if you make a carbonation, you've got to worry about 1, two shifts, so the molecule I chose here won't have that problem. The next step is going to be the bromine coming in and finishing the job. [drawing] So in two steps we have the electrophilic addition of HBr to an alkene.

Jean-Claude Bradley: Now by using ethylene, there was only one product that I could get from that, because it didn't really matter which way the HBr added. Let's take a look at an example where there is a difference. There are two different ways you can add it. If I have something unsymmetrical, like propane, then I can think that the Br could either go on the end of the molecule, or it could go in the middle. Let's look at both scenarios.

Jean-Claude Bradley: Now when I'm drawing the proton coming in on the alkene, you want to think of it almost like opening a door. The door will open either to the left or it will open to the right. If I put the proton on the left-hand side to show that the carbonation is ending up on the right, I'm going to be opening it up like this. [drawing] And I've got my carbonation in the middle. And then as a last step, we add the bromine, so then we get 2-bromo-propane. Ok the alternative is to do it the other way. [drawing] So now the carbonation would go on the end. [drawing]

Jean-Claude Bradley: Okay, so let's stop at this point right here. Which one of these two is more likely, do you think? [inaudible response] The one on the left? [inaudible response] Yes, it just comes down to everything we've seen before: primary, secondary, tertiary. So that's secondary, this is primary. So what ends up happening is that given that choice, you really don't get any of the right-hand side. All of the product will come from that secondary carbonation intermediate. [drawing]

Jean-Claude Bradley: So, you know, we can go through the mechanism every time and show, you know, I get secondary vs. Primary or tertiary vs. Secondary, whatever you want, but the pattern that actually comes out of that, and that has to do with where the hydrogen ends up going. If you notice in this case, as I'm adding HBr, the H goes where there are already more hydrogens. In other words, in this double bond on the left-hand side I have two hydrogens; on the right-hand side I have one hydrogen. So the rule is that in electrophilic addition, the H will go where there are already more H's. That's an empirical rule and that's called Markovnikov's rule. [writing]

Jean-Claude Bradley: So that's a way of explaining things without having to invoke the mechanism. I'll give you another example here. We want to apply Markovnikov's rule. Ok so if I have this compound, and I react it with HBr. So the first thing you need to recognize is that that is an electrophilic addition, that's the first step. Then you determine, how is that HBr going to add? You can go through the mechanism, but if you don't want to go through all that, you simply look at how many hydrogens are on each side of the double bond. On the top portion of the double bond, I have zero hydrogens. On the bottom portion, I have one hydrogen. So to apply Markovnikov's rule, we simply add the hydrogen where there are more hydrogens on the bottom. So I can draw the product then [drawing] by doing that. Okay, that's following Markovnikov's rule. Everybody clear on that? You're going to have a bunch of examples to do anyway, in the problem set for that. So Markovnikov's rule is pretty useful but there is one caveat, and that is if you have 1, two shifts, you're going to have a problem with applying Markovnikov's rule, so you need to be a little bit careful about that. But as long as you don't have any 1, two shifts, the rule will apply exactly like this.

Jean-Claude Bradley: So as I said?? [question from student, inaudible] Right, that's why I didn't finish that one, because I didn't want to get into that whole debate about the shift. I mean, basically what we're deciding at this point is just, where is the proton going to go? Is it going to go on the left or the right? So you're right, after we put it on there, other things can happen. Normally, though, you're not likely to get a shift a lot of the time, simply because you already have a selection of a more stable carbonation. But it is possible to design a problem where you will have going from secondary to tertiary afterwards. But typically because you have that initial choice, you're going to get probably the most stable carbonation, which is why Markovnikov's rule will apply almost all of the time. But that is a good question, and that is why I stopped at that point.

Jean-Claude Bradley: Okay, so we can call these products Markovnikov products, and that's what that means. [writing] If we can have Markovnikov products, then if you want to do the opposite, if I wanted to actually put the bromine on the bottom of the ring and put the H next to the methyl, that would be called an anti-Markovnikov addition. And you just simply can't do it with this particular approach, there's no way that just by adding HBr you can do it. But there is a trick, and that is if you add peroxide, you can actually have anti-Markovnikov addition. So let's take a look at that. [drawing]

Jean-Claude Bradley: Key things about this: the peroxide that I add does not have to be equimolar. I can have catalytic amounts of peroxide. I can have a trace. I think often in the problems, they write "trace" or something like that. It won't hurt it if you have more peroxide, but you don't need a full equivalent. The other thing about the peroxide is that it doesn't have to be hydrogen peroxide, it can be any R-O-O-R, where R can be any alkyl group. So sometimes you'll have dimethyl peroxide, or whatever. It doesn't matter, as long as you have a peroxide. What will happen is you will get the anti-Markovnikov addition. [drawing]

Jean-Claude Bradley: So anti-Markovnikov addition means the H will go where there are fewer hydrogens. There are fewer hydrogens on the top, so that's where the H will go, and the Br will go on the bottom. [drawing] Also this reaction will really only work with HBr, so you can't do it with HCl. If you wanted to put a chlorine in there instead, then you'd have to use some of the other knowledge you have, like you could do an SN2 reaction, on that alkyl halide, and put a chlorine, but you do need to remember that this only applies to HBr. This mechanism is actually pretty extensive, and it's going to be similar to another one we did previously. And I think that is where I will start the next class, at the mechanism for this anti-Markovnikov. That's it.

Transcription by CastingWords

Lecture 027: Alkenes 1

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:03:00 -0700

Lecture 027: Alkenes 1

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Jean-Claude Bradley: Just before we look at different ways of making alkenes, let's look at a bit of nomenclature. An alkene will be any compound that has a "Carbon=Carbon" double bond, as we've seen many times. And the way we name these is, if we just have a simple alkene, we start with the alkane. The corresponding alkane to this would be ethane, so I change the A to an E and I have ethene. So you can always do that, but a lot of the smaller alkenes have more common names. So this one would be the same as ethylene. But the proper IUPAC name would be ethene for this compound. Same thing for this, IUPAC name would be propene, but it could also be called propylene. When we have more carbons, then we have the possibility of the double bond in different locations. There's no reason to number the ethylene and propene, because there's only one place the double bond can be. Now we have four carbons - four carbons, so it would be butene - so I have to specify where the double bond is. I'm going to start counting from the side that gives me the lower number. Now so far the compounds I've drawn don't have cis/trans isomers, which we spent some time evaluating, so there's no other descriptor we need to add to the name.

Now when we get to compounds which do have cis/trans isomers, then we absolutely have to specify whether the methyl groups are on the same side or opposite sides. The only way we have to do that right now is using cis/trans, so for the top compound, we're looking for trans. How do we know it's trans? We have two groups that are the same on opposite sides. Another way of saying that is I can say that the methyl groups are trans and the hydrogens are trans, and that's unambiguous so we can say it's trans. And specifically it's trans-2-butene. Same thing down here, the two H are on the same side, or the two methyl groups are the same side, so this would be cis-2-butene. We also came across some examples where you did have cis/trans isomers and we couldn't name them. The best we could do was to say this group was cis, but that's not really a way of naming a compound if you're looking for it. So let's see what happens in those kinds of situations.

All right, so if I have a compound like this, it has four different groups on it - I, Cl, F, Br - this is ethene, it's name is actually bromo-1, chloro-2, fluoro-2, iodoethene. So that compound would be two different possibilities, because we do have cis/trans isomers. So we do something similar to what we did to R/S. We divide the alkene into two parts, and we determine the relative priority of each group. So on the left we have I and F, so the I has higher priority than the F. We use the same rules as for R and S. So that would be one and then 2. So now we're done with the left side, we forget about it and go to the right side. Between the Br and a Cl, the Br will be higher. So you're always going to end up with two possibilities, either the two high priority groups are on the same side or different sides. We can't use cis and trans for this because they have the condition that two same groups are on same or opposite sides. We don't have the same groups here. So instead we use an E and Z nomenclature. So where the two highest priority groups are on the same side, that would be Z. If they're on the opposite side, that would be E. So now you can name any alkene unambiguously. Now the E and the Z actually derive from German words so it's hard to remember how to remember. So one of the tricks we can use is if you look at the E, if you picture the top part of the E flipping back, now you can see they have to be on opposite sides. So that's kind of an unusual thing to remember, so you're likely to remember. It would be nice if the letters look the same, but it doesn't. It's the same with Z, it's got two groups pointing in opposite directions, but it's really on the same side. So that's basically all about how you name alkenes.

The other issue about alkenes that you need to worry about that we did discuss - the eight carbon rule, that is- in a ring, we can't have a trans double bond unless we have at least eight carbons. So this would be the smallest cycloalkene that you could make that would be trans, otherwise it's not possible. Okay, so this would be, if you wanted to name this guy, it would be trans-cyclooctene. And questions about that?

So let's actually get into the preparation. There are five methods we have to prepare alkenes. Some of these will be review, but we'll put them in one place: dehydrohalogenation. Let's break down the word. So halogene would be X (F, I, Br, Cl), hydro means H or water, in this case H, and de- means to remove. So dehydrohalogenation is an easy elimination reaction, so E1 or E2. We spent a lot of time on this, so let me just put one example. You need a base, typically to do E1 or E2. And you would get your alkene out. And there's the HBr that you lose. So we spent a lot of time on that, enough to say that you have to keep it in mind, because one of the things you're responsible for here - we're not going to look at the mechanisms of all of these, but you have five ways of making alkenes, so you could have a questions where I draw an alkene and that could be made from a bunch of different things. So even if you didn't see the mechanism, you should know five ways of making alkenes. So I could make an alkene from ethyl bromide.

Let's see how else we could make an alkene: dehalogenation. That's one we haven't looked at yet. So we break down the word: halogene could be X (F, Cl, Br, I), and de- means remove, so here we're removing only a halogen. The way we do that is by actually having two halogens, and there are two reagents we're going to look at, one of them is zinc, or you could use I-, and that's specific to I-, you can't substitute Br- or something, it's something special about the I itself. So with those reagents, you lose the two Br's and form an alkene. So it's not any dihalide that will do this, it's only dihalides that are vicinal. Vicinal means a 1, two relationship, so these two Br's are vicinal because they have a 1, two relationship. If they're on the same carbon, it's not vicinal, if they're another carbon away, it's not vicinal - they have to be next to each other. That's the only condition for this kind of dehalogenation reaction. Next we have the dehydration of alcohols. So dehydration means lose of water. Okay, we have just covered this today at the beginning of class. So if we want to lose an OH, we can't do that directly, we have to protonate the OH, then we can do E1 or E2. Typically you use a strong acid, usually sulphuric acid. You can use other acids, but this one have the advantage of not being nucleophilic very much. So if you're were to use something like HBr, you have the complication of the Br doing SN1 or SN2 on your compound. So if you don't want the complication of substitution reactions, you use something like sulphuric acid, which is an excellent acid but a poor nucleophile. So we actually looked that the mechanism of this earlier. Bottom lines if you end up with your alkene. You do have to be wary though, because you could have 1, two shifts. Because if you do an elimination by E1, you have to make sure there's not any carbocation that can rearrange. In this example there isn't any, but that is one small thing you need to worry about. And we revisit something we saw first week, when we look at reactions of alkanes, when we could break alkanes into smaller pieces, and some of those pieces were alkenes. So we're going to add this as a way of making alkenes. Use the same example as I used earlier. So propane in some catalyst, palladium, platinum, whatever, some heavy metal catalyst, and high temperature and usually high pressure, will actually break this molecule into smaller pieces. One of the pieces would be methane, and the other piece will be ethylene. The reason I get an alkene in this case if because if we break a C-C bond, we can't put a H there because we don't have enough H's, so you notice that breaks in such a way that we have the same number of H's on both sides. Industrially this is a pretty good way to make small alkenes like ethylene, but in the lab it's impractical. For one thing, the conditions are very harsh, so if you have anything else on this molecule on this molecule besides the alkane, this would destroy it. And it's not very specific, so if I have eight carbons, it would be random where it would break, so that's not good if you're making a specific alkene. But industrially it's very good because these are cheap processes, so everything has its place, and this is not going to have a place in the lab where you're going to be working. On the other side of that, we have a reaction that's much more expense, but is very specific, so if a chemist wants to make an alkene, this is one of the best ways to do it. This is called the Wittig synthesis. And what you need is an alkyl halide and a ketone or aldehyde. So let's start with methyl bromide. I'll take you through the whole mechanism; in the future I'll summarise it, but in the beginning let's go through it carefully. This is actually going to be an SN2 reaction, we have triphenylphosphene and methyl bromide. P is right under N, so it has a lone pair, still reasonable nucleophilic and is going to attack an alkyl halide the same way you would use and amine to attack it. So we end up with - these phi symbols are benzene rings, same as Ph or drawing out the whole thing. The reason we write triphenylphosphene like that is because it's really a pain to have to draw three benzene rings, so typically it's abbreviated like this. So just like the reaction of an amine with an alkyl halide, I have a positive charge on the P, and I've also lost Br- here. What we want to do is basically make the carbon that's attached to the P nucleophilic, so we're going to remove a proton from it. If I remove a proton from this, I put a negative charge on a C. Normally it would be difficult to put a negative charge on C, but here we have positive P here that would stabilise it a lot. You still can't use a mild base, but you can use a base that's available, and that base is typically N-butyl lithium. Now the way you want to think about N-butyl lithium is like this, you have a CH2- and a Li+. Now there is a covalent bond between the C and the Li, but it acts as if we have a negative charge on the C. So we draw the arrow, the CH2 is extremely basic, so it's going to extract a proton from the C next to the P. So now I have positive charge on P, negative on the C, and when you have that kind of situation, where there is a negative charge next to a positive charge, that is called an ylide. This is your Wittig reaction, we spent time preparing it. When we talk about the Wittig reaction, it's going to be the phosphorous ylide. So one way to draw the ylide is like this, with the + on the positive and the - on the carbon, and you'll notice that this follows the octet rule nicely. But I'm also going to draw it in a way where you might see it in problems, where the octet rule is not followed. So you can also draw it with five bonds on the P, and you don't have to draw a charge, but it's understood that if you're going draw a correct Lewis structure for this, you're going to draw this octet, but this is more convenient because you don't have to draw the charges. So I've created the Wittig reagent. The Wittig reagent is actually half of your alkene. The other half is going to come from either an aldehyde or a ketone. So let's say we just use an aldehyde as an example. The Wittig reaction will react with the aldehyde in such a way that the C=O will be removed, and exchanged for C=C. And a side product from the Wittig reaction proper will be triphenylphosphene oxide. What makes this a particularly useful synthesis is that it's extremely general to pretty much any alkene you can draw. So you'll have a problem where you have an alkene and try to figure out what original alkyl halide you could use, what original aldehyde or ketone you can use to make the alkene. And when you work backwards with the Wittig, is you just need to remember you stitched the double bond in the end. So you break it off, and you see what groups you need in the alkyl halide and what groups you need in the ketone or aldehyde. So that's something you're going to need to know fairly where; there are a couple of quiz questions on that alone. So that's pretty much it for the theory part of chapter 7, and we're going to practise this next time doing problems.

Transcription by CastingWords

Lecture 023: Alkyl Halides 3

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:02:00 -0700

Lecture 023: Alkyl Halides 3

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Jean-Claude Bradley: We have one part from chapter six left to do, which involves some of the rules with respect to minor and major products of elimination reactions so let's do one compound which encapsulates a lot of the concepts we've been working with.

Jean-Claude Bradley: We did this reaction between this halide last class and hydroxide or chloride and we did SN1 and SN2, now what we're going to do is E1 and E2 and again make sure we understand how to tell the major products from the minor. If I tell you to do this via an e2 reaction, then that means its second order, its an elimination, so you are going to get an alkene and its going to happen in one step just like the SN2, the E2 happens in one step. Now what you have to eliminate is HBr and to know which hydrogen is actually eligible to be removed, a good way to do that is to count to three from the bromine. So if I start on the Br I can count 1, 2, three and find the hydrogen atom, so i can do an E2 reaction and eliminate that hydrogen.

Jean-Claude Bradley: Is there another hydrogen in that molecule that we can do? The one over here? Yes, one of these other three hydrogen also can be eliminated. Is that it? Are there just two possibilities?

Jean-Claude Bradley: The first step in doing these problems is to determine how many different ways that we can lose HBr so, count to three from the halide and identify these 2.

Jean-Claude Bradley: I'm going to draw two mechanisms for each of these eliminations. In the E2 the base comes in, grabs the proton, we lose the C-H bond and form a carbon-carbon double bond where that arrow is going into and at the same time we lose a Br-. So E2 is always going to have those three arrows in that exact order. Now when you draw the products from that E2 or any alkene a good suggestion is always start to draw the carbon-carbon double bond and then draw the others around it, because the carbon-carbon double bond is the only one that had a rigid geometry and its a lot easier to start with the SP2 hybridized template than it is to try to fit it in when you draw the molecule. So if we're forming the carbon-carbon double bond then the molecule's going to look different, so that is the other reason why you have to be careful writing the product out. You have to determine what are the two carbons where the double bond is forming. I'm going to draw a yellow rectangle around them. That's where the double bond is going.

Jean-Claude Bradley: What you're asking is what's left on the left and what's left on the right. I remove the hydrogen and on the left two methyl groups are left. On the right I remove the bromine so what's left is a methyl group and a hydrogen.

Jean-Claude Bradley: Now one of the things we have to use is when we did exercises to figure out if alkenes have cis/trans isomers this is where you apply it. Basically if you have cis/trans isomers you have to write them out. This compound does not have that, it has two methyl groups on one side so there is only one product for this pathway. If you had cis and trans you need to put both in. This is one way to do the elimination. The other way, let's look at the methyl group and see what happens.

Jean-Claude Bradley: Same arrows, same pattern, three bonds from the bromine. Again, draw the carbon-carbon double bond template and look at what's connected to it. If I draw a yellow box around the carbon-carbon double bond, one side has two hydrogens that are left and on the other we have a hydrogen and an isopropyl group. We check for cis and trans and we see that again this compound does not show cis/trans isomerism, so we only have one compound. We're almost at the end of this. We identified all possible HBr patterns that can be removed by E2, we worked out these two solutions, we worked out all the products. Only thing left is to determine the major product and minor product. What is going to determine that is degree of branching of the alkenes? An alkene is said to be branched if it has anything other than a hydrogen is on it. So the top alkene has three methyl groups and one hydrogen, so it has a branching of 3. Does not matter how big the group is it counts as one branch. I would write here branching is equal to three.

Jean-Claude Bradley: What would be the branching of the lower one based on that definition? Just one, one alcohol group, all other hydrogens, has a branching of one. What we're going to use is Saytzeff's Rule?? the most highly branched. I said I would give you very unambiguous definitions for major and minor product in each reaction. In this reaction the products that have the highest branching are going to be major. If we compare we have a branching of three and a branching of one. The one that has a branching of three would be the major product in this case and anything else would be a minor product.

Jean-Claude Bradley: The branching would be any groups on the alkene that are not hydrogen, would count as one branch. These are probably the longest problems you'll have to do in this course, you need to take your time and there are several questions on them. I ask you to go through and draw out the problems and then I ask you questions about it. Some of the questions might be like how many total products? We have two products total here. Sometimes you'll have more than one major product. If you have two products with the same branching-I had another here with a branching of 3, then you'd have two major products. Basically no shortcuts you really do have to work out all of these products in their entirety to answer questions about them.

Jean-Claude Bradley: There is one more we haven't done yet, the nastiest one of all, the E1. Let's take look at this example and see what happens. The E1 and SN1 are similar; they start out the exactly same its just how they end up is different. The 1st step is always what? E1 and SN1. That's the one where we have carbocations. In order to have carbocations we have to first loose the halide. Always do the same in these cases. We are going to Br- Br- first.

Jean-Claude Bradley: We just made a carbocation so before we do anything else we need to worry about rearrangements. Is there a 1-2 shift possible in this case? This is a secondary carbocation, so the only shift we need to worry about are the ones that would create a tertiary carbocation. To find out what groups could potentially do the 1-2 shift We start at the carbocation and count to two; so I can go one, two and think about moving the methyl group over. If I move that methyl group I would have generated a secondary carbocation so it wouldn't work. I could move the hydrogen and it would make the same situation even worse in fact, I would go from secondary to primary carbocation. Only one left is 1-2 moving this hydrogen over, that would generate a tertiary carbocation. So that is the one we must do before we finish the problem. The 1-2 shifts you start in middle of the carbon-hydrogen bond and end up on the carbocation. Move the hydrogen and its two electrons over. So now we have the tertiary carbocation. Now we are finished doing the 1-2 shift; this is the same for SN1 and E1 so far. Now is where the difference comes in. If I had an SN1 the nucleophile would attack C+ and it would be done, in this case I'd have an alcohol because I'm using hydroxide. But we're doing an E1 so that means we're going to lose a proton instead and generate an alkene. Because there's going to be more than one way to lose proton from this I can get a lot more products. If you want to use a counting strategy to figure out what you could lose, start on the C+ count to two to see what hydrogen is left. Let's say in yellow I will circle the candidates here?? 1,2. I could lose that proton to generate an alkene. Recognize that this methyl group is the same as this methyl group. They're on the same carbon. It wouldn't matter which of these six hydrogens I lost, I would still end up at the same alkenes. Otherwise I can go to the right and count 1,2, I could lose a proton and make a different alkene going to this side. That's it, so there are two pathways we have to work out.

Jean-Claude Bradley: So I could lose a proton on the left methyl group, and if I do that the only thing I left to do to make the double bond is to move the electrons between the CH to between the CC. So again the E2 is really, you can think of it - the E1 is like an E2 done in two steps, sort of, except that you potentially having a rearrangement in between. But we're doing the same thing?? we're losing a bromine and losing a hydrogen. So to draw this product. First we draw our SP2 template. What do we have left? On the left we have two hydrogen left and on the right we have a methyl and an ethyl group. Now I do the same thing with the one on the right. On the left I have two methyl groups left and on the right hand side I have a methyl group and a hydrogen left.

Jean-Claude Bradley: We're careful to check for cis/trans isomer. Again, these two compounds don't have any cis/trans isomers. Total we would have two products from this E1. We want to determine major and minor, that is done in exactly the same way. I look at the carbon-carbon double bond, the number of alkyl groups on the top one is two so the branching is two. On the bottom we have three alkyl groups and one hydrogen, so the branching is three. Using Saytzeff's rule the alkene with the highest branching is the major product; that would be the bottom one in this case.

Jean-Claude Bradley: So that is basically it. This example is good because it has everything really - SN1, SN2, E1, E2, has rearrangement and major and minor products. That is it for chapter 6, any questions?

Transcription by CastingWords

Lecture 022: Alkyl halides part 2

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:01:00 -0700

Lecture 022: Alkyl halides part 2

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Jean-Claude Bradley: So last time we looked at the SN1 and the SN2 reactions. We also looked at what happens with chirality for SN2 reactions; we get inversion. So we're going to kick it off from there and we're going to look at solvent effects for these reactions. We talked about if I have a tertiary Alkyl halide, I'm more likely to have an SN1 reactions. If I have a primary I'm more likely to have SN2. That's just a tendency. For secondary you can have both easily. So it turns out, as a chemist, you have some control over that. By selecting a solvent you can favor more SN1 than SN2. So we're going to look at the solvent effects and we're going to try to understand them and won't have to memorize them. Let's look at doing an SN2 reaction first and two different solvents. SN1.

So if we do an SN1 reaction and we want to talk about solvent effects on speeding up or slowing down, remember that it's only the rate-limiting step we have to worry about. OK? Because anything else we do is not going to have an effect on the rate. So the rate limiting step in an SN1 reaction is the first step. It's that first step where you generate the carbocation. So say I use the same example that we did before, [xx] bromide. The first step is always the same thing, I'm going to lose the R minus and form a carbocation. So we're going to consider two different types of solvents, the very polar, hydrogen bonding solvent like methanol and we'll consider something that's less polar like chloroform. And we'll see if we can predict whether the SN1 reactions go faster in one solvent versus the other. If I were to view this in methanol and we want to find out if it's going to go faster; if there's going to be more stabilization on the right. If it's going to inhibit it we might see more on the left-hand side.

So let's go through the left and the right. On the left do we expect a strong interaction between methanol and the alkyl halide? It's just going to be dipole dipole and there's going to be solvation. We also want something that will dissolve the material we're doing. Ultimately, there's not any strong stabilization that I can draw here on the left but on the right however, I have charged species. When I have charged species, then it becomes very easy to bond with the methanol. So I can actually draw a hydrogen bond between this bromide and methanol like this. So normally, with a neutral bromine, I still have lone pairs but we never really draw a hydrogen bond. I have a bromine on the left but it's neutral so I just don't draw a hydrogen bonding in the bromine and the methanol. However, when I put that extra lone pair in, that now becomes negative so that's going to be strongly attracted to the slightly positive hydrogen on the methanol.

That's why we have hydrogen bonding on the right and not on the left. If I have this kind of bonding I'm going to get a stabilization of the Br minus. I could draw something similar for the carbocation but just for simplicity if I have an ion that's generated I'm going to get stabilization in a very polar or hydrogen bonding environment. That's the first conclusion here that in methanol the right-side is more stabilized. So if it's more stabilized that means it's going to be easier to form a Br minus and a carbocation in methanol than it would be in say chloroform. What we're predicting here is that for highly polar and hydrogen bonding solvents, the SN1 will be faster. So just to put my comparison here, in chloroform, CHCL3 we have less stabilization. The prediction is that in highly polar or H bonding solvents, SN1 is faster. So that's useful to know because if you wanted, for whatever reason, to get SN1 versus SN2 you could control that by just trying to get as polar a solvent as you could but still having the ability to dissolve your compound because that's one of the requirements of having the solvent effect work; your compound still has to be soluble. Any questions on this rationale?

OK, let's do SN2 and see if we can come up with something similar. For SN2, the rate-limiting step; the only step in the reaction has to be the rate-limiting step. So here we're going to look at what's on the left and what's on the right. And again, just to be consistent, I'll use my previous example; I think I used methyl bromide. So we're going to do methyl bromide plus Cl minus. This is SN2, Cl minus, methyl bromide. After the SN2 we lose the Br minus. We have those two arrows. Now I'm going to do the same kind of analysis. On the left hand side we would expect stabilization of anything with methanol. Same thing, we still have an ionic species so it's going to get stabilized. Also on the right we have a Br minus so we can expect stabilization in both places. We see stabilization on both sides so if we try to apply the same argument we did for the SN1, what do we predict would be the effect on the rates? You think it would stay the same? The thing is we have something else going on here.

If we look at the absolute energies that may be true, they both may go down to the same place but notice that if I have four hydrogens bonding to the Cl minus, it's tying up the lone pairs that are required to do the SN2 reaction. It's a different argument but it's kind of related. The effect of the methanol is to block one of the lone pairs that are required for the attack so the effect is the opposite. It's actually going to slow down an SN2 reaction. What happens on the right-hand side is not a concern at all because once the SN2 reaction happens, it's over and we go on to the next molecule. The fact that it's stable on the right really is not that big of a deal in this case. So we can say, it traps the lone pairs, which slows down the SN2 attack. So to write this more generally, we would say that highly polar or hydrogen bonding solvents slow down SN2. So the effect is in the opposite direction. That's good. It means now there's a reason why it's going to go faster and a reason why it's going to go slower in both cases.

So for the secondary alkyl halides now you can go one way or the other. Again, these are just tendencies. It's not so much that you're going to go from SN1 to SN2 but you might change the ratio and get fewer side products. And the problem set has a couple things on solvent effects; that's how we're going to rationalize that. Questions on that?

OK, it looks like we've covered everything we can on SN1 and SN2 so let's try to make sense out of this reaction. Here's a secondary alkyl halide and we're going to react it with Cl minus. I'm going to ask you to do it by SN2. So using the information about the mechanisms, we know that SN2 is a single step reaction and two electrons are going to start from the Cl minus. We'll attack the electrophilic and kick out the Cl minus. I didn't specify the chirality so we don't worry about it, but obviously we're going to get into an inversion here at the chiral center. In that transformation, we are going to expect to get 100% of the product. That is the case. So SN2 reaction goes exactly the way we've seen it before. Let's try the same thing and do an SN1.

I can't do this in one step so I'm going to have to do some work down here to go around. The first step in the SN1 reaction is the loss of the Cl minus and the formation of the carbocation. In terms of arrows, it's always the same, just one double headed arrow. And because I'm going to do a couple of steps here, instead of writing the carbocation plus the Br minus, I'm going to put the Br minus on the arrow but because I'm not adding Br minus, I'm losing it, I'm going to write minus Br minus. So that's a little shorthand when we have multiple steps and don't want to worry about side products. So we'd expect to get that and if we then follow what we're supposed to do at this step (this is the fast step), the step where the Cl minus attacks, we'd expect to get the same product that we did for SN1. However, when we do this reaction, that's not what happens. You don't get any of that product. The problem is that this carbocation that we just made can rearrange which is not the case in SN2. SN2 reaction as I said is the preferred reaction if you want to make carbon-carbon bonds because you're only going to get one product, it's going to be a clean reaction and you're extremely unlikely to get side products but when you have SN1 reactions, you have a lot of opportunity to get reactions that you're not looking for. The reason for that has to do with this carbocation.

What I'm going to say now is going to apply to any organic chemistry or any course you take in the future. Whenever you generate a carbocation, the first thing you have to look for is, can it rearrange to a more stable carbocation? We already looked at the stability of the carbocations. Tertiary is more stable than secondary, is more stable than primary. Same order as the radicals. The thing is the radicals didn't have a mechanism to rearrange. If I got a primary radical, it just stayed a primary radical, that's not the case for the carbocations. We had a secondary carbocation that I just drew here. If there's a way for it to become a tertiary carbocation it will. It can't do whatever it wants. There are rules in organic chemistry. One rule is when you have a carbocation you can do a 1, two shift. You can take a bond that is one carbon away and move it over as long as it gives you a more stable carbocation. So the group that I'm going to move is this hydrogen with its two lone pairs.

So I'm going to move the hydrogen and the bond over by one. The way that I'm going to show that (let me copy this molecule over so the arrows make sense) is by starting the movement of electrons at the CH bond and I'm going to end up on the carbocation. What that tells me to do is take the hydrogen and its two electrons and move it over to the carbon and don't leave any electrons behind. Now the carbocation on the right is no longer a carbocation, it's just a CH2 group. What I have left behind is no electrons so that carbon is going to be C plus. The driving force for this is that I generate a tertiary carbocation, something more stable than what I started off with. So we're going from secondary to tertiary.

Again, I can't move any bond I want, I have to move a bond that is immediately adjacent to that carbocation. This is called a 1, two hydride shift. There are many shifts that are possible. You can have 3, 3. You can have all kinds of things. The 1, two specifies that relative to the carbocation, the bond that's moving is second position. That's why you call it 1, two to specify they are adjacent, they have to be adjacent. It's a hydride shift because what's moving is not a proton. A proton is a hydrogen with no electrons. A hydrogen atom is a neutral hydrogen; a hydrogen with one electron. But this is a hydrogen with two electrons and that's called a hydride, which is an H minus. So you always do that whenever you make a carbocation, in any context. Now the mechanism resumes, so after we do the rearrangement we do the last step of the SN1 which is the attack of the nucleophile. I think we're using chlorine here and that just completes normally. That would be the only product that you would get from that SN1 reaction.

A common mistake here is to think that we are going to get a minor and a major product. Like I said, each chapter is going to have clear definitions about what's major and what's minor. Do not do any (see where the 'X' is here). Do not do any of this. Always do the hydride shift before any other product. The reason is the hydride shift is going to be faster the second step. Anybody have any idea why that should be. Why is a hydride shift faster than with Cl minus? It's closer. All you have to do in a 1, two shift is simply have the molecule rotated at the right angle and bingo, it happens. In order for the Cl minus to react the Cl minus has to find that carbocation so it could be traveling for quite a while relative to how long it takes to rotate. That's why the 1, two shift is always going to be faster than any intermolecular reaction that we can do. So remember that reasoning when you work these problems out.

So basically, that's it for SN1 and SN2 in terms of the complexity of the reaction. We've seen the effect from chirality, we've seen the effect of solvents and we've seen the effect from 1, two shifts. There's one other scenario I want to work out. What if we replaced this hydrogen, this hydride that moved with a methyl group? Is that a way we can avoid the 1, two shift? Let's take a look. So far, it's almost the same molecule. Again, I do Cl minus SN1 conditions. Let's take a look. First step will always be the same; we're going to lose a bromine ion. So now the question is do we do a 1, two shift? There's no hydride that can move here so the question is, can I move the methyl group just as easily as a hydride? That would make it easier, right? Because then you don't have to worry about hydrides but the reality is you can move any alkyl group. So yes, in this case the same thing will happen. It won't be a 1, two hydride shift; it will be called a 1, two alkyl shift but it's going to have the same probability of happening if it can make a more stable carbocation. This is a 1, two alkyl shift. So we go from secondary to tertiary. Ok, so that would be the only product in this reaction.

