Formal charges have their plusses and minuses. Har har.

One one hand, they’re an indispensable accounting tool. If a molecule bears a charge, it would drive us nuts (for nomenclature reasons) if we didn’t adopt some kind of system where a charge was unambiguously assigned to one atom.

In many instances, the formal charge on an atom is an “honest” expression of its electron density. We’re all familiar with the ions Cl(-), HO(-), CH3O(-),Br(-), Li(+) and so on. The formal charge assigned to these atoms truly reflects that these molecules bear additional positive or negative charge.

However! then there are the outlier cases. And these cause problems. From someone who preaches “opposite charges attract, like charges repel“, it’s important to know when to pay attention to formal charge, and when to ignore it.

“Formal” charge is called “formal” because it’s ultimately an accounting issue. It doesn’t take into account the true electron densities of a molecule, which are based on a combination of electronegativity and resonance.

When trying to understand a new reaction, apply electronegativity to understand electron densities , not formal charge.

For instance in the bottom two examples, the curved arrows, as drawn, would be showing the formation of an oxygen-oxygen bond. This doesn’t make sense.

When you apply electronegativities, however, you get a much better picture of the true electron density of a molecule. And this can help you figure out how a reaction might proceed.

Here are some other common molecules where formal charge can be a misleading indicator of electron density.

Keep this in mind, and you’ll have a much easier time of properly understanding how reactions work.

When learning any new reaction, I think you always have to start with the “what”. What bonds are forming, and what bonds are breaking.

After you answer “what”, then you can start asking “where” – as in , “where are the electrons of the reactants?” What areas are electron rich? What areas are electron poor? There’s going to be attractive interactions between electron-rich and electron-poor areas. Then, you can use electronegativity and resonance to figure out the potential reactions.

But even that only gets you so far. It helps you identify different possibilities for reactions, but doesn’t really help you with getting to the final result.   In order to get there, you need to be able to evaluate how, as in “how do the electrons move?”. In other words, “in what sequence do these bonds form and break?” And that will open up a new set of questions.

So “How do the electrons move?” will be the next question to address here, and it will involve a lot of curved arrows. Because (forgive me for repeating myself), reactions are transactions of electrons between atoms, and curved arrows are the accounting system that let us visualize those transactions.

Let’s go a little deeper into the details of that accounting system. Last time I mentioned that there were just three “moves” for drawing arrows (lone pair to bond, bond to lone pair, and bond to bond). However, if you consider that there are two types of bonds (sigma bonds and pi bonds) this gives you different “flavors” within those categories.

 First, there are arrows that form bonds (lone pair to bond)

The source of our electrons here is a lone pair of electrons, and this lone pair is accepted by a second atom. The result is the formation of a bond.

The atom that accepts the electrons has its charge become more “negative” by one, and the atom that donates the electrons has its charge more “positive” by one.

Secondly, there are arrows that break bonds (bond to lone pair)

The source of our electrons here is a bond, and the pair of electrons moves to one atom. This leaves the second atom bereft of electrons. The source of electrons can be either a sigma bond or a pi bond:

Again, the formal charge of the atom that accepts the electrons becomes more negative by one, and the formal charge of the atom which loses the electrons becomes more positive by one.

The third type of arrows are those which both break and form bonds. 

We can also break sigma or π bonds to form a new sigma or π bond.  Note that in contrast to the first two situations (which involve two atoms), these arrows involve three atoms. The atom in the “middle” sees its charge remain the same, while the atoms on either end see their charges changing by plus or minus one.

So what?

In trying to evaluate a reaction, we’ll often ask ourselves:  how likely is this bond to break, or this bond to form? One pathway to being able to answer this question is to have a good knowledge of experimentally measured bond strengths. And that’s OK, although somewhat limited because bond strengths measure homolytic cleavage, not the heterolytic cleavage we’re talking about here.

But there’s a much more powerful way to think about evaluating these “transactions” than simply knowing bond strengths.  Note how every single reaction involves changes in electron densities? In the last series, I talked about how to determine the electron densities on different molecules using electronegativity and resonance. Now, I’ll go into more detail about how these electron densities change as bonds are broken and formed - and this will have a huge impact on whether or not a reaction is favorable or not.

