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CHEM 125b: Freshman Organic Chemistry II
- α-Reactivity and Condensation Reactions
As in many synthetic procedures, an important challenge in ketone alkylation is choosing reagents and conditions that allow control of isomerism and of single vs. multiple substitution. β-Dicarbonyl compounds allow convenient alkylation and preparation of ketones and carboxylic acids. The aldol condensation, in which an α-position adds to a carbonyl group to generate a β-hydroxy- or an α,β-unsaturated carbonyl compound, can be driven to completion by removal of water. The Robinson annulation reaction is an important example of conjugate addition to α,β-unsaturated carbonyl compounds. α-Acylation of esters as in the Claisen condensation is a key step in the biosynthesis of fatty acids. Determining the constitutional structure of sugars posed a daunting challenge to early carbohydrate chemists.
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Freshman Organic Chemistry II
CHEM 125b - Lecture 36 - α-Reactivity and Condensation Reactions
Chapter 1. The Soxhelet Extractor [00:00:00]
Professor McBride: Since it’s going to take a minute for this to start up, I’ll show you a demonstration here. It’s a way to make tea. It’s a really clever device called a Soxhlet extractor. And it was invented in order to extract stuff from like bark or leaves or something like that. Something that’s very insoluble. So you could imagine to do it like tea. You put it in a filter paper, pour solvent through it, some of it dissolves, it goes on through, and then you have to concentrate the solution.
But there’s another way to do it, which is to use this thing here—
How are we doing? Let’s see if I can get the show up here.
So the idea is that the problem with that is that the stuff isn’t very soluble. It takes an enormous amount of solvent to get it out. So what was invented by this man Soxhlet—
and I’m having trouble here. It says audience. Why is this not on? Oh. Somebody’s unplugged it. Sorry for the confusion here. We’re technically challenged today. There we go.
OK. So what we have here is a flask down below that’s going to have the solvent, and not very much solvent. Let me get it like that.
And so we’re going to boil the solvent. It’s going to come up through this arm here, up to a condenser, and then drip into the tea leaves, or whatever it is we want to extract, held in a little cup inside here. And it’s solid down here, so this is just a cup.
But there’s this funny snakelike tube that comes from the bottom of this, up, around, down, and back inside, toward the flask. So watch what happens when I pour some liquid in here.
So if you just pour water through a filter, the water runs through very rapidly. It doesn’t have time to become saturated. So it’s inefficient in two ways. You’ve got to pour lots of water through, and it doesn’t get saturated on the way through.
This way, the liquid you distilled up drips down in there, sits with the stuff, which is held, remember, in like a filter paper inside here. And you can see that the liquid– see if I can get it so you can see. Did you see that the liquid went not only of course in here, where it’s with the tea leaves, but also through this tube and up that far? So it’s there. You can see the meniscus in the side arm.
Now, what’s going to happen when I pour a little more in? Drip, drip, drip, actually, of course, from a condenser. It’s getting right up to the top of that side arm. What’s going to happen? So it’s sitting there and becoming saturated. Then we go a little bit more.
What’s it doing? It siphons, right? But it sits there with the hot liquid until it gets to the top. Then it draws all that off. And then what happens? More liquid boils up and comes in and dissolves. It’s the same liquid, of course. It’s been distilled now. So it comes on up, and pretty soon– it’ll go a little quicker this time, because there are drops in the side arm. And it drains again.
So you use just a small amount of liquid over and over again. And there are no moving parts, right, except the boiling and the liquid. So it’s a really beautiful design of a clever apparatus, if you want to make tea efficiently.
Now this is starting up. And if I’m lucky, it will have backed up my most recent work. It’s got almost all of it. That’s good. See if it will com on here. Yeah. OK. Thanks for bearing with me.
Chapter 2. Alkylation Regiochemistry [00:08:44]
So a little more about the reactivity of acid derivatives, and then on to the sugar– glucose proof, one of the finest logical creations of the 19th century– or ever.
First, alkylation. We started talking about this last time, about the fact that an enolate has partial charge, both on the carbon and the oxygen. So if you want to put an alkyl group on by an SN2 type reaction– what’s the problem?