You notice how long it took to do the problem. That's how long it will take you on any quiz or exam. This is probably the longest problem that you're going to have to do. Just take your time. It's always the same steps you need to apply, just be careful about those issues. We're going to do a bunch of examples in the problem set also. This is basically it. So how do you know what group you could do a 1, two shift? An easy way to do that is to count the two from the carbocation so if I put my pen on the C plus I'll count to two and see what I fall on. So if I count one, two, I'm in the methyl group, so that methyl group could be a 1, two shift. Same thing, I could do one, two and hit all these methyl groups. We already spent a lot of time, so hopefully it's clear those are all the same shift so we don't repeat those. The hydrogen is directly connected to the C plus. I only count the one and I hit it. The only other group that I could move is one of the hydrogens on that methyl group. So if I start at the carbocation and count one, two - I could think about moving that hydride over but that won't happen because that would generate a primary carbocation. You should be prepared to do all the possible shifts and then determine if any is going to give you something more stable. You can generate something of equal stability; secondary to secondary, don't worry about it. I'm interested in seeing the major shifts so I'm interested in primary to secondary, primary to tertiary, secondary to tertiary or we also saw [xx] was stable. So anything that is much more stable.

In reality do you have secondary to secondary shifts? Yeah, you do and that's what complicates SN reactions a lot more; but what I want to see is that you show me the major shifts that are going to happen. If you have two tertiary products then you need to go both ways. You're going to have both products. Questions on this? This is just practice, you've got to get in there and do some problems. So far I've used Cl minus as the nucleophile but if you tried to do the quiz, I often used OH minus and there's a good reason for that. Let's take a look at a simple example of 2-bromopropane reacting with OH minus. And again, when reacting with OH minus I have to tell you what kind of reaction to do. Let's take a look at SN2.

OH minus is very similar to Cl minus; they are both pretty good nucleophiles. I have a negative charge on an electronegative atom; that makes it nucleophilic. Now, I'm going to do the standard SN2 reaction that we've done several times. So my only product would be 2-isopropanol. Similarly I could ask you to do the same reaction; OH minus and SN1. We've done this a number of times. Get a carbocation. Now you know, once you generate a carbocation, you have to look for 1, two shifts. This is a secondary carbocation. If we count to two from the carbocation, one, two, we end up on one of the end hydrogens. However, if I tried to do a 1, two shift I would go from secondary to primary so nothing will happen. So, we checked it and there's no 1, two shifts.

Now, we move on to the last step. In this case we also get isopropanol. So whenever you don't have a 1, two shift odds are you're going to get the same product for SN1 or SN2. The exception of course being with chirality. SN2 does inversion and SN1 will randomize the chiral center but aside from that they are pretty similar. So that would be the end of the chapter except that there are other reactions that will happen between OH minus and electrophiles such as this. There are two other cases. The alternative if we look at the SN2 reaction, OH minus is acting as a nucleophile which means it's coming, attacking with the carbon, and is displacing the leading group. But OH minus is also a pretty good base so if OH minus acts as a nucleophile, you'll get your substitution but if it acts as a base you're going to get an alkene from that. The difference is pretty minor arrow-wise. Instead of attacking the carbon, we're going to attack one of the hydrogens on the methyl group. Let's look at how that works.

It starts off the same way but it's going to attack a hydrogen that is three bonds away from the bromine. So as it's grabbing on to that proton the bond is breaking between the H and the C. So I start from the middle of the CH bond and end up in the middle of the CC bond. I can't stop there because I would have a carbon that has five bonds so in order to complete this, at the same time the bromine has to leave with two electrons. So if I follow everything it is telling me to do it's saying make the carbon carbon double bond between those two carbons and on the left side of the alkene I have methyl and a hydrogen and on the right side I have two hydrogens. This reaction is also second order. It happens all at once, just like the SN2 reaction but it's not a substitution reaction, it's an elimination reaction so instead of SN2 it's called E2. Two meaning second order. The other products we have here are water and Br minus. So if there's an E2, you can imagine there's an E1 and it's going to work in a very similar way...

So now I'm specifying that I want to do this by E1. That tells me, because it's an E mechanism, it's going to start off the same way as an SN1. Once again, any context where you have a carbocation, the first thing you do is look for 1, two shifts. We don't have any in this case so if we were to finish this reaction with an SN1 we would just clip on the OH minus, but we are not, we're doing an E1 so what happens instead is that we are going to lose a proton. The OH minus is going to come in and instead of going to the C plus it's going to go to this end proton and then we are going to get the double bond. So in this case E1 and E2 gave the same product and there is only one product in this case. That's basically the four reactions we cover in this chapter and that is the core mastery material and being able to take into account all of those situations, what happens with chirality, what happens with solvents, and being able to predict the products. On Friday, I will finish up the theory part and we're going to get into some problems.

Transcription by CastingWords

Lecture 021 Alkyl halides part 1

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 11:00:00 -0700

Lecture 021 Alkyl halides part 1

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Jean-Claude Bradley: We're going to start off on chapter six today, which involves alkyl halides. You're already familiar with alkyl halides; we've used them in the free radical halogination, that's how we make them, that's one way we saw how to make them. In this chapter, we're going to take a closer look at alkyl halides in detail, how to make them different ways than what we saw so far, and some of the reactions that they do, and then also we're going to look at some additional mechanisms that are completely different from the free radical halogination.

We're going to start off with nomenclature, just to make sure that we all know how to name these things. Again here, no big surprise, you've already been naming alkyl halides, because when you look at the products of free radical halogination you have to know how to name them, but let's just make sure we know two different ways to name alkyl halides. If we have a compound like this, you can use upac format, and you can say I have three carbons as the longest chain, so I have propane, the chlorine is on the second position, so this is 2-chlorl propane. What's an alternative? It'll only work for groups that have four or fewer carbons, though. If we consider the three carbons to be a fragment, then we can consider this to be an isyl propyl chloride. That's the other way that you came name these. It's a little bit less general because it'll only work with a fewer number of carbons, but you should be familiar with that also. Don't get confused by the fact that it almost looks like it should be a salt. Sodium Chloride is a propyl chloride, this is certainly not an ionic compound, it's just nomenclature. Really, it's all covalent. There's really not that much more to do about nomenclature because we've done so much already with alkyl halides.

I'm going to do the preparation, and then I'm going to look at the reactions of alkyl halides, but first the preparation. There are four ways we're going to look at making them. I have to list what we already saw, free radical halogination, we spent a lot of time on that, so I'm not going to give you all the details, I'm just going to summarize it with ethane. As an example, we use the bromination. You now know all the details about that reaction, you know what all the major products are going to be, you know what happens with chlorine, you know what happens with fluorine from the salt activity, so I'm just going to list it to make sure we're complete here.

Second, we have hydrohalogination of alkenes. In organic chemistry, there's a lot of really big words, don't get intimidated by that, just break it down, so if you see a new reaction, basically just take a look at what composes it, so in this word here, hydrohalogination, we have halogen, which we know what it is, it'll be an F, Br, Cl, or an I, and we have hydro. Hydro can sometimes mean water, and sometimes it can mean hydrogen. In this case it happens to mean hydrogen. We're doing a hydrohalogenation, which means that we're going to be adding a hydro halogen, so a hydro halogen would be something like HF, HBr, HCl, HI, of alkenes, so an example of that would be this. Reacting this with HBr, and I'm going to put the H on one side, and I'm going to put the Br on the other. What I've done is I've added HBr, and so this is a special case of a larger group of reactions that are called addition reactions. We add two things together, one plus one, we only end up with one molecule, and there's special rules for this reaction that we'll take a look at a little later, probably in the alkene chapter, but for now it's basically the important thing is that you know you have an alkene, you know you can add HBr, and you know you're going to get an alkyl halide, but again if you have different kinds of alkenes, you'll notice there's two ways you can add the HBr, so there's actually rules you'll need to follow for that, but for now in this chapter, we'll just look over this very generally, and we'll look at the details later.

The third, we have from alcohols. There's a couple ways to do this, but the way that I want to mention here is this. We start with an alcohol, and we react it with HBr, so we're going to substitute the OH for a Br, so we're basically going to swap the OH for Br, and the H will go on the OH, so we'll end up with water and an alkyl halide. Again, there's some rules for this, but we're going to leave that a little later.

The last one is from another alkyl halide. If we start with methyl bromide, and we treat it with Cl-, we can actually swap the Br for Cl, so that means the other product in this reaction will be Br-. A couple of things about this, on the arrow I'm putting Cl-, obviously I can't just add Cl-, I have to add a counter ion, but the counter ion is not important, because it's just going to be a spectator ion, so I may or may not name it. I can say that methyl bromide reacts with Cl-, I can say that methyl bromide reacts with NaCl, it's all the same, and depending on the problem you may see it differently, so just keep in mind that just because we don't mention a cation, there's still a cation there, just like there's a salt there that we don't list here. This here is a substitution reaction, because we're exchanging one atom for another, or one group for another, it's not like an addition where we take two molecules, put them together, and end up with one larger molecule, here we're just swapping a Br for a Cl, so it's a substitution reaction, and we're going to look at this one in detail in this chapter.

Of those four, what do I expect you to know? Again, we didn't look at the mechanisms for number two and number three, but I do expect you to know that those are two additional ways to make an alkyl halide, so you could have things like Ethyl bromide could be prepared from which of the following? Well, we saw that we could make it from an alcohol, so ethyl alcohol would be okay, we saw that we could make it from an alkene, so ethylene would be one possible answer, so that's at the level that I expect you to know it until we look at the mechanisms of these things in more details. Same thing for the reactions, that's how we make it, now what do you do with alkyl halides? They're actually really useful compounds, as we'll see why we're spending so much time on these mechanisms.

There are two reactions we can do. First thing is an elimination reaction, and that's actually just the reverse of what we did when we made an alkyl halide with an alkene and HBr, so I can reverse that process, let's say I start with this ethyl bromide, and all I really need to do is add a reasonably strong base. As an example, hydroxide, OH-, but many other bases, including amines will do the same thing, so basically what we're going to do is reverse the reaction, we're going to lose Br in this case, so we'll end up with the alkene, so there's the Br going backwards. This is another reaction that we'll look at in detail in this chapter.

Let's focus on what we're doing here, we're looking at reactions of alkyl halides, one of the things that we can do is we can create an alkene, another thing we can do with an alkyl halide is I can repeat the substitution reaction, so again if I have any alkyl halide, I can treat it with Br-, and I can do that swap again. Alkyl halides are actually really useful intermediates, these are just two of many many reactions we can do, they're really a good start, because these mechanisms that we're going to look at will be very general for lots of stuff that we're not going to explicitly do here. Let's say that we want to consider this reaction in detail the same way that we looked at the free radical halogination. What we want to do is look at the mechanism, how it actually goes from the left to the right, because if we don't look at the mechanism, if I draw another alkyl halide and I ask you for the products, you can take a guess, but you won't be able to predict with certainty. Once we knew the free radical halogination mechanism, you know what would happen just based on primary, secondary terstry[sp], and based on the stability of the radicals, so that's the level that we want to get at. One of the ways to understand the reactions is to do kinetics, to see how long it take for the two reagents to react with each other, so let's do that with this reaction. What we're going to do is I'm going to draw this a slightly different way, and I'm going to draw, instead of putting it on the arrow, I'm going to make it a little more clear. That's the reaction we're looking at, so if we look at the right of the reaction, one thing we can do is we can change the concentration of the reagents, and see what happens to the rate. So if we know what happens to the concentrations of these two, we can predict how long the whole reaction is going to take. The rate will be equal to a constant K, which you calculate experimentally by doing a couple runs, and it's found experimentally that it's related to the concentration of the methyl bromide, but it's also related to the concentration of Cl-. You guys familiar with these brackets? That should be straight out of first-year chemistry, right? So what's the order of this reaction? Second order. So the order of the reaction is, if it's dependant upon the concentration of two reagents, it'll be two, if it's dependant on one, it'll be one, if it's dependant on the square of one of the reagents, it'll be two, right? Basically this is a second order reaction and it tells us something very important, that basically if I double the concentration of the Cl-, the reaction will double, if I double the concentration of methyl bromide, the speed will double, if I double both of them, the reaction will quadruple. So if we're looking at the rate, if you remember the free radical halogination, it wasn't every single step that actually determined the rate of the reaction, right? There was actually one step in the mechanism that determined the rate, and that's all that you can really measure experimentally, is the rate of the rate-limiting step, so what this is telling us macroscopically, is that in the rate-limiting step, both methyl bromide and chloride are involved, right? Because if methyl bromide was part of the rate-limiting step and chloride wasn't, then chloride would have no effect on the speed of the reaction. So I'm going to rewrite this to say specifically what the conclusion is. So from these experimental results, we conclude that both methyl bromide and Cl- are in the rate-limiting step.

It turns out that this one is a lot simpler than free radical halogination, there's only one step, and it turns out to be the rate-limiting step, right? It works like this. The methyl bromide, we learned, was polar, it's got a polar CBr bond, so we know from that that the bromine will have slightly higher negative charge, and the carbon will have a slight positive charge, which is important. The Cl- has a bunch of electrons; specifically it has four on lone pairs. One of those lone pairs is going to be attracted to the slightly positive center on the carbon, so when we draw an arrow, I'm starting the arrow on the lone pair, remember when I start an arrow it always has to start at the source of electrons, and it has to end up where the electrons are going, so the electrons end up directly on the carbon atom, that's what they're attracted to. Now if I didn't draw any more arrows, then my carbon would end up having five bond, so-called texas carbon, you never want to have that, so what has to happen at the same time as I'm forming that bond, is I want to break a bond, I want to break the CBr bond, and the way I show that is by saying my electrons will start out at the bond, and will end up on the bromine, like that, so simultaneously, as the Cl- is coming in, the Br- is leaving; at no point does my carbon have five bonds, it's going to have three bonds plus two partial bonds, which is okay, we just can't have five full bonds on that carbon. The result of this is that I end up with a Cl, my hydrogen's, and Br-. That's the whole mechanism. This is a substitution reaction, and it's called a nucleophilic substitution reaction. Substitution we already saw, because I'm exchanging one thing from the other, and it's nucleophilic because, very differently from the free radical halogination where it was all free radicals, here we actually have charges, we have Cl- coming in, and there's actually two reactions we have coming in, it's not one electron reactions. So in this case we have one species that is seeking a nucleus, seeking a positive center, that would be a nucleophile, so in this case a nucleophile, which is generally electron-rich, is at least going to have at least one pair to be able to do the attacking, that will be the nucleophile. Phile means to like, nucleophile means to like a nucleus, the reason for that naming is that nuclei are positively charges, so you want to translate nucleophile to seeking positive charge, or slightly positive charge. On the other hand, I have the electrophile, that loves electrons, is looking for electrons, more specifically a lone pair, so that's why this reaction is called nucleophilic substitution, but we have to make one more refinement here, it's a nucleophilic substitution reaction that is second order, so if we want to completely name this reaction so every chemist would know what we're talking about, we would call this S for substitution, N for nucleophilic, and two for second order. That basically for now is what we need to know about the SN2 reaction. The electrophile is actually the whole molecule. The carbon would be the electrophilic center, it would be the point of attack, but if I asked you to point out the electrophile, it would be the methyl bromide. The full arrow means the movement of two electrons.

This is a little more useful, you know a little more about the reaction, and you can start to make predictions, but you will see that your prediction will fail very quickly. Let's take a look at another molecule and try to use the same mechanism. If I try to do the same thing with t-butyl[?] bromide, react it with Cl-, I can in fact substitute the bromine for the chlorine and end up with Br-. Now if I do the kinetics on this, though, something strange happens, I get a very different result than I did the previous time. The rate of the whole reaction, in other ways the rate at which I observe the t-butyl[?] chloride forming is equal to a constant times the concentration of t-butyl[?] bromide, and that's it. In other words, it doesn't matter what the concentration of the chloride is as long as I have some minimum amount, because if the concentration is way too low, obviously I can't do the reaction, but if I have a minimum, it doesn't matter what the concentration of the chloride is, which is very different from the previous reaction, so this reaction is not second order, this is first order, and it tells us that in the rate-limiting step, only the t-butyl[?] bromide is present, so that tells us we absolutely have to have more than one step in this reaction. Let's take a look at what could actually produce this, like I said we have to have at least one step, luckily we have two steps, so there's a lot less steps than free radical halogination. In the first step, what happens is the t-butyl[?] bromide, all by itself, will decompose heterolyticly[sp], so this arrow is saying take those two electrons from that carbon-bromine bond, and but both of them on the bromine, and leave none on the carbon. So if there're no electrons left on that carbon, I have to have a carbyl cation, I have to have the C+, and I have the Br-. In the second step, this carbyl cation will then react with this Cl-. Again I start at the lone pair on the chloride, and I end up on the carbon that's positively charged. So overall, it's basically the same thing, it's just that the SN2 reaction happens in one step, and this, maybe you can guess, would not be an SN2 reaction, it'd be an SN1, because it's first order. Everything else is pretty much the same, we have our nucleophile the same, Cl-, we still have the electrophile, which is the alklohalibe, the nomenclatures very similar, everything's very similar, but as we can see, this kind of thinking will cause a problem. In general, as a chemist, we want to do SN2 reactions, because we can see SN1 reactions have some pretty nasty side products that'll come from it, because we end up with this carbyl cation. That's the overall picture; we'll look at that in a little bit.

What's different between those two compounds that one is going to go by SN1 and the other is likely to go by SN2? It comes back to our primary/secondary tersery[sp]. One is a tersry[sp] alkyl halide, this one right here; this is the rate-limiting step. Again, to compare these two, we have a tesry[sp] alklhalide that's doing an SN1 reaction, and with alkyl bromide that was actually beyond primary, so I use those as two extreme examples, so what we can do is actually write out a little page. So we have a spectrum, and what we can say is that on the left hand side, we're going to have a tendency towards SN1. All the way over on the right side, we would have SN2. Now, in the case of secondary, you would expect that you would have competing SN1 and SN2 in a lot of cases. Now I have to stress that these are tendencies, so what that means is that if I have a compound, I can make you do an SN1 or an SN2 on any alkyl halide, it doesn't mean that it will be the dominant reaction, but still if I ask you what the tendency is, for example if I draw you t-butyl[?] bromide, you would look at it and expect it's probably going to undergo an SN1 reaction, but any question I can definitely ask you to do either one. You'll see that in the quizzes as we go through them. So why is it that we have this tendency? We can actually go back and understand that if we look at what the intermediates are. In case of the tersery[?] alkyl halides, you'll notice that your intermediate is a tersery[?] carbyl cation. Tersery[?] carbyl cations and radicals have a very similar stability. So that's one reason that basically your tersery[?] will favor SN1. In other words, if I had methyl bromide, and I tried to do an SN1 reaction, it would be a lot harder because that methyl cation is relatively unstable when compared to the t-butyl. The other reason has to do with sterikindrence[sp], which we looked at a little bit with Newman projections, with the cyclohexane folding, and it just means that two groups don't want to be in the same place at the same time. If I'm trying to do an SN2 reaction, you'll see that the nucleophile has to come in intimate contact with the electrophile. It actually has to get to that carbon before the reaction can happen, so if I don't have big groups that are around the carbon, it's going to be much easier to do that SN2 reaction, so if I try to picture doing an SN2 on t-butyl bromide, look how different it is... So relative between these two, there's little setrokindrence[sp] in the case of a primary or a methyl alkyl halide, so this would be a primary or methyl, and in the case of tersery[sp] the Cl- is going to have to negotiate those methyl groups in order to make it to the center to that molecule, and that can be difficult, so Cl- is bouncing around. The probability that they get right between those three groups and manage to meet the center of the carbon is far less than in the case of the primary or methyl.

Those are the two basic reasons, and they both go in the same direction, so they're additive, so there's basically two reasons why your primary or methyl alkyl halides will do SN2, but again you should be ready to do either one if you're asked to. When we do the problems, also we'll take a look at these. There will be two molecules, and you'll be asked which one will do an SN2 reaction faster or slower. So SN1 and SN2 have a lot of different consequences and one of them will also be with chirality, so what happens when we do SN1 and SN2 with chiral compounds? In order to understand what happens, we have to look at the details of the SN2 reaction, so I'm going to go back and draw a transition step for that, and then we'll see if we can see what actually happens. I want to draw everything in one line, I want to draw the nucleophile on the left and the electrophile on the right, and this is an SN2, and I'm going to end up with Cl on the left. Drawing a transition state, all we do is we look at the average of the left and the right, the energies here are pretty similar, so basically we can assume that the transition state is going to be in the middle, so your transition state is going to have an equal length of bonds, so Cl- comes in, attack, Br- goes out. I'm going to have a partial bond between the chlorine and the carbon in my transition state, and I'm going to have a partial bond between the carbon and the bromine as well, because I'm going from a full bond to zero bonds. What happens to the other three groups is kind of interesting because if you notice on the right, it's almost like they went over to the other side, and that's exactly what happens actually. Chlorine is coming in on the left, and this is an sp3 hybridized center, so what happens at that transition state, those three groups actually end up all lined up at the middle, so they could go from the right side, and flip flop over to the right side. One more thing about this transition stat is that we have to worry about charges, so the charge on the chlorine would be what in the transition state? We don't need to know exactly, but it's going to be somewhere between minus one and zero, so any transitions that you ever do, you do it the same way. You basically look at where it starts off and where it ends up, and in the middle we assume it's going to be something between the two, but we don't have to say necessarily that it's negative one-half, we just say that it's slightly negative, for the same reason that we say the bromine is slightly negative, but the carbon doesn't at all, because it doesn't go from plus to minus, it goes from neutral to neutral. So this is a transition state, and what happens is that those three groups in the middle flip from one side to the other. It's very similar to what happens to an umbrella when it gets inverted in strong wind, you've got this pyramidal kind of structure, and then it flip flops over to the other side, so this is called an inversion, named after the guy who discovered it, a Walden inversion. The fact that the three groups flip over for methyl bromide really doesn't matter because after you make the molecule it doesn't matter what the hydrogen's went through, you're still going to get the same compound. The time when a Walden inversion is important is when you have a chiral structure, because a Walden inversion will invert the chirality, so let's do a chiral molecule and see what happens.

So if we start out with enantiomerically pure 2-bromyl butane and we treat it with Cl-, and I specify that you do an SN2 reaction, then the Cl- will come from the back. An SN2 reaction is always a backside attack where your leaving group is coming off and the nucleophile is always coming directly behind, so what happens is we basically invert the chirality of that center. So we get a hundred percent conversion of that center. This is the first time something like this has happened, so far whenever we go through the mechanism, we end up with a mixture, this is different. This is also why SN2 reactions are so useful, because you don't get random products, you get 100% inversion, so biology uses this to its advantage. DNA is made like this, proteins are made using SN2 reactions because of the very tight control you have, and there really are very very few side products if any from SN2 reactions, so it's a really good reaction to understand carefully as an organic chemist.

Transcription by CastingWords

Lecture 018: Chirality part 3

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 10:58:00 -0700

Lecture 018: Chirality part 3

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Jean-Claude Bradley: I had a request for a question that I will start with.

[Background noise]

Jean-Claude Bradley: This is one of the ones that we skipped over. That I said I would be happy to revisit if there was a question about it. This is where we need to determine the different number of hydrogens and therefore the different number of products we would get from a free radical halogenation reaction.

Jean-Claude Bradley: So there are a lot of different hydrogens here. In fact, I hope I have enough colors for all of them. First of all the methyl groups, all the three Hs on each methyl group are obviously going to be the same, but the three methyl groups that are listed, they are all different. OK? Because the two methyl groups that are on the ring, one of them is going to be pointing up, one of them is going to be pointing down, and the ring is not symmetrical horizontally so all the three hydrogens, all of the nine hydrogens, are going to different.

Jean-Claude Bradley: In the red I have one population, and you have another population here. Then another population here. OK, so also we don't have a symmetry. This is drawn at an angle so you are kind of looking at the molecule on the side because of the two methyl groups that I just circled that are on the ring, that basically breaks the symmetry and so we have the hydrogens in the front being different than the hydrogens in the back. In green here I would draw a different hydrogen that would be different from the one in yellow.

Jean-Claude Bradley: And the same thing for the other two hydrogens that are on the ring. One of them is pointing up, the other one is pointing down. So those two will be different from each other, and the other four hydrogens will also be different from each other for the same reason.

Jean-Claude Bradley: We would have, and one more I forgot here, pointing out. OK, so we would have one, two, three, four, five, six seven, eight, nine, ten, eleven, twelve. Twelve different hydrogens. The consequence of that would be twelve different products, and let's see, how many major products would we have for this reaction? Let's say you did the reaction with bromine and the presence of light. We have twelve total products, how many are major and how many are minor? Yeah?

Student: One major?

Jean-Claude Bradley: One major, I can see more than that. You can see three? Well let's start pointing them out. So which one, which color would be one of the replacements?

Student: For major?

Jean-Claude Bradley: For major.

Student : [Unintelligible]

Jean-Claude Bradley: The yellow one?

Student : Yes.

Jean-Claude Bradley: OK, so let's see. I guess I almost have no colors here. So, this one is tertiary and so it will be major. So if you were to put a bromine there it would be one of the major products. Where else?

Student : [Unintelligible]

Jean-Claude Bradley: Which color?

Student : Right next to the green one.

Jean-Claude Bradley: Yeah. OK. So this one here?

Student : Yeah.

Jean-Claude Bradley: Yup. That one is also tertiary. And what else?

Student : The green one is also tertiary.

Jean-Claude Bradley: And the green one, OK. And everything else would be either secondary or primary, and so we would have twelve products, three major products and nine minor products. That's what happens when you start to break the symmetry, you get lots and lots of different products. Were there any other questions on that?

Jean-Claude Bradley: These three destructions are a little bit harder to do because you have to take into account perspective but it's really the same principle as doing the straight chain.

Student : [Unintelligible]

Jean-Claude Bradley: yea.

Student: The ones that you do with the straight box, because one is the axial and the one is the equatorial, is that why there are two different types like the ones that you drew over there.

Jean-Claude Bradley: Well it's a little bit hard to talk about equatorial and axial because this is not a chair confirmation.

Student : Ok, so the one pointing down [unintelligible].

Jean-Claude Bradley: What you are trying to draw, is that you have SP-3 hybridized centers at each point, so I'm putting a wedge to kind of make it obvious that it's coming out towards you. Because when you draw these on a two dimensional surface, you can have the optical illusion, some people may actually see it from a different angle and so I'm just drawing the wedge to show you that I mean that that one is actually coming out. In case you did not see the perspective the way that I was trying to draw it.

Student : So, the other ones, can you make it clearer for us, we could draw them as wedges with dashes? [Unintelligible]

Jean-Claude Bradley: Well the other ones are pretty close to the plane I would say, expect for the one that I drew the dashed wedge which is definitely going in the back. But otherwise, no, I don't think any other one, it would really?" I don't think there are any other ones that are clearly going in the back and clearly in the front. In the way that I drew this.

Student : So [unintelligible] symmetry. [Unintelligible]

Jean-Claude Bradley: When you have symmetry, right? If I didn't have the methyl groups then basically the green hydrogens would be the same as the yellow hydrogen. Because I could take the molecule, flip it over, and I couldn't tell the difference. So, anything else? Any other questions, actually, on chapters one to four?

Student: [Unintelligible]

Jean-Claude Bradley: Could I have quiet please? It's hard for me to listen to the question.

Student: Can you kind of explain how it's tertiary? I know it has to do with the carbon...

Jean-Claude Bradley: OK, tertiary. What I'm looking at when I am saying that this hydrogen is tertiary. Let's say I'm looking at a yellow one, I'm looking at the carbon where that hydrogen is connected and I'm counting how many carbons are directly connected to that carbon. Let me see. I'll use another color here. If I'm looking at the yellow one, I'm looking at that carbon and I count one, two, three carbons.

Student: [Unintelligible]

Jean-Claude Bradley: Yes. That's the definition of primary tertiary, secondary tertiary. Anything else?

Jean-Claude Bradley: All right. We had left off with, let's see, looking at compounds that have two chiral centers in them and we had talked about diastereomers and entromers, and I think the one thing that we didn't mention is how you would actually name these things. So let me actually draw back a Fisher projection. And, just to be different, we'll make a slightly different molecule.

Jean-Claude Bradley: So this is a Fisher projection, so once again I'm going to draw three lines, which means equivalency. So I am going to re-draw this in a way, because we just started to use Fisher projections, kind of to remind you what they mean. All of the horizontal bonds are coming out towards you. And the vertical bonds are in the back of the board when possible.

Jean-Claude Bradley: Clearly I can't put the central carbon-carbon bond between carbons two and three, both behind the board, because it is the same bond. So when you are actually looking at a Fisher projection that has multiple chiral centers in a row, what you're doing is it's actually curved. You can think about it that way. So I can't really draw it exactly the way that it would appear if you were to make a model of it. But when you analyze the Fisher projection for R and S, you would definitely think of the horizontal bonds as being behind the board, I'm sorry, the vertical bonds being behind the board and the horizontal ones being in the front. Think of this equivalency when you have two chiral centers.

Jean-Claude Bradley: If you wanted to name this molecule, the first thing we have to do is to figure out how many chiral centers we have. And we have two carbons that have four different groups onto them, so we have two chiral centers to name. When you name them, let's say I'm naming the first one. In fact let me put a different color for these. So I've got a yellow chiral center, and I have an orange one. So if I'm trying to figure out if the yellow one is "RS??, I'm not concerned that the other one is chiral or not. Ok? When I'm looking at the yellow one, I'm simply asking, "OK I've got four different groups??, I'm going to have a certain order or priority in those four groups. The lowest priority group is the hydrogen and the highest priority group between those three would be what?

Student : Bromine.

Jean-Claude Bradley: Bromine. That one is easy because it's bromine versus carbon versus carbon. So we're going to put a one next to the bromine. Now when we go to the next groups we have a CH-3, and the other group that we're looking at is actually this entire thing here, the left, this whole fragment that is the remainder of a molecule. So if I look at the shells for CH-3, I have a C followed by H, H, H. Whereas for the bottom group I have C. That C corresponds to that chiral center, and then that carbon will be connected to three other things: an H, a CL, and a carbon. Does everybody see that?

Jean-Claude Bradley: When I'm now comparing the other two groups, the two carbons from the first shell cancel out and what I'm comparing next is the highest atom from the second shells. The highest atom from the CH-3 would be an H, and the highest atom in the bottom fragment would be CL. So I'm comparing the CL versus the hydrogen, and, for that reason, the bottom group would be number two in priority, and the methyl would be third. As I go one, two, three, I find that I am going clockwise. Clockwise is normally R, but the lowest priority group is in the front, so this is actually S. So I have determined that carbon two has an S configuration. When you guys do this on a piece of paper what I very strongly recommend you do is to copy the molecule to analyze your second chiral center, because otherwise you're going to end up with a mess, especially if you don't have different colors, it's going to be truly a mess to try to figure out what's happening.

Jean-Claude Bradley: So I am going to do that to determine the bottom one here. Well in fact I can use the one on the right here, it's not a problem. If I'm analyzing the one on the right then the lowest priority group would be H, and, using the same analysis, chlorine would be number one, followed by this whole group here that has the bromine, and then the ethyl. I am going counter clockwise, counter clockwise is normally S, but the lowest priority group is in the front, so this is R, this one. When I want to name this compound, what I'm going to do is to specify, so this is one, two, three, four, five, this is pentane. I'm going to specify that the two position is an S. I would write 2S, 3R, and then I would name the compound. So that would be 2-bromo, 3-chloro, pentane. So that's how you can unambiguously name chiral compounds.

Student : 2-S or 3-R what determines what it is?

Jean-Claude Bradley: Well you have to number from the side that will give you the lowest number. If I number from the top methyl, that would be one, two, three. If I started to number from the other end it would be three, four, which you can't do.

Student : ok.

Jean-Claude Bradley: Ok. That's basically all there is to naming these things and we'll do some more examples in the problem set. Now we're going to look at reactions involving chiral centers.

Jean-Claude Bradley: The first one we will look at is creation of a chiral center. The reaction of interest here is this. What happens when I take butane and react it with light and bromine? What's the major product from that? We saw from the last chapter that the major product would be a substitution of the secondary position. But now what we didn't pay attention to in the last chapter is that when we do that reaction we create a chiral center, because now that carbon has four different groups. It didn't in butane. Butane is not chiral, it has no chiral centers, but that product has a chiral center. Knowing what we know about the mechanism of that reaction and knowing what we know about chirality now, can we predict what the products are going to be? Are we going to get the R isomer, the S isomer, are we going to get a mixture of the R and S? Or is it not possible to predict what's going to happen? When we try to predict consequences of reactions, we're going to use the mechanism we know. We're going to step through it and see what actually comes out of that.

Jean-Claude Bradley: To work out what happens in this reaction we are going to go through the steps. The first step is always the same. It is the reaction of light --the reaction of bromine with light-- to generate two bromine radicals. The second step, the first propagation step, we will have the reaction of the bromine radical with butane. I'm only drawing the hydrogens here that I'm going to be doing a bond cleavage with. The first thing that we need to think about is that this bromine radical is approaching the butane, we already know that it's going to prefer to react with the secondary hydrogens, but now we can see that there's actually a choice of two hydrogens that he can actually extract. Is it going to make a difference which hydrogen that it extracts at this point?

Student : [Unintelligible]

Jean-Claude Bradley: Let me draw the radical and we'll take a look at the question a little bit more carefully. When I draw this radical, what is the proper way of drawing it? In other words, what is the hybridization of that radical? Of that carbon? What do you think? We never actually asked that question.

