CHEM 125b: Freshman Organic Chemistry II
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Freshman Organic Chemistry II
CHEM 125b - Lecture 35 - Acyl Insertions and α-Reactivity
Chapter 1. Acyl Insertion of O, NH, and CH2 [00:00:00]
Professor J. Michael McBride: OK. We were talking about acid derivatives. We started last time talking about acyl insertions, although I’m not sure I mentioned the term. And then about reactivity at the position adjacent to carbonyls, alpha-reactivity and condensation reactions, which are related to that, which are chapters 18 and 19 in the Jones text.
So acyl insertion is to put an atom in the bond between the R and the carbonyl group, the acyl group. So you just move the R aside and put X in there. So X can be an oxygen, the Baeyer-Villiger reaction of ketones. It can be an NH group, so you can make an amine from a ketone. It can be a CH2 group inserted into the carboxylic acid.
So in the first case, you get a ketone transformed to an ester; in the second case, a ketone transformed to amide. In the third case, it’s an acid goes to an acid, but it’s an acid that’s longer by one carbon.
Can you think of any other ways of having an acid and making an acid that’s longer by one carbon? It has to be sort of a roundabout way. You could go from the acid to the alcohol, from the alcohol to the halide. Now you’ve got something you can make reactive. It’s reactive in itself, the halide. You could displace it with a carbon nucleophile. That would add one carbon to the chain. So it could be CN, right? And then we know you can hydrolyze CN to make carboxylic acid. So that would be one unit longer.
Or once you have the alkyl halide, you could make it into a Grignard reagent, or an alkyl lithium, and add CO2, which is a carboxylic acid again one way. So there are many ways, as we said last time, to skin the cat. But one way is this Arndt-Eistert procedure, which gives good yields.
So we were starting to talk about the Baeyer-Villiger reaction last time. And I’m going to write some more on the board here with this.
It’s supposed to turn that projector off. I think I’ll just turn it off and we’ll use the side ones for the rest of the lecture.
OK. So the Baeyer-Villiger reaction, as we said, starts with a ketone and reacts with a peroxy acid. Which is, of course, an acid. It’s got a pKa, I think, about 7 or something like that. So we can get the anion here.
We could add the acid and make the analog of a hemiketal, but instead… We don’t have a leaving group here that we can leave. It looks like the only leaving group is that. But there’s a fancy way to do it, which is, this can make these electrons of high energy. So we can use those electrons to displace the leaving group.
So now we have COR’ OR. So we’ve inserted that oxygen between the originally acyl carbon and the R group. So we have an ester, plus, of course, the carboxylic acid, which was a good leaving group, the anion of the carboxylic acid.
Now, I indicated here that we had different R groups, or could have different R groups, R and R’. The question obviously arises of which one migrates, which ester do you get?
Now, this is called migratory aptitude, and it comes in a reasonable order. H, tertiary radical, secondary, primary, and CH3. But I didn’t tell you whether this is increasing or decreasing in migratory aptitude.
So what is the group that shifts? Is it a cation, a radical, or an anion? What have I shown with my curved arrows? When R moves over, does it move without electrons, with one electron, or with a pair of electrons?
Student: A pair of electrons.
Professor J. Michael McBride: With a pair of electrons. So it’s R- that moves. It’s a methide. Or an alkide, shift, we say, like a methide shift.
So do you think methyl is the best or the worst at shifting, say, compared with a tertiary? Which should be the best? Suppose it were a cation that were shifting. Which would be best? What do you think? Liang, what do you think? Which is the best cation? Methyl, a primary, a secondary, a tertiary?
Professor J. Michael McBride: Can’t hear very well.
Professor J. Michael McBride: Yes, tertiary. So, it’s just the opposite with anions, So if it’s an anion that’s shifting, then one would predict that methyl would be the thing that would be easiest to shift, and tertiary would be the hardest. But in fact, H is better than this. And this is better, better, better, right? So it’s not the best anion that shifts. It’s the best cation that shifts. Doesn’t that seem curious?
