Lecture 002

2006-01-09T11:02:00.000-05:00

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Instructor: Jean-Claude Bradley

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So we are going to talk about aromatics today. Now we did discuss
aromatics in 242 at the end so I'm not going to be redoing all of
that. We're going to be looking at reactions that we haven't
covered so far, but I think it's important to review a little bit
of nomenclature. Okay, so we're going to be interested mainly in
the chemistry of benzene and its derivatives, so if we start to put
groups on the benzene ring, all right, like if we put a methyl
group on here, we could call that methylbenzene. We could also
look at it as being a derivative of methane, so call it
phenylmethane, and what would be another name for this one?
Toluene. So there's a few special names that I'm going to be
expecting you to know. Not too many, and I think we've covered
most of these before. Toluene. Okay.

So as you know, also, with aromatics, we have multiple groups on
the aromatic ring. There's a relationship between them, so if I
use two methyl groups, any two groups, there are three different
ways I could position them -- next to each other, removed by one
carbon, or at opposite sides of the benzene ring. So all these
compounds are called xylene. Dimethylbenzene is xylene, but we
have three possibilities. First one is the ortho, second the meta,
and then the para. So we would call these orthoxylene, metaxylene,
and paraxylene, and you can do that for any compounds that have
only two substituants. If you have more than two substituents, you
can't do that. You have to use a numbering system. So if you were
to use a numbering system in this case, because you start with a
ring, you can start numbering anywhere you want, so you have to
start numbering at one of the knuckle groups. In ortho xylene, I
put "one" on the top knuckle, then "two" would be where the second
knuckle is, so I would call that "one, two, dimethylbenzene." So
you can always name aromatics, no matter how many groups they have
on them, with the numbering system. But if they only have two
groups, then you can use the ortho-meta-para.

Now I don't want to spend too much time on this -- I know that we
already covered this, so this is just a reminder. There's a couple
of other compounds -- a benzene ring that has three methyl groups,
positions one, three and five. That's mesitylene. And the other
thing I want to remind you of is the difference between a phenyl
group, so if I have an aromatic ring, okay, if I have a
substituent, that's phenyl, if I have CH2R, then that's benzyl.
Okay, so benzyl has that extra carbon. Oftentimes those two get
confused, so just make sure you know the difference.

Okay, that's actually all I want to do in nomenclature. Any
questions?

Okay.

So, properties I'd also like to talk about briefly. So aromatic
compounds, as long as they don't contain electro groups, of course,
are going to be hydrophobic. So in that sense, they're pretty
similar to alkanes. You would expect van der Waal's interactions,
they should be soluble in alkanes, they should not be soluble in
water, and that's pretty much what you find unless you put some
electron-withdrawing groups on there. One thing that's special is
the relationship between melting points and the relationship of
groups on the aromatic ring. So let's say I have paraxylene and
orthoxylene, which one do you think would melt at a higher
temperature? Remember how to do that? This goes way back. Well,
what determines the melting point is the energy that is required to
separate the molecules, so the more orderly the solid, the more
difficult it will be to melt it. And when you have para
substitution, these are straight, so they're easier to pack, so
they would be higher melting. So this is the same logic we use
when we talk about alkenes. So I would write here that the
paraxylene would be higher melting.

Okay, so that's all I wanted to do for properties. I really want
to get to the reactions today, so we're going to learn a new
reaction. So we're starting with benzene, t-butyl chloride, and
aluminum trichloride. So this reaction will put the t-butyl group
on the aromatic ring. Okay, and because we start with just
benzene, and we were just adding one group, there is only one
product. We're going to see, of course, if you have a group
already on the aromatic ring, you could have multiple products, and
you need to figure out, you know, what they're going to be, but in
this case, it's pretty straightforward. There's only one product
in this alkylation. And this particular reaction happens with a
new mechanism. We looked at nucleophilic substitution, right? The
SN2 reactions, SN1 reactions. This one is not nucleophilic, it's
electrophilic, so it's seeking electrons. So this is called
electrophilic aromatic substitution. I'm going to do that on
another page because I'm going to go through the mechanism.