Jean-Claude Bradley: If the single electron is defined as a group of electrons it would be SP-3. But I never said that a single electron is considered a group of electrons. So, in fact, based on what we know, there's no way for you to know what the hybridization of that center is. It turns out that you don't count a single electron. Now that's something new that you know and that's going to have consequences. So that single electron is actually in the leftover P-orbital of that carbon. The radial is SP-2 hybridized, which is why I drew it this way. It's flat, 120 degrees, and the radical is in the leftover P-orbital that's on the top and the bottom of that flat structure. If you have an SP-2 center you can't have chirality at that point. In other words, I can't draw this. I can't draw the groups in any order that when I take their mirror image I can't superimpose them. And the reason is that I could take it, I could just flip it over, go from one flat structure to another flat structure, and re-order the group in any way that I want. This is important because that has a very significant consequence with respect to predicting what's going to happen here. So we know therefore, from this first step, that it makes no difference which one of those two hydrogens gets taken out. We're going to get the same radical, and, furthermore, we know that that radical is not chiral. There's only one possible radical from that reaction.

Jean-Claude Bradley: In the next step, the second propagation step. So I start with that radical. Let me start labeling this stuff here. This is SP-2, make sure we're clear about that, and we are going to react it with the bromine molecule.

Jean-Claude Bradley: OK, so, I told you where the electron is; it's in the P-orbital, which is either on the top or the bottom. So, when the bromine actually comes and reacts with it, there's now a choice: the bromine can attack from the top, or it can attack from the bottom, and that choice is going to determine the chirality of the product because at that point we are going to make a chiral product. We are either going to make R or we are going to make S; so, what determines whether you make R or S is the approach of the bromine relative to your flat radical. That's where the decision happens.

Jean-Claude Bradley: So, I'm going to draw here two possibilities. And of course the - actually sorry, I made a mistake, it's HBr. Second step we get the bromine radical so, basically, at this point it attacks from the top or it attacks from the bottom and we're going to get those two chiral centers and I drew them like this out of convenience. The only thing I did is I switched the H and the Br. I could have switched any two groups on that tetrahedral center, however, this is the easiest way to tell that these are mirror images and andromers, so when I can I am always going to draw andromers like this actually this is. Unfortunately I do not have an eraser on this thing. Let's try that. Ah, very good.

Jean-Claude Bradley: All right, thank you. So I have Br on the left, H on the right. So there are my two entromers, and the question we have to ask ourselves is "is there a reason that the bromine would tend to attack from the top versus the bottom?" I have a flat structure, perfectly symmetrical; there is not any conceivable reason why the bromine would attack from the top or the bottom, so you would expect to get just random attacks. We would predict, based on our knowledge of all of this stuff, (the mechanism, the geometry) we would predict that we would get fifty percent - exactly fifty percent of one and the other, and that's exactly what we get.

Jean-Claude Bradley: Fifty percent of the R and fifty percent of the S, so this ratio of getting exactly the same amount of R and S happens very often in reactions, and so this ratio has a special definition, it is called a racemic mixture.

Jean-Claude Bradley: It just means fifty percent R, fifty percent'or, I shouldn't say that. I should say fifty percent of each enantiomer. One of the consequences of a racemic mixture is because one enantiomer will rotate and polarize light exactly the opposite direction as the other, the racemic mixture has no optical activity - ok, so there's one consequence if you have exactly the same amount of each, it will not rotate plain polarized light.

Jean-Claude Bradley: Ok, any questions about the creation of the chiral center? Ok, lets move on to the next one which is reaction at a chiral center.

Jean-Claude Bradley: So, here's a molecule that's chiral - it has one chiral center and lets say we start with this molecule and we react it with bromine in the presence of light, so, if we look at the major products, we see that there is only one tertiary center so we would expect the bromine to go on that center.

Jean-Claude Bradley: However, that's the chiral center so the question is: are we going to get a hundred percent of this or are we going to get a racemic mixture, or are we going to get inversion at that center? So, these are the options that we have. That's the question. So, we're going to step through the mechanism to see if we can predict what the answer would be if we were to do that reaction.

Jean-Claude Bradley: OK, so, again, we go through the mechanism, generate our bromine radicals. So now there is no question about which hydrogen will be attracted. In the last one we had a choice between two, now there's no choice, there's just one hydrogen so now the fact that we know the hybridization of the radical is of great importance, because that would actually change the answer, because we know that the radical is flat, we know the radical is not chiral. It doesn't matter which enantiomer we start with, we are going to destroy the chirality at this step right here.

Jean-Claude Bradley: So, left hand side is chiral, right hand side is not chiral. Now the problem gets solved in the same way as the previous one. Now again I have my flat radical. I got the bromine coming in. It has an equal probability of coming in from the bottom or the top, so again I would predict getting my racemic mixture.

Jean-Claude Bradley: The only product we would have - a little bromine radical you could generate here, so this has some interesting consequences, because what it tells you is that you start off with something that's chiral, and once the reaction is finished you have a racemic mixture that is not chiral. What makes this interesting is that you could actually monitor this reaction in a polarimeter. As the reaction proceeds, you would see that plain polarized light would initially be reflected. As the chiral material gets destroyed, it would go to zero so as you put together all these tools, that's how you start to design experiments and figure out what's going on and test out your mechanisms. We couldn't do that in the last example because there was nothing chiral to begin with or end with but here we do, we destroy the chirality. Any questions about this one? Number 2? All right, number three is a reaction next to a chiral center.

Jean-Claude Bradley: So we're using this molecule which is a Fisher projection. We want to react it with bromine and light and we want to look at the major products, so the reason we put a phenyl ring and a t-butyl group and a methyl group is we want to make sure that the major product is at the secondary center, up top here, so nothing else will be the major product except for replacing one of those two hydrogens. So the question is going to be, of course, what happens the possibilities are as follows.

Jean-Claude Bradley: ok there are those two secondary hydrogens and we can imagine replacing either one of those, and the question we want to answer is: do we get one completely and not the other? Do we get a mixture of them? If we get a mixture can we predict what the ratio is going to be between them? So, let's analyze it the same way. Yes?

Student: Is it [unintelligible]

Jean-Claude Bradley: Ah, is it?

Student: [Unintelligible]

Jean-Claude Bradley: So, it isn't, right? So you stole some of my thunder there, because that was one of the questions at the very end. You're right, they are not mirror images so they are not enantiomers, and that is actually going to be a very important consequence when we start to look at the ratios of these things. Let's do the reaction.

Jean-Claude Bradley: The first step is to generate the radicals. Now normally when you draw a Fisher projection you have an intersection, you have a chiral center. In this case you'll notice that the one with the two H's is not chiral, but I drew it that way because it is going to become chiral. So you rarely would use that unless you had a good reason to, and in this case we have a good reason because that is where our chiral center is going to be and there is only one chiral center in this molecule. Ok, so the bromine radical comes in.

Jean-Claude Bradley: It has a choice between those two hydrogens. Let's ask the question: does it matter which of those hydrogens it takes off? It doesn't, and the reason is because the radical will be the same.

Jean-Claude Bradley: Ok, so the bottom part here is the same. Now the top part, we have to be careful how we draw it because it's no longer SP3 hybridized, so it's no longer an intersection in the fisher projection; it is now going to become flat, so I have to draw it flat, and I got a radical, so plus HBr so, the next question is going to be for step number 3. I'm going to take that radical, so I'm going to have to use a new page here.

Jean-Claude Bradley: All right, so now we are going to attract this, and what we said in the previous two examples was the radical is flat. It's not chiral, so when the bromine comes in it has an equal probability of attacking from the bottom or the top. Is that still the case here? Is there an equal probability of the bromine coming in the top or the bottom? Anyone have an idea about that? It is not exactly the same scenario. When your radical is flat it looks absolutely identical from the bottom and from the top. It's not chiral, remember that, it's not chiral. This thing is chiral, the chiral center may not be at the radical, but it is chiral, so that means that when its going to be, when its going to twist to its lowest energy confirmation, that is going to look different from one side than the other. You got a big phenyl ring there, you got a big t-butyl group that's going to block the attack more from one side than the other so because this radical as a molecule as a radical is chiral, I am not going to get 50-50 mixtures here. I do know that I will get a mixture, so I will likely get a mixture but I can't predict based on what we know right now what the ratio is going to be between the two. The best I can say in this example is that we get a mixture of diastereomers.

Student: [Unintelligible]

Jean-Claude Bradley: Well, it's something that actually may be very difficult to draw on paper in order to actually answer the question, you'd have to make a molecule model and twist it around in such a way that you could see that the steric interactions are minimized. Remember, you've got a lot of single bonds here; the Fisher projection is a convenience: it doesn't tell you the way the molecule actually looks in 3-dimensional space. It's just a convenience. All it tells you is that you have a chiral center down here, you have a radical up here. I'm not trying to show you that one of these is easier than the other. What I'm trying to explain to you is that it looks different from the bottom and from the top because the molecule is not symmetrical about its mirror plane. You don't have to understand exactly how it's different, just that in principle it's different because its chiral, it doesn't look the same way one way as the other so, that's basically it, we are going to move on to the problems now. Next class, I want to make sure, the test starts on Friday make sure that you read the syllabus and the FAQ because I have been getting some questions that have been causing me some concern. There are not a lot of things to worry about, but definitely make sure you know what rooms you're going to, and when the test is. I won't accept as an excuse that you didn't read the FAQ so you didn't know where the rooms are, but if anything is unclear let me know, and I will be happy to answer any questions.

[Sounds of people leaving]

Transcription by CastingWords

Lecture 017: Chirality part 2

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 10:58:00 -0700

Lecture 017: Chirality part 2

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Jean-Claude Bradley: We determined last time how to determine if a carbon center is chiral or not. It needed to have four different groups, if it has four different groups; we know we definitely have a chiral center. One thing we did not talk about was how to actually name these things, so that's what we're going to start off with.

When you're drawing a chiral center, you don't have to do it this way, but it really helps a lot if you use this kind of projection, where the horizontal bonds are coming out towards you, and the vertical bonds are in the back. I'm going to draw a chiral molecule at random here. The only thing I have to do is draw four different groups, and I'm guaranteed to have chiral molecule. I'll put an H here, I'll put an F, Cl, and a Br. We know immediately that this is definitely going to be a chiral molecule; it has a chiral center, and when we talk about chiral centers, generally put a little star, if you look through the chapter you'll see that that's the way that they're referred to.

If we wanted to have this compound and do something with it, we'd have to find it in a lab, we'd have to order it, we'd have to have some way of actually naming it, so what we're going to do is find a way to name all these chiral molecules so that you can talk about them. What you'll notice is that because we have four different groups, we can differentiate these atoms depending on what different groups they have and the order in which they appear. What we're going to do is actually prioritize the groups based on their molecular weight, so we're looking at the molecular weight of the atom that is directly connected to the chiral center; not the whole group, but just the atom. This is the simplest example, all I have is atoms, but later on we're going to have actually groups that have multiple atoms, and we'll have to deal with those a different way, but the first thing we're going to be concerned with is the molecular weight of each of these atoms. What you generally do is look for the lowest priority group or the lowest atomic weight group, or actually atomic number in this case, and put a box around it. So hydrogen is easy, because nothing is going to have a lower weight than hydrogen, so that's going to be your lowest priority. For the others, we're going to look at the highest priority and go down, so the highest priority between bromine, chlorine, and fluorine, is what? Bromine, and then chlorine. So we just go down our little periodic table that I asked you to memorize. Bromine first, then chlorine, then fluorine. You'll find that you will always be able to draw a circle that is going either clockwise, or counterclockwise, and in this particular case, we're going counterclockwise.

The definition to name these is that when you're going counterclockwise and the lowest priority group is in the back, this will be called S. 'In the back' means that it has the bond with the dotted lines that tells you it's in the back of the board. Something you'll notice about the way that we draw this sp3 hybridized carbon is that there's no bonds that are in the plane of the board, they're either in the back, or they're in the front. That's part of what makes this useful. When you have a group that's neither in the back or in the front, you have to start to do some things so that' you're able to put the lowest priority group in the back. If you draw your chiral groups like this, you're never going to have that problem. Counterclockwise is S, and when the lowest priority group is in the front, we're going to call that R, so let me draw the mirror image of this molecule...

This is just the diastereomer of the previous molecule. Lowest priority group is in the back, I put a square around it, you start up at bromine, then chlorine, then fluorine. We're going clockwise. That's the definition that you'll find to be consistent through all the organic textbooks that you're going to pick up. Any questions on this first part?

Let's do an example where the lowest priority group is in the front, and let's see if we can handle that. Let's say that I redraw this molecule, and this time I put the H in the front, the lowest priority group. So if you proceed to analyze this, there are two ways to solve this at this point. What you can do is, you can rotate the molecule so that the lowest priority group is in the back, and that's a little bit difficult for some people to do. Some people are really good at rotating three-dimensional objects, some of you don't. The method that I will give you will work for everybody. What you do is still determine everything the same way you did the other, put a square on top of the lowest priority group, and still order the other groups, so bromine is one, fluorine is second, chlorine is third... We are going counter clockwise, and the way that we talk about this is, counter-clockwise is normally S, but the lowest priority group is in the front, therefore the molecule is R, and the reason that I do that particular logic, is that again, if you look in any organic chemistry textbook, you'll see that the counter-clockwise always corresponds to S, and I want that association to be held constant so that you don't get confused. So this is counter-clockwise, so the logic is counter-clockwise is normally S, but the lowest priority group is in the front, so it is R.

So we'll do several more examples to see how this works. If you change the chlorine and the fluorine, then you would get clockwise, and then it would be S. Actually, if you ask a question like 'What happens if you switch the chlorine and the fluorine?' Any two groups that you switch, you will automatically create the mirror image, and so if you've already determined that this is R, then if you simply switch two groups, automatically the molecule that you draw will be S, you don't need to redetermine it. That will always be the case. And we will use that to speed things up when we do analysis.

All you want to do is switch the chlorine and fluorine. So now we're going one, two, three, now we're going clockwise. The lowest priority group is the same in each case. The lowest priority group you can think of is four, and you don't include that in the circle; the clockwise and the counter-clockwise only refers to one, two, and three. When I draw these, they're going to overlap the square, but I don't use the square to draw the circle.

So what happens when we have multiple atoms? First thing is you have a molecule here, you have 1sp3 center, and you have four different groups, so we can immediately say that this is a chiral center. Now we want to determine if it's R or S. We are going to put a square around the lowest priority group, and then we're going to compare the other three groups to each other. Remember that it is not the groups we are concerned with; it is the atoms that are directly connected to the center. Now in the case that it is F vs. C vs. C, it is clear that F is going to be higher priority than the other two groups, so F will be number one. Now, between the C and the C, those are equivalent at this stage, so what I need to do at that point is to use this concept of shells, which is I'm going to keep going outward into the groups until I find a difference between them, and at that point I'm going to make an assignment as to which is a higher priority, and one way to do that is to draw brackets, and to list the shells separated by commas. In the first shell for methyl, I have a carbon, comma, and then I ask myself 'What are the three things that are connected to that carbon?', and the answer is H, H, and another H, for methyl, so now I'm going to do the same thing for this CH2I group. The first shell is the atom directly connected to the chiral center; that will be a carbon, comma, three things are directly connected to that carbon, in this case it's H, H, and I. Okay, so when I compare these again, I'm not looking for averages or sums, I'm actually looking for the highest priority atom in each one of those collections. So in the first shell, carbon vs. Carbon, those two cancel out; in the second shell, the highest atom in H, H, H is an H. In H, H, I, the highest atom is an I, so the only thing I'm comparing is an I vs. An H, and so basically I and H, the I will win out. That will make the CH2I the second priority group, and the CH3 will be the third. So as I go around, I find that I am going counter-clockwise, and the lowest priority group is in the back, so that means it's going to be S.

That's basically all there is to it. Sometimes you've got to go 3, four shells until you find that difference, that just means you need to be careful and check what's going on, but certainly the largest group is not the highest priority group. Like in this case, the CH2I is much bigger than the fluorine, but it's not a higher priority. It even has an iodine in it; that doesn't make a difference because the iodine is in the second shell and not the first. That way all the chemists can agree on a standard, and now when you name things, all you need to do is put an R or an S in front of it, and everyone will understand what you're talking about.

One of the reasons that I used the representation that I did with the horizontal bonds coming out toward you and the vertical bonds being in the back, not only is it convenient for finding R and S, it's also an easier way to make the other projection that we call a Fischer projection. You remember the Newman projection? It's got a circle; it's got three bonds in the front, three bonds in the back. If you're not taught how to read a Newman projection, it doesn't make any sense; it's the same thing with a Fischer projection. Fischer projection is always going to be in the form of a cross, or a multiple of crosses on a vertical line, like this. In a Fischer projection, you never put a carbon in the middle, so it's not like a skeletal formula, you don't have a choice of putting a carbon in the middle, you always have to put it as an X, and you can't tilt it, so if you only have a skeletal formula, there's no rule that says you can't draw it at any angle you want. In a Fischer projection, you must draw it exactly like this, vertical and horizontal, and a Fischer projection, it is implied that the vertical bonds are behind the board, and the horizontal ones are coming out towards you. If we draw this molecule, and we say it's a Fischer projection, then it's really the equivalent of drawing it like this.

If you have one chiral center, it's not really that big of an advantage to use a Fischer projection. Where the Fischer projection comes in really handy is when you have things like this. When you have multiple chiral centers in a molecule, Fischer projections are very useful because it'd be very difficult to draw a molecule like this showing all wedges going in, going out, and this is used for a very specific application in biochemistry. What kind of molecules have lots of chiral centers? Not proteins, sugars, yes, carbohydrates. Carbohydrates are really a pain to draw unless you have something like a Fischer projection, where all you need to do is draw a line, put a couple of horizontal lines, and you're done, pretty much. So that's the kind of projection we're going to use usually if we have more than one chiral center. Any questions on that?

The definition of Fischer projection is that all the horizontal ones are in the front, and that's it. That's why we have to agree on what it means, otherwise I couldn't draw a molecule like I did with the bottom and have anyone know what it is I'm referring to.

In this case, there's no Sys or Trans because there's no alkene, and there's no cyclic alkene, so sys and trans has no meaning in this particular projection. The Fischer projection is just a convenience, it's not like the cyclohexane conformation, that's actually the way that the molecule looks. In a Fischer projection, the molecule will look very very different from this, actually.

Let's start to use these Fischer projections in a practical way. Let's pick a molecule that has two chiral centers in it. Actually the molecule I'm going to draw all the isomers for is 2, 3, dicholorabutane. If you take a look at the quiz, you'll see that there's some questions that are related to this kind of processing. You have 2, three dichlorabutane, the way that I drew it a the bottom is I didn't draw it with any regard to chirality. You can't tell anything, basically they're drawn flat, it's just to see the connection of the atoms. At the top left, I drew a Fischer projection. That tells you at no uncertain terms what the chirality is of those two chiral centers. I've got two chiral centers in that molecule, because I have two sp3 hybridized carbons that have four different groups attached to them; two chiral centers.

One of the things that you'll need to determine is how many different isomers can you have like 2, three dicholrabutane? Fischer projections are a great way to do that efficiently. What you do is you draw a Fischer projection, and you only draw a cross where you have a chiral center. I've got two crossing points. Then, you switch left and right, the two groups that are on the horizontal bonds, until you generate all of the possibilities. I've got two groups, two times two is going to be four, there will be four different ways in which I could draw this. I could draw the two chlorines on the right, or I could draw the two chlorines on the left, and when you start to do this with Fischer projections, you can't change what's on the vertical axis, so if you decide that you're going to put the methyl's on the top and the bottom, and the chlorines on the horizontal bonds, you can't change your mind after and now put the methyl's on the horizontal bonds, because you will not be able to analyze the problem properly. Initially, there's nothing really that tells you that you must put the methyl groups vertically. Convention says that you probably put the alkyl groups on the top and bottom, but that's certainly not an absolute. It wouldn't put a wrong Fischer projection had I put the chlorine on the top and the methyl on the horizontal ones, but once I start, like in the Newman projection; I'm committed to finish the problem that way. I have the two chlorines on the right and the two chlorines on the left, the other two possibilities are as follows. I could put one chlorine on the left, one chlorine on the right, and then the opposite. If you look at the number of combinations we can make here, that's it, there's no other ways that I can mix up those two chiral centers. The question now is how many different molecules do I actually have on this page? How many molecules here are mirror images on each other? There's two pairs, I drew them in such a way to make it easy for you to see that. That's another really really nice thing about Fischer projections; it makes it really easy to see if you have mirror images or not because the vertical groups don't change.

We have two sets of mirror images, and we know we have to be careful about mirror images, because it doesn't tell you that you have enantiomers, it tells you either that you have enantiomers or the same molecule. Is there a case here where we might have the same molecule? How do you know that the first one is the same molecule? By definition, you can superimpose it, but how can you tell, just by looking at it? Are you doing a translation, a rotation? With a Fischer projection, another really nice property is that you can rotate it 180 degrees by keeping everything in the plane. The rotation that we're talking about here is this. If I rotate, follow that arrow here, and rotate 180 degrees, the top methyl group will end up at the bottom, the bottom methyl group will end up at the top, and the two H's on the right will end up on the left, the two chlorines on the left will end up on the right, and because you're not flipping through the plane, all the bonds in the back are staying in the back, and all the bonds that were in the front are staying in the front, you don't have to worry about that. The only kind of rotation you can do with a Fischer projection is the rotation that I'm drawing here. Don't take anything in and out of the plane, just rotate it, turn it upside-down basically. And if you do that you can see that I can superimpose the structure onto it's supposed mirror image. Everybody see that?

If you try to do the same thing to the other two, you've got a problem, because although the methyl groups match, the Cl group that is on the right here would end up on the bottom left, so if I rotated that fourth structure on the right, it would end up being the same thing. I could rotate it forever and never generate the third structure. That's a real advantage of using the Fischer projection.

What we know from drawing these, then is that we have actually three molecules. The first two possibilities that I drew are actually the same thing, that counts for one, and then we've got two, and then three, so one of the things that you need to be able to determine is how many different compounds do we have for 2, three dicholrabutane, the answer is three. There ought to be a couple of quiz questions to help you do that for some additional molecules.

The other thing that we have to do with this is name the relationship between the molecules that I drew. So the two molecules on the right that are mirrors of each other, but which are not the same molecule, are enantiomers. That's the definition of a enantiomers, that they are mirror images, so that definition doesn't change. Remember that the term diastereomer is a relationship, so a molecule is an diastereomer of another molecule, or I can say that those two molecules are enantiomers. So the one new things we have here is I have a molecule that is different chirality-wise from another one, but they're not enantiomers. The second structure I drew, and the third structure are not mirror images of each other, however, they have the same connectivity, so that's a new relationship, that's a new word, those two are diastereomers.

There'll be consequences of that fact. What we know from enantiomers is that enantiomers will rotate in polarized light, and they'll rotate it in opposite directions. We know that the third and the fourth compound have the same melting point. Sometimes I might ask you the question with a couple things that you have to integrate, like I can ask you do these compounds three and four have the same melting point, and to answer that, you have to understand that they're mirror images, that they're enantiomers, and one of the properties of enantiomers is that they have the same physical properties, so they have the same melting point. What we also know is that the first compounds and the third or the fourth compound, the diastereomers, those are not mirror images, and so they will not have the same physical properties; they won't have the same melting point, they won't have the same boiling point, they'll have different solubility's, they have different diapole movements... They're similar, they're pretty close, usually, but they're not going to be identical, and for that reason you can chemically separate diastereomers much more easily that you can enantiomers, so if you were to do a reaction and you had mixtures of those three compounds, it's generally much more easily to separate diastereomers from each other, because you could do things like crystallization or distillation, things like that, they would have different properties. You would have to use some other tricks to separate the enantiomers.

There's one more piece of nomenclature that we have to look out. The compound on the left, that I drew the mirror image, is that a chiral compound? It's got two chiral centers; however, its mirror image is super imposable upon itself, so although it has two chiral centers, it is not chiral. It's a little bit counter-intuitive at first, but if you think about it, this is how it's possible. Because this is a special molecule that has multiple chiral centers, yet it is not chiral itself, that's called a meso isomer. It's called a meso isomer, or meso compounds, but that's a special term you use whenever you have this kind of situation. Meso is not a relationship term, it's not like enantiomer, it's just a name, so I can say what is the meso isomer for this compound, and this is it.

Transcription by CastingWords

Lecture 016: Chirality Intro

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 10:57:00 -0700

Lecture 016: Chirality Intro

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Jean-Claude Bradley: Chapter five is about chirality, where we're going to be looking at mirror images of molecules.

So, if we start with methane. There are actually a couple of different ways that I can represent a tetrahedron on a two dimensional surface. We saw one of them when we did the dipole moment, and I used a particular way to represent it because it was easy to draw vectors, and to see how they added up or canceled out. But in this chapter I'm going to use a different representation that will serve me better, and that's something I'll try to use consistently, whenever I want to talk about chirality, and that's if you have two bonds, the vertical bonds are in the back of the board, and the horizontal bonds are coming towards you. It's a really convenient way to draw an sp3 hybridized carbon. OK, so there are no bonds in the plane of the board, they are either all coming towards you or away. If you have a model kit that you want to prove to yourself that is the case, you'll see very quickly that a tetrahedron really works out like that.

The reason that it's more useful to use that in this chapter is that we are going to be talking about mirror images, so if I want to take the mirror image of this molecule, we can draw a mirror plane and reflect the atoms across it, and I can see that it's really easy to do that. The hydrogen that is closest to the mirror on the left side will be closest to the mirror on the right side, the one that's farthest away will be farthest away, and the top stays on top and the bottom stays on the bottom.

So by taking the mirror image of CH4 I can see that it's identical. I can take that mirror image, and I can see that obviously it's the same on the left as the right. So CH4 does not have a difference in it's mirror plane, there;s no difference at all.

Lets see, as we start to add complexity to this, how that's going to change.

So if I add a single chlorine, I have chloromethane, and then I reflect everything around it, I can see that again it's exactly the same thing, the left and the right are absolutely identical. I can say that I can superimpose the mirror image on the original molecule, so there is no difference in the mirror images of these two molecules.

So if I add a fluorine to this, I can use the same exercise and see that now the mirror images are also the same.

Now when I add a bromine to this, now something interesting happens. Because when I take a mirror image of that compound on the left I can't superimpose it with it's mirror image. In other words I can't rotate it in any way to make all of the atoms line up. If I try to, for example, rotate it, turn it upside down, to make the Br and the H match up, I'm going to have a mismatch between the Cl and the F. So any two atoms that I line up, are mismatched with the other two. So these are clearly different, they are different molecules, and these are what we call chiral, they are chiral molecules.

We have a new kind of isomer here, they are absolutely identical except in a mirror plane, so that's going to have special consequences that we are going to investigate.

There are tricks to figuring out if a molecule is chiral or not without always having to draw a mirror plane. One of the things you'll notice is that in all the four examples I did, this is the first example that I had four different groups on an sp3 hybridized carbon. So it turns out whenever you have four different groups on an sp3 hybridized carbon, you're going to have a chiral center, and the molecule will be chiral. So that's four different groups on sp3 carbon.

[inaudible question]

Jean-Claude Bradley: OK, let's take a look at that..So what do you want to do, the top and the bottom are that same?

So, are those two molecules superimposable? They are superimposable because I'm allowed to rotate any way I want. And another really convenient way of drawing molecules this way, is that it enables me to rotate 180 degrees without moving groups in and out of the plane. So if I take the second molecule that I drew, and then if I rotated it 180 degrees, what will happen? The top H will go to the bottom, the bottom H will go to the top, and the F that's on the right would now go to the left, and the bromine would go to the right. So in this case, you could have predicted that easily by seeing I've got two hydrogens, I've got two groups that are the same, therefore this is not chiral, end of story.

[inaudible question]

Jean-Claude Bradley: Well the way that you drew this projection, you can't flip the groups in and out of the board, because then you're never going to be able to superimpose them. So really the only thing that you can do is to start to rotate, and keep everything the same, then your bonds are going to line up every time.

Anything else?

So that's basically a little introduction to chirality. So let's also give a definition here. When I do have a chiral molecule- remember the definition, that a chiral molecule is one where you cannot superimpose it's mirror image. Now when I'm talking about determining a chiral center, a chiral center is one in which I have four different groups on an sp3 hybridized carbon. So those are the little subtleties, and we'll see actually how those subtleties will play out when we have multiple chiral centers. But the only definition that counts for a chiral molecule is taking the mirror image, that's it. If you can't superimpose it it's definitely chiral.

Ok, so now we have a set of isomers that are new. They are different from the structural isomers, they are different from the geometric isomers that we saw with the cis and trans. They have a special relationship, these two isomers, they are called enantiomers. So to have an enantiomer all I have to do is have a mirror image of a molecule that is not itself. That's the biggest challenge when you're doing these problems is, if you have two molecules drawn and they are mirror images they can either be enantiomers or they could be the same molecule and you haven't rotated it carefully to make sure that it's not the same molecule.

Enantiomers have very similar properties, because if you think about it, if the only difference is in a mirror plane, then all the bond lengths are going to be the same, all the bond angles will be the same, and therefore all of the energies are going to be the same. So if I have one enantiomer and it's pure, and I'm determining a melting point or a boiling point, it's going to have absolutely the same properties as it's enantiomer. Because that's it, if they are just living in mirror worlds they are going to have the same melting point, the same boiling point, and a lot of the same physical properties.

So they are going to have the same properties unless they are interacting with something else that's chiral, and that happens actually quite a bit. It's not going to happen if I take a Bunsen burner, heat is not chiral. But there are some things that are chiral. For example, a lot of drugs have chiral centers, and if you inject one enantiomer, it's going to have a completely different effect than another enantiomer. The reason for that is that we are chiral. All our proteins are chiral, everything, basically, that's biological is based on some chiral building blocks. So when you have the molecule that's coming in and it's got to fit into a receptor, that receptor is chiral. So that's why it;s very rare that you are going to have enantiomers with drugs that are going to have the same effect, because they have to fit into a receptor. So there's one area that they are definitely going to be different.

Another area that they are going to be different is when they interact with plane polarized light. So let me draw a little picture here of how we set up this experiment.

So if we take light, and we pass it through a polarizing filter. If you have sunglasses it's probably a polarizing filter. It's basically a filter that has tiny grooves that are all oriented in one direction. And what happens is that, just random white light is composed of an electric field vector and a magnetic field vector, it's going up and down, and it's going up and down in all directions. When you have a polarizing filter, it filters out only those waves that are going in one direction, that are going up and down. So if I have a filter like this, all the lines are vertical, if the light is undulating in other directions it's not going to make it through, so I end up with light that is called plane polarized light after it passes through a polarizing filter.

Now if I have a solution that contains a chiral compound that just contains a single enantiomer, that light is going to interact with that compound by turning either to the right or to the left. So this is a solution... I have a chiral molecule, I have a solution, it goes through and now it either goes to the right or to the left. And then, we're going to hit another filter. So in this example I rotated the filter to the right by a certain angle. So if the light let's say got rotated by ten degrees. As I rotate this filter, when I hit ten degrees I'm going to get maximum output of the light. If I'm perpendicular to the direction of the plane polarized light, it's going to turn black. As I turn it closer and closer to the maximum angle it's going to become lighter and lighter and I can tell when actually it matches up. SO if it rotates it by ten degrees I can tell that by having a second filter and seeing how much I have to rotate the filter to let through most of the light.

All right, and then I have here a detector. So if you want a prettier picture than this take a look at you book, there's a nice drawing of all this, but this is just the concept, Ok? You've got plain polarized light, you have your solution, it will rotate it by a certain amount, and I know that by the detector that I put here.

So what we can say about chiral molecules and enantiomers, if we go back here, we can say that enantiomers will rotate plane polarized light in opposite directions. And by the same amount, and there will be the same concentration on the same equipment.

So if we have one enantiomer that rotates negative ten degrees, we know that it's mirror image will rotate plus ten degrees. That's a property that all enantiomers are going to have. Now you can't tell by looking at a molecule whether it's going to rotate it to the left or to the right, or by how much. We can't tell that, but we do know that they will definitely rotate it by some amount.

So how do we actually standardize this so that we can compare... If you make a molecule and you want to see if it's the same thing that someone else has made you can have your polarimeter, you can put it in your polarimeter and it will rotate it by a certain amount, but if you don't do the experiment the same way how can you compare your results? So in order to standardize all this so that any chemist will be able to get the same results we need to quantify certain aspects of this experimental design.

So the first thing we have to worry about is C which is the concentration, and then we have L, which is the length of the tube in which you have your solution, and then basically you have the rotation that you measure, so you have the angle here. What I can do is write a formula here. So alpha OBS is your observed rotation on the machine, and the alpha that's inside square brackets, that's called specific rotation. The specific rotation is the parameter that is going to be listed in books. So it doesn't matter what your equipment was like, if it had a different path length, all that is accounted for in our formula and everyone will get the same number.

Basically what it tells you is that if you double the concentration, you're going to double the angle of the rotation, which would make sense, and if you make the path twice as long you're going to double the rotation, so it's all linear and it all adds up.