Student: Are you rehearsing timings?
Professor McBride: Pardon me?
Student: Are you rehearsing timings?
Professor McBride: Oh, thank you. One more time into the breach here. Jon is on the button about whether I’m rehearsing the timings.
OK. So the question is whether you react with the carbon or the oxygen when you do SN2, when you react with methyl iodide. Now, you could do a calculation and see where the HOMO is, right? The HOMO looks like that. It’s a pi orbital, obviously, with one node, in addition to the p orbitals that it’s made of. And that’s big on the carbon. So that would favor reaction on the carbon. And the attack would be from the top, because it would come in on top of the p orbital of the carbon.
On the other hand, if you look at the electrostatic potential, the energy a proton would have on the surface of this molecule when it’s in contact with it, it’s not the most favorable at the position of the HOMO. It’s –143 kcal/mole for a proton that would be there.
But a proton here is –170 kcal/mole. And notice that it not only favors the oxygen, but it’s a sigma attack, rather than a pi attack. So you could imagine either of these positions attacking the methyl iodide.
Now, in fact it turns out– so that if it attacked the carbon, it would attack from the top, as shown up here on the right, out of the plane. If it attacked the oxygen, it would attack in the plane. And in fact, the major product is attack on carbon with methyl iodide. And it’s a negligible amount of O-methylation.
But it’s not always that way. If you have a different reagent, a different electrophile to be attacked by the nucleophile– for example, silicon– displace chloride– then it turns out to be just the other way. It goes on oxygen, 85% yield, and a negligible amount on the carbon.
So this is obviously a finely balanced thing, and it’s not something that you’re going to be able to predict. This is a part of lore, which things go where.
Now, if you use a strong anion [correction: base] to extract a hydrogen adjacent to the carbonyl to make the enolate, its pKa is 12 units higher, so that the equilibrium lies completely to form the enolate anion.
But one could imagine two different enolate anions, where a proton’s been taken from the right or from the left. So there’s another question about possible reaction mixtures, when you react this with methyl iodide. And if it’s done with two equivalents of methyl iodide– this was done by Professor House at MIT– he found that 41% of the product alkylated on the CH2 group, and 9% alkylated on the group that had a methyl group attached to it already.
But these aren’t very good yields. There was also 22% of recovered starting material. So the question is, where did the proton come from that went on to the recovered starting material?
Now, this was done by adding methyl iodide to the enolate solution. So there’s just a little bit of methyl iodide there that reacts and makes the ketone, one of these two ketones down here. But then, additional enolate could abstract a proton from one of these, and that could become the new enolate. That could be where the proton came from to make the starting material. And in fact, 21% of the product has a second methyl added, and 6% of the product has three new methyls added to the compound. So it’s a real dog’s breakfast. It’s not the way you want to do this kind of thing. It’s a very inefficient reaction.
There are other ways to do it, like for example, use lithium diisopropylamide, another really strong base. This is done at very low temperature. And then the people who did this put in manganese dichloride in a complex with lithium chloride, again at –78°, and they made a different enolate anion, the one with manganese as the counterion.
And then they added to this 1.3 equivalents of benzyl bromide and a different solvent, the N-methylpyrrolidone. And did that at room temperature.
Notice this is lore. This is a lot of fine description of exactly what solvents, what temperatures, and so on– how you do it. And under those conditions, they were able to get 88% of this single product after distillation. Right? So it’s the pure stuff. It’s possible to do these alkylations as a very high-yield reaction.
Now, this is what’s called an “Org. Syn. Prep.” When there’s a very valuable technique that people would like to be able to repeat, it’s submitted by the people who created it, and checked by people at this Organic Syntheses operation, and then they publish it. And then you can believe that if you try it, you’ll be able to do it here, because it’s been checked. So things that are an “Org. Syn. Prep.” have been very carefully done, and all these conditions are there. But it would be very difficult ahead of time to figure out exactly all the tricks you need to do it. But you can look it up in Organic Syntheses.
OK. LDA can also be used with carboxylic acids. So we want to make the enolate. We want to extract that red proton. Can you see any problem involved when you treat this with a strong base? Megan? What’s the name of this starting material? What kind of compound?