Electrophilic aromatic substitution. So there are a couple of
steps here. The aromatic ring is not involved in the first step.
The first thing that happens is, we're going to react aluminum
trichloride with the alkyl halides. So the aluminum doesn't have
any electrons left on it, right, it's what we call a "lewis acid."
It's going to receive the electrons, and the way it's going to do
that in this case would be like this. The t-butyl chloride would
lose the chlorine, and it will make the aluminum negative, and it
will leave behind a carbocation. So this is really just another
way to make a carbocations, and everything you learned about
carbocations is applicable here. So that gives us the t-butyl
cation, and an aluminum that has four chlorines around it -- a
tetrachlo aluminate. So that has a negative charge on the
aluminum. And as we'll see, the important thing here is not so
much the aluminum trichloride and the t-butyl chloride, it's that
we generate a carboction. If you can generate a carboation
by any means, you can do this electrophilic aromatic substitution.
So we'll see a few other ways of doing that.

Now that we have the carbocation, we're going to react it with
the aromatic ring. So I'm drawing all the hydrogens on the benzene
-- if you don't do that, you'll see that it gets a little bit
confusing as to what happens. So the carbocation comes near the
aromatic ring, and it has a positive charge, so it is, by
definition, electrophilic. So that's where the electrophilic part
of the mechanism comes from. It will be attracted to the pie
electrons in the aromatic ring. So what will happen is that you
will actually break the aromaticity, and the electrons will jump
in, make a signal bond between the aromatic ring and the cation.

So the reason I suggest whenever you do these problems to leave to
hydrogens on is because it becomes easy to forget if the carbon has
a hydrogen or not once you break the aromaticity. Okay, so you'll
notice that in this step here I haven't removed any hydrogens, so I
have to put hydrogens on all the six carbons. Now I have swung
that double bond open, okay, and I put the t-butyl cation on the
top to show that that's where it comes in, so that means that one
carbon now becomes SP3 hybridized, and I have a hydrogen and a t-
butyl group. And I have a carbyl cation now at the other position
on the ring. So you'll notice that this is not aromatic, right,
we looked, last term, we need a continuous ring of p-orbitals, we
don't have that here, because we now have an SP3 hybridized carbon.
So obviously this is not, you know, something that is energetically
very favorable, but once you do form it, the reaction goes in
another direction, so over heating this thing, you're going to be
driving the reaction. Okay, but this is not a downhill process at
this point, because we're losing the aromaticity.

Okay, so what happens now is, either I can go back, or I can lose a
proton and go forward. So if I'm going to lose a proton by an
elimination reaction, we're going to need a base. The base we have
here is the tetrachloral aluminate that we made in the first step.
So we're just doing the reverse of what we did in the first step.
Now we're actually starting the arrow at the ALCL bonds. This is
just an acid-base reaction just like every other elimination
reaction we did last term, and we eliminate. And we regenerate the
aromathicity, and that's a very favorable reaction. So what are
the products of this? Well, I'm going to redraw all the hydrogens.
I've now regenerated the double bond at its positions and I've just
made t-butyl benzene. The other thing that I've done is I've
regenerated aluminum trichloride, and if you follow what the
arrow's doing here, the CL is joining with the H, so I've generated
HCL. Okay, so overall in this reaction, I am not consuming the
aluminum trichloride, so the aluminum trichloride is a genuine
catalyst. You don't need one equivalent of it to get this to work,
you just need a catalytic amount, so that's one thing to keep in
mind. And you're generating HCL. So you don't have any base here
to absorb the HCL, so basically, as the reaction progresses, it
gets very acidic. Okay, so that's one thing you can expect from
that reaction.

Okay, so this is an example of an electrophilic aromatic
substitution, actually, I have to label one more thing on here.
This intermediate, where the aromaticity has been broken, is called
the sigma complex. Okay, so that happens when you have an SP3
hybridized carbon as part of your intermediate, like this. We call
that a sigma complex. Okay, and this reaction is called a Freidel-
Crafts reaction. So it is a subset of electrophilic aromatic
substitution reactions. It happens whenever you use an alkyl
halide. Okay? So we're going to see other examples of
electrophilic aromatic substitutions that don't involve alkyl
halides -- they won't be Friedel-Crafts reactions. So with Friedel
-Crafts you try to make alkyl benzenes, or alkyl aromatics.

Questions?

Okay, so let's complicate this a little bit. Let's try to predict
what happens if we again take benzene, aluminum trichloride, and
1-bromyopropane. Okay, in the last example we started with t-
butyl chloride, we ended up with t-butyl benzene. So what do you
think the product of this reaction would be? Benzene and propyl
chloride. One propyl chloride. No memories from last term? Well,
we're going through a carbocation intermediate, okay, so
everything you learned about carbocations comes back now. So if
we draw out what happens initially, okay, the aluminum trichloride
comes here, and we'll do a similar thing, okay, so I'm kind of
approximating this a little bit, you probably don't end up getting
a full carbocation from this, but it will behave in that way. So
aluminum tetrachloroaluminate, and we end up with the carbo
cation of propane on the one position, so you get a one-two shift.
When I introduce carbocations, I told you whenever you come
across them, they will always do one-two shifts for anything else,
but here's another example. So before this has a chance to react,
it'll do a one-two shift.