It’s very useful to understand  how likely it is, for example, that an O–H bond will break so that the pair of electrons end up on oxygen as opposed to hydrogen. Or whether it’s more likely that a lone pair of electrons on oxygen or nitrogen are more likely to form a bond with a proton (H+).

If you understand the factors that stabilize (and destabilize) negative and positive charge, you will gain a great deal of insight into why (and how!) certain bonds break and form, and others do not.

P.S. Today marks the two-year anniversary of this site: Feb 20, 2010, “Orgo Hacks” was born.  I think the site got 20 pageviews that month, but it’s grown like a little weed.  The first two weeks of Feb 2012 the traffic was greater than that for all of 2010. Thanks to everyone for reading!

Grignard reagents are very useful carbon-based nucleophiles and will readily form carbon-carbon bonds with carbonyl compounds. Here the reaction of Grignard reagents with ketones is described, along with several examples and a detailed mechanism.

Have a great weekend!

I can’t claim credit for this idea but unfortunately I can’t remember where I read it either.

Here’s a helpful and quick way to make use of a study partner.

Let’s say you’ve got a long list of reactions that you’ve learned, but you need to practice your synthesis skills. Here’s what you can do.

Both you and your partner can write out a plausible sequence of reactions that you know.  It can be two or three steps – or even longer (or shorter) if you like. Include the reagents. Keep it relatively simple. Don’t worry about making it hard. Stick to what you know. If you’re unsure of anything, don’t do it.

I’m just thinking something simple like this:

This is your copy. Now draw out the starting material and the product, and give it to your partner.

It looked easy when you saw all the answers in front of you. Looks a little harder now, doesn’t it?

By doing this not only will you get practice in drawing out reactions in the forward direction, you’ll learn about how to think backwards too.

Here’s an example of a possible Org 1 version.

By the way, Ian Gould’s site at Arizona State is a great resource for synthesis problems.

If you think of electrons as the currency of chemistry, reactions are transactions of electrons between atoms. Just like double entry book keeping was developed to formalize how financial transactions  are recorded, chemists have developed their own convention for showing transactions of electrons between atoms. It’s called the curved arrow formalism. Previously I covered how we apply the curved arrow formalism to drawing resonance forms. Here, I’m going to show how we can extend it to show reactions. The same principles that applied to resonance forms apply here to reactions, except that unlike resonance forms we’re dealing with actual reactions, not components of a resonance hybrid.

The purpose of the curved arrow is to show movement of electrons from one site to another. Electrons move from the tail to the head. Most of the arrows you’ll see have a double-barb at the head, representing the movement of a pair of electrons. (there are also single-barbed arrows depicting the motion of a single electron; those are covered in detail elsewhere, and will be covered in a subsequent post).

Bottom line: there are only three legal moves you can do with curved arrows. Other than a few problematic examples, every reaction you will encounter in Org 1/ Org 2 can be described using a combination of these three “moves”.

This is just like the three moves for drawing resonance curved arrows! However, unlike drawing resonance forms – which only involve changes in π bonds – the bond here in question can be either a single (sigma) bond or a π  bond.

Curved arrows are also a way of tracking changes in formal charge:

• Since a pair of electrons are being donated from the “tail”, the atom at this site will have a formal “loss” of one electron, making its charge more positive by 1.
• Also, since a pair of electrons are being accepted at the “head”, the atom at this site will have a formal “gain” of one electron, making its charge more negative by 1.

Here are three general examples of each transaction. There’s a second layer of analysis that can be done here (sigma bond versus π  bond) but that can be saved for later.

The first example shows the simplified example of a lone pair on a naked hydroxide ion going to hydronium ion (H+).(Note 1). The arrow shows the formation of a bond between O and H, with electrons from the lone pair on the oxygen. Note the changes in formal charge: the “tail” become more positive, and the “head” becomes more negative.

The second example shows the reverse reaction: dissociation of water to give H+ and hydroxide ion. Here, the arrow shows the breaking of the O-H bond, to end up as a lone pair on oxygen.

