 Good morning. Good morning. So today we finish up chapter 24. And as I said last time, chapter 24 really brings us just to the height of a lot of aspects of organic chemistry. It gets us thinking about mechanism. It gets us thinking about synthesis. It gets us thinking about retrosynthetic analysis. The work in chapter 24 is the basis of a lot of synthetic organic chemistry. It was the enabling technology that allowed chemists to make steroids for the first time. Chemistry made the New York Times for helping in the war effort back in World War II being able to synthesize cortisone that was needed as an anti-inflammatory. And this is really your entree into that beautiful chemistry. So what I'd like to talk about today is the material in chapter 24 that we didn't talk about. And I'm doing things in a slightly different order that I see as more logical. We're going to start by talking about the Robinson annulation reaction that builds upon the ideas from the last lecture. And then we're going to move into related carbonyl chemistry of the Claisen reaction. Now normally when I start I don't recap, but I'd like to recap three reactions that we learned last time and in a moment you'll see why. So we learned about the Michael reaction, sometimes called the Michael addition. And your textbook has taken a lot of the color out of things by calling everything a reaction and I've kept with that in my titles here. But you'll probably hear me and other people referring to it as Michael addition. So in the Michael reaction the example that we had, we had several of course, but one of the examples that we had which is fine, we took something that was going to make a nucleophile. Not too basic a nucleophile either. I said that typically we're talking about things where the conjugate acid, right? So this is a conjugate acid of an enolate. Where the conjugate acid of an enolate is not too basic where it's only moderately basic. This particular example, ethyl acetoacetate has a pKa of 12. And I said generally less than or equal to 15 pKa we go through Michael addition and generally greater than 25 we go through one, two addition. By the way, another name for Michael addition or Michael reaction is one, four addition. So in a Michael addition we generate an enolate or generate a nucleophile of some sort and we have an alpha, beta unsaturated carbonyl compound very, very broadly defined. Alpha, beta unsaturated ketone, alpha, beta unsaturated ester, even other things in the broad carboxylic acid family like an alpha, beta unsaturated nitrile. And in the presence of something that allows us to make an enolate here, in this case one of our two favorite bases we've been seeing again and again is sodium ethoxide in ethanol. We end up with a one, four addition of the enolate to generate a new enolate. This is all in your notes before for the mechanism so I'm not going to rewrite it out but I want you to at least process things in your head. So we end up with a conjugate addition of the enolate to generate a new enolate and then protonation to give a ketone and we see we get a one, five carbonyl compound. So that was one of the things we learned last time. Another reaction we learned last time was the aldol reaction and we learned it in lots and lots of different forms. I started off by teaching the directed aldol reaction. I said chemists like to have control and so in a directed aldol reaction we specifically make an enolate first and then allow that enolate to react with a ketone or aldehyde. So the example that I gave was cyclohexanone, one of the several examples and we treat it with LDA and THF. LDA is a very strong base conjugate acid diisopropyl amine as a pK of 40 and then we add in the second step we add a ketone or aldehyde. I said this is a very general recipe and I used isobuteraldehyde in this particular example, two methyl copanol and then we carry out an aqueous workup and I wrote in one case I wrote aqueous ammonium chloride and another case sort of generic H3O plus what you'd get from throwing a weak, throwing an acid into water like HCl in water. And so the product in this case was a beta hydroxy carbonyl compound and the last class of reactions we learned really was a variation on the aldol reaction. So aldol reactions can occur under many, many conditions. They can occur with strong base like lithium diisopropyl amide. They can occur with a weaker base, a moderately strong base like sodium hydroxide. And so the other example, these are all sort of archetypes was the aldol reaction with dehydration. That's where we lose water in the end. So here the mechanism is you generate an enolate. It adds into the carbonyl compound on workup. You protonate the aldolate, the oxyanide. So aldol reaction with dehydration generally occurs under conditions where you've mixed components together, either an aldehyde and base, an enolizable aldehyde and base or a ketone and