Student: Carboxylic acid.
Professor McBride: Why do they call it a carboxylic acid?
Student: It gives up protons.
Professor McBride: Because it gives up protons. Is that the proton it gives up?
Professor McBride: No. It gives that one up. So you don’t get the enolate.
However, this is a strong enough base that it’s possible to pull off a second proton and in fact get the enolate, which lacks the [correction: insert “second”] negative charge that’s supposed to be there. And then that can react with an alkyl bromide at the more nucleophilic, less stable anion center, and give a good yield of the alkylated acid.
Chapter 3. Alkylation of b-Dicarbonyl Compounds [00:16:35]
It’s also even better, of course, to do dicarbonyls, which are more strongly acidic, so you don’t need these really strong acids [correction: really strong bases].
So you use ethoxide to do this. It’s acidic enough that you could imagine using hydroxide. But you don’t use hydroxide to do this. You use ethoxide. Can you guess why you use ethoxide rather than hydroxide?
As always when you begin to get complicated reactions, and this is not a very complicated one, there are several ways things can react. And this is not… as in the last case, where you could pull off the wrong proton. In this case, the base can not only pull off the proton, it can also act as a nucleophile. And what would it do as a nucleophile? Jack?
Student: Attack the ester.
Professor McBride: Hydroxide would attack the ester, right? OH– would come on, –OEt would come off, and you would convert the ester to an acid, which would react with the strong base, hydroxide, and that would be the end of the line. Because you don’t have this really super strong base that could pull off two protons.
But if you use ethoxide and it attacks here, who cares? One or the other of the ethoxides is going to come off, and you’re back where you started again. Right? So then it’s possible to wait until you pull that proton off.
OK. So that compound is a beta-ketoester. It’s got a ketone in the second carbon from the ester group.
OK. So you make the anion and react it with an R with a leaving group, and now you treat it with hydroxide. Right? Because what you were really after was getting this acid with an R group on it. But you don’t use OH– for that.
Or you could use two esters to make the thing acidic. Do exactly the same stuff and get the R on, now with two acid groups. And of course, it would be possible to do it a second time, and put a second R group on while it’s still an ester, before you add the hydroxide.
OK. So you can put alkyl groups on these beta-ketoesters or malonic ester. The CH2 with two carboxylate groups on it is malonic acid. OK. Now, you might want those compounds.
Or we already, in fact, saw using cyanoacetic ester. You remember when we saw that? We used it where R was this, R with the leaving group. It was the Mitsunobu procedure that we went through and showed that that also can hydrolyze the CN group, the nitrile, as well to give a second carboxylate group.
But mostly, people aren’t after this. What they’re after is protonating those to give the beta-keto acid or the malonic acid.
Why did I draw it that way? Do you remember?
Student: Because in that structure the H is shared.
Professor McBride: Pardon me? I can’t hear well.
Student: The H is somewhat shared between the two Os.
Professor McBride: Right. The H, of course, would be hydrogen bonded to this oxygen, but it can do more than that. Remember from when we were talking about the Mitsunobu reaction? We can do this, which results in decarboxylation when we heat it.
And those are enols, which of course can isomerize to the related ketone, or on the second case, into an acid. So the real purpose of using this beta-ketoester or the malonic ester is to put the R group on, and then get these compounds. So it’s another way of getting to what was effectively that enolate, except having the second carbonyl on made it easier to get. You didn’t need to use those really strong bases.
So you can you can synthesize an alkylated acid or an alkylated ketone, alkylated in the alpha-position, by using these dicarbonyl compounds.
The one on the top is called acetoacetic ester synthesis. Because when this was still an ester here, this was CH3, acetoacetic acid. So it goes by way of that to get the ketone. Or this one, as we said, was malonic ester that we started with to get the acid.
Notice that these are ways of adding to the R group two carbons. Either with a ketone– I mean, it would be another carbon, of course, if it were a ketone– or two carbons here to make an acid.
And so when one’s thinking about synthesis, you try to think about how to put more carbons on. Remember the Arndt-Eistert procedure, we talked about with diazomethane, or cyanide, displacing chlorine from an R group, or making a Grignard out of the alkyl halide, and then adding CO2? Those all add one carbon to the chain. But this thing adds two carbons, at least, or more, in the case of this, when you have an R group there.