Note the changes in formal charge: at the “tail”, hydrogen goes from “sharing” to “lacking”, thus losing an electron. At the “head”, oxygen goes from “sharing” to “owning”, gaining an electron. We adjust the charges accordingly.

(Why does it break this way? Rule of thumb: bonds generally break so as to put the electrons on the atom that will best stabilize them. Oxygen, being more electronegative than hydrogen (3.5 vs. 2.2) will better stabilize the additional electrons).

Finally, an example of a bond breaking to form another bond. Here, the arrow shows the breaking of a π bond between C1 and C2, and the formation of a new C–H bond between C1 and H.

Again, the atom at the “tail” (C-1) becomes more positive by 1, and the atom at the head (H+) becomes more negative by 1.

There’s a little problem with this type of curved arrow: the identity of the atom that is forming the new bond (C2 in this case) is somewhat ambiguous. For this reason, some instructors (myself included) occasionally draw an additional “dotted line” to remove the ambiguity. Others have developed a “bouncy arrow” technique.

So to summarize, this “accounting system” lets us not only account for the bonds that are formed and broken during a reaction, it also lets us keep track of the charges. This is really useful! If you’re given a molecule with these “curved arrows” drawn on it, it’s a lot like a computer program. The arrows give you precise directions on what to do in order to obtain the product.

Note 1: . I say it’s somewhat artificial because in reality, each of these will be accompanied by a “counter ion” of opposite charge).

Reaction Friday is back!!! Today’s video is on a commonly encountered reaction for making alkenes, the oxymercuration of alkenes using Hg(OAc)2 and water, followed by reduction with NaBH4.

Thanks to all of those who’ve made helpful comments on making videos.

First, an explanation. This happened. So things at MOC have basically  been on hold for a few weeks. Life is now returning to normal.

The following discussion has nothing to do with the above, other than the fact that my dad’s best friend wryly observed of him: “When he opened his wallet, moths flew out”. I suppose I share this trait with regard to a lot of things.  But I do not scrimp on can openers. This is why.

About 10 years ago when I was living in Montreal I was very poor, and outfitted my kitchen with items available from the dollar store. Including a can opener I paid exactly $1.14 for (in Canada, they hit you hard with sales tax).

After taking this small financial hit, life was somewhat better. I could now eat canned Ravioli. However this state of affairs did not last long. Within the next week my $1.14 can opener had opened (poorly!)  about three or four cans before flying apart while trying to pry open an 800 mL can of Bravo spaghetti sauce. So I was back to where I started: with no working can opener.

This time around I decided I wasn’t going to screw around, so I went to the local Canadian Tire and pulled out $6.99 .  It was more painful to do this, but the can opener I received was of superior quality. And it did not fail me in the subsequent years that passed (although it did get a tad rusty).

So if we compare these two situations we get this:

Now what does this have to do with chemistry?

Well, it resembles a situation that occurs in the addition of strong acids (like hydrochloric acid, HCl) to certain dienes such as butadiene. Two products are possible: the 1,2-product and the 1,4-product.

Two notes:

•  With butadiene, the addition of HCl can lead to two possible carbocations. Note that the carbocation on the top is secondary while the carbocation on the bottom is primary (they are both stabilized by resonance, however).  Since carbocation stability increases with substitution, the first carbocation is more stable. In other words, less energy is required for its formation, so there will be a lower transition state, and it will occur faster.
•  However, when you look at the final product here, the 1,2 product has a monosubstituted alkene (one carbon substituent) while the 1,4-product has a disubstituted alkene (two carbon substitutents). Alkene stability increases with the number of carbon substituents! 

The reaction diagram looks like this.

Each reaction begins with diene A, and then progresses through a transition state (B) to form carbocations ( C ) , which undergo attack through transition state D to give final product E. The height of each step in the process is related to its overall energy.