But the reason is that it doesn’t break completely away to become R-. Its electrons, which have been boosted in energy by having this charge next door, are being shared with sigma* here.
So in fact, those electrons have to be given to something else in order for the rearrangement to take place. If it were an alkide shift in which the R pulled away to become anion, and then attacked sigma*, then the best anion would do it. But that’s not the way it goes. In fact, exactly the opposite. Because it actually is giving up electrons in the process of shifting. These electrons are being given to sigma*. So it’s the best cation migrates fastest, which seems sort of funny, when you say that it’s an alkide shift.
One other feature is that if the group that’s migrating is chiral, then, since you’re just moving these electrons over, it’s not any kind of backside attack. So the R in the product will retain stereochemistry.
OK. This is the Baeyer-Villiger reaction. So that inserts an oxygen.
Then we said that the Beckmann rearrangement– notice that these are old. Remember, the Baeyer-Villiger was done at the same time that Gomberg was there, essentially, 1900, roughly. The Beckmann rearrangement is 1886. This was done by one of the few organic chemists who was also a physical chemist, Beckmann. He invented a very, very sensitive thermometer, as well, and worked in the same department with Ostwald, whom we mentioned before.
But the Beckmann rearrangement puts a nitrogen in. So it starts with a ketone, like cyclohexanone, and inserts a nitrogen in the bond between carbon and oxygen. And the reagent that does it is hydoxylamine, and then treated with acid.
So we already talked about what amines do with ketones. They give imines. And it’s especially good to do for amines, you remember, that have unshared pairs that are interacting unfavorably. So you lose water and generate what’s called an oxime.
Now, the Beckmann rearrangement happens when this is treated with strong acid. You could imagine it protonating the nitrogen. It probably does that. But when it does it, it comes off again. But it can protonate the oxygen, as well.
And now we have the leaving group. But N+ doesn’t look like such a very good cation. So in fact, what you use is the electrons here. These electrons shift up, so it’s very much like the R-, or R+, if you wish, shifting in that case.
Pardon me. This is a C. I’ll write it explicitly. And we have the N here, and a double bond here.
And notice that since these electrons left this carbon, there should be a positive charge on this carbon in the formula. Right?
So since this is done with water present, we can put water on there.
Well, actually, let me draw a resonance structure of this first. Notice that it’s hard to pull these electrons away to leave this, what would appear to be a vinyl cation, right? But remember that the nitrogen has a pair of electrons in its pi system. Have I got that right? Yeah. It’s got another pair of electrons. These electron are displacing this, so they’re coming in backside to that orbital. But there’s another pair of electrons on the nitrogen, and at the same time that these electrons are leaving from this carbon, these electrons can go toward that carbon. And if that happens, we get a resonance structure of this thing which has the positive charge on the nitrogen. Right? So it’s not as unstable as you might have thought.
But at any rate, when we add water and lose a proton, we can get this, which is of course related allylically, if we just shift the proton, to this amide. So we’ve indeed inserted NH into the carbonyl-to-alpha-carbon bond.
Incidentally, this particular amine, this particular lactone– a lactam, pardon me, a lactone is a cyclic ester, a lactam is a cyclic amine, or amide– and this particular one is called caprolactam, and it’s of some commercial interest. Because if you hydrolyze it, it becomes this amine, which, of course, can cyclize, the amine can form an amide with the COOH. But it can also form polymers where the amine attacks the carboxylic acid of another molecule. So instead of a ring, one can then get a chain, and you can get a very, very long chain.
This is called nylon. The polymer– I’ll write it this way– is called nylon-6. So the 6 means that there are six carbons in the chain between the amide groups.
Now again, we could wonder, in a case where we have different Rs here, which are migrates. That is, which one gets next to the nitrogen.
Now, one might think it would be something like this up here, where the best cation would move, because its electrons trends are being shared with the sigma*, that’s allowing the water to leave. But it’s actually not that. Because you’ll notice, there are two different isomers of oximes, and if you’re careful, you can separate them. And it turns out that in this one, R migrates, but in this one, R’ migrates, which doesn’t surprise you, because it’s a backside attack. Displacement of the water. So it inserts, the nitrogen inserts, in the anti bond, the one that’s away from the OH.