And remember adding a nucleophile to open an epoxide also adds two carbons. So if you’re in the business of organic synthesis, these things become second nature. I need to add this, aha, I would use that reagent, or that reagent, or that reagent.
Chapter 4. Aldol Condensations [00:22:12]
OK. Then one of the most important reactions is the aldol condensation. It was discovered first with acetaldehyde, which, treated with base, makes the enolate. But it can react with itself. And then when you protonate the O–, you get this compound, which, in the case of acetaldehyde, this particular one with a methyl group here and a CH2 group here, has the name aldol. So this is called the aldol reaction. It makes aldol. But you can make it with other groups here as well, not just with the methyl group. And so generically they’re called aldols. Obviously the name is aldehyde and alcohol, aldol.
But it can be dehydrated. Lose the OH and an H down here to give the unsaturated aldehyde.
Now an aldol is a beta-hydroxyaldehyde. And this is an alpha,beta-unsaturated aldehyde. So these are things that when you’re looking at something you want to make, and you see, aha, it’s a beta-hydroxycarbonyl compound, or it’s an alpha,beta-unsaturated carbonyl compound, you think, the aldol reaction.
If I assigned you a zillion problems to do that, that would become second nature to you. If it turns out that that’s important to you, you will do a lot of problems with it, and it will become second nature to you.
Now, I’ve shown it here base-catalyzed, but it can be acid-catalyzed as well, and I would expect you to be able to figure out how the mechanism would work with acid catalysis.
Now, the equilibrium for this is favorable, because this isn’t very crowded with H and H here, and H’s down here, and H here. But if you start with a ketone instead of an aldehyde, then the product is more crowded and less stable. So if you start with acetone instead, and make the anion, and try to do this reaction, and make the beta-hydroxyketone now, you find that the equilibrium is unfavorable. So you’d like to figure some trick, if that’s the compound you want, that would allow you to get over there.
One way is not to start with the ketone, but to start with the enolate, so you’re much higher in energy. So these things we talked about before, about using really strong bases to pre-form the anion, those are ways to do this. But if you want to deal just with hydroxide, there’s a way to do it.
Now, we already went through the demonstration of showing how you make tea by pouring hot water through a tea bag in a funnel. But a much more efficient way is to use this Soxhlet extractor. So the extractor looks like this you boil liquid down below, and you condense it up above, and it drips down into a paper cup that’s holding the tea or whatever it is. And it begins to rise up, and when it gets up to the top and finally goes over the top, it drains down, and then the whole cycle starts over again.
So that’s a much more efficient way to do– to extract to make tea. But it’s also a way to make aldol because of what you put in the cup.
So the liquid you’re going to use is acetone. The liquid is acetone, so it’s distilling up, sitting there for a while, and then draining back, and distilling up again.
Now, remember that it takes a catalyst to make the aldol. A base is what we’ve shown here. So suppose you had a base that was only in the cup. Then the acetone will come and sit in the cup and reach equilibrium, if it’s sitting there for a while with the base, which has just a little bit of the aldol in it, because the equilibrium lies to the left.
But then it drains down into the pot, where there’s no base, so it can’t go back again. But the dimer, of course, is much heavier than the starting material. So it doesn’t distill. The acetone goes back up, drips through again, makes a little bit more, it drains in, up a little more, a little more, a little more. So all you have to do is put in that cup up above a base which is not soluble in acetone. Barium hydroxide is such a thing. So that’s a really fancy way to make an aldol when it was unfavorable. Using a Soxhlet extractor.
OK. Now it’s possible to do a crossed aldol condensation where the enolate comes from one carbonyl compound, and the carbonyl that gets attacked comes from another one. In particular, suppose you start with what’s called acetophenone. It’s a methyl phenyl ketone. Make its anion. And it attacks– if it were to attack, itself, with methyl and phenyl, the equilibrium would be very unfavorable. It would be way too crowded, right? So this one won’t work. That’s an unfavorable equilibrium.