a nonenolizable aldehyde in which case you have a crossed aldol reaction. Then the example that I started us out with was the parent aldol reaction. It was acid aldehyde with a base and I said historically calcium hydroxide had been used but we'll just think of this with sodium hydroxide and heat for this reaction. And actually if you do this it gives kind of a mediocre yield of 50 percent yield. And so this reaction, the aldol reaction with dehydration is very similar to the aldol reaction here. You in an equilibrating conditions generate an enolate. That enolate reacts with another molecule of aldehyde to give a beta hydroxy carbonyl compound but under these equilibrating conditions you generate another enolate that then undergoes elimination of hydroxide. So mechanistically those are things you've all seen before. Oh, by the way just while I'm on this topic of color. So you'll also hear this reaction referred to as the aldol condensation. A condensation is a reaction where you lose a small molecule so when you have put two molecules of acid aldehyde together here we're losing a molecule of water so it's a condensation reaction. The Claisen reaction which we'll learn about later you're losing a molecule of alcohol which is why it's sometimes referred to as the Claisen condensation. All right, so I want to put these three different reactions together now in sequence. In the sequence I've talked about and the reaction we're going to learn about is the Robinson annulation. And I'll start with a specific example of this particular ketone, 2-methylcyclohexane 2,4-dione. And I'll say in general we can say, let's say in general a ketone or a 1,3-dicarbonyl compound. We're going to mix it with an alpha-beta unsaturated carbonyl compound. In other words, something very much like this over here. I'll start out, and this is kind of the archetypal example, I'll start out with this ketone. I suppose it's IUPAC name is 3-butene 2-one. Every practicing organic chemist who does a Robinson annulation will call this methyl vinyl ketone, a nickname because you have on one side of the ketone a methyl group and on the other side a vinyl group. If you really want to sound in the know you could call it MVK. Anyway, this is an archetype for an alpha-beta unsaturated carbonyl compound. I'll write that in shorthand, alpha-beta unsaturated. And I'll say more specifically an alpha-beta unsaturated ketone because we're going to need that. Okay, and again we're going to mix this with some base. Honestly, base in the real Robbins, you're going to kind of get a baby exposure to the Robinson annulation. In the real Robinson annulation, base is kind of a sticking point and it's not always as easy as you just mix things together with the first base you try. We're going to in this particular example use sodium hydroxide and water, sort of one of our favorite conditions. But I'll say a suitable base here. If we have time I'll even give you an example of an acid-catalyzed Robinson annulation reaction. But okay, here's the real beauty. So we mix these two components together, our carbonyl compound, ketone or 1,3-dicarbonyl compound, something that can form an enolate and an alpha-beta unsaturated carbonyl compound, something we've seen as a Michael acceptor in the presence of base. And these two components come together in a series of reactions to form, whoopsie, these two components come together in a series of reactions to form a cyclohexenone. And you look at that and you say, my God, how did that occur? Obviously there's a lot of chemistry going on here. This particular ketone happens to be important. It's called the Velen-Mischer ketone and it's used as the basis for the synthesis of some synthetic steroids. Synthetic steroids are the basis not only of cortisone, as I mentioned before, but for example, birth control pills. So these are very important compounds. Okay, so what's going on here is we're going, oh yeah, and so let me, since I've been writing all of these archetypes, a ketone or 1,3-dicarbonyl and alpha-beta unsaturated ketone, let me write out the key thing here. The key thing is that we're forming a cyclohexenone. In other words, the key thing, the new feature that we've added is a six-membered ring containing a carbonyl group and cyclohexanone but also a double bond, so a cyclohexenone. All right, so what's going on here, we're going to break this out in just a moment, is three different reactions in sequence. So the first reaction is a Michael reaction or Michael addition. The second reaction is an aldol reaction and the third reaction is a dehydration. And if you want to remember something, just come up with a crazy way of remembering it and you look at this and you say, how in hell does that occur? That's crazy, that's mad, we're just going to go ahead and remind ourselves, M-A-D, the Robinson annulation is a mad reaction. All right, but I'm not mad, so I will