• When temperature is low, there is enough energy to form the 1,2-product –  and that’s it. The product ratio is determined by the reaction rate (i.e. the height of B). Such a reaction is said to be under kinetic control.
•  When temperature is high, there is enough energy to form both the 1,2- and the 1,4-products. Furthermore, the reaction to form products is reversible (i.e. there is enough energy to go from E → D and thence to A. In this case the ratio of products is determined by the relative thermodynamic stabilities (i.e. the height of E). Such a reaction is said to be under thermodynanic control.

So it’s a little bit like buying can openers. It’s easiest to get the money together to buy the cheapest one,  but the more expensive one is often the more stable product overall.

As Benjamin Graham said,  “Price is what you pay: value is what you get. ”

——–

Notes:

1. The can opener analogy doesn’t go into the “reversibility” part of things. However if you had a large population of people who had to make the identical decision, the population of people with deluxe/crappy can openers should likewise segment according to cost and reliability as a function of available resources. Note that if the product is too stable, nobody ever buys a replacement and the company goes out of business. That’s where planned obsolescence comes in. 

2. Not all additions to dienes follow this pattern. It is important to be able to evaluate the relative carbocation stabilities and alkene substitution patterns independently for each given diene. Don’t automatically assume that the “1,4″ product is always the most stable (try cyclopentadiene, for instance). For a good time, also try 2,5-dimethyl-2, 4 hexadiene. 

3. When you buy an item from the dollar store with moving parts, caveat emptor.  

… as will regular posting – next week.

Organic chemist Brian Coppola of the University of Michigan recently won a $275,000 teaching award from Baylor University. The press release and his presentation are in the link below.

Chemist Brian P. Coppola Wins Prestigious Teaching Award (University of Michigan)

The hour-long talk is extremely good. Coppola is a master of highlighting the connections between chemistry and aspects of everyday life, especially language.

The addition of HBr to alkenes is the focus of today’s video.

Fast forward to the fifteen seconds at the end for maximum awkwardness. I am an idiot when it comes to this stuff. Is making videos supposed to be brutally fricking hard? I don’t know how Gary Vaynerchuk and Salman Khan do it.

Thanks for bearing with me during this painful learning process. If you’re an expert on making videos (but not organic chemistry) and want to set up a mutually beneficial tutor-barter arrangement, drop me a line.

Loyal reader Sandy writes in with a comment about memorization:

I wish I had understood much earlier that there is a place for memorization in Orgo – like the reagents and what they contribute to a relationship, what reactions they are used with, etc. We were heavily discouraged not to memorize for the class – and I think it hurt my study plan. What we should have been told is “understand the mechanism – movement of electrons”, but you must memorize reagents, format of reactions, etc. I know he had good intentions, but he was not clear. It was at this point that I began to struggle. I did make an A, but spent the entire time between Orgo 1 and Orgo II reviewing and making sure I got the information I needed in my head.  This was the single most difficult issue I faced – trying not to memorize.

Great point.

No matter what anyone says, here’s where memorization helps: learning vocabulary, names of reactions/reagents/solvents, abbreviations, nomenclature, functional groups, and the bond-forming/bond breaking pattern of each class of reaction.

In short: vocabulary, conventions, and results obtained by experiment. Anything that answers the question “what?”is something that can be memorized. 

Where memorization doesn’t help as much is in answering the question “how?”

• how would you use [this data] to determine [this property of a molecule] ?
• how do you synthesize [this molecule] from [these molecules]?
• how will changing [this property of a molecule] change the [property of this reaction]?
• how will [this molecule] react under [these conditions] compared to [these conditions]?

See what’s going on here? We’re changing two variables at once, and the number of  possibilities (and things to memorize) increases exponentially. This is where applying concepts just becomes more efficient than memorization. I think this is what’s behind instructors’ advice to “not memorize”: it’s sound counsel.

In the best organic chemistry courses, you’re taught to apply the concepts you’ve learned to solve problems. So one piece of advice for students : ask the “how” questions a lot.* (“How does this reaction occur” is a subtly different question than “what’s the mechanism of this reaction”!).

* “Why?” is also a good question to ask, although the answer can sometimes be unsatisfying. (example “Why do we use [this reagent] for this reaction?” “Because it’s convenient, reliable, and cheap”!]