OK. So that’s the Beckmann rearrangement.
And then the Arndt-Eistert reaction, which is the homologation, the one that makes an acid one unit longer. So you have a carboxylic acid, and insert CH2 in this bond.
And the way that one goes is first, to react with SOCl2. What will that do to the acid? SOCl2, remember that reagent?
Student: Switches OH out for Cl.
Professor J. Michael McBride: Replaces OH with Cl, the same as it does in ROH. So we have an acid chloride. And now, nucleophiles can do association-dissociation and displace the chloride. And the particular nucleophile that’s used in this case is called diazomethane. It’s a very poisonous and unstable compound. It’s a good idea not to work with it. It’s one of these interesting molecules that you can’t draw a Lewis structure without putting charge in it, which is sort of a tip that it might be reactive.
But you can draw not only that resonance structure, but also this one. The nitrogen in the middle has to be positive, but the negative charge can be written on either end of the molecule. It’s probably better to put it on nitrogen than to put it on carbon, but the reactivity is of the carbon.
So these are the electrons that come up. And then loss again, so substitution at the acyl carbon, as we’ve been speaking about.
So now we have CH2N+ triple bond N. Which loses a proton to give R, O, CHN2– OK. I’ll draw it this way. Have I done it right? Yeah. OK, good. Plus minus– which is called a diazoketone.
Now, it doesn’t surprise you that diazoketones, when heated– and it’s often done in the presence of silver– lose nitrogen to give an unusual compound, because there’s just two bonds to that carbon. It’s like a carbene. So it has a vacant orbital and an unshared pair.
And now you can see how we’re set up for the rearrangement. These electrons can shift just as they did in the Beckmann rearrangement; just as they did in the Baeyer-Villiger reaction. But at the same time, these electrons can come here to make a double bond. So we have O=C=CHR.
And notice that this, if we trace through this carbon here, became this carbon, became this carbon, became this carbon, and it’s now this carbon. So we inserted that carbon between the carbonyl carbon and the R group.
Now, we’ve seen this kind of compound before. It’s an acid derivative. It’s a ketene. So it can add water, as we’ve discussed before, to give RCHCOOH, which accomplishes what we wanted to do. CH2 now, because of the H from the water.
OK. So that’s the Arndt-Eistert procedure, which lengthens a carboxylic acid by one unit.
Chapter 2. α-Acidity [00:25:26]
I want to speak a little bit about– the rest of the stuff we’re going to talk about with acid derivatives is reaction at the alpha-position. That is, making an anion at the alpha-position and using that. So like an enolate.
But first we should look at the acidity of these compounds. So I’ll just use the slide to talk about that.
OK. So the pKa of acid aldehyde is 17, right? Acetone, we talked about this before, is slightly less acidic. The anion isn’t quite as stable. 19, two orders of magnitude in the equilibrium constant.
Now, the ester is another five orders of magnitude less acidic. Right? The equilibrium constant is 105 smaller. Why is it that the ester is less prone to form the anion by losing that red proton than a ketone is?
I could make the opposite argument. I could say oxygen is electron withdrawing, which would make it easy to form an anion. So there must be something else that goes in the opposite direction that’s even more important. What is it? Ruoyi?
Student: The unshared pair on O.
Professor J. Michael McBride: Right. That’s right. What allows you to form the anion at this position? In these cases, it’s stabilization by mixing with the LUMO, *pi. But in this case, the LUMO is always already being used to stabilize the electrons of the oxygen.
In fact, you could make the same argument here, in terms of hyperconjugation. That the pi* here is being used to stabilize the electrons in the CH bond. So when that vacant orbital, the pi*, mixes with the occupied orbital, whether it’s CH bond or, even higher, the unshared pair of oxygen, then they mix, and the vacant orbital becomes higher and less able to stabilize the anion on the carbon here.
So that’s understood.