So you'll recall one-two shifts originate, and count to two from
the carbyl cation, one, two, and whatever you hit there you can
move, if it's either an alkule or a hydrogen, more specifically, a
hydride, will move by one position, if you will generate a more
stable carbyl cation. And in this case, certainly, you go from a
primary to a secondary carbyl cation, so that will happen. Okay,
so here is the secondary carbocation, and now it proceeds exactly
the same as it did the other problem we did. This now is
relatively stable. It won't do any more shifts. So this aromatic
ring will open up again in the same way, and will form a sigma
complex with an isopropyl group on the ring. And again, put those
hydrogens in so you don't get confused where the carbyl cation is.
All right, there's the sigma complex, and now we're going to lose
the proton. And the base in this case will be the tetrachloro
aluminate we just made, so we generate isopropyl benzene from this.
Okay, and of course, the HCL and the aluminum trichloride is
regenerated.

Okay, so that's basically one thing to watch out for. Make sure
you verify, you know, if you can get the one-two shifts, make sure
you do them before drawing the product out. Again, the reason why
the one-two shift happens before anything else is because it's
intramolecular -- it doesn't have to wait until two molecules bump
into each other. Everything that's needed to do one-two shifts
just has to twist in the right conformation, and that's it. It
gets done. That's why it's quicker.

Okay, so there are other ways to make carbyl cations. Aluminum
trichloride and an alkyl chloride is just one. We can also use
alchohols. So in this reaction, I am reacting benzene with
isolproponal and boron trichloride. Boron trichloride is another
potent lewis acid, and you have an alcohol on here. That alcohol
will make a complex with the boron trichloride. So that's just an
acid-base reaction, the lewis acid-base. So what do we get out of
this? Well, we first have to activate the alcohol. So the three
chlorines are very electron-withdrawing, so this makes it very
easy. There are two long pairs on the oxygen, the alcohol. One of
them will grab onto the BF3. So we're just going to follow what
the arrows are telling us to do. Notice that we didn't break any
oxygen-hydrogen bonds, so we have to keep that in, still have one
of the long pairs, and BF3. On here, the boron now has four
groups, so it's got extra electrons, it's going to be negative.
The oxygen will be plus. So the idea here is that we make the OH
group a good leaving group. It's kind of similar to the tosylate,
actually, if you look at it that way. And we generate a carbyl
cation like that. So we have an isopropyl cation. And the other
product here is the OH on a boron and two chlorines. Three
chlorines.

Okay, so the important thing is that we generated the isopropyl
cation, now we can do the standard Friedel-Crafts reaction, which
I'm not going to repeat every single time we do this. All right,
we go through the sigma complex, and then, well actually, I'll
repeat it just to make it clear what the base is in this case,
because it is a little bit different. Okay, so we're going to open
up the aromatic ring again. I'm doing this in the same position
that I did the other two problems. Isopropyl group, still have a
hydrogen, and a cation inside the ring, and this is the sigma
complex. So now this boron compound will come in, and kind of like
the tetrachloraluminate, except in this case, it's chlorine that's
coming off. So we end up with isopropyl benzene, and the other
compound would be HF, and then we have the BF2OH.

Okay, so once again we generate the Friedel-Crafts reaction.
Notice in this case that the boron, the lewis acid, is not a
catalyst. You don't regenerate it at the end. You actually
consume it. So it's better to think of it in terms of promoting
the reaction, but it's not a catalyst in the true sense of the
word, like aluminum trichloride.

Okay, so there's one more way we can make carbyl cations, and
that's from alkenes. So if I start with propene, in order to
convert the alkene into a carbyl cation, there's something you
should recognize. That's the first step of a Markovnikov addition,
right? So if we use certain acids we can actually create a carbyl
cation ion from this, and what's used here will typically be like
HF. Okay, so what's the product from here? Well again, we're going
to generate a carbocation first, and it helps to draw out all the
atoms in this case. HF, strong acid, will protonate that, and if
you remember how Markovnikov addition works, the first ion will go
at the end of the ion, where there's more hydrogens, because I'll
generate a secondary carbocation, as opposed to a primary, if I
put it in the middle. I won't repeat all of that stuff. I'll just
remind you. That's why it goes at the end. Okay, so we've made a
isopropyl cation, and there's no one-two shifts that are possible
for this, right, it would give a primary carbyl cation, so we stay
with that, and then in the next step, we make the sigma complex.
Sigma complex with cation inside the ring, and an SP3 hybridized
carbon. And again, we have the isopropyl group. Okay, so the base
in this case would be the F-. So we regenerate to aromatic
structure, and the final product is a propylbenzene again, and HF.