The 25 refers to room temperature, it's 25 Celsius and the D is the D line of sodium because we need to specify a certain wavelength that we are measuring. The specific rotation will vary depending on the temperature a little bit and it's going to depend on the wavelength you use to do the measurement. SO to make sure everything is taken into account we usually write down the D line of sodium.

So obviously if the specific rotation of one enantiomer is plus ten degrees its enantiomer will be minus ten degrees. That's all that it means.

So when we say that a compound can rotate plane polarized light to the left or to the right we say that it has optical activity, and a compound that will rotate the light to the right will be called dextrorotatory, and the opposite would be laevorotatory. So in any pair of enantiomers you will have one that will be dextrorotatory and one that will be laevorotatory. Sometimes when you see the naming of the compounds dextrorotatory will be denoted as a small d, laevorotatory will be a small l. You can also see a plus in parenthesis to show that it rotates plane polarized light to the right, and a minus. So those are ways of actually specifying the properties.

So I think this is a good place to stop, and I'll hang around if you want to discuss anything.

Transcription by CastingWords

Lecture 015b: Test 1 makeup review

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 10:57:00 -0700

Lecture 015b: Test 1 makeup review

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Jean-Claude Bradley: This is a review session for the make-up starting immediately after class. I'm going to go through your questions first. After we go through that, if there's any time left over, I will look at the most commonly misanswered questions. That might give you some feedback as to what concepts that most of you are having trouble with. Again, the same thing, once the make-up period starts, I won't be able to answer your questions. So this is the best time to do it.

I have one question here that is the reaction of 2-methyl butane with bromine. This would have to be in the presence of light and we're looking for a minor product. The answers that are listed here are 1-bromyl pentane; 2-bromyl 2-methyl butane; 2-bromyl 3-methyl butane; 2-bromyl pentane; and, of course, none of the above.

The first thing to do obviously, is to translate the words into an equation, and then to figure out what all the minor products are. So 2-methyl butane, that's the first step. That's going to react with bromine and light. It's not a question about mechanism, so the only thing we want to concern ourselves with in this problem is how many hydrogens do we have, and how many different products, and which one of those products is going to be the minor product.

So I'm going to circle the different hydrogens. How many do we have in this case? Well, let's start by circling them. Let's start from the right-hand side of that molecule. I have a methyl group, so those three hydrogens are going to be one group. Now, is the methyl group on the right the same as the methyl group on the left? I think we, in fact, did this exact problem in class before. The six hydrogens on the left are different than the right. Why? Because it's not symmetrical. I have to start counting from the left-hand side and not the right, because the methyl group is on the second position. That's all I can circle for one group. Now I move down the chain, I have a C-2 group, that's clearly different from any of the other hydrogens Then I have that tertiary hydrogen, CH, that's certainly different from any others, and then finally, I have the six hydrogens at the end. In fact, we have four different products in this reaction.

Although I'm doing one question here, I'm actually doing several at the same time. Because there were several questions, one of the questions asked was how many different products. So the answer is four. One of the questions might be which of the following is a major product? So I'm going to answer all those questions simultaneously. There's a question?

Student: [unintelligible]

Jean-Claude Bradley: The one on the left? Are you asking why those two methyl groups aren't different?

Student: Yes.

Jean-Claude Bradley: Because those two methyl groups are identical and they're on the same tetrahedral carbon. They're on the same carbon. If I add three methyl groups, those three methyl groups would be the same. If I had two chlorines, those chlorines would be the same. That's one way to look at it. Another way to look at it, is if we put a bromine on there, and we name the compound, it'll be the same compound. You can think if I count through this molecule, why should I start counting at that methyl instead of at this methyl? There's no reason. Any way that you can look at it that it makes sense. I like to think of it as if you've got two identical groups that are on the same sp3 hybridized carbon, then they're the same. Sometimes the way you draw it might be a little strange, but that way that's always going to be true.

Four products. Now when we're looking at major products versus minor, the way we define this consistently is we're going to look for tertiary first. Tertiary are more stable. So far, we've only looked at tertiary in this class. Basically we have one tertiary center. We have one secondary center. Again, when I use the term secondary, I'm saying we have a secondary carbon, or I can say we have two secondary hydrogens, but those would generally give rise to secondary alkyl halides. Then we have two different groups, which are primary. Is everybody clear on that?

Student: [unintelligible]

Jean-Claude Bradley: The hydrogen that circled tertiary. In order to figure out what it is, you look at the carbon on to which it's connected. Then you ask the question how many carbons are connected to that carbon? The answer is three. You're always really thinking about the carbons. Think of it that way. Any other questions on that? Any molecule that I've brought you, this is the same process.

This exact question asks about the minor products. Again, we define in this chapter, that the tertiary would be the major, and everything else will be a minor. We would actually have three minor products in this case. You may have come across a question that was almost the same, but with a different selection of options. The same question in this case can have three correct answers. There are three minor products. What we're going to do, is look at what those products would be. I can put the bromine on the left side. I can put it on the secondary carbons. And also I can put it on the right-hand side, if I'm only interested in minor products.

Let's take a look at our options. The first one really has nothing to do with this. First of all, it's a pentane. Hold on a second, let's draw this out. Right. 1-bromyl pentane would be five carbons in a chain. We do have five carbons in the molecule, but it's branched. So, it's not going to be 1-bromyl pentane. 2-bromyl 2-methyl butane. What that one is really talking about is the major product. 2-bromyl 2-methyl butane. The question is about the minor product, so that's not the answer in this case. The next one is 2-bromyl, did I reapeat that? Three. 2-bromyl 3-methyl butane. I see that in the second product I drew I have 2-bromyl 3-methyl butane, so the answer for this particular one would be C.

Whenever you do these problems the most important part is to write out the equation and make sure that the products you draw are drawn correctly and then go through and systematically compare these each. Yes?

Student: [unintelligible]

Jean-Claude Bradley: I'm not sure I understand. Are you asking me why is it that the 2-bromyl 2-methyl butane is not a minor product?

Student: Yes.

Jean-Claude Bradley: 2-bromyl 2-methyl butane, which I can draw down here, this is a major product. It's a major product because it arises from the most stable radical, from the tertiary radical. I'm not sure how else to explain it. Basically, it's the definition that we use in this chapter, we basically going to name our major products just from being from the most stable radicals. In this case it's only one of the products. I actually covered probably a dozen questions with this, that were related.

We have a Newman Projection question here. Three methyls in the back and three ethyls in the front. The first thing is we'll just affix the groups. We draw a tmplate. Why am I drawing it staggered? It's a conformation question, it doesn't matter, it's just easier to draw a staggered conformation and than an eclipse. Ultimately, we just want to position the groups. Then I randomly put the groups, the three methyls in the back, so my back three carbons are here. Three ethyls in the front. That's what that Newman Projection would look like. We want to retranslate that. I think the question is about naming it. I'm going to retranslate that, and I'm going to do the reverse, I'm going to draw the axis of the Newman Projection in a yellow rectangle. Then we're going to fix the groups. So, it's really the same thing now. You've got the three methyls on the left and the three ethyls on the right.

Now I don't have any options for the answers, but we can go ahead and name this thing. We're going to pick the longest chain, so you notice on the left that I've got three methyls, it doesn't matter where I start from, I might as well start from the middle one, on the right also, we've got three ethyl groups so they're all going to give you the same length chain. If I number this guy, I'm going to number it from the side that's going to give me the lower numbers. I find that my longest chain has five carbons, so it's going to be a pentane. What is it in addition to a pentane? It's got 2-2-dimethyl and 3-3-diethyl, so if we want to name it alphabetically, we would say that we have 3-3-diethyl 2-2-dimethyl pentane. If the order is different, that shouldn't prevent you from answering the question because you're going to see from the list of options given, I don't know what the options were in this particular case, but you would be able to draw out from each of the options given and if it correlated with the molecule that you drew, it would be correct. The key thing is to take your time and draw out your options. Use up all the time to do review. A lot of you are finishing really quick, and are not doing that well, on the make-up take a little bit more time to review, to make sure everything is correct, that you have the right number of carbons and everything. Do I have any other questions, these are all the paper questions that I have. Yes?

Student: [unintelligible]

Jean-Claude Bradley: That's the question we just did, that was the first question. The minor product of the reaction of 2-methyl butane with bromine. Like I said, you may have had a slightly different selection of answers, but from those answers you would see one of those three products and if you don't, it would be none of the above. Okay, anything else?

Student: [unintelligible]

Jean-Claude Bradley: Because in chapter five we covered enantiomers and diastereomers, I'm telling you not to worry about counting enantiomers and diastereomers, because that's not the way we learned it in chapter four. There are actually more products than the way we do it in chapter four, if you count mirror images and diastereomers. The way you're doing the problem is exactly the same way you did it on the quiz in chapter four. All you're doing is looking at the different kinds of hydrogens.

If the question is where we are reacting pentane with bromine in the presence of light, how many products can you have? Let's take a look at the number of hydrogens. What do you think, how many different kinds of hydrogens in this molecule?

Student: Three.

Jean-Claude Bradley: Three. Let's start with the ends, we have a methyl group, so those three hydrogens are the same, but unlike the first problem we did, this molecule is symmetrical. There isn't any reason that that would be the one position. If I go on the left side, that just as easily can be the one position. Another way to look at it, if I put a bromine on the right, it would be called 1-bromyl pentane, if I put it on the left, it's still called 1-bromyl pentane. Same color here, six hydrogens are the same. I continue to make that argument, if I go inward, the CH2, that's the two position if I start to count from the right and the same thing on the left. Then we end up in the middle, and those would be different. The total number of products would be three, and let me do a little more work on this in case we have a similar question. Those are the only three products we can make, 1-bromo, 2-bromo, and 3-bromo pentane. How many minor products do we have? Let's label primary, secondary, and tertiary, and let's see. The ones at the end are what?

Student: Primary.

Jean-Claude Bradley: Primary, and these are secondary, and the middle ones?

Student: Tertiary.

Jean-Claude Bradley: Not tertiary. Look at the hydrogen, how many carbons is that connected to? Two, so that is also secondary. Based on our definition of major and minor we have how many minor? Just one. The 1-bromo would be minor, and the other two would be major. In other words, if you do this reaction you can expect an even distribution of the two middle ones, or a significant amount of the two middle ones. Yes?

Student: [unintelligible]

Jean-Claude Bradley: When you're thinking about major versus minor, the major will be the products that come from the most stable radicals. We have no way no way of distinguishing between the stability of those two secondary radicals, so we're going to assume they're the same stability. If there were a tertiary radical, then that would be the major product, but we don't, so we have to look at the most stable that we have. And that's secondary. Yes?

Student: [unintelligible]

Jean-Claude Bradley: Okay, let me make sure we don't have any more questions on this. You have a question on this?

Student: [unintelligible]

Jean-Claude Bradley: Well, the first one is minor because it comes from a primary center. The primary radicals are less stable than the secondary radicals.

Methyl cyclopentane reacts with bromine in light. The first thing is we're going to make sure we draw all the hydrogens and not miss any and then we will circle the different ones. The easy ones to spot are the three Hs on the methyl. Those are clearly different from anything else. The hydrogen right underneath the methyl, there's nothing else that looks like that, so that would be the second one. Now you've got to be a little bit careful, as we go around the ring, there may be some that are the same. If I go down and look at the two hydrogens that are on the carbon next to that, are they the same? They're different. They're different because one of them is cis to the methyl group and one of them is trans to the methyl group. They're not the same relative to each other, but if I go on the other side of the ring, then those would be the same. If I circle the top hydrogen in blue, then those two would be the same. That's why we're not consideringenantiomers and diastereomers, because that would complicate this further. The way we did these problems, those two are the same. The ones that are underneath are going to be the same as each other. Then we go down to two carbons away. Same deal, same side, opposite side.

One of the questions you may have had from this is how many different products do we have from methyl cyclopentane? We have six products in this reaction. To determine the major ones versus the minor ones, we have to determine primary, secondary and tertiary. The Hs on the CH3 would be what? Primary, and this H, tertiary. It's connected to three carbons. Then as we move down the chain, secondary, in fact, all of these are going to be secondary. Again, using our definition of major and minor, we have a tertiary, so the tertiary will be the major product, if I put the bromine there. Anything else is going to be minor. There are five possibilities for that question. We could have cis 1-bromo 2-methyl, trans1-bromo 2-methyl, whatever. Those would be all the minor products in this case. Yes?

Student: [unintelligible]

Jean-Claude Bradley: In total we have six? One major product and five minor products. Let me at least draw the major product. The major would be 1-bromo 1-methyl cyclopentane. All right, what else?

Student: [unintelligible]

Jean-Claude Bradley: The question is this, 2-2- 3-3- tetra-methyl butane reacts with bromine and light in this case. How many products do we have from that? Let's write out the formula for that. Here you have the butane. We have at the two position, we have two methyl groups, at the three position we also have another two methyl groups. This looks awful familiar, I think this may be straight out of one of the problems. The point is, we have to determine how many different products we have. So, how many different hydrogens do we have in this case? Let's start from the right side. Clearly these three hydrogens are the same. What about the other methyl groups? The one that's here and one that's here. They're the same. Remember that rule? They're on the same tetrahedral carbon, so those would be the same. If I keep going on the other side of the molecule, you would notice this molecule is completely symmetrical. There's no reason I would number it from the right side versus the left side. The nine hydrogens on the left are also the same. There would be only one product from this reaction. It would be 1-bromo 2-2-3-3-tetratmethyl butane. No matter where you put the bromine, it's going to be in the one position.

Student: [unintelligible]

Jean-Claude Bradley: Because all of the methyl groups are the same. You can see that the three on the right are connected to the same sp3 center, so those three methyl groups are the same. You see the right side of the molecule is the same as the left side, so there's perfect symmetry there. Let me use another argument. Let's say we actually do put a bromine in there, and let's say we try to name our molecule so it's actually one product. Let's say you became convinced that hydrogen was different from the one on the right. What I would do now, now that I think that I have two different compounds, I would try to name them. If I try to name them, the bromine has got to be on a position on the main chain. When I count my longest chain, I'm going to include the bromine in it. If I start from that position I find that I have a butane. I would have a 1-bromo 2-2-3-3-tetramethyl butane. Now you go into the compound on the right, I have a bromine and I have to include that bromine as part of the main chain and I have to find the longest chain. Now my one position is the same, so I would end up with the same name. If you're really confused, you can name the thing to see if it comes out to the same name, alternatively if you use the strategy I just described, where are the groups that are on the same tetra-carbon are the same, and if there is a perfect symmetry, then each half has to be the same.

Student: [unintelligible]

Jean-Claude Bradley: When I say symmetry, I did refer to a plane. If you have perfect symmetry on all sides, there is only going to be one product, like cyclo-hexane, cyclo-pentane, methane. Any time you have perfect symmetry, there's only one bromo-cyclo-hexane. There's only one bromo-benzene. When you start to introduce a symmetry into the product, if I put a methyl group on a cyclopentane, like we did up here in the last problem, if I have cyclopentane, there's only one product, because it looks all the same. As soon as I introduce a methyl group, I now break the symmetry. Although I have symmetry vertically here, I have a mirror plane, that's why the left-hand hydrogens are the same as the right-hand hydrogens. However, horizontally, I've lost my symmetry, so now I've got six products instead of one. It's not fair to say if you have symmetry you have only one product. You have to have symmetry along all planes to have only one product.

Student: What's the difference between [unintelligible]

Jean-Claude Bradley: A halide is fluorine, chlorine, bromine, iodine. Mono means one, so basically what it means is don't put more than one halogen. When we say mono halogenated, and it happens to be bromine, it's the same as mono brominated. It just means standard conditions. Standard questions that you did on all the quizzes that we did on all the problems. What else?

Student: [unintelligible]

Jean-Claude Bradley: The question is how many Newman Projections? 2-methyl propane. If I didn't specify which two carbons, C1, C2, but even if I hadn't, I could've asked the question without specifying, because you have no choice, there's only one way to do this one. I would have to pick two carbons, C1 C2. We're going to draw two projections for that. There's many ways to draw that, I just chose to draw it like this, the two methyls in the front. What I'm going to do is I'm going to rotate the back. Because I've got three identical groups it's a little bit easier to see the problem that way. I'm going to rotate the back clockwise. If I continue to rotate the back clockwise, I will just get back the first one. Basically, it's three of the same groups. The only thing I can really do is just draw them staggered or draw them eclipsed. But overall, I only have two conformations. That's the answer. If we had a different product, we could end up with more projections, but in this particular case, we only have two.

Student: [unintelligible]

Jean-Claude Bradley: We have a cyclohexane conformation question. What's the cyclohexane?

Student: [unintelligible]

Jean-Claude Bradley: Trans. We do these problems always the same way. We draw out the two conformations, the templates at least. You have your choices as to where you want to put them. I suggest you always put them in the front because it's easier to draw. We have 1-ethyl and 4-methyl, so if I count along here I look at four. The four position would be all the way across and I have to be careful to make it trans. The way that I put the ethyl is underneath the ring, so I want to make sure the methyl is on top of the ring, which is easy to see from the axial position. All I have to do is to draw the other conformation is to just flip each one. This one is a lot harder visually to see that they're trans because they're both equatorial, but they are definitely trans. In this case, which one would be more stable? Left or right? The right one. We want to put as many of the groups as possible on equatorial. Sometimes, you have to make a tradeoff, in this case we don't. Both groups are either equatorial or axial, so it doesn't matter what the groups are. The right one is going to be more stable.

We want to note here that the ethyl is equatorial, the methyl is equatorial. If you had this question, there's a number of different possible answers, but if you have an answer where the ethyl is equatorial and the methyl is equatorial, that would be correct in this case.

Student: [unintelligible]

Jean-Claude Bradley: If you could have them all equatorial that would be good, but sometimes you can't. Sometimes you can have one equatorial and one axial. In that case, you're going to pick the structure where the largest group is equatorial.

Student: [unintelligible]

Jean-Claude Bradley: Do you see all the bonds I drew straight up and down? That's axial. The ones that are kind of horizontal are equatorial. I'm not a great artist, so if you want to see perfect angles, look at your book. It's much easier to see the axial ones because they're definitely exactly up and down..

Student: Why would one be axial and one equatorial?

Jean-Claude Bradley: Because they're trans. Actually, that's what I'm testing you on, is that you're able to see that when they're trans they're on opposite sides of the ring. Trans is just opposite sides of the ring. So methyl is underneath and ethyl is on top. They're both trans, it's the same molecule. Are there any questions about the make-up or anything else related to it?

Student: [unintelligible]

Jean-Claude Bradley: All the same rules apply for the make-up. Make sure you read the FAQ carefully, find out which rooms you're supposed to do it in and all that stuff, you don't have as much time, I think. Does anybody know how long it is?

Student: [unintelligible]

Jean-Claude Bradley: It's always going to be more questions, I'll put more questions in, in the same amount of time.

Student: [unintelligible]

Jean-Claude Bradley: I will take the second score, no matter what it is. This is an opportunity if you really messed up the first one, it's an opportunity for you to do better, but you don't want to take it lightly. It's just an opportunity. You don't have to tell me you're going to do it, you just do it. Everybody can do it if they want to. But, by the way, the average, do you see the average when you log in? The average is about 81 percent, so if you want to relate yourself to that, that's what it is. You have a question?

Student: [unintelligible]

Jean-Claude Bradley: The same thing, 90 minutes.

Student: [unintelligible.

Jean-Claude Bradley: Well, I think it's four days, you've got to log in and see what period of time. I think the next make-up, I can't give you as much time. But this one, you've got plenty of time. Anything else? I wish you luck, if you're doing it again.

Transcription by CastingWords

Lecture 015a: Review for test 1

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 09:59:00 -0700

Lecture 015a: Review for test 1

View Lecture

Jean-Claude Bradley: I'm going to start with a question I have from an online student on the quiz for chapter three, there are two questions here, and they have to do with the most stable conformation of cis, 1-ethyl, 2-methyl cyclohexane, so we're looking at quiz chapter 3.

Whenever you're doing a problem that has to do with conformations of cyclohexane, you have to draw out the two conformations, then figure out which one is more stable, and then answer the question depending on what the options are going to be, so there's no shortcut to doing this question, you really have to go through each one of these steps.

I'm drawing the two conformations, and I'm drawing the axial bonds, the equatorial bonds, for each one of the templates. We're looking at 1-ethyl, 2-methyl, and it's advisable to use the carbons that are facing towards you because you have a little bit more room to use, so I have to put an ethyl group, and the methyl group will be one carbon away, and it's cis, so it has to be on the same side of the ring. If I drew the ethyl group, under the ring, position one, then in position two, I have to put it on the bond that is closer to the bottom of the ring, and that will be this position here, and there we have hydrogens.

To draw the other structure, all we do is flip each one of those groups, so now the ethyl group will go equatorial and the methyl group will go axial. Between these two, let's say that we call the first structure A, and B, which one will be more stable? B. Why? The big group is equatorial. You'll notice that I can't put both groups equatorial. If I could, then clearly I would, but the molecule just can't flip in such a way to put both groups equatorial, what you have to do is look at the least disruption, or the least spheric hindrance, so that will be the one where the largest group is equatorial. The largest group is equatorial, and therefore structure B is more stable. Yes?

Student: What if it is cis 1-2 dimethyl? Would there be any difference between the two?

Jean-Claude Bradley: No, then there wouldn't be any difference between the two in stability, they'd really be the same conformation and energy, so there would only be one conformation.

That's the first step, so you see how long it took me to do this? You can't do these in two seconds; you have to take the time to do the problem. Once you've come to this point, now you look at the options that you have. In the most stable conformation of this compound, the methyl group is axial, so is that true? Yes, that's true. The other options are, the methyl group is equatorial, no. All the other ones basically don't describe the properties of this, so you can't really do these without drawing them out, so in this case the answer would be A, the methyl group is axial.

The next one is the most stable conformation of trans 1-ethyl, 2-methyl cyclohexane. The same relationship between the methyl and the ethyl group, but this time it is trans, if I put the ethyl group under the ring, the methyl has to go on the other side and I just flip these. Now, between A and B, which one would it be? B. This one is easier to tell actually, because it doesn't matter where the groups are. One of them has both groups equatorial; the other one are both axial. B is more stable.

Now we answer the questions. The methyl group is axial, no that's not true. The methyl group is equatorial and the ethyl group is axial, no, that's not true. The methyl and the ethyl group are axial, that's not true, the methyl and the ethyl group are equatorial, that will be true. I'm just identifying where the groups are, so in that case, it would be D.

While we're on the topic, any other questions on cyclohexane conformations?

I have a question here on the Newman Projection of butane, end butane. Which conformation would be the highest in energy?

For a compound like ethane, propane, butane, we did all of those, so when we talk about conformations of butane, we know that it's between C-2 and C-3 unless specified otherwise. If I give you any other molecule I will definitely tell you between which carbons you're going to do the Newman Projection.

To translate the butane into a Newman Projection, we draw circles, and we've got three groups in the front, three groups in the back. You arbitrarily select which three groups are going to be in the front or the back, so let's say that the left side will be in the front. I have a methyl group, I have a hydrogen and a hydrogen, so those three groups are going to be the groups that I put in the front. It doesn't matter where you put them initially. Then I have three groups that're going to be in the back. Again, I'll have a methyl and hydrogen and a hydrogen. The question here asks for both the highest and the lowest energy, so let's draw all the conformations and we can pick out which is which.

When you're determining the various Newman projections, you want to keep either the front of the back constant, it doesn't matter which, let's say in this case we keep the back constant, so all my structures are going to start off looking like this. Then I'm going to rotate clockwise, you can do it either way, but I like to do it clockwise. We're going to rotate that until we reach another conformation. The next conformation we're going to reach is when they methyl group is on top of the hydrogen. Then we continue to rotate, keeping the back constant. Finally, we can draw one more, where the two methyl groups are on top of each other.

In this problem, I think your selections are anti, gauche, totally eclipsed, partially eclipsed, so why don't we label them, and that'll be general for any question. The one where the two methyl groups are opposite is anti; the one where the methyl group is on top of a hydrogen is an eclipse conformation, specifically for n-butane it is partially eclipsed, that's the name of it. Then we have a staggered conformation where the two methyl groups are next to each other, that's gauche. Finally, we have the totally eclipsed conformation.

The one that's going to have the highest energy will be the one that has the highest spheric interaction between the two methyl groups. The highest energy is always going to be an eclipsed conformation, but here we have two eclipsed conformations, so we're going to look at the one with two large groups on top of each other, so that will be our highest energy. Another way of saying highest energy is most unstable, you can think of it that way. The one that has the lowest energy, or that is the most stable, will be the one where the two groups are as far away from each other as possible, so that would be the anti. I think I've covered every aspect of this question. Anything else on that?

Student: [unintelligible]

Jean-Claude Bradley: Yes. Only n-butane can be named specifically with these terms, anti, partially eclipsed, gauche, totally eclipsed. What I can say for any conformation is that the two methyl groups are anti, or that the two methyl groups are gauche, or that the methyl group is gauche to a propyl group. There should've been a couple of questions that used that kind of reference, especially when I ask you to find the most stable Newman Projection on a bigger molecule. I'll say the ethyl group is gauche to the methyl group or something like that. I don't have a question in writing on that, but I can certainly do one if we need to.

Now I have a question on the number of Newman Projections. How many different Newman projections can be drawn for n-pentane using C1-C2 axis? The C1-C2 axis will be this in yellow. Now we do it the same way we always do it; we start with a circle. We're going to have three groups attached, let's say to the back carbon, and then there'll always be three groups attached to the other carbon. That would be a hydrogen, a hydrogen, and an n-propyl group.

If I want to figure out how many different groups I have, what I have to ask is, if I keep rotating either the front or the back, how long before I regenerate the same projection? You'll notice that one of them have three groups that are the same, so hopefully, it will be obvious that I can only draw one more. All I can really do is draw the eclipsed conformation of that. Let's say that I keep the front constant. If I kept rotating the back, I would just go back to the first structure, so the answer to the question is two. That's pretty much it for this number of projections, you just count them.

Next question is actually draw a Newman Projection. If you have three hydrogens in the back, one methyl in the front; I have to have three groups in the front and three groups in the back. I don't know how the question was originally posed, but let's say that it means one methyl and two hydrogens in the front, so three Hs in the back, and one methyl in front. There are a number of things I can ask you about, this might be a question where I'm asking you what molecule is it, so you work backwards. Three Hs in the back would be like this, and in the front, we have a methyl and two hydrogens. Again, working backwards, the yellow rectangle will correspond to two of the carbons that consist of the circle. On one of those carbons you're going to have H-H-H, and on the other carbon you're going to have H-H-methyl, so the molecule that has that Newman Projection would be propane.

Next I have a reaction of propane with bromine gives how many monobrominated products? We're looking for monobrominated, which just means that you only have one bromine, which is the standard the way we've always done this problem; we've only put one bromine on the alkene. So, monobrominated or monohaloginated would mean the same thing, just put one halogen on there.. This is where we have to identify how many different hydrogens we have. We've done this problem before. We've got six hydrogens at the ends that are the same, then we have the two hydrogens in the middle which are the same, so there're only two different kinds of hydrogens in this molecule. These would be the two products. The question doesn't ask about major or minor, but we can do that as well. The first one would be 2-bromopropane, and the second one would be 1-bromopropane. The one that would be the major one would be from the secondary carbon. Anything that's not major is obviously going to be a minor product. I think that's about all that we can do with that question. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Okay, I can do hexane. Any other questions for propane?

Student: [unintelligible] major or minor group?

Jean-Claude Bradley: When we look through the radicals, in order to make 2-bromopropane, you have to go through a secondary radical, and to go through 1-bromopropane, you have to go through a primary radical, so secondary radicals are more stable than primary radicals. Without having to draw the mechanism every time, the way that you know that is you look at the molecule, and you have to be able to answer, the red hydrogens are primary, the green hydrogens are secondary. If I had tertiary, the tertiary would be the major products. If I had two different tertiary hydrogens, then I would have two major products. Anything else, primary or secondary, would be major products. That's how we've consistently used the terms major and minor in this chapter. When we do other chapters with different reactions, I'll be very specific about what is a major product and a minor one, but in this chapter, this is definitely how we are defining it.

I can do hexane, let's see what happens. What do you think, how many products do you think you have?

Student: [unintelligible] two primary.

Jean-Claude Bradley: Let's start with the primary ones. This is a symmetrical molecule, so I can't tell one end from the other, those six hydrogen in red would be the same, and you're right, they would be primary. Now all the other hydrogens are secondary, but they're not all the same. We need to determine the different hydrogens, so how many different hydrogens are left?

Student: Two.

Jean-Claude Bradley: Right. Let me use a different color here, green. Those two are different, but they're really the same as the two hydrogens on the other side. Those green ones are all the same. Then let's put some blue ones, and the blue ones will be the same, so the answer is we would have secondary hydrogens, primary hydrogens, we would have a total of three products. How many major products?

Student: Two.

Jean-Claude Bradley: Two. Exactly, because we have two secondaries and no tertiaries. Two major, one minor.

Student: Why, exactly, are those two secondaries different?

Jean-Claude Bradley: Let's draw the products, because if I draw the products and the products are the same, then the hydrogens would be the same, so let's see. If we exchange the red hydrogen, we would end up with 1-bromohexane. That would be the blue hydrogens being replaced, so do you see that 2-bromohexane is different from 3-bromohexane? That's where you get your three products. Let's say that you couldn't really tell, and you drew it, when you name the molecule, you should end up with the same name, if you start numbering from the side that would give you the smallest number.

Next, what is the intermolecular force between dimethylamine and water? First thing we look at for intermolecular force is we look at the strongest force, and we have hydrogen bonding between those two. We need to have HxH, and in amine we have an NH, and in the water we have an OH, so there's actually a couple of ways to do hydrogen bonding between these two. Just one of these ways would be through the nitrogen, so the answer is H-bonding.

Next I have bromomethane and methanol. The first question is, can we have hydrogen bonding between those two? Methanol has an OH, it can hydrogen bond to itself, but it can't hydrogen bond to bromomethane. Bromomethane doesn't have a nitrogen or an oxygen that it can bond to. The first question is answered no, cannot hydrogen bond. Do we have dipole-dipole, is that possible? To have dipole-dipole, we need to have dipole moments in both molecules. Well, let's figure it out, do we have a polar bond in bromomethane?

Student: [unintelligible]

Jean-Claude Bradley: Carbon bromine, right. In fact, that's how we learned about dipole moments.

Bromine is more electronegative, so that's the only bond we have in there, therefore it adds up to a positive dipole moment for the molecule, so it's polar. Do we have a dipole moment in methanol?

Student: [unintelligible]

Jean-Claude Bradley: Carbon oxygen?

Student: And the oxygen.

Jean-Claude Bradley: Yes, and the oxygen-carbon. It's towards the oxygen. That oxygen is an sp3 hybridized center, let me redraw this in a way we can more clearly see. So, oxygen is sp3 hybridized, we have dipole moment from the hydrogen to the oxygen, from the carbon to the oxygen, and we can consider through the lone pairs. We have four arrows and they're all pointing towards the left, so they definitely don't cancel out. We would say that methanol would definitely be polar, so we have two polar molecules, so dipole-dipole would be the dominant intermolecular force.

Those are all the written questions I have. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Resonance structure. It will have consequences in respect to figuring out whether the molecule is polar or not. It will have consequences with respect to figuring out the average charges on atoms. If you don't consider resonance, then SO3 looks like it's really polar in this resonance form. It also looks like these two oxygens are charged -1 and the other oxygen is charged zero, but, in fact, that's not the case. SO3 is not polar and the charge on each oxygen is minus two-thirds. That's a consequence of resonance. If you don't take into account resonance, then you won't get the right answers for those kinds of questions.

By considering all three resonance forms of SO3, we see that we get three arrows, and the three arrows cancel out. It's more likely to be questions that'll get you into trouble if you don't know understand resonance, rather than a direct question about resonance.

Student: [unintelligible]

Jean-Claude Bradley: The charge of the oxygen will be, in this case, each resonance form contributes equally, so one of the resonance forms has -1, one has -1, and one of them is zero, so the average charge on the the oxygen will be minus two-thirds. The average on sulfur would be +2, because it doesn't change. What else?

Student: [unintelligible]

Jean-Claude Bradley: Will it have dipole-dipole bonding?

Student: Yes.

Jean-Claude Bradley: It's not polar, so no. But you could get hydrogen bonding, with water, for example. Anything else?

Student: [unintelligible]

Jean-Claude Bradley: Acids and bases, we didn't do too much with that. Basically all I did was I defined a Lewis acid and a Lewis base.

Student: [unintelligible]

Jean-Claude Bradley: The test is going to be similar in format to the quizzes, that's the whole point of the quizzes. It doesn't mean there might not be a couple of questions that are a little different, but if we haven't spent a great deal of time on a particular concept, then it's unlikely that I will have many questions on it. The acid and base thing, again, is because we're going to use that concept throughout the Lewis acid and Lewis base, we just didn't spend that much time on it. Are there any questions about the format of the test?

Student: [unintelligible]

Jean-Claude Bradley: How many questions are there? There are 25 questions, which you can know just by looking at the test that's been posted for several weeks. It tells you how many questions, it tells you how long it is, it tells you when the start time and the end time it; that's under 'quizzes and tests'. That tells you what the tests are, so you know pretty much the format and everything.