show you how this occurred. All right, rather, by now we have written curved arrow mechanisms for every one of those reactions I just erased. We've written a curved arrow mechanism for a Michael addition. We've written a curved arrow mechanism for an aldol reaction and we've written a curved arrow mechanism for the base-promoted dehydration. So I'm not going to write out the curved arrows, I would expect you to be able to do this, but I am going to think us through all of the intermediates. So we have an acidic carbonyl compound or cyclohexane dione. We have a base, the most acidic thing is the cyclohexane dione. So what we end up generating first is an enolate. We're under equilibrating conditions, so reactions. So I guess I will write here to help us keep track. This is the M step, the Michael step of our mad reaction. So we're going to generate an enolate, and I am like so, and that's going to act as a nucleophile in our Michael addition reaction. We talked last time about different representations and keeping resonance structures straight in your head. If it's hard for you to think about pushing all those arrows, the arrow down from the carbonyl, from the double bond to the alkene, then just go and think of the other resonance structure here. If you want to go ahead and think about it this way, we can just say, well, I'd rather think about this resonance structure. I don't generally draw this resonance structure because it's a more minor resonance structure. The more major contributor has the negative charge on oxygen, but this is a fine resonance structure as well. Regardless of how you want to think about your resonance structures, in the Michael addition reaction, your enolate is going to add to the alpha-beta unsaturated carbonyl compound to give you a new enolate in the second step of the reaction, and you've seen that. You've drawn your curved arrow from the nucleophile, from the source of electrons to the electrophile to the beta position of the Michael acceptor. We saw that yesterday on Tuesday. Now, under equilibrating conditions, we have water. If you did the reaction, sodium hydroxide and water, you'd have ethanol. If you use ethanol, another perfectly good solvent, sodium ethoxide and ethanol, or you can even mix it up and do sodium hydroxide and ethanol. Protons go on, protons come off. PKA of a ketone is on the order of 20, which means the enolate is pretty basic. PKA of water is on the order of 15.7. PKA of an alcohol, if you're using an alcohol, would be 16 or 17. Protons come on, protons go off. And so such ends are Michael addition reaction. We're now, you've formed a product that has formed another, that's a 15-dicarbonyl. And remember, we've learned that you can recognize a Michael addition product because there's a 15 relationship. One, two, three, four, five. It's a 15-dicarbonyl compound. All right, that's our Michael step. We've learned that the aldol reaction can occur in intermolecular fashion and intramolecular fashion. And we've learned that in equilibrating solvents and hydroxylic solvents like water or like alcohol, protons come on, protons go off. Every possible position here can form enolites. Some of them don't lead anywhere. In other words, every possible alpha position can form enolites. Some of them don't lead anywhere. Protons come off, protons go back on. Some of them do. I'm going to generate the enolate that's going to take us on in our next step. So we can imagine if you want to write a curved arrow mechanism, your base, your hydroxide, pulling off the alpha proton, pushing in electrons to form a double bond, pushing up electrons onto the oxygen to form an enolate. Now that enolate is poised to form a six-membered ring. And we said six-membered rings like to form. We've seen aldol reactions occur intramolecule to form six-membered rings and five-membered rings. So in our next step, that enolate attacks the proximal carbonyl group to give an aldolate like so. And again, I'm not drawing curved arrows. You should know by now how to draw your arrows down from the oxygen to form an oxygen double bond, oxygen-carbon double bond, from the double bond into this carbonyl carbon and up onto the oxygen to form this aldolite. In aqueous solvent, protons go on, protons come off. And so the last step, we have lots of protons, whether we're in water or we're in alcohol as our solvent. The last step is protonation and thus our aldol reaction has proceeded. Whoops. The last step is dehydration. By now, we should be pretty well set up to see it. And the reaction conditions, protons come on, protons go off in all different positions. But if we pull off the proton next to the carbonyl here, we're set up to form an enolate that can undergo elimination. And elimination is going to give us a stable conjugated enone. So again, I'm just going to go ahead and draw intermediates, not curved arrows. And so you can imagine forming