But suppose instead you use benzaldehyde. Now it’s not hindered because it’s an aldehyde. So you can make the product with benzene here, benzene here, and H here by this crossed– two different reagents, this one and this one– aldol condensation.
Now, it’s clear that this one is not going to react with itself, with its own ketone, because the product will be too hindered. But why won’t the enolate of that react with itself? It would be much less hindered. This is a trick question. Why doesn’t the enolate of this carbonyl compound react with the aldehyde? Ellen, what would the enolate of this be? How do you make an enolate?
Student: Take off the H.
Professor McBride: Where do you take the H from?
Professor McBride: From the carbon that’s adjacent to the carbonyl. Right? So that the pi* of the carbonyl will stabilize the anion. So what’s the story here?
Student: Oh, there’s no H there.
Professor McBride: There isn’t any. It’s a benzene ring. There’s no hydrogen on that carbon. So this can’t form an enolate, and this can’t react with itself. So they cross over without complication. So that one can’t form an enolate. And you can then lose water to make the alpha,beta-unsaturated ketone.
Chapter 5. Conjugate Addition and Robinson Annulation [00:29:41]
Now I’ll just go through this, which I did on the board last year and took a picture of. This is the compound we just made on the previous slide. It’s an alpha,beta-unsaturated ketone. And it undergoes what’s called conjugate addition. That is, you expect a nucleophile to react with the carbonyl carbon, the π*.
But if you have a situation like this, cyanide could react not only there, but also here. And if it reacts here, and then you put on a proton on the O… you’d have the enolate anion once the cyanide is attacked up at the top left here. So there’s the charge here, charge here.
So you could put the proton either place. The more stable place to put the proton is down here, so you have a carbon oxygen double bond, rather than to put it on the oxygen, so you have an enol. We know that carbonyls are more stable than enols.
So this is the product you get from that. And I think it’s safest to call it conjugate addition, because you add to a double bond that cyanide would not normally add to. Remember, alkenes are usually attacked by electrophiles, not nucleophiles. But this one is special, because adding here, you generate a stable anion, the enolate.
OK. So that’s called conjugate addition. Or sometimes it’s called 1,4-addition. So here’s the one, two, three, four, position. You add a hydrogen there and cyanide here, that would be 1,4-addition to the thing. The product, of course, would be an enolate, but the enolate will become a ketone. So that’s just the name that’s given this, 1,4-addition. But conjugate, I think, is better.
OK. Now, if you react with the Grignard reagent, phenylmagnesium bromide, it does the same thing cyanide does. It undergoes conjugate addition, and you get a 96% yield of this. And if you use copper, it turns out that that sometimes helps to get conjugate addition, if you add it.
But if you use, instead of phenylmagnesium bromide, phenyllithium, then it doesn’t do the 1,4-, or the conjugate addition. It adds, as you would naïvely have expected, to the carbonyl group.
So obviously, there’s a lot of subtlety involved here, and a lot of lore. I would certainly not have expected you to know– I would expect you to know that you can add here, and having thought about it a little bit, I would expect you to think that you can add here. But as to which way it goes, and why a Grignard reagent goes here, and why copper helps it, while phenyllithium goes here– that, I wouldn’t expect you to know. That’s the lore that comes from being in the business a while.
But when you’ve been around, you know what to do. You know to throw copper in, if this is the kind of product you want.
OK. And the same kind of thing happens with hydride addition. So if you have this cyclohexenone, and reduce it with lithium aluminum hydride, the hydride does what you expect it to do. It adds to the carbon of the carbonyl, and you get this unsaturated alcohol, cyclohexenol, 97% yield.
But if you use a different source of hydride, the BH4– instead of AlH4–, then you get only a 59% yield of this product, and you get 41%, the rest of it is the material that’s lost the double bond. So the double bond has been reduced, too.
Now, BH3, remember, adds to double bonds. But BH4– doesn’t. And when BH3 adds, it adds an H, and also the boron, and that’s the way you get an alcohol, remember, the hydroboration/oxidation. But BH4– is different. It doesn’t react with double bonds. And if you take this material and react it with sodium borohydride, it doesn’t do that. So that’s not where this came from.