Student: I understand the times [unintelligible]

Jean-Claude Bradley: If you have any suggestions on how to make the FAQ clearer, I would appreciate it because you just walk in and do it; you have four days to do it. There's no need to reserve or tell me when you're going to do it.

Student: [unintelligible]

Jean-Claude Bradley: No. I get that question so much, I'm really curious why that's being missed. Question nine on the FAQ gives you four rooms where you can do it in. It's pretty clear, four rooms. What else?

Student: [unintelligible] sigma bonds and pi bonds [unintelligible]

Jean-Claude Bradley: Why we did the hybridization?

Student: [unintelligible]

Jean-Claude Bradley: You basically need to understand, like for ethylene, it's anatomy. It has double bonds, but when we labeled stuff, we said this is a sigma bond, this is a pi bond, you need to know there's one pi bond in ethylene. You would need to know how many sigma bonds there are, you would need to know that when you draw those two lines indicating double bond, one of those is a sigma bond and one of them is a pi bond. You do need to know that, I'm not going to ask you to draw a picture obviously, but you need to understand that and to understand the consequence that it prevents the rotation between the two. Yes?

Student: [unintelligible]

Jean-Claude Bradley: Entropy? Yes, there's a lot of other stuff in the chapter that I didn't talk about. The syllabus basically gives you a checklist of the topics that I cover, so I don't think I talked about entropy once, so I really wouldn't worry about it.

Student: [unintelligible]

Jean-Claude Bradley: We definitely covered that, and that's part of the syllabus. All the functional groups, we drew them out. Yes, so you definitely need to know the differences between the functional groups. Any other questions? Gong once, twice; all right, good luck on the test!

Transcription by CastingWords

Lecture 014: Free Radical Reaction Part 2

noreply@blogger.com (Jean-Claude Bradley) — Mon, 22 May 2006 06:57:00 -0700

Lecture 014: Free Radical Reaction Part 2

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Jean-Claude Bradley: Okay, let's get started. So we're going to continue on Chapter 4.

[student chatter, settling in]

Jean-Claude Bradley: All right, so this is the reaction we're looking at, free radical halogenation. What we did in the last class was basically look at the steps that were involved in this reaction so we know now that this does not happen in one step, it happens in multiple steps. Let me just reiterate those because we're going to need them to redraw the energy diagram.

So the first step is that we're going to take bromine and then in the presence of light we're going to generate two bromine radicals. In the second step, the bromine radical will react with the CH4. Then in the third step, the methyl radical reacts with the bromine. Then finally, we have three possible termination steps. I'm not going to list all of them here, just one of them. Termination being the combination of any two radicals.

Ok, now the key thing with this is if we want to understand the energy diagram, is we have to look at how the bulk of the products are formed. That means that the initiation step is a very very small part of this whole process, it just happens for one out of every 10, 000 or 100, 000 molecules. The termination step is also an extremely minor part of any products being formed. So the only two reactions that we are going to look at in the energy diagram will be propagation steps, steps two and three.

In the energy diagram we will ignore one and four. Does anyone have a question with that or is that clear?

Okay, so lets start off then, we're going to have two steps. The first step will be the first propagation step. It will be the bromine radical plus methane. So we're going to start off here, and I'm doing step two only here: bromine radical plus CH4. And the end result of step two will be the formation of HBr and CH3 radical. Okay, so it's going to look a little strange, the way it is here, because I'm going to have products then I'm going to start at the same place then I'm going to do the second propagation step. So in the second propagation step what do we have? We have methane radical plus Br2. And that is going to go to CH3Br and a bromine radical.

So what I've done basically here is I've filled in a little bit of what was missing in the middle. What we had done initially was we determined from the left hand side to the right hand side was an 8kcal/mol drop, so that's still true but now we see that in between there we have another step which is actually a higher energy than where we're ending up. Let me draw this energy curve. It will look something like that. This energy curve is two steps. The first part is propagation step two the second part is propagation step three. If you remember a little bit about how these energy diagrams work, we have multiple components here.

The first thing that I'm interested in is this distance here, what's that called? Activation energy. That is the energy that's required to go over the top of the energy hill to make it to the products. The higher the energy of activation, the slower the reaction is going to be.

So the first arrow is energy of activation for step two. Now the enery of activation for step three is going to be much smaller, it will be on the other side of this.

Knowing where one of the energy of activation is much much higher than the other that tells us that the rate of the entire reaction is going to be dependent not on both of these steps but is only going to be dependent on the slow step. The step that has the higher energy of activation is going to be the slow one, which is also called the rate limiting step.

What we can tell from this diagram is that step two is the rate limiting step. That's going to be really important when we look at some additional reactions that have multiple steps because if you know what the rate limiting step is, then you can control the rate of the whole reaction. Because when you do this reaction, you don't actually usually measure the intermediates. You don't measure the methyl radical, you don't measure what's happening in the first point. What you are measuring is basically the rate of the production of your two products.

Greater production of CH3Br, HBr, and if you know the mechanism though, then you can start to control that reaction. You can change conditions, you can speed it up or slow it down. But the key concept here is that no matter what you do to the second step in terms of making it faster or even a little bit slower will have no effect whatsoever on the rate of the entire reaction. Okay, so that's the concept of rate limiting step. Any questions?

Okay, there's one more part of this diagram that we need to point out. And that's at the very top of these energy hills we have a structure that's called what? You remember what this is called? Transition state. Which is not a molecule, it's basically a configuration of energy and geometry that is the highest possible energy, the most unstable part of this entire curve. So this is called a transition state. We have one for step two and we have another one for step three. The other thing you have to learn how to do is to estimate what is the geometry of the transition state going to be. That's one little trick you're going to have to use in this reaction and in some additional ones we will do later.

In order to predict transition state, what you need to understand is that atoms are going to move from left-hand side to right-hand side and they don't move by jumping around. They move in a continuous way. You can guess what that transition state is going to look like by just using common sense and thinking what's in between what's completely on the right and what's completely on the left.

Let me use number two as an example. We start with Br radical and CH4. We start off like this and start off on the left with that. And on the right what I'm going to do is keep the position of all the atoms the same that don't move. So if you look carefully at what the products are like the only thing that changes here is I move one of the hydrogens over from the right to the left. In the end I move that hydrogen over and on the CH3 group I now have a radical. If we want to figure out what the transition state looks like in between those two, you basically take an intermediate view.

A transition state will be drawn with square brackets around it. The bromine on the left is uncharged, the bromine on the right is uncharged, so the bromine in the transition state is likely not going to be charged, that should be your first guess. So I would put a bromine. And on the left there are zero bonds between the bromine and the hydrogen and on the right there is one bond between the bromine and the hydrogen. So we can guess that in between those two there's going to be a partial bond between the bromine and the hydrogen.

So I'm going to put the hydrogen between those two but there's something else I'm going to have to worry about before I put the hydrogen. There's a full bond between the carbon and hydrogen on the left and there's zero bond between the carbon and hydrogen on the right so I will again have a partial bond between the carbon and the hydrogen. So when I'm done here I will put that hydrogen. The other Carbon hydrogen bonds are unchanged.

There is no change in charge there's no change in bonding between the left and the right so I know I can just leave those things intact like this. So the only thing I haven't put here is the hydrogen and partial bonds. If we don't know the energies of the reaction, we could take a guess and just put the hydrogen in between the carbon and the bromine. That would be a good guess. We don't know exactly where it's going to be so if we don't know the energies, then we'll just put it randomly in the middle. However in this case we do know the energies. Let me go back to this diagram.

We know that there is a huge energy of activation, goes to a top, then it has just a little bit more to go before it goes to the products. So based on that, would you put the hydrogen closer to the bromine or closer to the carbon?

Student: Carbon.

Jean-Claude Bradley: You think closer to the carbon? Does the transition state look more like the products or more like the reactants? Let's go back. Just on a picture level, the height of this peak, is it closer to the reactants or the products? Definitely the products. So an easy way to think about this is the transition state is going to look almost like the products. And what do the products have? They have a full bond between the bromine and the hydrogen and have no bond between the hydrogen and the carbon.

So the hydrogen has migrated from the right to the left. In the transition state it's almost completely connected to the bromine. So when we put the hydrogen, now that we know the energies, we'll put it a lot closer to the bromine than the carbon. My little dash here is a partial bond. The hydrogen is closer to the bromine because the transition state is more product-like.

The transition state is more product-like just because it's closer in energy. This is called Hammond's postulate. Let me write that out. Hammond's postulate basically says that the transition state will look like the species that it is closest to in energy. Hammond's postulate is one of those things that if you try to learn it by looking at the definition you're just going to confuse yourself. It's much easier to understand by looking at the picture first.

So just to make sure everybody's on the same page here, any reaction you can do this with, but you have to know the energies to be able to apply Hammond's postulate. If you don't know the energies, you just can't apply it and that's why you would just guess and put everything in the middle cause you don't know. When I talk about energetically closer, I'm just simply looking at is it closer to go down the right-hand side of this hill or the left-hand side.

Student: [xx]

Jean-Claude Bradley: Well, I'm telling you that. I have to tell you what the energies are for you to be able to do this, but yeah, if I didn't tell you what the energies were you really couldn't do the problem. Anything else on that? Any questions?

So I will let you do step three as an exercise. See if you can draw the transition state for step three.

[phone rings, silence while students work on problem]

Jean-Claude Bradley: We looked at how to brominate methane. All we've really looked at is that you know this reaction is kind of general, because we called it the halogenation of alkenes, we didn't call it the bromination of alkenes so you know that you can probably do it with other halogens and you know that you can probably do it with a lot of other alkenes. But the one thing you haven't learned yet is how you can predict the products you're going to get. So let's take a look at a simple example, which is propane.

So if you treat propane with bromine in the presence of light, and if you're not used to this way of writing, if I put a reagent on the arrow, it's the equivalent of saying propane plus Br2; it's just more convenient sometimes to do it that way. So if I react it with bromine in the presence of light what we know is that we're going to swap a hydrogen for a bromine and generate HBr. So, the key question now is how many different products can we make from propane. Because there's more than one place that I can put the bromine on here.

Let me see what the answer is. Three? Let's start with the first one. Where's the first place I could put the bromine? At the ends? ok so let's start with that. Where would be another place? On the middle carbon. Anyplace else? The reason we spent so much time going through sigma bonds, and all these compounds are the same; this is the reason, because now we have to apply this stuff. So I gave you a couple of the tricks to find out if compounds are the same or not.One thing you can do is just name it. One is 1-bromo propane and the other is 2-bromo propane. If you draw anything else, you'll come up with either 1-bromo propane or 2-bromo propane; however you draw it.

Okay, so, that's the first thing you can do here is determine how many products you'll have; if you attempt the quiz here you'll see that that's a common question. Now, we also have HBr that is resulting from this. So HBr will be kind of a side product here. When I ask you how many products you get I'm only interested in the organic products. Be very clear about that. Propane will give you two products only. Two possible products.

Okay, so if we try to predict the ratios of these two, there's a couple of approaches that we could take. What would be an argument? Could you argue that you'd get more of one than other? Is there any argument that you can make?

Student: It depends on the stability of the radical.

Jean-Claude Bradley: We didn't learn that yet; so, you're right, it depends on the stability of the radical. Let's say you didn't know that. Just based on the information that we have so far, how could you guess rationally?

Student: [xx]

Jean-Claude Bradley: There might be a hindrance issue, so that the central carbon would be more hindered. Let's say though that there's no hindrance issue whatsoever. It's purely a statistical effect. Let's say that the probability of substituting any hydrogen in this molecule is exactly the same. What would be the ratio that I would expect of those two compounds? Fifty-fifty? It's not fifty-fifty.

Student: Wouldn't it be the first one because [xx]

Jean-Claude Bradley: Right, so how many chances do I have? Well, I've got six hydrogens that would give you 1-bromo propane, I only have two hydrogens that would give me 2-bromo propanes, so I would expect to get this in a ratio of three to one just from a statistical expectation. I expect three to one, just by counting the hydrogens.

When I do the reaction I find that that's not at all what I get. I get the second one at over 99%. Just using common sense and just the tools we have looked at so far we did not have enough information to do this problem properly. So we need to understand some additional properties of these intermediates, and as you said, it's going to be related to the radical stability between these.

Let's think about what's happening in these two competing reactions. So let's go back to making at 1-bromo propane and we're going to step through the steps of the free radical halogenation and see what's happening exactly.

[phone rings]

You should become very familiar with this mechanism. We're going to go through free radical halogenation. The first step is always going to be the same, don't forget that you need light to do this first step. Now the bromine radical is going to approach the propane. And it has a choice. This is where the decision gets made as to what product you're going to make.

If it goes on one of the six end hydrogens, I'm going to get HBr and an n-propyl radical. And then that n-propyl radical will react with bromine. And then we're going to have termination steps and we can have three different termination steps; I'm not going to bother to write all three of them, we don't need that really to understand the problem so here's just one of the three.

Again, the key thing to remember here is what happens at the second step. Once we're at the second step, that propyl radical is not going to change, it's not going to move around.

[phone rings]

So that means if it's at the tip, it's going to stay there and always go to step three. So we can think that the outcome, the product is decided at step two and is decided because we made an n-propyl radical.

I could redraw this whole thing to show you how to get to the isopropyl radical but it would be the same thing. At step two, I would have the isopropyl radical and that would commit you to making 2-bromo butane. So, if we're not getting a statistical ratio here, then clearly it's got to be an energy difference. As was stated earlier, that reason is that the n-propyl radical is much less stable than the isopropyl radical.

So let me draw that in the next slide here. That's the comparison you need to make whenever you do these problems. You figure out the radicals that are involved and you determine their relative stabilities. So this will be the isopropyl radical, will be more stable between these two.

How do we know that one radical is more stable than the other? What's the key difference between these two?

Student: [xx]

Jean-Claude Bradley: What is the type of carbon?

Student: [xx]

Jean-Claude Bradley: That's the key thing, exactly. This is a secondary radical and the one on the left is primary. So you don't have to memorize that isopropyl is more stable, what you need to understand that will work for any instance is that there is an order of stability for radicals.

And that radical stability will be - tertiary is more stable than secondary is more stable than primary which is more stable than methyl radical, which is even beyond primary. It doesn't have any carbons connected to it.

We're going to stop our analysis at that point, you don't need to understand further why is it that those radicals are more stable, the key thing that you need to understand as base knowledge is that here's your order. Okay?

All right. So we're going to do a bunch of examples where you're going to be comparing -- you're going to have a molecule, you're going to have to compare different sites on a molecule and try to predict which is going to be your major product and which is going to be your minor product. Let's go back to this problem that we did with propane and try to do it in a little bit quicker fashion here.

So the first step is to determine how many different products you have then the second step is to determine your major product.

You don't have to go through the mechanism every time, that was just to explain to you why this happens; this is secondary alkyl halide and this is primary. The secondary will be your major product and the primary will be your minor product. So in this chapter the definition of the major product will be the one you get the most of and its going to be related to the order of the carbon. So if you have two products that are both secondary carbons, you would have two major products and everything else will be a minor product.

The other thing you'll notice is that I've been using bromine consistently through this. That's not an accident. Basically, this will work for bromine but it doesn't work so well for other halogens. Let's pick again the example of propane. This is what we just looked at. Do the bromination, we saw that we get above 99% of the secondary and less than 1% of the primary.

If we try to do the same thing with chlorine, the first point is that we do in fact get all the same products, we'll get the 2-chloro propane and the 1-chloro propane. But in this case, the ratio is going to be very far from an overwhelming amount of the secondary bromine. And the reason is that the reactivity of the chlorine radical is much higher than the bromine radical.

So what's happening is that when the bromine radical is coming along and is about to grab a hydrogen off of the propane, it's not too reactive so it has chances to bounce off the propane many times until it reaches just the right angle, just the right energies, and it's going to make the more thermodynamically stable radical. When the chlorine radical is coming along, it is so reactive that it's going to grab pretty much the first thing that it bumps in to.

And because we have more hydrogens that are primary than hydrogens that are secondary, its going to skew the results a lot. Basically, you cannot predict major products and minor products for chlorine, so remember that when you're doing problems. However, you still get the reaction. Ok so that would be the case for fluorine and chlorine. Those would be way too reactive to be selective.

The problem with iodine is that most of the time it doesn't react; it's not even reactive enough to even do the reaction so that's why bromine is just in that sweet spot. It's just reactive enough to do all the reactions that we want but it's not so reactive that it's not selective.

Let me write out here an order of reactivity. So fluorine radical would be most reactive, followed by chlorine, bromine, and iodine radical, where the iodine is not reactive enough, and what follows from that is the selectivity. The selectivity will be: bromine is more selective than chlorine which is more selective than fluorine. I'm leaving the iodine out of it because if it doesn't react then selectivity is meaningless.

Those are the two concepts that you need to add to the rest of this theory.

Last thing I want to touch at in this chapter is make sure everything is clear about these radicals. Okay, so a radical is not charged, that's a very common mistake that I see. The radical is not charged. Why is it not charged? Go back to the basic theory of the first week. Carbon normally has four electrons for charges, there are exactly four electrons here. This does not satisfy the octet rule, which is one of the reasons it's really unstable but it does exist. It exists fleetingly, but it still exists long enough to do the reaction. But remember that it's not charged.

You can have a CH3 with a charge; if I have a lone pair of electrons on that carbon, I would have five electrons for charges, carbon normally has four, so this would have a charge of -1, and that's called a carbanion. Anytime you have a carbon with a negative charge on it it's a carbanion. The only other possibility is if I have no electrons on that carbon left, now I have three electrons for charges, and so this is going to be a carbocation.

Okay, so make sure you understand the differences and why it is that we don't put charges when we do free radical halogenation mechanism.

So this is a pretty good place to stop, we're going to go right into the problems next class.

[general noise of students packing up and leaving]

Transcription by CastingWords

Lecture 013: Free Radical Reaction Intro

noreply@blogger.com (Jean-Claude Bradley) — Fri, 19 May 2006 12:29:00 -0700

Lecture 013: Free Radical Reaction Intro

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Jean-Claude Bradley: We are going to take this entire chapter to look at one single reaction, and that's something that we looked at briefly, the halogenation of alkanes. When we looked at the reactions of alkanes, this was one of the three. As we will see, as you move on through organic chemistry, you'll see that reactions fall under certain types. This particular type is called a free radical chain reaction; you'll understand why it has all those terms in its name after we're done here.

This is something that's specific to halogenation of alkanes, but it's also general. Once you learn the free radical chain reaction, then you can understand a lot of other chemistry that does not seem related at first, but actually is.

I'll just remind you of the example that I likely gave, which was if you take methane and reaction it with a halogen, let's say bromine, and shine light on it, you're basically going to swap the hydrogen for the bromine, I'm going to put the bromine on the CH3 group and we end up with HBr from that.

That's useful on one level; you do know what the product is from methane reacting with bromine. Because we covered this in the alkane chapter you could probably guess that propane will also reaction with bromine, but you really have no means of predicting for different products that you would get. For methane, there is only one product; I can only put the bromine at one place. For longer alkanes, there's all kinds of different places I can put the bromine. Without knowing the mechanism, you really don't have any predicting power to be able to even guess, rationally, as to what kind of products you're going to get.

By looking at the mechanism, it's pretty powerful because you'll be able to predict how the free radical halogenation will work, not only on molecules you haven't seen before, but on molecules that have never been made. That's the whole point of organic chemistry, to make stuff that hasn't been made before and predict how it's going to behave.

Let's start to analyze this, based on some stuff that you already have seen, which is let's look at the energy of this reaction, at least get a big picture view of it. To look at this we're going to consider bond association energy, which is the energy it takes to break a bond. There are two ways hat we can break a bond, and we have to be specific about which one we are going to use. If I bring a bond, the bond has two electrons and here are two ways that I can split those up. I can either electron on each hydrogen, so each hydrogen ends up with one electron. To do that, I'm drawing these arrows that have not a full arrowhead, these are half arrows. That means movement of one electron.

What we saw previously with the curved arrow formalism was the movement of two electrons. If I wanted to show that in the bottom example, all I would do is use a full arrow. That would basically tell me to take the two electrons in that bond and put them on one of the hydrogens and leave no electrons for the other hydrogen. These are two ways that I can disassociate; does anybody remember what these are called?

Student: Homolytic and heterolytic.

Jean-Claude Bradley: Right. Homolytic, homo means the same, so they get the same number of electrons. Heterolytic, different. The one that we're going to be concerned with, to do these calculations, is going to be the homolytic. On page 134, there was a whole list of bond association energies that are relevant to this chapter, that's what I'm going to use to do the following calculations.

In order to look at that, which is going to tell us how much energy, if energy is being released or absorbed by the reaction. We are going to draw only the bonds that are being broken or formed in that first reaction. I'm going to draw a CH4 like this, CH3-bond-H. Then another bond that is broken is a bond between the two bromines. All I need to do to figure out the energies of this reaction, is I'm going to write down the energies for the homolytic dissociation. For the CH bond in CH4, we have 104 kilocalories per mole. For the bromine, we have 46. For the CH3-Br, 70 and then 88. The higher the number, the stronger the bond is. You can see here very quickly, the bond between the two bromine atoms is a lot weaker than the CH bond in methane, which is partially why bromine is so reactive.

To calculate the energy of the whole reaction, we're basically going to sum up the bonds that are broken, minus the bonds that are formed. That's just basically (104+46)-(70+88), those are all kilocalories per mole, we end up with -8 kilocalories per mole. What does that tell you about the reaction?

Student: It's giving off energy.

Jean-Claude Bradley: Right. It's giving off energy when you have a negative number. It's not a huge amount of energy, but still, it's going to be definitely liberating heat as you do this reaction.

The first thing that it tells us, if we look at the energy diagram, if we start off with CH4 and bromine on the left, we end up with CH3-Br and HBr on the right. The only thing that we really know is that we are going to have an eight kilocalorie per mole difference between the starting material and the products. What we don't know is the landscape between those two, what actually happens energetically between those two. That's what we're going to look at now. We're going to fill in this diagram as we do the mechanism.

You want to think of this in a logical way. What happens is we're going to take a container and fill it up half with CH4 and half with bromine. The first thing is that we know that without light, nothing happens. I could have bromine and methane coexist and nothing will happen. Obviously, light is going to be involved in the first step of this mechanism. The very first thing that happens, methane is pretty inert to light, but bromine is photosensitive, so the bromine is going to dissociate, like that. That requires light. H nu is traditionally the symbol you use for light in organic chemistry.

Again, all of the electrons, because we're going to use these arrows extensively through the rest of the class. The half arrow tells you it's moving only one electron. Whenever you break a bond, using only half arrows, you're going to need two of them every single time, because I have to remove two electrons to break a bond. The first thing that this tells us is that we now have something new in the mixture, and that is a bromine atom. When I write Br dot (Br.) what I actually mean by that - I'm only going to do this once because it's a pain to draw it like this every time - is the full correct Lewis structure for a bromine atom is a bromine with seven electrons around it. It's neutral; bromine only has seven electrons. Now that we've done the first couple of weeks, you know what it should look like. We are going to abbreviate that by just drawing Br dot, but there are three lone pairs there that exist.

We're going to follow this down logically, to a generated bromine radical. There are only two things that can happen, the bromine radical can reaction with either Br2 or it can reaction with CH4. It probably does reaction with Br2, but it will just break up again into another Br2 and a bromine radical. That's not something that we care about, what we care about is the reaction between the bromine radical and the methane. You don't really even have to memorize this, if you just think about it logically, about what gets added.

Now, I have something new and there's only one logical thing it can do and that is to reaction with the methane. It's going to take three arrows to do this. The bromine radical is going to form a bond with the hydrogen. We're going to make HBr at this point. In order to form a bond between the bromine and the H, it's going to take me two electrons. Bromine only has one electron to donate, so the other electron has to come from somewhere. The only place that it can come from is between the CH bonds. When I'm making a bond, I'm taking two electrons and bringing them together in space. I'm going to depict that by putting one half arrow in space, between the Br and the H, and the other one will come from the bond.

If you think about the meanings of the arrows, you're not going to make mistakes. If you just look at this, it should be obvious that this is wrong because I can never just remove one electron from a bond. What would I have, where is the other electron? That electron has to go somewhere. Where is that going to go? It has to go on a carbon. It can't go on a hydrogen because the hydrogen is fully bonded now to the bromine. It's not going to go off into space. You have to put it someplace and the only place it can go is on the carbon. I'm going to, again, start the arrow on the middle of the bond, and end it up on the carbon atom itself. That will give me BrH and CH3 radical, so this is CH3 that's uncharged. How do you know it's uncharged? Go back to the basics that we learned in the first week; there are four electrons around the carbon for charges. Carbon only has four electrons, so it's not charged.

With this step, we have now generated two new species. We now have HBr and we have CH3 radical. Now, there are a number of things that can happen here. First thing you'll notice is that we generated the HBr, which was one of the two products of the reaction. That's actually how that HBr forms.

The next product that we have to make is the CH3-Br. That's going to be a reaction between the methyl radical and some more of the bromine. The flow of the arrows is going to be completely identical to number two. I'm going to take the radical and put it into space between the C and the Br. I take the other electron from between the two bromines, because we're going to break that bond and have it join the other one. Again, if I leave it like that, I have to show what I'm doing with the other electron in that bond, so I have to put it on the other bromine, just like I did in number two. There are lots of arrows here, but it's the same pattern over and over again.

I've just now created CH3-Br and a bromine radical. I already have the bromine radical in here, so the only thing new is the CH3Br. This is actually an interesting reaction because I now have a bromine radical. That bromine radical can go back into step two and react with another methane molecule. I have here basically, the way that it's drawn, an endless cycle. For every step two that I don't I make a bromine radical, for every step three I do, I can go back and do step two again. Steps two and three keep going in a cycle. It doesn't go on forever, for one, I have a limited amount of material here. Presumably, after all the methane and bromine is gone, then the reaction is over. More practically, this will only go around ten thousand or a hundred thousand times before it stops. Anybody have an idea why it might stop?

Student: Because the reactants will eventually run out?

Jean-Claude Bradley: No, I'm saying that way before I run out of reactants.

Student: Some bromine might [unintelligible] and Br2.

Jean-Claude Bradley: Right. As long as I use up the bromine in step three and step two, I can keep going until I run out of reagents. However, if I trap that bromine radical, then it's over, and I have to start over. I have to go back to step one and generate more bromine radicals. That step, where you take two bromine radicals, or generally any radicals, in steps two and three, if you combine them, you now have a termination step. That's why it's not infinite.

I'm going to draw steps four, five and six. Step four could be the reaction of two bromine radicals. Step five could be a bromine radical and a methyl radical. The third possibility is I could have two methyl radicals combining. These are very straightforwardly just two half arrows because that's all I have. I just have two single electrons; they come together in space somewhere and they create a bond.

Basically, we have all the steps of the mechanism, there are six total steps, but they're divided up into categories of reactions. The last three are the termination steps, because they stop steps two and three. Steps two and three are called propagation steps and number one is called initiation. You'll notice here, light is only needed in the first step. If you were to flash a brief pulse of light, what would happen is you would generate the bromine radicals, then you get steps two and three, like I said, ten thousand or a hundred thousand times, it would go over, and then it would be over.

The thing you have to understand about the relative importance of each one of these steps. Why is it that steps two and three can go on for tens of thousands of times? That has to do with the probability of two radicals finding each other. After I do step three, I go back to what's in the mixture. I have bromine radical, and around it I've got 50 percent methane and 50 percent Br2. The probability is it's going to find a methane molecule way before it has a chance of finding another bromine radical. It's all a concentration thing. That's why the termination steps are so rare.

Consequently, when we look at the energies of these things, a common misunderstanding when doing this is that step five produces one of the products, CH3Br, but it does so in an insignificant amount. You're only going to have one part in every hundred thousand methyl-bromide molecules made this way. It simply does not come into our calculations of the energies. It's important that you understand that because it really is a big distinction when we do the energies. The terminations are just aberrant behavior in the system.

What we are going to do next time is we're going to look at the energies of those steps, and then make predictions.

Lecture 010 Transcript

noreply@blogger.com (Jean-Claude Bradley) — Fri, 19 May 2006 12:20:00 -0700

Lecture 010 - cycloalkanes
Speaker: Jean-Claude Bradley

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Jean-Claude Bradley: So we're going to resume the Newman projection.
We left off at propane last time, so I'm going to do the Newman
projections for butane.

So when we look at butane, we've got four carbons. So we're looking
at the Newman projections for butane, and when we look at that,
there's actually two different Newman projections we can do with this
one because I could draw a box around the C-1 to C-2, or I could draw
a box between C-2 and C-3. So, with butane, we're going to look at
C-2 to C-3 because it gives you more different confirmations. But
generally, with molecules that have more than one possibility, I will
tell you which two carbons to do the Newman projection around. So,
if we're doing it from C-2 to C-3, then we draw the box.

Next step, we go and draw the circle, and it doesn't matter which one
you put in the front and which one you put in the back; as long as
they're consistent with all the projections, it'll work out. So, I'm
going to start first with a staggered conformation, they're a little
bit easier to draw. And I'm asking a question, if I decide to make
the left carbon the front, the question is what are the three things
that are connected to that carbon? A methyl, an H, and an H. So
that's what I'm going to put in the front of the Newman projection.
A methyl, an H, and an H.

Now, why did I put the methyl on the top instead of on the right? It
doesn't matter where you start, because when you generate all the
Newman projections, you're going to generate them all, regardless of
where you start. Okay, so the important point is just to put them
down there somewhere. For the right carbon, that'll be the one in
the back, and again we have a methyl, a hydrogen, and a hydrogen.
So, I'll put the methyl at the bottom arbitrarily, and then I'll put
the other two hydrogens.

So just like what we did for propane, I'm going to keep either the
front or the back constant, and I'm going to rotate the other part.
So in this case, let me make the back constant. So I have to put the
methyl on the bottom like that. And now I'm going to rotate, and I
generally rotate clockwise, so I'm going to rotate this clockwise
until the methyl group is on the top of the first hydrogen that it
comes across. You want to go to the center, as much as possible.
Okay. So, we have now the methyl on top of the hydrogen, and you
notice the other hydrogens on top of the other groups.

Now we want to continue rotating until we generate a duplicate
structure. So I'm going to keep rotating until the methyl group is
around the four o'clock position. Now it's really important that if
you decide you're rotating one side, then you don't change that.
After you draw the first projection, you have to maintain that
throughout all the Newman projections, otherwise you're going to get
confused and you're going to come up with the wrong counts for them.
So you'll notice here that the back here is going to remain unchanged.

So, now the two methyl groups are close to each other, so you can see
that the third Newman projection is different from the first one,
right? So now if I continue to rotate, I can draw one more. Okay,
again the back is constant. And now I'm going to show that the two
methyl groups are on top of each other. Okay. So, if I were to
continue to rotate clockwise, the methyl group would show up here.
Well, that actually is the same thing as the third structure, and as
I continue to rotate, I will basically create duplicate structures as
I go along, so these are the four conformations that are important
for butane, that's it.

Now, they have names, and the first one, because the methyl groups
are on the opposite sides, it's called Anti. So that's the Anti
conformation. Now, when the methyl is on top of a hydrogen, this is
called an Eclipse conformation, but we have to distinguish between
this eclipse and this eclipse, so we call this partially eclipsed.
Someone ran out of imagination when they went to the eclipse
structures, unfortunately. The staggered ones, though, are still
pretty interesting. When the two methyl groups are next to each
other, that's called Gauche. And the final one you might guess from
the partially eclipsed, we go to totally eclipsed when the two
methyls are on top of each other.

Okay, now we have to make the distinction between the name of the
conformation and the class of the conformation. When I refers to the
anti and the gauche, both of those conformations are staggered.
Okay? However, because we have more than one staggered conformation,
staggered is not the name of the conformation; it's simply the class
of the conformation. So whenever you have a conformation where the
front group is inbetween the back group, that's always going to be a
staggered conformation, but you'll have to be more specific, so
someone will know exactly which one you are referring to. So we use
staggered as a class, and of course we can also use eclipsed as a
class. Okay, so that's the nomenclature of the master here.

The other thing is that when we use the term gauche, anti and butane,
we can specify, Draw me the gauche conformation of n-butane, but I
can also say, that the two methyl groups are gauche to each other. I
can say that the two methyl groups are anti to each other in this
structure. So I can use that term also not just as a name, but to
describe the relationship in a Newman projection, and I'll be doing
that when it's relevant.

So what I want to do now is draw an energy diagram for this as we
rotate through. So the anti has the two methyl groups as far away
from each other as possible, okay? And so you remember that if you
have nothing that is attracting two groups, they'll naturally be
repulsed from each other, so the farther away I can put those groups,
the lower the energy. So we're going to predict that the anti will
be the lowest energy. The two eclipse structures, they're the groups
that are absolutely as close as they can be, and specifically for the
totally eclipsed, I have the two methyl groups in the closest
position, so that one we predict will have the highest energy in the
diagram. Okay, so I'm not going to give you specific number; I just
want you to be able to gauge the shape of this thing.

So if we do the same thing in an energy diagram that we did for
ethane, let's start with the lowest energy, being the anti, and as I
rotate I'm now going to go to an eclipse conformation. The eclipse
conformations will always be at the top of an energy hill, right?
Because as you reach the peak where they're on top of each other,
that's the most unstable that the rotation will take you, so I'm
going to go up here some more, like this. So again, maintaining the
back constant. All right, so that'll be the top, and as I continue
to rotate the methyl group, I'm going to go back to a staggered
conformation. Specifically the gauche.