Now, notice that a cyanide group, as we said before, a so-called nitrile, has much the same character as a carbonyl. In fact, this has the same acidity as the ester does.
Now, in an amide, so this amide, has a pKa of 18. Does that make sense to you? What would you have predicted? Ayesha? If the aldehyde is here, when you have a bond next door you get a little bit of raising the vacant orbital, so it’s not so good. Unshared pair on oxygen is even more important and makes it not acidic. What would you expect if, instead of oxygen here, if I had nitrogen? How about its unshared pair?
Student: The nitrogen pair would be less shared because it has a larger nuclear charge.
Professor J. Michael McBride: Aha. You’ve got everything right except the periodic table. Nitrogen comes before oxygen, so nitrogen has less nuclear charge than oxygen, so it’s unshared pair is higher in energy. So what would you expect?
Student: It would be more shared.
Professor J. Michael McBride: It would mix even more with pi*. pi* would go up even more. And it should be less acidic, it should be greater than 24.
And the reason is that that’s not what ionizes. What ionizes is a hydrogen from the nitrogen. So the amide is a little bit like a carboxylic acid. So it’s not comparable with these others, actually.
But if you put two R groups on the nitrogen, so it doesn’t have a hydrogen it can lose, then indeed it’s less acidic than the ester, which is what you predicted once you get the periodic table right.
Now, you can do even better if you have two carbonyls. So notice that all these are weaker acids than water. So if you put them in with hydroxide, the equilibrium will have the proton on them, not on the water. So you can, especially in the cases up here, they’re close enough to water– water’s about 16, 15.7– they’re close enough that there is some anion at equilibrium, a significant amount, so you can do reactions with it.
But mostly, it’s not in that form. Mostly it’s in the protonated form.
But if you have two carbonyls, then it becomes quite acidic. And we spoke before, when we were talking about the Mitsunobu reaction, about active methylene compounds. Methylene groups, CH2 groups, that are activated by two adjacent pi* orbitals, so that then they become much more acidic. This one has a pKa of 5.
And if we have two methyl groups, it’s like going from here to here. The pKa is 9. In fact, it looks uncannily good for that, because if you got two orders of magnitude by putting on one methyl group, from H to methyl, here when you put on two, you get four orders of magnitude.
That’s spurious in how good the agreement is. Because you see, if you put ester on, instead of the methyl, as we did here, ester instead of methyl, the pKa goes up, as it did here. But here it went up five orders of magnitude. Here the acidity went up only two orders of magnitude. So they’re really not additive the way we were talking about, but it’s in the right direction.
If you have two esters, then it’s another two orders of magnitude. If you have nitrogen, it goes up five orders of magnitude for the reason that we– in fact, compared to this one, it went up seven orders of magnitude. Over here, compared to the ester, to the nitrogen, it went up six orders of magnitude, so that one looks reasonable.
And finally, if you have two nitrile groups, it’s still quite acidic. And notice that all these are stronger acids than water, except the one here, which was deactivated by the unshared pair of the nitrogen.
So if you put these with base, they’ll be in the anionic form with hyroxide or alkoxide.
OK. So you can get these anions from the simple compounds– a small amount at equilibrium in strong base– and for the doubly activated compounds in high concentration, if you put strong base in there. In fact, if you use an even stronger base, like lithium diisopropylamide– so amide, remember, is a confusing name, because it means both the functional group O=C-N, but it also means a nitrogen anion, and you just have to know from context what you’re talking about. So anyhow, this is a nitrogen anion. So it’s a strong anion. The pKa is 36. So if you were to take this that you don’t get very good formation of, pulling off this hydrogen, the one that’s much deactivated, but you put it with this very strong base– this one has a pKa of 25, but this one has a pKa of 36– so you have an equilibrium constant of 1011, favoring formation of the anion. So you could make 100%, essentially, or just one part in 1011 not that way, if you use this very strong base.
So sometimes, even these alpha-anions that are difficult to make, if you use a really strong base, you can get them all made. So you have complete formation, even of the amide enolate, and not just a little bit of equilibrium.