Where it came from was having the first addition be conjugate. So hydride added to this position, the end of the double bond that’s conjugated with the carbonyl, as in the top left there. That made this enolate, which then got protonated in this position to give them the ketone instead of the enol.
And now that carbonyl can be reduced by borohydride, the same way this one could at the beginning, and now you get the 41% of that. So you must initially have gotten almost equal amounts of reduction of the carbonyl group directly by 1,2-addition and conjugate addition, 41% conjugate addition, which then led to that.
So again, this is lore. And if we drilled you on that, you would know something. You would be able to have quick recall of it. Having seen it, I expect you to recognize things like that. But I don’t expect you to know all the details. In fact, I think, no one knows all the details. This is all sensitive to what temperature, exactly what salts you use, what solvent and so on. So it’s a complicated thing. Or adding the copper.
OK. Now a very interesting reaction is called the Robinson annulation. Annulation means making a ring. And this was invented in 1935, and here’s the beginning of the paper where it was invented by Rapson, a graduate student, and Sir Robert Robinson, who got the Nobel Prize for synthetic organic chemistry.
So it’s experiments on the synthesis of substances related to sterols. So you remember steroids, these hormones, are six-membered ring, six-membered ring, six-membered ring, five-membered ring, with various lettuce hanging around it. So there are lots of these different things.
And the 1930s through the 1950s and ’60s was the golden age of steroid chemistry. And remember, that was the basis on which conformational analysis was invented. In 1958, Barton, who was working in this field, started looking at which side, axial, equatorial and all that, that was all done in the context of steroids.
But what Robinson says here is, “The methods for synthesis of 1,5-diketones”– So here’s a ketone. One, two, three, four, five. So a 1,5-diketone. “The methods for synthesis of 1,5-diketones have been explored in this laboratory for two reasons. One, it was thought that pinacols such as (I) might furnish the characteristic angle-methyl ketone group (II) of estrone…”
So estrone is a female hormone, right? And it has that five-membered ring up in the top-right, has a ketone there, and what’s called an angular methyl group adjacent to it.
So how are you going to synthesize that? So he thought that that could come from this via a pinacol rearrangement. We talked about the pinacol rearrangement. Remember, what’s going to happen is you would protonate here, lose water, cation, methide shift. So now you’ve got the methyl there from this five-membered ring.
So you need that compound, if you want to be able to make it that way. Where do you get a pinacol? You get it from pinacol reduction. So you would get it from this forming the bond across there. And we talked about the pinacol reduction, where you put an electron on here, an electron on here, the two free radicals couple, and then you have O-, C-C bond, O-. And they get protonated and give the pinacol.
OK. So that would be the pinacol reduction. So that was what he was hoping is a way to try to make this five-membered ring with that methyl group, which is hard to put in there otherwise, “by intramolecular change,” which we would now call rearrangement.
And a second reason was “the diketones themselves, III, might be dehydrated with the formation of cyclcohexenones.” So you could make a six-membered ring with a double bond carbonyl.
Now, what do you notice about that double bond? He says you might be able to go from here to here, so these double-shafted arrows are thinking backwards. It’s called retrosynthesis. What could something come from? The ultimate arrow will be this going to this.
But when you see this, you think of this. Why? Why, when you see this functional group, do you think of this diketone? You look at that functional group, and what would you call it? Can you give a name to that? Kate, can you think of any names for functional groups there? What’s this functional group, the C=O?
Professor McBride: Ketone. And this one?
Student: An alkene.
Professor McBride: An alkene. So you could call it an enone. Right? But then you wouldn’t know where the ene is. So you have to say where the ene is relative to the ketone to get that functional group, to know that they’re conjugated. So if you start with the ketone, it’s the alpha and then the beta position. So it’s an alpha,beta-unsaturated ketone.
And there’s another name for alpha,beta-unsaturated ketones. They’re the dehydration product of aldols. So the aldol reaction, where you make an anion here, it attacks that carbonyl to make OH, and then you dehydrate it– he says dehydrate it here– lose the OH and an H here to make the double bond. So that’s the reaction we just talked about. The aldol reaction.