So, if I want to try to guess where the energy could be, we know that
it's going to be at the bottom of an energy well, but it won't be as
low as the anti, because it actually is closer, the two methyl groups
are closer in the gauche than it is in the anti, so we're going to
predict somewhere in between. And then finally, when we have the two
methyl groups on top of each other, that'll be the highest energy.

If I continue the rotation, all I'm going to do is draw the mirror
image of this, so this would keep going on and on as I rotate. You
can actually calculate what the various energies are. Why is that
important? Because in this case, if you know what the energies are,
you can actually predict what percentage of the molecules are going
to have a certain conformation, which is important especially from
ecology, when drugs fit into receptors, you need to know what they
look like at the temperatures you're going to be using them, and
that's where this comes from, originally. I think that's all I
wanted to do with Newman projections. Do we have questions?

Let's talk a little bit about cycloalkanes. So, we're just talking
about molecules that have just carbon and hydrogen and that are in a
ring. The smallest ring you can make is three, so I can have
cyclopropane. Generally, three-membered rings are less stable than
larger rings, but it doesn't mean that they're impossible to make, so
cyclopropane is a gas, and you can buy it, and it's pretty stable,
but it's going to be less stable than cyclohexane, for example. But
when you're drawing this thing, you're drawing it flat, and each one
of these centers is sp3 hybridized, so whenever you draw a ring
that's an alkane, you're going to have one bond coming towards you,
and one bond going out. So if I want to draw the hydrogens on this
thing, I would do it like this.

Okay, and you'll notice that you can't rotate around those bonds. So
if I were to replace one of the H's with a methyl group, I find that
there are two possibilities. I can either have the methyl groups on
opposite sides of the ring, or they can be on the same side. So
those are two different compounds because I cannot rotate around this
sigma bond. All right, if I have an open-chain alkane, I can rotate
around any sigma bond I want, I have free rotation, and that's why
you have different conformations, but here we actually have two
different compounds, because they're locked in the same position,
just like the double bond locks a molecule in position. So we would
name this thing trans-dimethyl-cyclopropane. Or trans-1, 2.

So as I go from cyclopropane to cyclobutane to cyclopentane, if you
make molecular models of these things, you'll find that it's pretty
much going to stay flat. You're basically going to have a really
flat structure. With cyclopentane, you're going to see that it
buckles a little bit, but it's still really very close to being flat,
but when you hit cyclohexane, it's definitely not flat anymore.

So if we want to look at the conformations of cyclohexane, because
it's not flat, it's going to get bent in certain ways, and we want to
talk about the certain ways in which it's bent, because we absolutely
have to know what these molecules look like, depending on what we
want to do with them, so for cyclohexane, we're going to look at it
from the side. So this is side view of cyclohexane, and the reason
that it bends like that is because you have 109.5 degree angles, and
you can't make a flat hexagon that has those angles. So you find
that basically it can bend like this, or it can bend like this. And
those two conformations have the names of what objects they resemble;
this is called the chair conformation, and this is called the boat
conformation. And whenever you have a comparison between the chair
and the boat, the chair is more stable. Okay, so that means that if
I have a solution of cyclohexane, there are going to be more
molecules that are going to look like the chair than the boat, but
those two interconvert all the time. It's just that we want to know
what the most stable conformation is. If I just have cyclohexane,
there are only two conformations, but if I start to put some situates
on there, we're going to find that we sometimes have more than one
chair conformation.

Let me give you an example with methylcyclohexane. Okay, so here's a
skeletal formula of methylcyclohexane. That doesn't tell you at all
what the molecule looks like, because it's drawn flat here. What we
want to know is what does methylcyclohexane look like the most when
it's in 3D? So the first thing is we don't bother with the boat
conformations because the boat is going to be less stable, so what we
want to do is compare the chair conformations of methylcyclohexane
and see how we can analyze if one is more stable than the other.

Now to do this, you have to get used to drawing this little chair
side-view, so if you've managed to draw this, you'll find that you
have edges that are pointing either upwards or downwards, so let's
say this one here on the left, this is definitely pointing down, so
I'm going to draw a vertical line pointing down, like this. Exactly
vertical. This wedge here is pointing up, so I'm going to make a
vertical line pointing directly up. This wedge here is pointing
down, same thing. This is pointing up, I go up. Those are the easy
bonds to draw because they are exactly vertical.

The other bonds we have to draw are a little bit harder because all
the angles here are 109.5 degrees, right? So in order to do that I'm
going to draw an angle that's a little bit larger than 90, so I'm
going to do something like this. So not exactly horizontal, you want
to make that angle a little bit more than 90. I'm going to go around
the molecule and do that everywhere. Okay, so I strongly recommend
that you put the vertical bonds first because they're easier.

So I have to put the methyl group somewhere, and you'll notice
there're really only two kinds of bonds here, I have the bonds that
are going up and down, and I have the bonds that are kind of going
horizontally, so there's only two different places I can put the
methyl group. So on the left structure I'm going to put the methyl
group on the bonds that go up and down, and that of course means that
all the other things are hydrogen's. Okay, so the methylcyclohexane
is flipping around, and the other chair conformation I want to be
able to do for that. The easiest way to do that is learn to draw
this template once and for all, and then we're going to put the
groups on it in a different position, so once again, I'm flipping all
of the vertical bonds, and then the horizontal ones. So if I put the
methyl group on a vertical bond, now I want to put it on a horizontal
one. It doesn't really matter which horizontal one I put it on, but
I'm using this carbon because it's more accessible to me in terms of
my ability to draw a methyl group on it. It's easier than the bonds
that're in the back here of this projection, so that's the only
reason that I'm using this particular carbon. And the rest is loaded
up with hydrogen's.

Okay, so now I've drawn the two chair conformations of
methylcyclohexane. The difference between the two in terms of
stability is that when they methyl group is on one of these vertical
bonds, it's actually pretty close to these two hydrogen's. That's a
1, three interaction in cyclohexane. So the bigger the group, the
worse that that interaction is going to be, but still with methyl
it's significant enough that the molecule wants to put that methyl
group far away from the ring, so when it puts it on these horizontal
bonds, it's no longer close to these other hydrogen's.

Okay, so we know then that the structure with the methyl group on the
horizontal bond will be more stable. Okay, so these horizontal and
vertical bonds have names actually, the ones that're going straight
up and down are called axial, and the ones that are going near
horizontal are called equatorial. So equatorial should be easy to
remember, near the equator, horizontal. So in general what we want
to do is put the large group in the equatorial position. So if
you've seen the quiz for this material, you'll see that there's
questions like that, or a certain derivative of cyclohexane in its
most stable conformation, where's the methyl group? Well, in its most
stable conformation, they methyl group is equatorial, that's how you
answer those kinds of questions. Now when you only have one group,
it's pretty easy because there's only two choices, equatorial or
axial, but let's take a look at what happens when we have more than
one group on the cyclohexane ring.

So I'm going to again start with the skeletal formula so we know what
we're dealing with. So I have 1-methyl, 2-ethylcyclohexane here, and
notice that they're on the same side. Up and down will have nothing
to do with axial or equatorial, and you'll see when I draw the
picture. The up and down will have to do with whether it's coming
from the top of the ring or the bottom, and that's a little tricky
the first time you see that, so I'm drawing this skeletal formula so
we keep in mind what exactly we are trying to draw. I am looking for
the most stable conformation, so I ignore the boat. I draw my two
templates. So I can start ending where I want, it doesn't matter,
but I like to use these carbons in the front, because I have more
room to draw things with, but it really doesn't matter where you
start, you'll end up with the same answers. So I'll just arbitrarily
put the methyl group over here.

Now I have to put the ethyl group, where? I have to put it on the
next carbon, so the next carbon would be this one right here, for
example. Now, they're on the same side, so let me actually draw it
over here. In one of them, the methyl group will be equatorial; in
the other one the methyl group will be axial. The question is, where
is the ethyl group going to be? So what you're looking at is the side
view of the cyclohexane. You'll find that there's always one bond
that's higher than the other, so those two yellow dots that I've put,
those are both on the same side of the ring, they're both underneath
the ring, whereas these two dots are on top of the ring. So if I put
the methyl group on the yellow dot, I have to put the ethyl group on
the yellow also, and then I have hydrogen's everywhere else. When I
draw the other conformation, I flip both of those groups, so now the
ethyl group that was axial will now be equatorial at this position,
and the methyl group that was equatorial is now axial. Both of these
are the same compound, they're two different conformations of the
same molecule.

So now we want to determine which one is going to be more stable.
Well, you'll notice that, in each case you have one that's
equatorial, and one that's axial. We want to put all the groups, if
we can, equatorial. We can't do that, because we don't have one of
the conformations that does that, so we want to pick the conformation
that has the biggest group in the equatorial position. That would be
this one here. So even though we have the methyl group that's axial,
the ethyl will take over.

[question from student]

The question is, if there are cysts, are they both going to be axial
or equatorial? It depends on the position. For a 1, two
relationship, the cysts will have one axial, one equatorial.
However, if I have a 1, three relationship, then if they're cysts
they're both going to be axial, or both of them will be equatorial.
So in other words, don't try to memorize it or anything, just learn
how to draw the picture, then put them and see what it turns out to
be. It's the only way you really can do these problems.

Okay. Did I cover everything here? Do we have any other questions?

So these questions take time. I certainly cannot do them in my head;
I have to draw these out, so you definitely will have to. I'm going
to ask you questions about this, then I'm going to look at the
picture. So for example, this one, the methyl group is axial in the
most stable conformation. You couldn't possibly know that unless you
went through this process of drawing them out to find out exactly
what is the lowest conformation for the whole molecule. Okay, so
you've got some practice with the quiz, and some of the problems.

Lecture 007: Alkane Introduction

noreply@blogger.com (Jean-Claude Bradley) — Fri, 19 May 2006 10:26:00 -0700

Lecture 007: Alkane Introduction

View Lecture

So we're going to resume the polarity issue and talk about intermolecular forces. So we spent a lot of time building up since last week to this. Lewis structures determining the shape and determining if the molecules were polar or not.

The reason for that is now you can use that information and predict some relative properties such as melting point and boiling point. In order to do that you need to understand what forces keep molecules of the same compound together.

Okay, so there are three major forces, and we're going to start with the most powerful force and go down to the weakest.

The first force is called hydrogen bonding. It occurs whenever you have an X-H... :X pattern like this, where X is nitrogen, oxygen or fluorine. So you need some fairly electro-negative elements to form a hydrogen bond.

So an example of this, the example you are probably most familiar with, is water. So water basically has an X-H component. The oxygen is the X in this case, and we find that we can line up another water molecule next to the first one. Through the lone pair of the oxygen, on that second molecule we can form a hydrogen bond.

So that hydrogen bond is always drawn from the lone pair to the hydrogen. It is basically that lone pair is being shared a little bit between the oxygen and the hydrogen. So it is not a full bond. It's a very partial bond, but it has a strong enough that that's the first thing we need to consider when we talk about intermolecular forces.

There are three main forces and you go in that order. So if you find that your compound has hydrogen bonding, then you are done. The major force holding water molecules together is hydrogen bonding- the strongest force. You don't need to go to the other ones because they will always have it, but it will be less strong and so it won't contribute as much to the intermolecular forces.

Okay, so that's the strongest. The next force here, number 2, would be dipole-dipole. And you will get this when first, you have no hydrogen bonding so you go and ask, "Is there any dipole-dipole interaction?" Whenever you have a polar compound then yeah, you will have dipole-dipole interaction.

Let's pick something we've already used before, chloromethane. And you recall this has a dipole. There is a polar bond from the carbon to the chlorine making the chlorine slightly negative and the carbon slightly positive.

So again, I'm using the delta symbol. That's not a formal charge. Okay, a formal charge would just be a plus and a minus. This is a delta or a slight minus and a slight plus.

So the way the dipole-dipole works is the two molecules line up anti-parallel, like this. So in blue here I can show that it still comes down to a plus minus attraction, but it's partial plus to partial minus.

Okay, so that's how dipole-dipoles line up and that is why chloromethane will be attracted to itself. The stronger the interaction, the harder it will be to pull those molecules apart. So it's easier to pull chloromethane molecules apart than it is to water molecules, for example.

Okay, the third one, which is the default force, is the Van der Waals interaction. So I'm not going to go into detail here about the differences here with London dispersion and all that. It's basically hydrophobic interactions.

When you don't have hydrogen bonding, when you don't have dipole-dipole, the only thing that would be left would be the Van der Waals or dispersion forces. So let's pick an example here like methane. So methane has no significant dipole moment. So you would say that would be the force responsible for the intermolecular attraction.

So this is the weakest force, which means that it will be easier to pull apart two methane molecules than two chloromethane molecules or two water molecules, of course.

Okay, so that's what we're getting at in terms of talking about the forces. Let me give you a specific example of how you would use this and the type of questions you might come across.

You may be asked to predict which one would have a higher boiling point, ethanol or dimethylether. So drawing them in a condensed form is not terribly useful because we can't tell exactly what's happening. So I am going to redraw those molecules in their full Lewis structure with the geometry and everything correct, well at least correct to the oxygen.

Okay so I'm trying to draw the tetrahedron for the oxygen on ethanol. So that will have two long pairs. And let's do the same thing for the ether. So the lone pairs are important because we're going to need them if we are going to worry about the dipole moment.

I didn't draw, for ethanol, every tetrahedral structure, you know for the CH3 and the CH2. The reason for that is that there is no significant dipole moment between carbon and hydrogen. So it doesn't really matter what they look like, there won't be any vectors to draw there. The important part of this molecule is the oxygen, because there are significant dipoles and you have to add them up, at least to confirm that they don't cancel out.

So you're going to go from plus to minus. All right so from the carbon, towards the oxygen. Carbon and hydrogen have very similar electro-negativity. So if you predict that there will be a dipole moment between carbon and oxygen, you would also predict that this would be the case between the hydrogen and the oxygen. So I'll draw another arrow going in that direction.

And for oxygen and nitrogen you typically take into account the lone pairs, drawing a vector from the oxygen or the nitrogen through the long pairs like this.

Okay, so we have a tetrahedral structure that has four arrows on it. The key thing that I want to point out to you is that they definitely don't cancel out. Right? Everything is going up.

So I'm not trying to get you to draw the exact place just to be able to draw it well enough to understand whether or not it cancels out. And clearly in this case it doesn't. So let me draw the overall vector up and to the right in this case.

And for the dimethylether we're going to do a very similar analysis. Draw two vectors from the carbons to the oxygen and then through the long pairs. Okay, so once again we have a dipole moment for both of these molecules, ethanol and dimethyl ether.

Now, now that we have this information. Let's step through these forces. For the first force, what we want to ask is "Is there hydrogen bonding?" So, is there hydrogen bonding in ethanol? What do you think? It's possible.

So again, let me draw this out. Now we're looking for an X-H X pattern. Okay, so in this case the X-H will be O-H. We also have CH3, and CH2. All right and we have the other oxygen. Again worry about the lone pair, because you need the long pair to do the hydrogen bonding.

Draw the hydrogen bond like this.

Okay, so in being able to draw this X-H X pattern, we confirm that we have hydrogen bonding. So the answer to the first one. We stop there. It will be hydrogen bonding. It will be the major force.

Okay, so do we have hydrogen bonding for the ether? We have an X. Right? And we have hydrogens, but that's not enough. We need to have a situation where we have an X directly connected with a covalent bond to hydrogen. And we don't have that here. The oxygen is only connected to carbons so there is no hydrogen bonding for the structure on the right.

Okay, so we then go to the next question- "Do we have dipole-dipole attraction?" Yeah, if you have a dipole moment, you'll always have a dipole-dipole. So the answer to the one on the right, to the question "What is the dominant force?" would be dipole-dipole.

Okay, but I worked out the fact that the ethanol has a dipole moment as well. It also has dipole-dipole attraction, but it's not important because it is overshadowed by the hydrogen bonding. Okay, so that's why I have specified the dominant, not the only.

Okay, so with this information now we know the intermolecular attraction for ethanol will be higher than for dimethyl ether. Okay, so it's going to take more energy to separate two ethanol molecules, to take them from a liquid state to a gas state, than the ether molecule. So we predict that we have a higher boiling point for ethanol on the left.

Okay, and the boiling point for ethanol is 78 Celsius, whereas the boiling point for diamethyl ether is -25 Celsius. Okay, so there is almost a hundred degree difference between the two. So hydrogen bonding is a really important force, as you can see. It has a huge impact.

Now we can't compare any two molecules when we do this comparison because you know what happens if I have a big molecule on one side and a small one on the other and they have a different groups on them. Well you can't really do the comparison in that case.

The only reason we can confidentially say that ethanol has a higher boiling point is that they are the same size, or roughly the same size. So if that criterion is met then you can do all the comparisons for their respective boiling points.

Does anyone have any questions on this?

Okay so melting point is the same principle. When you are melting a compound, you are basically separating molecules, except that in this case you are separating them from a solid state to a liquid state, but it is the same principle. So the higher the force the higher the melting point, as well.

And we'll do some examples of that. I gave you some problems to do and we can see that later.

Okay, the other physical property I want to cover is solubility. Ok, where like dissolves like. Okay, but let's think a little bit about what it means for one compound to dissolve into another.

It's true that generally like will dissolve like, but what happens in the situation where two compounds don't dissolve. If we look at it from the situation of intermolecular forces, what you are comparing is "If I replace a molecule, interacting with itself, is that going to require energy? Is it going to give off energy? Or, is the energy going to be pretty much the same?'

Okay, so let's do a couple of examples and see if that becomes clear.

We want to predict if water and methanol will be soluble in each other. There's methanol. There's water. So, let's go a little bit more quickly over this. Methanol is really very similar to ethanol. Its dominant intermolecular force will be hydrogen bonding. With itself, it will hydrogen bond.

For water it's the same thing. Its dominant intermolecular forces will be hydrogen bonding as well. So let me draw a little water molecule bonding to it self with a hydrogen bond through the lone pair. The same thing will happen with the methanol.

So if you mix these two together, when one dissolves in the other, there is going to be an interaction. So the question we now want to as is: "What is the interaction between the methanol and the water? Can they hydrogen bond to each other?"

Well yeah, because one will have an X-H and one will have an X.

So let me draw the mixture. Here's the methanol. And here's the water.

So the dominant intermolecular force between methanol and water is also hydrogen bonding.

So what it means is that I am replacing hydrogen bonding with hydrogen bonding. So I'm not really losing that much in energy. Therefore, we would expect those two compounds to be soluble in each other. And they are. In fact, they're miscible, meaning they're soluble in all proportions.

Now let's take another example of two compounds that are different from each other. So let's look at, for example, CH4 and water. So water, we just determined, was hydrogen bonding. And for methane, does it have hydrogen bonding? No. Does it have dipole-dipole? No, because it doesn't have significant dipole, therefore it has to be Van der Waals interactions.

So what will be the force between methane and water? I have a methane molecule. I have a water molecule. Is there any force between the two?

Well, there's always a force. Okay. You can never have no force. So it's not hydrogen bonding. It's not dipole-dipole. Because for dipole-dipole you need two dipoles. So the water has dipole, methane doesn't. So the only thing left is Van der Waals.

So yes, there actually is a positive interaction between the two and it's attractive. So for the mixture, the dominant force will be Van der Waals.

So what you might think is; "Well if it has an attractive interaction, what's the difference? Why doesn't methane dissolve?"

And the reason for this is that in order for one methane molecule to go into the water, it has to displace a water molecule, to do that it has to break a hydrogen bond. If it didn't have to do that, there would be no reason for methane not to go into the water because it's an attractive interaction. So that's basically the reasons that compounds are not soluble in each other. Okay, because you're replacing a strong force with a weak one. So we would predict that these two would not be soluble, and they're not.

Okay, now solubility is a graded kind of thing. You can have miscible compounds that are soluble in all proportions like water and methanol. I mean methane has a measurable solubility in water, it's just so small that we say that it's insoluble. But keep in mind, with all of these things, there is a certain amount that will dissolve. It's just that we have to draw the line as "is it a reasonable amount?" And in this case its not.

What I want to take a look at now are functional groups. So to make things simpler for chemists, instead of considering each molecule different from another, chemists have figured out that certain kinds of functionalities on compounds give similar kinds of reactions. So you don't have to memorize reactions for every compound in organic chemistry. You can put them into classes. And for compounds that have never been made before you can predict how they are going to behave, pretty much. Okay, so that is what the purpose of functional groups are.

We are going to take a look at maybe a dozen functional groups that I'm going to expect you to know and be able to recognize. And there are going to be some examples that I'm going to ask you what functional groups are in a molecule.

So if you have any questions this is a great ask because you might have a few questions.

The first one is not really a functional group. It's kind of the default. If you have no functional group, you'll have an alkane. So an example of an alkane will be CH4, methane.

And later on we're going to take a look at some of these functional groups in greater detail. Okay, so right now I just want to list them, but we're definitely going to look at alkanes shortly.

The next one is alkenes. So alkenes have carbon-carbon double bonds. So I'm going to list the example ethene or ethylene. It has a carbon-carbon double bond with four hydrogens. So you can have an alkene if you have different kinds of groups on the carbon-carbon double bond. So if you have alkyl groups, aromatic groups, whatever, but it doesn't change the fact that this will still be an alkene.

Sometimes if you put another group on, it will change the functional group. And I'll let you know when that is, but in the case of alkenes, there is not much you can do to it to turn it into something else.

The same thing for alkynes, which is whenever you have a carbon-carbon triple bond. So just as an example, here I'm going to put a propyne. And there at each end, there could be a hydrogen, there could be an alkyl group, could be an aromatic, it doesn't really matter, it's not going to change the fact that you would call that an alkyne.

Alcohols, methanol is the one that we've talked about before, CH3-OH. So you'll have an alcohol whenever you have an R-OH pattern. Where R is an alkyl group, an aromatic, whatever, it's going to be an alcohol. But R can't be certain things. If R is a carbonyl group for example, then you no longer have an alcohol, you have a carboxylic acid. So you can't put anything you want under R, and that's why I'm saying, if you come across something and you're not sure, definitely ask me. But any alkyl group will certainly stay as an alcohol.

An ether would be similar to an alcohol, but instead of having H, you would have another alkyl or aryl group. For example diamethyl ether. So in this case we have R-O-R'. This is Where R and R' are alkyl or aryl groups.

So an aryl group would be like benzene. Benzene, benzene ring. So one thing that R can't be is a hydrogen because if R or R prime are hydrogens then I have an alcohol. That's the kind of limitations that we have to put on these functionalities.

Aldehydes. So aldehydes will have this format. C double bond O, -H. And you can, you can also have a hydrogen. So formaldahyde H C double bond O, H is also an aldehyde. And aldehydes will have the general formula R- C double bond O H where R can be a hydrogen, it can be an alkyl group, or an aromatic group. I don't think we can change the functionality. If you have a carbon at R, you're pretty much going to have an aldehyde.

Ketones are related. In this case you don't have a hydrogen on one side. You have C double bond O and you have two alkyl or aryl groups on either side. So general formula would be R, C double bond O R'. And here R and R' can't be hydrogens. The have to be an alkyl or aromatic group.

A lot of this might seem arbitrary. You know, what difference does a hydrogen make? And it really comes from the reactivity. It turns out that aldehydes are similar to keytones, as we will see, but they have enough difference that it's worth it to call them a separate functional group. Okay, for example aldehydes oxidize really easily into carboxylic acids. Keytones don't do that.

The naming comes from how they behave.

Carboxylic acids. So an example is acetic acid. Basically that is what is in your vinegar. It's just a solution of acetic acid. It has the formula R, the general formula R -C- double bond O, OH. So here in our the example. If instead of C double bond O we had a CH 2, this would be an alcohol. By putting a carbonyl group, it makes a different functional group.

Okay, if it didn't change the functionality, the behavior of the compound, we call these alcohols. But this change actually makes these compounds, as their name says, very acidic compared to alcohols. They have all kinds of other reactivity.

Acid Chloride. Acid Chlorides are very similar to carboxylic acids. In fact, you make them from carboxylic acids. They have the general formula R,C- double bond O Cl. And these are very reactive compounds. You'll see that we use them as reagents. They'll react very easily with some compounds.

Esters are similar to Carboxylic acids. We still have the C double bond O, O pattern, but now we have to have another alkyl or aryl group on the end of the oxygen. So the general formula is, R- C- double bond O, O, R', where R' here can't be hydrogen, it can be an alkyl or aryl. Okay, so in the case where R' is a hydrogen, then I don't have an ester, I have a carboxylic acid.

Next we have amides. So amides will have C- double bond O and then a nitrogen. And there's all kind of things we can put on the nitrogen on the other side of the carbonyl group. So we can generalize that into R, C double bond O, N and then R' and R''. And here you can have hydrogens, or alkyl, or aryl anywhere. So that won't change the fact that it's an amide.

Don't confuse amides with amines. Amines are similar. They also contain nitrogen, but there's no carbonyl group. So trimethylamine is an example. So in general R, R', R" can be anything. It can be hydrogen, an alkyl or an aryl group.

Okay, questions?

Well I'm sure you come across some things when you start to do the problems. You're asked to identify different functional groups in molecules so we'll see what you run into when you get to that.

Well I do want to talk a little bit more about the hydrocarbons, go into a little bit more detail about that.

Okay, so hydrocarbons basically mean is that all you have is carbon and hydrogen as the name suggests. But from that, there are different kinds of arrangements.

We can have an aliphatic. Aliphatic just means that we have an alkane. We can have aromatic, which means that you have compounds similar to benzene. And we'll get to that in another chapter. Or we can have alkenes, which we saw previously, which have carbon-carbon double bonding. So what we want to do is to look at the aliphatic hydrocarbons or the alkanes in this section.

Okay, so what you need to learn how to do here is to count to ten, in organic chemistry because those first ten names of compounds you are going to use as prefixes for any molecule that you want to name later.

Okay, so we're going to learn, for example that one carbon, example CH four is called methane. So the key ingredient here is the meth that means one carbon. So whenever you come across meth, you know it's one carbon. You know that this is methane and that it's an alkane because it has the -ane ending. But if I put a different ending, for example methanol, I know that it's an alcohol that only has one carbon. So that's what I mean by counting to ten in organic chemistry.

Two carbons, we have ethane. Three carbons propane. Four carbons, butane. Now when we hit butane, there is actually more than one way of connecting those carbon atoms together as we saw when we did one of the problem sets. So in order not to create millions of compounds here, I'm only going to draw the compounds that have a straight chain like this.

So if you start to do branching, then you'd have to use another terminology, but if I want to specify the butane that has all the carbons in one line, I put a small n in front of it, which means normal.

So if you see butyl group and it doesn't have the n, it is implied that it is the normal, that it will be a straight chain.

Five carbons. So in this case, we have a pentane. Six carbons hex, hexane. Seven carbons we have a heptanes, When we reach eight, we have octane. Nine, nonane, and finally ten decane.

Okay, so a lot of these you won't really have to memorize, because we're going to use so much. You know heptane, octane, nonane, decane you'll probably use a little bit less, but you're certainly going to come across the first ones a lot more. And that just tells you how many carbons you have.

Okay, so I had mentioned that there are different butanes that you can make. So if I connect the carbons in a straight chain I would have n butane. And if I make a branch structure, like this you go to isobutane. Okay, now with four carbons there are two ways of connecting them together.

With Three carbons, unless you make a ring out of them, there is only one way you can connect them together. But there's more than one place that you could link to that three-carbon fragment. So, although there's only one propane, there are two different ways that we can connect a propyl group to another subsistent.

So these are going to be fragments, not molecules. Okay, so if I write CH2, CH2, CH3, and put a line here that CH2 has to have one more thing connected to it but the group, outside of that connection, is a fragment. And that's called the n-propyl fragment.

Okay, now if I connect to the propyl group through the middle carbon, like this, now I have an isopropyl fragment. So again, these are not molecules. You have to add something to them to make them real molecules.

So an example would be, lets say I put an alcohol group. So if I put the alcohol on the straight chain group that would be n-propanol. Now if I add something else, let's say a chlorine, to the isopropyl, that would be isopropylchloride.

There are other ways of naming these compounds and we'll get to that. We'll get to that when we get to alkylhalides in detail. But basically isopropylchloride is a real molecule. Isopropyl is just the fragment. There are a couple of these fragments that I'm going to ask you to memorize. I'll ask you for three carbons and for four and then we'll stop.

There are names for the five carbons, but they're not used as much and you know, I don't think it will help you that much. But definitely butyl, you'll need to know.

So let's start with the one you already know n butyl. So, I can link to four carbons in a line by going at the end, or by going at one of the internal carbons, the second carbon. So if I go to the end it's n butyl, if I go to the second carbon it's sec butyl. By rearranging the carbon atom in a branch structure there are again two ways that I can link to that. If I link to it by one of the end carbons like this, that will by isobutyl. And finally I can link the branch structure through the middle carbon and that's called tert butyl, or t. Same thing, you can just put t butyl.

Okay, so with these tools you can construct a lot of the names of the simpler molecules. Questions? Okay.

So, another thing you have to be able to name on molecules is whether your carbons and hydrogens are primary, secondary, tertiary, or quaternary. So, let me pick an example. I'm going to draw propane and if we look at a carbon, and we and we want to ask whether it's primary, secondary, tertiary, or quaternary, the question is how many carbons are directly connected to that carbon. So, in this case, it's just one, so that would be a primary carbon. If I look at the middle carbon it's connected to two carbons, and that would be a secondary, and the bottom one, just like the top one, would be primary.

So we're going to be referring to, you know, a primary center and a secondary center. You always figure it out by counting the number of carbons on that particular group. Now, when I refer to the primary center I'm also going to be referring to the hydrogens on here. So, these three hydrogens on the top, we're also going to say that they're primary. So, the designations for primary will go for the carbon or it will go for the hydrogens as well. Okay, and that's something that you'll need in order to predict the outcomes of reactions. So, if I draw isobutane, this carbon would be what? Primary, secondary, tertiary...primary? Right?

Because I look at it and ask how many carbons are directly connected to it and the answer is one. And the carbon in the middle...it's got one, two, three carbons attached to it, so it's tertiary. I'm going to write these out too. Okay, so the other example is quaternary. So if I have this compound, I have four carbons connected to the central carbon, so that would be quaternary. Okay, so that's basically all there is to primary, secondary, tertiary, and quaternary.

And again, just by using the terms you're going to get to know them pretty well.

Okay, so we're talking about alkanes, let's look at some reactions of alkanes. So the first one is combustion, probably the most important reaction. Example, CH4 + O2 would be C02 and water. So, you get combustion whenever you burn something, and you burn something by combining oxygen with it to generate a lot of energy. So, that's the basic source of our energy and if that didn't happen, we couldn't drive cars, couldn't heat our houses and all of that stuff, so it's a very important reaction.

Now I'm not going to go into detail on this reaction, I will go into detail on some reactions, but not this one. So in this case, all you're really expected to know is that you generate carbon dioxide and water, but you don't need to know how. Okay? Also this reaction doesn't necessarily go all of the way to carbon dioxide. If you have limited combustion you can go to carbon monoxide.

That's also part of combustion. So this basically gives you some sort of an idea, you know, of how combustion works. The key point is that combination of oxygen with the alkane to generate heat.

The second one is cracking, where we take a long carbon chain and break it down into smaller pieces. So, a simple example of cracking would be starting with propane and then reacting it with some catalyst, like palladium or platinum, and generally this is done at high temperatures. What you do is you break the molecule, so I can go from a three carbon fragment into a two carbon fragment and a one carbon fragment.

I'm going to make an alkene in this case. Okay, so I've broken down propane into ethylene and methane, the key thing being that I have reduced the number of carbons. This is not something that's necessarily terribly useful for propane, but it's more useful for long chains.

So for example if you have fifteen chains you want to break them down to pieces of eight to make octane, which is used in gasoline, for example. So that's how a lot of the octane is made, it's not just found in the ground like that. Oil is found. Oil has longer chains, those longer chains are cracked into smaller pieces, and that's how we make most of the octane.

Student:Why is it not ethane.

Professor: Well the reason it's not ethane is I don't have enough hydrogens. Again, cracking is a very complicated subject, yes, you can take hydrogen and do the cracking, or if you do it without the hydrogen you're going to end up with an alkene. But both of them are considered cracking. Any time you break a long chain into smaller pieces.

Okay, next we have halogenation. Example, CH4 reacts with bromine in the presence of light. Okay, so I'm using hυ (H nu)to say that I have light, that's the common abbreviation in organic chemistry. I'm going to write this once. Okay, so what we're doing is we're going to swap a bromine and a hydrogen. So this is called halogenation because I'm using a halogen to do an exchange reaction like this.

Now this reaction we're going to look at in very great detail. So here you're just expected to know that this happens but we're going to look at it in enough detail that you can predict what's going to happen with ethane, propane, benzene, does it still work?

But in order to be able to do that it takes time to go into that in detail. Okay, so there's not a lot of reactions here. Primary alkanes are pretty inert. So there's really not that much you can do to them. These three reactions are probably the most important. Okay, the one thing we want to worry about in alkanes is how do we predict the shapes they're in, not reactions, but how they twist.