And noticed that it’s a hindered base. It’s not dimethyl or diethyl. It’s diisopropyl. And you want it to be hindered, because otherwise, this H isn’t so very easy to attack. It would also be possible to attack the carbonyl group, pi*. Add the nitrogen to this carbon, and lose the other nitrogen. You don’t want that to happen, so you use a hindered one that can only get at the hydrogen out there. Doesn’t attack the enolate.
Chapter 3. H/D Exchange and Racemization via Enol or Enolate [00:34:38]
Now, when you make these things, you have a nucleophilic alpha-carbon. That’s why you make them. Carbon, remember, in the ways we’ve talked about it mostly, is an electrophile, R-halogen. The pi* of C=O, our low LUMOs. Here’s a carbon that’s a high HOMO, the unshared pair here on that alpha-carbon.
So this allows you then to form carbon-carbon bonds with the other kinds, the electrophilic carbons, by using that unshared pair to form a new bond.
OK. So let’s look at some examples of that. And it could be done either with acid- or base-catalysis via, in the acid case, enol, and in the base case, enolate anion.
And you know you know how to do this. I’m going to just zip through it quickly.
So a proton comes in. In the acid-catalyzed case, you get this cation, which is very stabilized, because of the unshared pair on oxygen next door. But that can lose a proton to generate an enol.
But now, in the case of hydrogen-deuterium exchange, it’s interesting. Because if D comes up, we can reverse the reactions. We could have reversed it this way, and put a proton on. But if you have a deuteron available, then you can do it this way, and go to there, and now lose this D+. And what we’ve done is exchange, in the alpha-position, deuterium in place of hydrogen. So that’s the acid-catalyzed mechanism, which doesn’t surprise you.
You can also do it with base catalysis. You can have base come in and take the proton, generating an enolate anion, which of course has this resonance structure.
And now we have this, as we had here, a nucleophilic carbon. We also have a nucleophilic carbon here. So we can, in the enolate, we could bring in deuterated water. So either acid or base allows you to exchange those protons. And if there’s more than one proton in that alpha-position, you can exchange the others, as well.
Now, notice that you can also racemize in this way. Because if the other things attached to this carbon were A and B, you can go through exactly the same scheme– acid or base. But there’s now configuration here. So if the H is in front of the plane here, in this case, when the H+ comes back on, it could come either from the front or from the back. Because this thing is planar, and this thing is planar, not chiral. So we’ve lost the memory of whether the thing was right- or left-handed. So it could come in from the back just as easily as from the front, and then we racemize.
So either by deuterium exchange, or by racemization, you can measure the rate at which enolate is formed and destroyed.
Chapter 4. α-Alkylation-Halogenation [00:37:53]
You could also do alpha-halogenation. And it’s exactly the same scheme again. The only difference is, once you get the enol or the enolate, you react it with a halogen. I’ve chosen to show iodine here. You can do it with chlorine or bromine as well. So you can make an alpha-halo ketone. Nothing special here.
Now, the product we made was an iodomethyl ketone. But we could also do the reaction with that. If there’s another hydrogen here, we could do the acid-catalyzed or the base-catalyzed, and get a diiodo compound.
Or can we? In fact, in this case, only one of the pathways works. And the reason is that the iodine, the halogen, is electron withdrawing. So it gives positive charge to that carbon– partial positive charge. That means that if you went up here, you would be putting two positive charges next to one another, which isn’t so great.
So in fact, the acid-catalyzed pathway disappears after you’ve added the first iodine. So you can put one in, one iodine, one chlorine, one bromine, but you can’t put the second one in, because you can’t, with acid catalysis, get the enol.
But if you have a positive charge there, that’s all the better for losing the proton and getting the enolate. So this is even faster, the second one.
And what do you think happens next? If you have another hydrogen there, right? – so there’s the base catalyzed one works– but now you can do it with a diiodomethylketone and make a triiodomethylketone. OK? So that’s great.
What happens next with base? Well, you can’t have anymore hydrogens on that one. So if you brought in hydroxide at this stage, it would attack the carbonyl. But when it does that, you could imagine the OH coming off again, although it’s not such a good leaving group. There’s a better leaving group. What’s the better leaving group?