Okay, so now you've seen three different ways to make carbo
cations, and they all go by the, you know, very similar mechanism
to the Friedel-Crafts.

Okay, so I mentioned that Friedel-Crafts is only one example of
electrophilic aromatic substitution. Let's take a look at some
other examples. So another thing we can do is halogenation.
Again, I'm going to start with an unsubstituted aromatic. I'm
going to start with benzene, and we're going to react it with
bromine. Now, if you remember from the alkene chapter, when you
pick an alkene, you react it with bromine in the dark, it'll add a
cross double bond. But we never did that with an aromatic. Okay,
so if you simply mix bromine with benzene, nothing will happen,
because the aromatic stabilization is too much to enable the
bromination. But if you add to this a lewis acid like iron
tribromide, then you can get this to actually do an electrophilic
aromatic substitution and exchange a hydrogen for a bromine. And
you also generate HBr from this reaction.

Okay, so we're going to go through that mechanism. Turns out it's
similar to the Friedel-Crafts reaction. Again, it's the same kind
of concept -- the catalyst will react with the bromine to form a
complex. It's not a carbyl cation in this case, though, but iron
tribromide is just like aluminum trichloride and boron trichloride
for all lewis acids. Here's the bromine -- Br2 -- and you get a
lewis acid-base reaction like that.

So what does this look like? You end up with a positive charge on
the middle bromine, and a negative charge on the iron. Okay, so
you're not really generating the analog of a carbocation. You
want to think of this as activating the bromine, making it a little
more electrophilic. So the rest of this proceeds very similarly to
the Friedel-Crafts reaction, except we're not putting an alkyl
group on, we're putting a bromine. Okay, so I'm going to put it on
the other side this time. Okay, we're doing kind of like a, sort
of similar to an SN2 reaction. We're attacking the neutral bromine
and dumping the electrons on the Br+. So that will generate
another sigma complex. So I've put the double-bonds in that
position, so I need to put the bromine on the top. Hasn't lost any
hydrogens yet.

And now in the next step, we just have to lose a proton to
regenerate the aromaticity, and what do we have that we just made?
Well, in this step here, where we reacted the benzene with the
activated bromine complex, we generated iron with four bromines,
negatively charged. Tetrabromoferrate. So that is the base in
this case, and we get the same kind of elimination at the end that
we did for the Friedel-Crafts. All right, and what's left over is
iron tribromide. Well, we generated HBr and iron tribromide, which
is, now this is catalytic just like aluminum trichloride was in
Friedel-Crafts because you regenerate it, so you don't need a full
equivalent. You do need, of course, a full equivalent of bromine,
but that's not the catalyst.

Okay, so we're going to look at a couple more electrophilic
aromatic substitutions. Okay, actually, let's see here. Let me
just list the other halogens you can do this with, and then I'll
stop. Okay, so I won't go through the whole mechanisms here
because it's really very similar. You can use chlorine gas and
aluminum trichloride and draw exactly the same kind of mechanism
that we did with iron tribromide and bromine, and make
chlorobenzene. Okay, so when I'm drawing these reactions I'm not
going through all the details of the mechanism, okay, I might not
write all the side products. What I care about in this reaction is
the generation of chlorobenzene, but of course, you also generate
HCl. So that's how you put a chlorine on. Now if you want to do
the same thing with iodine, you can't really use a similar kind of
concept. In this case you have to use nitric acid and molecular
iodine. And that will make iodobenzene. I don't want to go
through the whole mechanism here, but it involves an iodonium ion.
So it involves an I+. Okay, you have to make that intermediate.

And finally, for fluorine, you want to make fluorobenzene, this is
really not a good way to do it. Fluorine is too reactive; you
can't really control the reactions, so you can't make it through
this mechanism. We're going to see other ways of making aromatic
fluorides, though, later.

Okay, so when we come back on Friday, we will resume with nitration
and sulfonation reactions.

CHEM243 transcripts

Lecture 002