So let's consider methane first. So methane has a carbon that's sp3 hybridized, and has four sigma bonds, right, and what we know about sigma bonds is that you can rotate around a sigma bond. Now hydrogen is just a sphere. When I rotate it, it still looks like a sphere. Okay, so basically, although I have sigma bonds, where rotation is possible, it doesn't change the shape of the molecule at all. So methane you really just have one confirmation, just the way that I drew it here.

But if I look at ethane the situation is different. So I'm drawing ethane showing two tetrahedral structures connected to each other. Again, if I look at the carbon hydrogen bonds, if I rotate around the sigma bonds nothing will happen, but if I rotate between the two carbons, then actually it will look very different.

The way that I drew the molecule here, remember that a bond that is neither a wedge going up or down is in the plane. All right, so the way that I drew this, this hydrogen is in the plane with this carbon, with this carbon, and this hydrogen. But if I rotate around the sigma bond, I can move the hydrogens away from each other.

So, it's really difficult to draw that using this kind of representation. Okay, so what we do is we're going to learn a new representation, it's called Newman projection that will let you see what the confirmation of an ethane molecule is like.

So, it works like this. So, we want to do the Newman projection for ethane. First thing we do is we write it out, and then draw a circle. That circle represents the two carbon atoms and ethane that I just put a box around, so this circle represents to two carbons.

And what we're doing is we're looking at the molecule from the side, like this. So if you look at the molecule from the side, you're going to see three hydrogens close towards you and three hydrogens far away. The hydrogens that are close to you we draw lines all the way to the middle like this. Okay, so the lines go all the way to the center of the circle, and it kind of makes sense because you should be able to sort of see these bonds to the carbon that's closest to you. To represent the carbon that's in the back we draw lines that go to the periphery of the circle like this. So, this hydrogen in blue would be the same as one of the hydrogens over here, and hydrogens in purple would be one of the hydrogens in the front. That's how we're labeling these.

So, the reason that these are useful for confirmation is that it's very easy to turn that sigma bond now. All I have to do is keep either the front or the back constant and rotate the other part. So for example if I keep the front constant, and it really doesn't matter which you do, and now I'm going to rotate that bond clockwise, so this hydrogen is now going to move over until it's right behind the hydrogen in the front, and of course they all move like that.

Now I want to say that they're exactly on top of each other, you can't of course draw them on top of each other so whenever you draw a hydrogen that's real close, like this, it implies that they are exactly on top of each other. That's the convention, okay?

When they're in between, it's very easy to draw that, because they don't overlap. So these are the two extremes in confirmations. The hydrogens are next to each other or they are on top.

So each confirmation will have its own name. The one where the hydrogens are next to each other will be called staggered, and when they're on top of each other it's called eclipsed. So, those words should kind of make sense. Eclipsed, it's like a planetary eclipse, go on top of each other.

Now there are actually an infinite number of confirmations here because I can have any angle between the front and the back, but for convenience sake we only name the extremes of that, so we name the staggered we name the eclipse, but you have to understand that there are really an infinite number. So, if you want to get an idea what that looks like, we can draw an energy diagram, to show how the energy changes as we go from staggered to eclipsed.

Let me go back to this diagram here. Which do you think would have a lower energy, be more stable between these two? The staggered or the eclipsed? What's the reason that the staggered is more stable? More space?

The way we say that in organic chemistry is it has less steric hindrance. Steric just means filling up space. It just means the two groups don't want to be near each other, because they take up space. That's exactly right. It turns out that, according to the picture, they are farthest away from each other when they are in the staggered confirmation. Okay, so we'll call that a steric interaction, or actually steric hindrance would be a little bit better in this case.

Okay, so that will be the reason that that will have lower energy.

Let's go back to our energy diagram then. Let's start at that confirmation. So, as we rotate we'll reach a point where we reach the eclipsed structure. And if we continue to rotate, we'll find that we wind up in the same position as we did originally.

So, we can draw the energy diagram like this. So we have energy versus angle. And of course I can keep spinning this forever, so I'm basically going to have something that looks like a sine wave, it's just going to be up and down like that. And, the importance of this is that you can calculate it, you can run a computer program that will tell you exactly what the energy will be at any given point.

The reason that's important is that it's these energies that are going to determine that shape of a molecule, any molecule. Now, for ethane, that's it. There's only one sigma bond that you can rotate around.

But, for a complicated molecule, like a drug for example, that has lots of sigma bonds, they can rotate in all kinds of different angles. So you need to basically draw these energy surfaces and figure out where the lowest energy point will be, you know, at the temperature of your body, basically. So you can predict that the molecule will have this shape, which is important because it's the shape of the molecule that determines if it will fit in a receptor or not.

It's important, because it enables you to predict the confirmation, the most stable confirmation for a molecule. All right, let's do one more here. If we go from ethane to propane how does that change the Newman projection? The thing about the Newman projection is that you can only do two carbons at one time. So you have to decide which carbons you're going to do.

In the case of propane, it doesn't matter. If I take the left two carbons or the right two carbons, it'll give me exactly the same thing. So, I shouldn't have to tell you which carbons to use. If there's more than one possibility, I'll tell you, you know, do C2 to C3, so that will tell you between the second carbon and the third carbon. Okay, so I'll need to specify that, but in this case we don't have to because there's only one possibility.

So that's what we start with is a box between the two carbons. And, what you'll find with all these Newman projections is that you'll have three things in front, three things in the back. You notice here that it's three hydrogens over here, and on the right side it is two hydrogens and a methyl. So, to translate this, again, draw a circle, three lines, pointing to the middle. And it doesn't matter which you put in the front and which you put in the back, okay, but if you start to put one in the front or in the back, you need to continue that in all of your projections. So, you can't change your mind once you start. That's really important.

So let's say I decide to put the methyl group in front. I'd have methyl, and then H and H. In the back I would have three things, H, H, and H. So I started with a staggered confirmation, just because it's easier to draw, but once we start we can then draw all of the other confirmations. If I keep, in this case, the back constant, and I rotate the front clockwise, this is what will happen. Here's the back, rotating clockwise. The methyl group will now be on top of this hydrogen back.

Okay, so again, I have a staggered and an eclipsed confirmation. It looks pretty similar to ethane, basically. If I continue to rotate this, I'm going to put the methyl group between two hydrogens and that is identical to what I started out with. So there are only two confirmations here, two extreme confirmations. The staggered and the eclipsed.

So, the energy diagram will look very similar to this, but most likely the difference in energy will be higher, because methyl group is larger than the hydrogen, so the effect will be more pronounced in terms of the steric hindrance. Okay, so the next one we have to do is butane, and there's a lot more confirmations in there, that's going to take a little more time, so I think for today we'll stop here.

On Fridays we have a little more time reserved. It's a great time to come see me after class. If you wanted to talk in smaller groups I'd be happy with that.

Lecture 003

noreply@blogger.com (Jean-Claude Bradley) — Sun, 9 Oct 2005 09:12:00 -0700

Instructor: Jean-Claude Bradley, Drexel University
View Lecture

1
Chem. 241 - Lectures 3
1 DISCLAIMER: This text is being provided in a
rough-draft fashion. Communication Access
2 Realtime Translation (CART) is provided in order
to facilitate communication accessibility and may
3 not be a totally verbatim record of the
proceedings.

8 LECTURE 3
9 Okay. So what we're going to do
10 now is look at especially covalent bonds and
11 how we can figure out how atoms are
12 connected together to form more complicated
13 molecules. So we're going to learn how to
14 draw Lewis structures. You probably already
15 know how to do this, but I'll show you a way
16 of doing it so that you can find the
17 solution to the Lewis structure very quickly
18 for more complicated cases.
19 Okay. So I'm going to show you
20 that method and that method is on the first
21 blog. Right here, Lewis dot approach to
22 molecular structure. So we're going to
23 apply this today and you should be able to
24 do any Lewis structure after that.
25 So I'm going to start with the
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11
Chem. 241 - Lectures 2 and
1 same example, hydrogen, since we already
2 know the answer. We're talking about Lewis
3 structures. So if we follow this little
4 algorithm that I gave you, it has six steps.
5 The first step is to compound the total
6 number of electrons available, and that's
7 where you pull out that valence periodic
8 table, all right?
9 So carbon is four, nitrogen is
10 five, oxygen is six, fluorine is seven, neon
11 is eight, and hydrogen is one. Okay? So
12 that tells us how many electrons we have
13 available. We have 1 plus 1. So I'm only
14 talking about valence electrons.
15 Now the next step is to figure out
16 this number 2: Find out how many electrons
17 are needed. When I refer to "needed," I'm
18 talking about the octet rule. And the octet
19 rule states that you want eight electrons
20 for each carbon, nitrogen, oxygen, and
21 fluorine and related elements. So sulfur,
22 phosphorous, iodine, all those little other
23 elements that we drew in that little
24 periodic table, the same thing. They all
25 need eight electrons because of the octet
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12
Chem. 241 - Lectures 2 and
1 rule. Hydrogen only needs two.
2 So what we're going to try to do
3 is figure out how we can have a molecule so
4 that every atom satisfies the octet rule.
5 Okay?
6 So we need -- well, I have two
7 hydrogens, and each hydrogen needs two
8 electrons. So that's 2 plus 2 is 4. Okay?
9 So that's the octet rule part.
10 So the next step is, we figure out
11 how many electrons are shared. And then we
12 divide by 2 to get the number of bonds in
13 the molecule. So this should make some
14 sense.
15 I have two electrons available;
16 however, I need four. So I'm going to have
17 to share a certain number of electrons. How
18 many? Exactly 4 minus 2. I need to share 2
19 electrons. By definition, a bond, a single
20 bond, is two electrons.
21 So I divide 2 by 2, and I come up
22 with 1, okay? So for this example, it
23 actually takes a lot more time than the way
24 we did it, you know, earlier, where we just
25 drew the electrons.
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13
Chem. 241 - Lectures 2 and
1 You see, when you do molecules
2 that are far more complicated than knowing
3 the number of bonds in the molecules, it is
4 critical to being able to solve the problem
5 quickly. So I'm going to be doing examples
6 that are increasingly more complex until we
7 cover examples that you should be able to do
8 anything.
9 So knowing that we have one bond,
10 the next instead of is you write out the
11 atoms. In this case, we have two. There's
12 only one way to link two atoms together with
13 a single bond, and that's like that.
14 Sometimes we will have a choice
15 and we will talk about how we decide how to
16 put the bonds. When you only have two
17 atoms, you don't have a choice and you have
18 to put it like that.
19 So it looks like we're done, but
20 there's actually a couple more things to
21 worry about in order to make sure that we
22 have a correct answer.
23 The next thing we're going to do
24 is number 4 here. After placing the bonds,
25 complete the octets. What I'm doing here
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14
Chem. 241 - Lectures 2 and
1 is, I'm looking at each atom and counting
2 how many electrons are around it. If I look
3 at the left hydrogen, I count all of the
4 electrons until the bond. So I have two
5 electrons that are surrounding that hydrogen
6 for the octet rule. Well, hydrogen needs
7 two electrons so, I don't have to add any
8 electrons if I had some things that were
9 missing. We're going to do that in the next
10 example that we do.
11 So, you know, by symmetry, the
12 other hydrogen also has two electrons. So
13 we have satisfied the octet rule.
14 The last thing you have to worry
15 about is counting electrons around the atoms
16 and placing the charges. So when I'm
17 counting for charges, I only count one
18 electron per bond because charges have
19 nothing to do with the octet rule; charges
20 have to do with on average how many
21 electrons are around each hydrogen.
22 So if I count one electron per
23 bond, I have one electron on each hydrogen
24 for charges. Hydrogen normally has one
25 electron; so, therefore, it is not charged.
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15
Chem. 241 - Lectures 2 and
1 So it is really important to do
2 that last step because if you get everything
3 else right but you don't get the charge
4 right, you will have a wrong answer, and
5 it's going to lead you to wrong solutions
6 for some of the problems we're going to do.
7 So when we come up with a charge,
8 you will see how that works. But in this
9 one, it doesn't have a charge so we're
10 pretty much done.
11 If you had an issue with the with
12 the charges, you can always recheck your
13 molecules to make sure that the octet rule
14 is satisfied everything. So that number 6
15 is just check up on it.
16 So any questions on that? We're
17 going to do a bunch of example so is if it's
18 not all clear now, it will be all clear.
19 Okay. Next example is fluorine,
20 F2. So we're going to have available, need,
21 share, and then bonds. Always the same
22 thing.
23 Okay. So how many do I have
24 available in this molecule? Each fluorine
25 is what? Seven. So how do I know that?
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16
Chem. 241 - Lectures 2 and
1 This is where you go back to the valence
2 periodic table. You look up fluorine --
3 seven, and that's it.
4 So we have seven, but we have two
5 fluorines. So that's 7 plus 7=14. How many
6 do we need? 16. How do I know that?
7 Again, I look here, and it tells me that for
8 each carbon -- nitrogen, oxygen, fluorine --
9 I need eight. So 8 plus 8 is 16.
10 I'm going to share 16 minus 14=2
11 electrons. And divide that by 2 and find
12 out that I have one bond.
13 So I have two atoms. I have to
14 connect them together and make a molecule,
15 so I have to at least put the bond between
16 the two atoms, right?
17 So now if we stop here, this is
18 incorrect, because a Lewis structure a
19 complete Lewis structure, satisfies the
20 octet rule, and all the valence electrons
21 are shown and all the charges are shown, so
22 we're not done here.
23 We're going to look at the next
24 step. So after placing the bond which we
25 just finished, we complete the octets.
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17
Chem. 241 - Lectures 2 and
1 So the fluorine has, for octets,
2 two electrons around it, so it's actually
3 missing six to be able to come up to eight.
4 So I'm going to add six electrons around the
5 fluorine, but I'm going to do it in pairs.
6 So two on the top, two on the left, two on
7 the bottom. Okay? So I'm completing the
8 octet rule at this point. And by symmetry,
9 the fluorine on the right, I have to add six
10 electrons.
11 Okay. So the next step is, we are
12 going to count and place the charges. So
13 we're going to count the electrons per
14 charges. So when I count the charges, I
15 only count one electron per bond, but I
16 count all the other electrons.
17 So one for the bonds, two, three
18 four, five, six, seven. So for charges, I
19 have seven electrons around the fluorine.
20 Fluorine normally has seven electrons;
21 therefore, it's not charged. So I don't
22 have to put a plus or a minus anywhere
23 Okay? So it's not very
24 complicated, but it's one step above
25 hydrogen. There's something new in that
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18
Chem. 241 - Lectures 2 and
1 example.
2 Oxygen is the next one, O2. So
3 available, need, share, number of bonds to
4 be filled out.
5 So oxygen is in the column of six
6 valence electrons. So that's 6 plus 6=12.
7 How many do we need? So we have, you know,
8 any atom that's listed here, oxygen is
9 listed. It needs eight. So that would be
10 8+8=16.
11 Share is 16. So we're going to be
12 sharing 16-12=34. 4 divided by 2=2.
13 Okay? So it's useful to know that
14 there are exactly two bonds in oxygen. So
15 let's see how we are going to put them.
16 Well, if you only have two atoms,
17 it's not really very difficult. There's
18 only one place to put them, and that's
19 between the two atoms. So the first bond
20 links the two atoms together. But I have to
21 put two bonds total, so the only place I can
22 put it is here.
23 Okay. So now I'm going to go a
24 little bit more quickly with each example,
25 okay?
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19
Chem. 241 - Lectures 2 and
1 For each oxygen, now we're going
2 to complete the octets. How many electrons
3 do I have to put on? Four. I have four
4 electrons per octet already there. I only
5 need four more.
6 So, again, putting them as pairs,
7 I put the four electrons like that. And by
8 symmetry, I will do the same thing with the
9 oxygen on the right. Okay, now we have the
10 eight electrons per octet, so it's
11 satisfied.
12 Next thing we're going to count
13 the charges. One two tree four, five six.
14 Again, it's only one electron per bond, so I
15 have six valence electrons -- I have six
16 electrons per charge. Oxygen normally has
17 six electrons, so it's uncharged.
18 So, once again, with the simple
19 examples, you know they're not going to be
20 charged. Now it's complete. That's the
21 full Lewis structure.
22 Okay. Let's do nitrogen next.
23 N2. Available, need, share, and number of
24 bonds.
25 Okay. So going back to the table,
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20
Chem. 241 - Lectures 2 and
1 nitrogen is in the column with five valence
2 electrons. So I have five plus five
3 available. Okay? Available valence
4 electrons. Nitrogen is one of the atoms
5 that requires eight to satisfy the octet.
6 So 8+8=16. I have to share 16-10, which is
7 6, which tells me that I have 6 divided by
8 2=3 bonds. So once again, it's useful to
9 know that going in.
10 I only have two atoms. So there's
11 only one place I can put those three
12 bonds -- between the two nitrogens.
13 Now I'm going to satisfy the octet
14 rule. I have six electrons on each nitrogen
15 for the octet rule, so I'm only missing two.
16 Put them in a pair like that. Opposite for
17 the triple bond. By symmetry, the nitrogen
18 on the right would have two electrons.
19 Counting for charges each one of
20 the bonds contributes one electron so
21 there's one, two, three. And then we count
22 all the electrons. Four, five. So we have
23 five electrons. Nitrogen normally has five
24 electrons. So, again, uncharged.
25 Okay. Let's start to put more
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21
Chem. 241 - Lectures 2 and
1 atoms and see how this works out. So if I
2 try methane, CH 4, the same thing -- we are
3 going to look at available, how many we
4 need, how many we share, and then the number
5 of bonds.
6 So the carbon is four. And each
7 hydrogen is one. So to make it very
8 explicit here, 4+4x1, that totals up to 8.
9 So how many we need is going to be
10 8 for the carbon, and this is where people
11 often make a mistake. Hydrogen is 2, it
12 requires 2. So I have 4 of them, and that's
13 4x2=8+8, total of 16
14 So I'm going to have to share
15 16-8, or 8 electrons. Number of bonds is 8
16 divided by 2=4. Again, not a bit surprised
17 for this molecule because I'm starting
18 something simple, so we can build up to it.
19 We have carbon. I have 4 bonds
20 and I have 4 atoms to connect it to so
21 there's only one way to do that. Hydrogen
22 can't have more than one bond, so the only
23 thing I can do is put the four bonds around
24 the carbon. So how many electrons does
25 carbon need to complete the octet in this
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22
Chem. 241 - Lectures 2 and
1 case?
2 It's already full. It already has
3 eight electrons. So there's nothing to add.
4 For charges, I have one, two,
5 three, four. Carbon normally has four, so
6 it's uncharged so we're done here.
7 Carbon dioxide, CO2. Available,
8 need, share, and number of bonds.
9 So carbon is going to need four.
10 And each oxygen is 6. So 6x2=12+4=16. So
11 we're going to need the carbon. And the
12 oxygens each require 8. 8x3=24. So we have
13 to share 24-16, so we have to share 8
14 electrons. That tells us we have 8 divided
15 by 2, and that's 4 bonds. So we have four
16 bonds and we have three atoms.
17 So now we have more than one way
18 of solving this issue, right? We can join,
19 you know, the three atoms in a triangle. We
20 can do all kinds of things like this. So
21 this is where you have to use heuristics to
22 tell you what is likely to be the most
23 stable form of this.
24 Generally, when you have one atom
25 followed by, you know, a whole bunch of
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23
Chem. 241 - Lectures 2 and
1 another one of oxygen, you're going to put
2 that single atom in the middle and surround
3 it. So if I follow that, it's going to work
4 95 percent of the time.
5 We will see maybe a couple of
6 exceptions to that, but that's pretty good
7 heuristics to follow. So following that, I
8 would put the carbon and then I would put
9 the oxygens on either side.
10 Okay? So what's the minimum I
11 need to hold the structure together? One
12 bond on the left, one bond on the right.
13 But I have to do something with all the four
14 bonds that I know are in that molecule.
15 So, you know, again, here I could,
16 you know, draw a bond between the two
17 oxygens and maybe make a triangular
18 structure, but as we will find out a little
19 bit later, three-membered rings are unstable
20 usually. So that's not going to be your
21 first choice.
22 So if I can't make a triangle out
23 of it, the only thing I can do is to put
24 double bonds. So I'm going to add C double
25 bond 0 on the left, C double bond 0 on the
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24
Chem. 241 - Lectures 2 and
1 right.
2 Okay? So now I'm going to
3 complete the octets. So I add four
4 electrons on the oxygens, and the carbon
5 already is satisfying the octet rule.
6 Looking at charges -- one, two, three, four,
7 five, six electrons around the oxygen.
8 Oxygen normally has six, so that's
9 uncharged. Carbon -- two, three four,
10 carbon, it only has four, so it's also
11 uncharged. So that's complete, the CO2
12 Okay? So let's look at BF3, boron
13 trichloride. Available, need, share, number
14 of bonds. So available, let's go back to
15 the table here. Boron has three valence
16 electrons. And the chlorine has seven.
17 That's 24, right?
18 How many do we need? How many
19 does fluorine need to complete the octet?
20 Eight. How many does boron need to complete
21 the octet? Boron is not listed in the list
22 of atoms to complete the octet rule, so
23 that's one thing you have to watch out for.
24 Boron will not satisfy the octet
25 rule, so you simply cannot use this
-----------------------------------
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25
Chem. 241 - Lectures 2 and
1 procedure. You're going to end up with a
2 molecule. If it doesn't satisfy the octet
3 rule, that doesn't mean you can't form a
4 molecule. It just means that it will not
5 satisfy the octet rule, so you don't do
6 this.
7 I'm not going to give you problems
8 that are terribly complicated with boron, so
9 you can pretty much figure out what has to
10 happen. The boron has three valence
11 electrons and it's bonding with three atoms.
12 So there's really not much else for it to do
13 except take each one of those valence
14 electrons and use it to form a covalent bond
15 with each fluorine like this.
16 So notice that in the structure,
17 which, if I wanted to show the bonds, this
18 would still not be complete because I have
19 not shown the long pairs on the fluorines.
20 So you'll notice with this
21 example, the borine follows the octet rule,
22 but the boron only has six electrons, and
23 that's just the way it is, okay? So if it's
24 a boron, it just doesn't satisfy.
25 Let's look at ammonia, NE3. So
-----------------------------------
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26
Chem. 241 - Lectures 2 and
1 nitrogen is five. NE hydrogen is 1. So we
2 have 8 electrons available. We need 8 for
3 the nitrogen and 2 for each hydrogen.
4 3x2=6+8=14. We have to share 14-8, which is
5 6. So we have 6 divided by 2, so it's 3
6 bonds.
7 So each hydrogen cannot have more
8 than one bond to it, so the only way I can
9 hook this together with the nitrogen in the
10 middle is to use up my 3 bonds. And now if
11 I go to the next step, I have to complete
12 the octets. Each hydrogen is satisfied.
13 The nitrogen only has six
14 electrons. It's missing two, which I will
15 put. And for charges, one, two, three,
16 four, five. Nitrogen normally has five
17 electrons, so uncharged.
18 Okay. Let's look at carbon
19 monoxide, CO. Available, need, share, and
20 number of bonds. Available, we have four
21 from the carbon and 6 from the oxygen, and
22 that gives me 10 total. We need 8. Plus 8,
23 that's 16. So we have to share 16-10, which
24 is 6. That means we have 6 divided by 2, so
25 3 bonds.
-----------------------------------
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27
Chem. 241 - Lectures 2 and
1 As we start to get into more
2 complicated examples, it becomes more and
3 more useful to know how many bonds we have.
4 There's only two atoms. I have to
5 put the three bonds between the carbon and
6 the oxygen like that. So the next step is
7 to complete the octets. The carbon has 6
8 electrons around it already. It's only
9 missing two 2. Same thing for the oxygen.
10 Now, if we look at charges, for
11 the carbon, one, two, three from each bonds,
12 four, five. But carbon normally has four
13 electrons, so it has an extra electron.
14 Electrons are negativity charged; therefore,
15 I have a formal charge on the carbon of
16 minus 1. So I don't have to write 1. If I
17 just put a minus, it implies I have one.
18 The oxygen -- I have one, two,
19 three, four, five electrons. Oxygen
20 normally has six. It's missing an electron.
21 So if I'm missing a negative charge, I go to
22 positive. So it has a plus 1 charge. I
23 just put a plus on it.
24 Now, after you're finished placing
25 charges, the total charge in the molecule,
-----------------------------------
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28
Chem. 241 - Lectures 2 and
1 you draw has to be the same as the one in
2 the problem. So my total charge in CO is
3 zero. I have minus 1 and plus 1, so that
4 cancels out. That has to happen. So if it
5 doesn't happen, you made a mistake
6 somewhere, okay?
7 Okay. Now let's get some other
8 kinds of examples. SO, one of the oxides of
9 sulfur. Available, need, share, and bonds.
10 So sulfur is 6. Oxygen is also 6.
11 That's 12 electrons. We need 8+8=16. We
12 have to share 16-12 which is 4. That means
13 we have 4 divided by 2, or 2 bonds. So I
14 only have two atoms. That makes it pretty
15 straightforward. I have to put the two
16 bonds between the sulfur and the oxygen.
17 Completing the octets. Sulfur is missing 4.
18 Oxygen is missing 4. And if we count the
19 charges, we will find six electrons on the
20 sulfur. Okay?
21 * * *
22
23
24
25
-----------------------------------
(Not a verbatim transcript.)

Lecture 002

noreply@blogger.com (Jean-Claude Bradley) — Sun, 9 Oct 2005 08:59:00 -0700

Instructor: Jean-Claude Bradley, Drexel University
View Lecture

1
Chem. 241 - Lecture 2
1 DISCLAIMER: This text is being provided in a
rough-draft fashion. Communication Access
2 Realtime Translation (CART) is provided in order
to facilitate communication accessibility and may
3 not be a totally verbatim record of the
proceedings.
4
* * *
5
6 Lecture 2
7 Okay. So we're going to start with
8 electronic configuration, so this will be a bit of
9 a review for you, but you have to have this
10 material fresh in your mind to actually see how we
11 will implement it. If you remember all this, just
12 bear with it a little bit.
13 You're going to be applying two rules
14 that you should be familiar with the -- the Pauli
15 solution principle -- and, by the way, the
16 software I'm using to write it on, I'll make that
17 available to you as a PDF, so you don't have to
18 worry about copying everything down because it
19 will be available in an easily printable format.
20 So the Pauli solution principle has to
21 do with the fact that only two electrons can
22 occupy one orbital.
23 And the other rule we're going to need
24 to be able to use to deal with electronic
25 configuration of the first ten elements is the
-----------------------------------
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2
Chem. 241 - Lectures 2 and
1 Hunds rule, where electrons like to be unpaired if
2 possible.
3 Okay. How many of you remember these
4 two rules? Okay, good. So we're going to apply
5 these. And we're basically going to, like I said,
6 fill up the first ten elements. So we're going
7 from hydrogen to neon. And these little lines
8 that I'm drawing are the orbitals, and those will
9 be the orbitals where we have to apply the Pauli
10 solution principle and the Hunds rule.
11 So as we go from the top to the bottom,
12 we're adding one electron per each element. So
13 hydrogen has one electron. And let me use a
14 different color here. We can show that electron
15 with a half arrow either pointing up or pointing
16 down. The direction of the arrow has to do with
17 the skin of the electron, which we don't have to
18 get into; it's just that you need to know that
19 this can either be up or it can be done.
20 So when we move the helium, we have two
21 electrons. Apply the two rules, put the helium.
22 So following the two rules, we put the first
23 electron there and the second electron. It
24 doesn't have another orbital of the same energy,
25 so the other electron can't remain unpaired so
-----------------------------------
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3
Chem. 241 - Lectures 2 and
1 we're going to be forced to pair it. And the way
2 we demonstrate that is by drawing a down arrow
3 next to the up arrow. So now we have two
4 electrons in the orbital, and it's filled up so we
5 can't put any more in there.
6 What is the name of this first orbital?
7 1S. So obviously, I'm going to have to draw more
8 lines here. I'm going to draw all the lines for
9 all the elements, and I'm leaving them blank until
10 I fill them up.
11 So continuing on with helium, it has
12 three electrons total. So the first one goes in
13 the first orbital. And the second one goes in the
14 same orbital, different direction. And then the
15 third one we put in the next orbital. So the
16 first orbital is 1 and the second orbital is 2S.
17 So now we continue with beryllium.
18 It has four electrons. Now I just filled up
19 the 2S orbital, so when I go to boron, that
20 has five electrons and I now have to use
21 these other three orbitals that are of the
22 same energy, and those would be the 2P
23 orbitals.
24 So I have 2PX, 2PY, and 2PZ. And
25 there's no particular reason for that order;
-----------------------------------
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4
Chem. 241 - Lectures 2 and
1 it's just distinguished between them.
2 They're absolutely the same energy. So that
3 like electron, then I have to put it in the
4 2PX orbital like that
5 Okay? So now we have one more
6 electron -- one, two, three, four, five --
7 and now we have three orbitals that are the
8 same energy. So the electrons will be
9 unpaired if possible according to the Hunds
10 rule, so we will put them as unpaired. So
11 I'm going to have one in 2PX and one in 2PY,
12 for example.
13 With nitrogen, I have one, two,
14 three, four, five, six, seven. And, again,
15 I'm putting another unpaired electron
16 because it's the same energy. All the 2P
17 orbitals are the same energy.
18 Oxygen -- one, two, three, four,
19 five, six, seven. And now the eighth one,
20 now I'm forced to pair it because I've run
21 out of orbitals of the same energy that are
22 vacant. So I'll put in the 2PX, for
23 example.
24 With fluorine, we fill this up
25 where I'm going to have to pair the electron
-----------------------------------
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5
Chem. 241 - Lectures 2 and
1 and the 2PY orbitals.
2 And, finally, neon has all of the
3 two P orbitals completely filled up with
4 paired electrons, okay?
5 So this is something I'm taking a
6 bit of time to go through because we will
7 need this when we look at molecular orbitals
8 that is based on this theory. So for now,
9 this should be a refresher.
10 The other reason I go through this
11 is because I want to point out a concept
12 which is valence electrons. When we look at
13 making molecules from collections of atoms,
14 those atoms are connected together with
15 bonds. And it turns out that not all of the
16 electrons in the atoms participate in that
17 bond formation; only the valence bonds
18 participate. So I'm going to put a square
19 around the valence electrons.
20 So, for example, for hydrogen and
21 helium, well, those are all the valence
22 electrons; that's all it has. But when I go
23 from lithium to neon, the 1S orbital is not
24 a part of the valence electrons.
25 So I'm going to put a square
-----------------------------------
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6
Chem. 241 - Lectures 2 and
1 around just the two orbitals, the 2S and 2P
2 orbitals.
3 So those are called valence
4 electrons. And so if I ask, you know, how
5 many electrons does carbon have? what I'm
6 really interested in is the four electrons
7 that are in the valence.
8 So generally when we talk about
9 how many electrons each atom has, we're
10 referring to valence electrons as we count
11 them
12 Any questions on this?
13 (No questions.)
14 Okay, so the next thing I wanted
15 to with these valence electrons is rewrite a
16 little useful periodic table for you with
17 the valence. The periodic table. So these
18 are the elements that we're going to be most
19 concerned with in this class: Hydrogen,
20 boron, carbon, nitrogen, oxygen, fluorine
21 phosphorous, sulfur, chlorine, bromine, and
22 iodine. We can put even lithium and sodium
23 on the left.
24 The reason this is a valence
25 periodic table is that we're going to divide
-----------------------------------
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7
Chem. 241 - Lectures 2 and
1 this up into columns and only worry about
2 the number of valence electrons. So
3 hydrogen has one valence electron; boron has
4 three; carbon, four; nitrogen and
5 phosphorous have five; oxygen and sulfur
6 have six. And the halogens -- fluorine
7 chlorine bromine and iodine have -- have
8 seven.
9 Okay? So those elements are going
10 to behave in pretty similar ways. For the
11 two elements have the same number of valence
12 electrons, there will be differences in how
13 they react, but there will be a lot of
14 similarities as well as to how they format
15 bonds with other atoms.
16 So this is something that, you
17 know, you should keep handy when you're
18 doing problems because you're going to need
19 to use it constantly.
20 Okay. So keeping this in mind,
21 let's now talk about actually making
22 molecules from atoms. Well, let's consider
23 the simplest thing we can consider, which is
24 which is two hydrogen atoms coming together
25 to make hydrogen gas, H2.
-----------------------------------
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8
Chem. 241 - Lectures 2 and
1 So hydrogen has one electron, and
2 I'm getting the Lewis structures here, so
3 this should look pretty familiar. I put
4 hydrogen and then I put a dot, or sometimes
5 we use an X. But the point is, you show
6 that there's only one electron.
7 If I want to show two hydrogen
8 atoms coming together, I can show one
9 electron and the other hydrogen. And then
10 when they come together, I can draw the two
11 electrons on top of each other between the
12 two hydrogens.
13 And what I've done here is, I've
14 created a bond, but there are two kinds of
15 bonds that we're interested in. This is a
16 covalent bond because the two atoms are
17 sharing the electrons. Neither hydrogen has
18 those electrons entirely to itself; it has
19 to share both of them, so that's why it's
20 called a covalent bond.
21 So another way that we can draw
22 the covalent bond is to draw a line. So
23 that's how you join two hydrogens, with a
24 line. Every line you see corresponds two to
25 two electrons. So we will be doing that
-----------------------------------
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9
Chem. 241 - Lectures 2 and
1 most of the time because it's more
2 convenient than drawing a whole bunch of
3 dots. So we need to draw dots when we don't
4 have bonds; but otherwise, we can just draw
5 a line like this.
6 Let's consider another case
7 lithium fluoride. Lithium has one valence
8 electron. Fluorine has seven valence
9 electrons. But in this case, instead of
10 sharing the electrons, the lithium actually
11 completely gives up its electron, and the
12 fluorine completely takes it, so that's
13 different.
14 If the lithium gives up its
15 valence electron, it now has lost a negative
16 charge, and so it's going to be plus,
17 specifically plus 1. And the fluorine now
18 has eight electrons around it. Fluorine
19 normally has seven electrons, but it has an
20 extra one and so it has an extra negative
21 charge. That's how we know that it's
22 negative.
23 Okay. And you notice that I'm
24 drawing the electrons in certain places and
25 I'm putting them in pairs. We're going to
-----------------------------------
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10
Chem. 241 - Lectures 2 and
1 actually find out why I'm doing that, you
2 know, in the next class. But for now, just
3 note as to how the electrons go. They don't
4 go randomly around the atoms; they go in
5 pairs and they're separated by certain
6 angles.
7 * * *