CI3. With three halogens on the carbon, it’s a sufficiently good anion that it can leave. So we can unzip in that direction, one of these association-dissociation mechanisms, and get that. Because –and then, of course, you want to transfer the proton– the pKa of the HCI3 is about 14, so it’s more acidic than water is.
OK. Now that’s neat. Because that compound, iodoform– remember, CHCl3 is chloroform; this is iodoform– but whereas chloroform is a colorless liquid, iodoform is a yellow solid. So this provided, in the days before there was NMR– like if you had a methyl ketone now, you’d look for a methyl singlet shifted down field at a certain position. We know from NMR how you would identify it, and in the carbon NMR, you’d look at a particular place. Right? But in the days before there was spectroscopy to identify these things, how would you would know if you had a methyl ketone? You reacted it in water with base and iodine, and see whether you get a bright yellow solid.
So this is a picture [fumbling with projection controls], and you can see, here are two test tubes. This one had a methyl ketone in it, that one didn’t. This one has a yellow precipitate. So that was a handy way of identifying methyl ketones in the days before spectroscopy.
Now, you can do this alpha-halogenation with other carbonyl groups, as well. For example, carboxylic acids. So we’ve seen before that you can make an acid chloride in this way. And since I had it animated, I decided to put it here so you can review it, if you want to. And that’s how you get the acid chloride.
But one of the most charmingly named reactions is the Hell-Volhard-Zelinsky reaction, sometimes called the HVZ reaction. But if you remember Hell-Volhard-Zelinsky, that’s another one you can impress your roommates with. So it was Hell did it first, and then Volhard, and then Zelinsky, at about two year intervals flanking 1885.
So the Hell-Volhard-Zelinsky– I forget whether I’ve got this animated here or not– involves PBr3 and bromine. So the same way that SOCl2 can make an acid chloride from an acid, so can PBr3. So the HVZ reaction. Carboxylic acid reacts with PBr3 to give the acid chloride, or bromide. But the acid bromide is very good at forming an enol. And if you have bromine present, bromine can attack the double bond, because the cation will be stabilized with oxygen next door. So then when you lose the proton, you have alpha-bromination, bromination adjacent to the carbonyl group.
So this was easy to do, because the acid bromide is good at forming the enol. But now it looks like you would need to keep making the acid bromide in order to continue doing this reaction. But in fact, this can be catalytic. Why can it be catalytic? Why do you need only a little bit? You need a full mole of this, of course. Because this can react with acid. So it displaces the bromide, and we have Br, so that the brominated product, but now in the form of an acid anhydride. A different acid derivative.
But we also have Br minus, which can attack here and then this comes– yeah. What have I done wrong? Oh, I’m losing it. Yeah. Right? And then back here. Because I didn’t have the Br, I wasn’t thinking right. So now we have the acid, plus the other product you can see is Br. So we traded which one was acid bromide and which one was carboxylic acid. So now this can give this, and this is a cycle that goes around. So you only need this to get it started, to get the first bromide, and the subsequent bromides can come from that. And the actual product, then, ultimately is– remember, H+ came off here– so the ultimate product is the alpha-brominated acid.
Chapter 5. α-Alkylation [00:46:50]
I was hoping I’d get further than this. I’m going to go– well, let’s see. Yeah, I’ll go just this slide.
So alpha-alkylation. We promised before that the real advantage of having these carbon nucleophiles is the possibility of putting an R group on there. So you do this with– see, what do I have here? So you use a base, you get an enolate from your ketone or aldehyde. And now, this you react with something like methyl iodide, so an electrophilic carbon, sigma*. And you can make a new carbon-carbon bond, alpha. So the same way we could put bromine in alpha, or hydrogen in alpha, we could put methyl in alpha.
But there’s a problem here. Because we also have the possibility that it could react here. So “Which nucleophile wins?” is this question.
I think I probably should let you go now, and we’ll go on to resolve the Perils of Pauline the next time.
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