So then you need this compound. You need that diketone if you’re going to make it that way. And how will you make it? You’ll make it by forming the red bond, attacking this double bond here. What kind of reaction is that? Luke, you got an idea? If you made an anion here, that would be an enolate anion. Pulled an H off there. OK? So you can make the anion here. And then you want to attack a ketone, so you think of making a bond from here to here. But this is making a bond from here to here. Did you ever see anything like that?
About eight minutes ago, we talked about conjugate addition. Addition to the double bond adjacent to the carbonyl. Because if the anion attacks here, then you have a minus charge on this carbon adjacent to the carbonyl. It’s an enolate anion.
So this is an example of conjugate addition. So what you would need, then, this would be conjugate addition to this ketone. And that ketone, before you make the red bond, is methyl and a C=CH2, vinyl.
So that’s methyl vinyl ketone, which is a very important reagent for making these six-membered ring compounds.
So the idea of making this ring compound by way of methyl vinyl ketone, attacking a ketone, is called the Robinson annulation procedure. The way of forming a ring. So a base to make the enolate, and then methyl vinyl ketone. So it takes this into two six-membered rings adjacent to one another with a carbonyl down here, which is just like what’s in the bottom left-hand corner of these steroids. Six-membered ring with an oxygen pointing down that way and another six-membered ring.
So the Robinson annulation is just the ticket for forming this thing, and they often have a methyl group in that position. So you see, these are the four carbons that become these four carbons in the product. So to do that, you’d need an enolate at this position doing a conjugate addition to attack here. So this carbon becomes that carbon. Then you form an anion here at that position, an enolate, it is. And beyond the carbonyl, there’s one more methyl group. This one. But once you’ve got that, then once you’ve got the methyl group here and still a ketone here, then you can have the anion of that methyl group, the enolate, attack the carbonyl, an aldol reaction, dehydrate, and you’ve got this thing.
So that all happens at once when you use methyl vinyl ketone as the ketone. Except you have to be sure that you’ve got the right anion here, that the anion is, at that position, the enolate that you’re starting with, and not at that position. And we’ll address that on Friday, how you make sure you’ve got the right enolate.
And Woodward– because on Friday, we’ll be talking about Woodward’s synthesis of the steroid cortisone, where he used Robinson’s method. We showed this picture before last semester, and they’re not acknowledging one another’s presence here. Both of them were quite… Each of them, I’m sure, was quite confident that he was most important organic chemist of the 20th Century.
And what’s especially interesting is here. You see the double bond is not drawn, but this is that cyclohexenone, the product from the Robinson annulation.
Chapter 6. Claisen Condensation and Fatty Acid Biosynthesis [00:44:17]
OK. Then there’s one called the Claisen condensation, which is now an ester. But all these are the same. You make an enolate, and you react here with itself. So in fact, when I made these slides, I could use exactly the same slide and just plug in one group different each time. So as you look at these, you’ll be able to organize them clearly in your mind.
So here now is an ester. Why do we use RO– rather than hydroxide? Because we don’t want to make the acid. We want to keep the ester.
So we have RO–, we talked about that before, make the enolate. And now this one comes over and gets attacked.
So it starts the same way an aldol does, if the starting material were an aldehyde or a ketone. But it’s different because you have a leaving group on here. So when it comes back in, that leaving group can leave. So instead of getting an alcohol from the group that gets attacked, you get a carbonyl group.
So it’s a beta-ketoester. Instead of beta-hydroxyketone, beta-hydroxyaldehyde– that’s the aldol reaction– it’s a beta-ketoester, because you had a leaving group. So it’s not an alcohol, right?
Now again, this equilibrium is unfavorable, if you have more complicated groups on here. But at first, it looked like the alkoxide was catalytic. We used it here to make the enolate, but then we got it back again here, so it looks like a catalyst that just cycles around and around. You don’t need very much of it.
But this beta-ketoester is quite acidic. So it actually “eats up the lye,” as in saponification that we talked about last time. And that’s good, because the equilibrium here might have been unfavorable. But once you do this very exothermic reaction, taking up one of those hydrogens, then the reaction is favorable again.