Lecture 001

noreply@blogger.com (Jean-Claude Bradley) — Fri, 7 Oct 2005 12:38:00 -0700

Instructor: Jean-Claude Bradley, Drexel University
View Lecture
1
9.26.05 Organic Chemistry 241
1 DISCLAIMER: This text is being provided in a
rough-draft fashion. Communication Access
2 Realtime Translation (CART) is provided in order
to facilitate communication accessibility and may
3 not be a totally verbatim record of the
proceedings.
4 * * *
5 INSTRUCTOR: Okay. So this is Organic
6 Chem 241.
7 Quiet, please. Thank you.
8 So for this class you received some of
9 my emails -- I'm going over where the information
10 is. Some of you have responded -- oops.
11 Okay, so you've received my emails about
12 where the class lectures are, things about
13 podcasting. For this class today, I will actually
14 go through pretty much everything that is
15 available to you and show you how to get to it,
16 answer questions, and then you should be clearer
17 at the end of this.
18 The first thing you is, you'll notice
19 that I sent you to a wiki, Chem 241, and one
20 student actually asked me why I don't do that on
21 Web CT. Well, this is an open course, meaning
22 that it has to be available to students from
23 around the world. So by putting the information
24 in the wiki logs, it gets indexed by Google, and
25 people can find the information a lot better. If
-----------------------------------
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2
9.26.05 Organic Chemistry 241
1 it's inside Web CT, basically when the class is
2 over, nobody gets the benefit. That's why this
3 information is in an open format.
4 The other reason that I bringing you
5 back to the wiki is that it's really -- think of
6 it like a table of contents. If you don't know
7 where something is, come here and take a look at
8 one the eight possible headings. If you ever get
9 confused, look back over here.
10 So the first thing that I want to point
11 out here is what you have to do in this class to
12 get points. A lot of the stuff that I will give
13 you is voluntary. You don't have to do the
14 quizzes, you don't have to play the games or do
15 any of it, as long as you score well on the the
16 two tests -- the exam and the final.
17 But the reality is that if you don't do
18 the quizzes, if you don't play the games, or you
19 don't do what you have to do, you're unlikely to
20 do well in the class. Make sure you understand
21 the distinction as to what you have to do and you
22 don't.
23 Now, basically, there's a one percent
24 extra-credit assignment, which I will get into
25 later when I go into the syllabus. That's the
-----------------------------------
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3
9.26.05 Organic Chemistry 241
1 only other place where you can earn extra points,
2 that 1 percent.
3 Aside from that, I don't take attendance
4 so there's nothing like that. You could take the
5 entire course on line fully. The lectures are
6 already recorded and you can actually listen to
7 the entire course right now.
8 Let me just take this one step at a
9 time. We will start at the syllabus. This is
10 pretty classic here. You have the courses and
11 objectives. So this course is the Introduction to
12 Organic Chemistry. We will look at molecules,
13 predicting their properties -- just basic stuff
14 like melting points, boiling points. We will look
15 at some reaction mechanisms. In the later -- in
16 the second half, we will look at SM1 and SM2
17 reactions. So pretty much the most important
18 reactions. We'll be looking at 242 and 243 later
19 on, which will be more complex stuff.
20 The textbook -- and, by the way, please
21 interrupt me with any questions. That's why I'm
22 going through this, so that you can bring up
23 issues.
24 The textbook is -- you can use a number
25 of them. I recommend that you use the Wade
-----------------------------------
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4
9.26.05 Organic Chemistry 241
1 edition five. But if you use edition four or six,
2 that's absolutely fine. It's just that the
3 problem numbers will change slightly. I think I
4 gave you the problems for editions four and five.
5 If you have six, we will see what the slight
6 differences are. Otherwise, that's fine
7 How many of you actually got the
8 book?
9 Anybody using another organic
10 text?
11 So you should do -- you know, the
12 beginning of organic chemistry is pretty
13 much the same everywhere. So if you have
14 another organic textbook and you're not sure
15 of it, just ask me, but the primary
16 information that you need to be able do the
17 problems and understand my class are from
18 the lectures and the problems that I solve.
19 So that has be your primary source.
20 You can just think of the book as
21 basically just additional reading material
22 to go a little further than what we do in
23 class. But, clearly the lectures will be
24 the primary source.
25 So the grading is pretty
-----------------------------------
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5
9.26.05 Organic Chemistry 241
1 straightforward. We will have two
2 one-and-a-half-hour exams, so 100 points
3 each. We a final of 200. Giving you total
4 of 400. That will be -- you know, the
5 percentage there is going to determine your
6 grade down here. Over 90 is A, over 80 is a
7 B, over 65 will give you a C.
8 Now, there might be a slight curve
9 to this but it will always be in your favor
10 if there is one.
11 There's an FAQ to this class. And
12 I will take a good chunk to go through the
13 whole FAQ because most of the pain in this
14 class is caused by not reading that
15 document. You have to know where to go to
16 take the tests, when you take the tests,
17 things like that. So if you're not clear
18 about that, that could be really upsetting
19 when you actually run into a problem. So we
20 will take the time to go through that.
21 The makeup test policy. I give an
22 automatic makeup for every test, but it has
23 more questions in it, all right? And I will
24 take the better of the two scores.
25 The idea behind that is that you
-----------------------------------
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6
9.26.05 Organic Chemistry 241
1 take the first test, and I expect you to
2 study for it properly to be able to do well.
3 You should be aiming to do well on the first
4 test.
5 But if, for whatever reason,
6 things don't go well on the first test, you
7 have an option to prove that you know the
8 material immediately following that with a
9 test makeup. And we will use the time in
10 class to basically go over reviews as to why
11 you didn't understand a certain concept.
12 But there's a penalty for that in
13 that you will have a lot more questions in
14 the same amount of time. So I think that
15 overall is fair for everybody.
16 Now, lecture archives are divided
17 into different sections. The theory part is
18 actually in the class blog, which we will
19 look at shortly. But the problems, the
20 problem solutions are all in Web CT. The
21 only reason for that is that there's a
22 copyright issue. I can't show the solutions
23 to the problems because Prentice Hall isn't
24 happy about me doing that, so I have to hide
25 it and put it in Web CT. But otherwise,
-----------------------------------
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7
9.26.05 Organic Chemistry 241
1 everything else is on the blog
2 Okay. Grading policy we went
3 through.
4 The extra credit. If you're
5 interested in doing that, we will talk more,
6 come and see me. To give you an idea about
7 of what this is about, up to 1 percent of
8 the grade -- so it's not a huge percent, but
9 there was like six students last term who
10 would have gone from B to A or from C to B
11 if they had just done the one percent extra
12 credit.
13 You know, it's not so much that
14 everybody in the class will do it, but it
15 should be an incentive if you're interested
16 in doing the project.
17 But the primary reason for doing
18 the extra credit is really for your own
19 benefit so that you can integrate the
20 material better.
21 And the way that it works is that
22 you go out and find a reference in a
23 peer-reviewed journal that has something to
24 do with a reaction that we do in class. We
25 will look at stuff that has to do with free
-----------------------------------
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8
9.26.05 Organic Chemistry 241
1 radicals. A good example of that is the
2 ozone layer and how it gets destroyed.
3 It doesn't have to be long; it
4 could be just a couple of sentences, but I
5 want to see a picture of the reaction you're
6 doing and a link to the original article,
7 and that's it. If you do that, then you get
8 full points usually.
9 Now, the earlier that you put it
10 up, the more time that I have to give you
11 some feedback on it. So there are two dates
12 here that are important -- November 4th and
13 November 25th. On November 4th at midnight,
14 I will go in and grade all the assignments
15 that have been put in.
16 And there will be no extensions
17 for any reason on these two dates. So if
18 you're interested in doing this, I suggest
19 that you do it as quickly as possible so you
20 have plenty of time to work out any issues.
21 If you post something and there are things
22 that are wrong in there, I can give you
23 feedback, or students in the class can give
24 you feedback so that you have time to
25 correct it before the final grade. So
-----------------------------------
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9
9.26.05 Organic Chemistry 241
1 there's two of those for the full 1 percent.
2 You have the benefit of accessing
3 previous examples of this. If you go back
4 to the Chem 242 class where they did this
5 assignment, what you will see are basically
6 examples like this.
7 You can go through -- there's
8 about 40 of these in that particular blog.
9 And you can read my comments to see if
10 that's a good way to do it or not a good
11 way. So here, if you look at the comments.
12 And this is one that's full mark.
13 Some are not going to be full marks and I
14 will explain why.
15 So if you have time, you can go
16 through that and take a look at, you know,
17 what is a good reaction to select and what
18 isn 't.
19 So there's all kinds of things
20 here. I think some of the more interesting
21 things, for example, were the Breathalyzer
22 tests the students did, the actual chemistry
23 behind it. There's a lot of very
24 interesting things that you're able to find
25 out and link to.
-----------------------------------
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10
9.26.05 Organic Chemistry 241
1 I think that's pretty much all I
2 want to say about this.
3 The only other issue about the
4 extra credit is that we will be using games
5 in this class. For example, we're using an
6 unreal tournament. So If you wish to create
7 an unreal tournament map of chemistry, that
8 can also be your 1 percent. Again, come
9 talk to me about it once you find out more
10 about how this works.
11 The material covered, at the
12 bottom of the syllabus, we have different
13 sections of what we will be doing. We will
14 start off with going back to introductory
15 chemistry, looking at electronic
16 configuration, and we will pretty quickly
17 reach doing Lewis structures, all right?
18 Now, there's a way that I teach
19 you how to do Lewis Structures. You can do
20 pretty much any structure very quickly. I
21 spend a lot of time on that. If you listen
22 to lectures, you will see how it works. If
23 you haven't, definitely get into there and
24 start to listen to the lectures and do the
25 problems for Chapter 1.
-----------------------------------
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11
9.26.05 Organic Chemistry 241
1 Chapter 2 is just an extension of
2 molecular orbitals, so we have hybrid
3 orbitals. And that's just basically how to
4 determine the geometry of the molecules once
5 you have the Lewis structure.
6 Now moving on to hydrocarbons,
7 which is the simplest form of
8 (indiscernible), and that's carbon and
9 hydrogen.
10 And then we will look at our first
11 reaction, which is the halogenation of
12 alkenes, so the reaction of bromine with
13 ethane. We will go into detail and find out
14 exactly how that happens step by step. And
15 then we have Test 1.
16 So, you know, you can listen to
17 the lecture at your own pace, but just make
18 sure that you are completely way ahead of
19 time for the Test 1.
20 Now, the Web CT -- I don't know if
21 you know there's a migration where we're
22 going to the Vista edition of Web CT. So
23 right now, everybody is getting transferred
24 over. I think this class is still not quite
25 done. As you will see when we go back to
-----------------------------------
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12
9.26.05 Organic Chemistry 241
1 the wiki, you can actually already do all
2 the quizzes and do everything you have to in
3 the prior version of Web CT if you log in as
4 guest. This week sometime, you will be able
5 to log in as yourself
6 That pretty much covers the
7 syllabus.
8 Any questions?
9 Okay, the FAQ. So the FAQ is
10 actually a blog, and it should become
11 apparent to you why I chose to make it as a
12 blog. I have one common FAQ for all
13 classes, so Chemistry 241, 242, 243, we all
14 share this common blog.
15 The other advantage is that if
16 there's a new question that I pose, if you
17 subscribe to this blog correctly, you will
18 see a new entry in it. It's basically
19 another way to keep updated about the class.
20 And I do expect you to subscribe to this
21 because I will be following most of the
22 posts.
23 So the way blogs work is they're
24 in reverse chronological order. So FAQ 21
25 is the most recent FAQ. If you want to read
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9.26.05 Organic Chemistry 241
1 a blog in chronological order, you just
2 scroll down to the bottom. So here's FAQ 1.
3 How does podcasting work in this
4 class? How many of you have heard of
5 podcasting? How many of you already
6 subscribed to podcasts?
7 Well, I think if you learn
8 anything from this class, you will learn to
9 use podcasting, which is very useful in all
10 kinds of different areas. There are a lot
11 of lectures now being put on podcast and on
12 iTunes, and they're free and available to
13 you, okay? So this is a resource that you
14 should at least be aware of.
15 Podcasting means you subscribe to
16 a feed and you get audio files automatically
17 copyed over onto your computer or your iPod
18 if you have one. You don't need an iPod.
19 In fact, I listen to stuff on my laptop, but
20 you can synchronize with it and listen to
21 lecture that is way.
22 The problem with the iPod is that
23 there's no video information, so I provided
24 the PDF's as well. When you go into iTunes,
25 you see a lecture an MP3 and PDF for each
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9.26.05 Organic Chemistry 241
1 lecture.
2 How do you do this? Well, I give
3 you information here on iTunes. I've
4 created tutorials that are 30 to 60 seconds
5 long. You click on that. We have gone back
6 to another section of the wiki.
7 This is how you subscribe to
8 iTunes. There's a screen cast. Let's take
9 a look at that. You can come back here
10 anytime and replay this. So this goes
11 through iTunes. After you download it, you
12 basically go to the music store, pick
13 podcasts, pick higher education, and then
14 you can sort however you like this.
15 The name of this class is CHEM241,
16 organic chemistry. Click on subscribe and
17 basically click on the little triangle to
18 get alll the backup results. They will just
19 be automaticically there in iTunes for you.
20 Okay, so that's one way of actually
21 accessing the class material.
22 You can also use other pod
23 catchers like iPoders is another one, but I
24 think iTunes is pretty good for the things
25 we try to do in class.
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9.26.05 Organic Chemistry 241
1 So we have other tutorials here.
2 If you want to use iPoder, there's a screen
3 cast. If you want to subscribe to the class
4 blog, we'll take a look at that.
5 Notice at the top of all my blogs
6 is a little button for subscribing -- click
7 through that, subscribe to it. What will
8 happen is that on the left-hand side of blog
9 lines, whenever there's a new posting of any
10 kind, whether the FAQ or the 241 class, it
11 will appear in bold on the left. So that's
12 something that you definitely need to do as
13 soon as possible.
14 How many of you have actually
15 subscribed through blog lines for this
16 class? Not too many. Again, that's
17 something I can help you with. If it's
18 still confusing after you look at the stream
19 casts, you know, the next time we meet we
20 can look at that.
21 Any other questions on podcasting?
22 Let's look at the next question.
23 When are the tests and the exams? When you
24 are able to log into Web CT -- again, that
25 should be sometime by the end of the week.
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9.26.05 Organic Chemistry 241
1 When you go to the quiz and test section,
2 you will see dates associated with each
3 quiz. The quizzes don't end, so they have
4 no end date, but the tests have a start date
5 and end date.
6 Typically, the first test is
7 around the end of Week 4, start of Week 5,
8 so you want to plan to have all of that
9 material covered that I've talked about up
10 to the first reaction done well before the
11 end of the fourth week.
12 Now, you will take these tests at
13 Korman. There are certain rooms. That's
14 actually the next question here, where they
15 are. You basically lock into one of these
16 Korman rooms, sit down, log onto Web CT, and
17 take the test.
18 So we use video surveillance to
19 make sure that it's you actually taking the
20 test, but it gives you the flexibility of
21 walking in pretty much anytime within a
22 four-day period. So you can take the test
23 at your own convenience.
24 Now, that means that you also have
25 to know when Korman is actually open. So
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9.26.05 Organic Chemistry 241
1 make sure that you look at the hours. I
2 gave you a link to know when they have
3 scheduled downtime or scheduled classes.
4 The last day is what I call a
5 "no-excuse day." Make sure you understand
6 that. If you wait until the very last day
7 to take the test, I will not accept any
8 excuse for you to get an extension. So plan
9 to do the test anytime before the last day.
10 The reason for that is at the end
11 of the last day, I open up the tests so that
12 you can see which questions you got right or
13 wrong, okay? And I'm not going to delay
14 that because somebody chose to take it on
15 the last day and had a problem. That's
16 really important.
17 What happens if you miss
18 something -- if you plan to do it on the
19 last day and missed it, too bad, take the
20 makeup. That's another reason for you to
21 take a makeup.
22 So we just went over this, which
23 Korman rooms.
24 Now, what material is covered. I
25 showed you in the syllabus. At the bottom
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9.26.05 Organic Chemistry 241
1 of the syllabus, there were items of what we
2 were going to cover, some of the reactions.
3 That's basically a checklist of how I make
4 up the tests. So you want to use that to
5 make sure you've actually covered everything
6 we're doing in class. And that's -- I try
7 to tell you as specifically as possible what
8 I cover.
9 Are the tests open book? Well,
10 they are. You can bring anything you want
11 into the test room, but don't mistake that
12 to think that you can do last-minute
13 studying in the test room, because there's
14 not enough time.
15 Organic chemistry, at least the
16 way I teach it and test it, is very much
17 like math. You can all the examples in the
18 world in front of you, but if you don't
19 understand addition -- you know, I will give
20 you an example of adding two numbers you've
21 never seen before.
22 So you can have all the examples
23 in front of you, but If you don't understand
24 addition, you can't do the question.
25 So organic is the same way. I
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19
9.26.05 Organic Chemistry 241
1 give you examples of reactions, make sure
2 you know how it works, and then ask you a
3 question about a reaction that you haven't
4 seen before.
5 That's why it's not a problem for
6 organic tests to be open-book. It might
7 lower your anxiety a little bit if you think
8 you might forget a few things.
9 Question 6, we already talked
10 about this. When is the grading going to be
11 released. Well, actually immediately after
12 you take your test, you will know what you
13 did, but you won't know which questions you
14 got wrong.
15 So at the end of the testing
16 period, after the last day, I'll let you
17 know which questions you got right or wrong
18 but I won't tell you what the correct answer
19 is.
20 So you still have to work to
21 figure out what it is that you didn't
22 understand about each question
23 So is Web CT available at all
24 times? Well, it should be. But it's not.
25 Sometimes there are problems. Most of the
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20
9.26.05 Organic Chemistry 241
1 problems are fixed within like ten or
2 fifteen minutes. Occasionally, there might
3 be longer problems. This is another reason
4 why you don't want to wait until the last
5 day, in case there is a problem. But Web CT
6 also schedules downtimes. So if you were
7 going to practice your quizzes over the
8 weekend, it's important that you do it on
9 that particular day. You might want to
10 check what the scheduled downtimes are.
11 How do you resolve technical
12 problems? Well, it depends on where you
13 are. If you are on campus, probably if you
14 just walk over to Korman and ask the
15 consultants, they are well trained to help
16 you with any technical issue you come
17 across.
18 If you're at home, here's the
19 phone number. Of course, if your Internet
20 goes down, you won't be able to get the
21 phone number.
22 Also you can do it on line. If
23 you click on that link, it will take you to
24 the support. They're usually really
25 responsive. Depends on where you are.
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9.26.05 Organic Chemistry 241
1 Honestly, I've never had a situation where
2 we had a technical problem that we couldn't
3 resolve quickly. As long as you give
4 yourself enough time to do everything,
5 you'll be okay.
6 I haven't tested this out yet. I
7 think we should be okay.
8 What does it mean that a classroom
9 is reserved? What you will see in the blog
10 sometimes is that I will say, you know, the
11 Korman 111 is reserved from noon to 3 on
12 this particular Saturday. What that means
13 is that there should be no other students in
14 there. I sometimes do that if I can. But
15 it ensures you that you will have a quiet
16 time where you can actually take the exam.
17 You don't have to take it during
18 the reserve time. Some people are not
19 disturbed by other students in there doing
20 other things. If you are, you might want to
21 schedule it at a reserved time. It will be
22 in the blog, that information.
23 Why don't you get the correct
24 answer after the tests are graded? Well,
25 basically, it's your responsibility to
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22
9.26.05 Organic Chemistry 241
1 integrate the material I'm going over. It's
2 my job to help you understand it, but it's
3 your job to tell me how you tried to solve
4 the problem so I can help you.
5 If I simply tell you what the
6 answer is, you will do exactly the same on
7 the makeup as you did on the first test.
8 That's really the kind of
9 interaction I want to have with you most of
10 the time. It's you coming to me. You know,
11 if you're not getting to the correct answer,
12 show me the steps you're taking, then I can
13 show you exactly which step where you're
14 taking the wrong turn.
15 So that's why I don't give you the
16 answers. That would be too easy.
17 So why are there more questions on
18 the makeup with the same amount of time?
19 Again, you're supposed to be considering the
20 first test as being the one where you do
21 best at. So, you know, you're going to be
22 penalized if you don't get the material
23 right away.
24 But you can still completely make
25 yourself up. There are extra questions, but
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9.26.05 Organic Chemistry 241
1 honestly, I don't think I've ever had a
2 situation where the student couldn't
3 complete it. It's just a little bit more
4 intense, but it's definitely doable. It's
5 not an unrealistic number of questions.
6 Question 12. Why do I get e-mail
7 notices iabout my exam room and time that
8 conflict with the times in Web CT? Because
9 you won't be taking the exam in a normal
10 classroom but in one of the Korman rooms in
11 exactly the same way as you take the tests.
12 These are all automatic. If you get an
13 alert from SAS, just ignore it
14 STUDENT: (inaudible).
15 INSTRUCTOR: Which exam?
16 STUDENT: A common exam that supposed to
17 be at 8 o'clock in the morning, aren't you
18 supposed to tell us? (Inaudible.)
19 INSTRUCTOR: That's news to me. I
20 haven't heard of that before.
21 STUDENT: It's on the schedule here
22 (inaudible).
23 INSTRUCTOR: Well, it's the first time
24 I've heard about it so I will try to find out more
25 about it.
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24
9.26.05 Organic Chemistry 241
1 So the exam period will start on the --
2 it's usually on the Friday right after the last
3 time that we get together. And you have maybe, I
4 think, until Wednesday or Thursday. So you have a
5 big chunk of time to actually schedule your exam.
6 Is there a curve? If there is one, it
7 will be in your favor.
8 Now, can you take this course if it
9 conflicts with another course? Yes, you can,
10 because the way the course is structured, I expect
11 that you should be able to do very well if you
12 take this class completely on line.
13 The last time -- all the classes I run
14 this way, we end up with an attendance rate of 10
15 to 20 percent. So as we move through the term,
16 the vast majority of you find it doesn't do any
17 good to come here to listen to me talk if you have
18 a screen cast available, which is just as good or
19 better, because you can pause it and go back and
20 forth.
21 Basically, you definitely can take this
22 class fully on line, and that has not been a
23 problem at all. We will get together, of course,
24 for workshops to go over material with, but I
25 won't be covering any material in the workshops
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25
9.26.05 Organic Chemistry 241
1 that is not explicitily covered in all the
2 lectures. So you won't be be missing anything key
3 if you don't come to class.
4 Question 15. How do I find test
5 averages? They will be posted on the blog like
6 any other announcements.
7 What's the best way to study organic
8 chemistry? First of all, keep up with the
9 lectures. As soon as you can, try the quizzes.
10 They're designed in exactly the same format as the
11 test and exam questions are.
12 Part of what you do with the quizzes is
13 going over the material. But the other thing you
14 do is look at the format of the questions, making
15 sure you don't have a misunderstanding about the
16 format.
17 So it's important for you to do the
18 quizzes even though they don't count.
19 STUDENT: (Inaudible.)
20 INSTRUCTOR: Yeah, the quizzes don't
21 count. In fact, I did make them count one term
22 and there weren't any more students taking them.
23 So basically, it's unwise not to do them, but you
24 don't have to do them.
25 A question in the back?
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9.26.05 Organic Chemistry 241
1 STUDENT: (Inaudible.)
2 INSTRUCTOR: Yeah, you have an infinite
3 number of tries with the quizzes, and that's just
4 the Web CT. You don't have to wait for it to be
5 activated; you can go directly into the existing
6 version of it and log in as a guest and take it.
7 If there are quizzes that you don't
8 understand, find the part that covers the topic
9 and listen again. As you're more familiar with
10 the way the lectures are archived, there's an MP3
11 audio, a PDF, and screen cast.
12 If you're looking for any material in
13 class, the quickest way to find material is to
14 look at the PDF, see roughly where it is, so like
15 if it's Page 3 out of 9, you know it's a third of
16 the way through. That's the quickest way by far
17 to find information. You don't have to listen to
18 the whole screen cast if there's only one part
19 you're interested in. You should be able to fine
20 any information you want immediately.
21 If you do that and still don't
22 understand, make a list of those concepts, and
23 when we have workshops, 1we can work on that.
24 One.drexel.edu is slow sometimes, but
25 what can you do? Two ways of accessing Web CT,
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27
9.26.05 Organic Chemistry 241
1 and one is through 1.webct.edu, which makes you
2 log in a couple times. If there's a problem with
3 1.drexel and you don't have access to Web CT, you
4 can directly access Web CT.
5 The link I'm giving you here is
6 webct.drexel.edu, and this address is valid for
7 the previous version of Web CT. The new version
8 of Web CT, as soon as I know the link, I'll put it
9 up. If you click on this link, you can actually
10 log in as a guest.
11 What do you do if your computer crashes
12 during the test? As long as you keep clicking
13 save, nothing will happen. You could even go to
14 another computer, restart it, and it will
15 remember. Obviously, click "save" every time you
16 make a selection.
17 How do you save the flash file? I've
18 only had one student ask this. If you're
19 interested in doing that, you can go through this.
20 If you're not, that's okay, it's not important.
21 As long as you have a connection, through the
22 Internet you will have all the screen casts.
23 Do you have to use Cite U Like? Maybe
24 you can think about if you would like to do it or
25 not and how you would like to structure it. I can
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9.26.05 Organic Chemistry 241
1 answer these questions if more detail. The Cite U
2 Like is a website that forces you to have the
3 proper reference when you're citing peer-review
4 journals look at the example in chemical 242. We
5 don't have to go into the details about that
6 today.
7 How do you subscribe to an RSS feed?
8 RSS stands for "really simple syndication," and
9 it's basically a way for people to automatically
10 get updates from various RSS feeds. Some are like
11 news; a lot of people get news like that.
12 Everyone who actually maintains a blog has an RSS
13 feed you can subscribe to.
14 There are two different kinds of feeds
15 relevant here. The one RSS is the one you text
16 for the class; that's where you use blog lines.
17 The other RSS subscription that's important is for
18 the pod cast.
19 I use the same blog. Not everybody
20 does, but I do it for simplicity. That's why you
21 need a pod catcher to get the files, and you need
22 a normal aggregater to get the text.
23 Again, if this isn't making any sense to
24 you at all, bring your laptop and we can set you
25 up. That's basically the idea.
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9.26.05 Organic Chemistry 241
1 I've linked here all the things we need
2 to subscribe to and have shown you the tutorials.
3 STUDENT: (inaudible).
4 INSTRUCTOR: When you subscribe to.
5 STUDENT: Blog.
6 INSTRUCTOR: To the class blog?
7 STUDENT: Yeah.
8 INSTRUCTOR: No, it's completely
9 anonymous. So anyone can subscribe. There's no
10 password.
11 All right. So there's a lot of
12 questions. You can come back to this if you're
13 unclear on any of this.
14 I have a little bit of time left here.
15 The third entry here is the actually class blog.
16 There are two things you will find here. Mainly
17 the arrest curve lectures and the class
18 assignments that you do. If you do one of the
19 class assignments you will be putting it in the
20 blog so you have to get me to add you as a number
21 if you do the assignment.
22 Right now, all there is really is just
23 the lectures. So again, it's reverse
24 chronological order. If you want to go to the
25 first lecture, you scroll down. What you see in
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9.26.05 Organic Chemistry 241
1 the first entry is the Lewis approach to molecular
2 structure this is what I go through in the first
3 lectures.
4 And the classes are numbered so it's
5 from 1 to 39. The only one not in there right now
6 is class one because that's this class I'm doing
7 now. It's being recorded now and archived. But
8 all the other lectures in lecture format are
9 actually already there.
10 So you can look at the PDF. Again, the
11 streaming screen casts that's in the same flash
12 format.
13 Scroll down. See this little cursor
14 here? you can move it anywhere inside of the
15 class. Okay?
16 There's also the PDF. This is going to
17 crash my computer. Acrobat is a little finicky
18 sometimes. So the PDF is basically all of the
19 screens that are in the screen cast that are just
20 available as separate static products.
21 So if you were looking for particular
22 material, I suggest you start with the PDF. And
23 let's say you were interested in this valence
24 periodic table. So you notice this is Page 3 of
25 4. So roughly, I would go, you know, somewhere
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9.26.05 Organic Chemistry 241
1 about three quarters through the screen cast to
2 try to find where I talk about that. That's the
3 easiest way to find something.
4 And the other one is the MP3. When
5 you're accessing the blog in this way, this is not
6 a pod cast; this is just a website you can go
7 through. And technically you wouldn't have to
8 subscribe to the podcasts or the blog, but if you
9 don't, you won't know when anything has been
10 changed.
11 So the reason you want to use the RSS
12 subscription is to know that something has changed
13 so that you don't have to keep checking the whole
14 website, okay? But you can certainly just go to
15 this blog and download, you know, the file
16 individually.
17 Now, if you use iTunes to do that, this
18 is what it will look like. So in iTunes, you see
19 here the Lecture 38, Lecture 38 PDF. So the audio
20 and the file will actually be here. That's how it
21 shows up on here.
22 Let me just give you a taste of what the
23 quizzes look like in Web CT. Any one of you can
24 do this now. You're not dependent on
25 (indiscernible). Let me see if...
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9.26.05 Organic Chemistry 241
1 This is the course menu, and one of them
2 will be quizzes and tests. It will show up "begin
3 quiz," and then you can just take it.
4 So here I'm testing you on the Lewis
5 structures, testing you on the charges on the
6 individual atoms, which, as soon as you listen to
7 the first lecture, will make more sense.
8 This is where you can come to test
9 your knowledge. And, again, the tests will
10 be very similar to this.
11 Okay. So really the only thing
12 that's in Web CT is the quizzes and the list
13 of the problems from the book.
14 Okay. Let's try to wrap this up
15 here. You see here, there's a button for
16 workshops. Give me a couple minutes,
17 please. Quiet.
18 So the workshops are preplanned up
19 to a certain point. I want to be able to
20 spend time with you that's most productive,
21 so that means I want to work with the things
22 you're having the most trouble with.
23 I'm out of town next week so the
24 next class when we meet together is going to
25 be October 5th, on Wednesday. And I have
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9.26.05 Organic Chemistry 241
1 your schedule, so subscribing to the class
2 RSS feeds, FAQ podcasts, how to update the
3 wiki, the electronic configuration in the
4 Lewis structures.
5 So I will give you additional
6 problems to work on. You can work either by
7 yourself or with a team. So when you come
8 in, sit next to people you want to work
9 with. And as you go through the problems, I
10 will circulate and help you try to find out
11 where you are having difficulty with each
12 problem.
13 But anything is fair game here
14 STUDENT: You don't have a time for
15 these workshops, do you?
16 INSTRUCTOR: It's the class time. Yeah,
17 this will be the last formal lecture that we have.
18 STUDENT: Wednesday and Friday won't be
19 formal classes?
20 INSTRUCTOR: All the formal classes are
21 already recorded, so those are now assignments for
22 you. You are expected to watch the archived
23 lectures. So then when you come to the workshops,
24 I expect that you've looked at the lectures, tried
25 to do the problems, and now we will try to figure
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9.26.05 Organic Chemistry 241
1 out what you have a problem with.
2 You don't have to come to every
3 workshop. If what we're going to be talking about
4 isn't relevant to the problems you're having, you
5 don't even need to come to the workshops.
6 You can e-mail me a request with a
7 topic, and I will add it as we have time.
8 Answer your question?
9 STUDENT: (Inaudible.)
10 INSTRUCTOR: This week? No, I'm out of
11 town this week.
12 STUDENT: These dates are the only time
13 there will be a workshop.
14 INSTRUCTOR: In the coming future.
15 STUDENT: You can refer to this page for
16 any date of the workshops?
17 INSTRUCTOR: Exactly.
18 * * *
19
20
21
22
23
24
25
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