Now here’s nature’s version of the Claisen condensation. They use the special ester. Instead of oxygen, it’s got sulfur here. But it does exactly the same thing. It makes the anion, it attacks, you get a beta-ketoester. It’s a thioester. And coA is coenzyme A, so it’s a special R group that’s attached to make it an ester.
If you’re interested, that’s the R group. There’s the SH that’s going to be that thing there. But that will be for another day.
So how does nature get around the fact that this equilibrium is unfavorable, that it would come back? It puts an extra CO2 in. So instead of starting with this being CH3, it starts with an extra CO2 group on it here. So it’s a dicarbonyl compound. Easier to form the anion.
Now, you might think, if that’s a negative charge, it’s going to be hard to form the anion. But in the enzyme where it’s held, this one is not ionized. I’m not sure exactly how it’s tied up, but anyhow, it helps make this one acidic, having the carbonyl next door.
And then you get this product. So it’s malonyl, not acetyl. And then the equilibrium is driven by losing the CO2. So that’s how nature does Claisen condensations.
Why? Because it wants to store energy, right? So notice it’s put this acetyl group on the original acid, or ester. It’s formed this new bond where it attacked and had a leaving group. So then nature reduces it using the reducing agent, NADH, that we talked about. So it becomes an alcohol. And then it dehydrates. It’s a beta-hydroxycarbonyl compound. Just like aldols, it dehydrates. And then it gets reduced again to put H2 in here.
So we have this product. We started with having just one methyl group, with just two carbons here, and now we have one, two, three, four carbons there.
And notice what you can do. You can take this ester and plug it in up above, and put another two carbons on when you come around to get this. Then you can plug it in, put another two carbons, another two carbons. So you can make a long chain of reduced carbons ending in an acid, or an ester, to start with.
So these are fatty acids. That’s how you store the energy of fuel to burn, as these successive CH2 groups that can be oxidized. But the fatty acids that occur in nature are all even numbers of carbon, not odd numbers. Because it makes them this way, and always adds two carbons at a time to the chain.
Chapter 7. Carbohydrate Structures [00:49:36]
I’m going to say just two words now– more than two, but not many more than two– about carbohydrates. So carbohydrate was named early on. I was going to look it up and find out. It must have been in the early 19th Century. Once they could analyze them and see they were CH2O taken n times. So carbohydrates. A hydrate of carbon. Right?
Now, Couper in 1858 had the idea of tetravalence. And once you had that… a number of sugars had been observed, but no one had any idea what structure was, so it was hard to talk about them. You could talk about some of their properties, but not, obviously, about their structure.
So if you have carbohydrate, Couper would put two more bonds on the carbon. So you could make a chain of these, say, six.
But this is unsatisfactory. You’ve got to do something with those two bonds on the end. You could make a carbon ring, a six-membered ring, cyclohexane with OH’s all around. But that’s not what you do. What Couper did in his very first paper, you remember, about structure, he gave a structure for glucose, and the structure he gave was this.
What did he do with the extra bonds? How did he modify this chain in order to get his structure? Well, of course, remember there are double oxygens where in fact there should be a single oxygen. That’s because of the– But what else? He took water and put H on one end, OH on the other.
And in fact, you can do that. This is a gem-diol. That’s an aldehyde that’s been hydrated. That’s one way to do it. Another way to do it is to take– this is what got lost when my computer crashed. So this one, notice, took the HOH that was here, took the H’s away, put one H up here, the other H down there.
And this one took the two H’s away from this one, put one H up there, the other H down there.
So these, then, are sugars, but who knows which one is which? And this is a page from van ‘t Hoff’s– that was 1858– 1874, stereochemistry. What makes things optically active? Van ‘t Hoff, in his 1874 paper, says these compounds are optically active: Glucose, levulose, lactose, mannitol. Because he’s got four different things on this carbon, the four that are in the middle of the chain, right?
Or mannitol, the one where you’ve added H to each end of the aldehyde. Then you still have these things. Or derivatives of mannitol, or the diacid where you have an acid at each end of the chain. All these things were optically active, right? That’s part of this.
So there were a number of sugars known by that time. But the really fine work began to be done in the 1880s by Heinrich Kiliani and Emil Fischer. And that will be a treat for next time.
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