this enolate. So, and then the enolate pushing electrons down and kicking out hydroxide to give our stable alpha, beta unsaturated carbonyl compound. And so this reaction actually in many, many cases works pretty well. And the two components come together in 50 or 60 or 70 percent yield to give a new cyclohexenone. The answer questions at this point. Stereo, ah, great question. What's our stereochemistry? This compound is chiral here. What do we get? One enantiomer, the other enantiomer, the R, the S. Both of them in the absence of something chiral to direct the reaction, we get as much of one enantiomer as the other. And one of the big areas of organic chemistry these days is being able to direct chemistry to get a single enantiomer. Why? The steroid hormone receptors in your body are all handed. If you want a birth control or you want an anti, a birth control drug or an anti-inflammatory, you want the handedness of the drug that's going to fit into the receptor properly. You don't want to be taking twice as much drug where half of the drug can be inactive or can be have a bad effect. In the case of the drug thalidomide, which was given for, I believe, morning sickness in pregnancy in the 50s, the children ended up being born with birth defects and many other drugs have had this particular, you know, various bad effects. In this case, the enantiomer of the desired drug causes birth defects. All right. So, what's a cutting edge area of chemistry is being able to make just one enantiomer. It turns out that in this particular aldol reaction, the amino acid proline can catalyze this to give as a base, well, actually through an enamine pathway, which I might give you in the discussion section, to go ahead and generate just one enantiomer preferentially. So remember when you read about CBS reagent for reduction and I said this is just giving you a taste of this whole area of asymmetric synthesis. Likewise, in the asymmetric synthesis of the Velen Miescher ketone and related ketones, a little taste of it is that a use of a chiral base or in this case a chiral amine can give rise to a single enantiomer. Good question, very important. And this becomes more important when you have multiple stereocenters in the molecule. So imagine for a moment now, let's say for a moment that we had a second methyl group at this position. I'm not going to draw it in. Then you'd be saying, wait a second, now we can get two diastereomeric products. I wonder which one forms. And then you might think, well, let's see, the one methyl can go equatorial. Maybe we'll have thermodynamic control. And that kind of gets beyond the scope of the course. But very, very big and important ideas and you're just getting a tiny taste of them in this introduction to one in this part of carbonyl chemistry that as I said is sort of a gateway to tons and tons of really important organic chemistry. All right, I want to, since this is such a big thing, I want to give us another example, many but not all of the examples do involve 1,3-dicarbonyl compounds because quite frankly, it's easier to generate the enolate and to do the Michael reaction when you have a 1,3-dicarbonyl compound. So I'm going to give us in a forward sense another 1,3-dicarbonyl compound, a beta-keto ester. We're going to see this beta-keto ester later in today's lecture. And what I want us to see is a Robinson annulation with a slightly different ketone and get a little faster and a little better at our thinking our way through this reaction. So I'm going to start by drawing the Robinson annulation product. Now in this case, I am choosing specifically not to use sodium hydroxide. Why don't I want to use sodium hydroxide? Why am I using sodium ethoxide? We have an ethyl ester in the molecule and if we were to go ahead and use sodium hydroxide, we would attack and lose and saponify the ester group and then end up with the carboxylic acid which would actually decarboxylate under the reaction conditions. So in general, if you have an ethyl ester you want to use sodium, well, an ethyl ester you want to use sodium ethoxide in ethanol. If you had a methyl ester say you'd want to use sodium methoxide in methanol. And our Robinson annulation product is this and again it's a cyclohexenone and I want us to think it through now a little bit faster. I want us to learn how to see these reactions forward and backwards. So let me show you how Johnny and I and Kim probably look at this and take it apart retrosynthetically. So Johnny and Kim and I look at this and say, ah, this is a cyclohexenone and I know cyclohexenones can come from Robinson annulations and I know that in this reaction we can break, we can think backwards about forming the alpha beta unsaturated bond but also the bond that's between the third and fourth position so the beta gamma bond to the carbonyl of the cyclohexenone. So we can break this molecule apart in one giant retrosynthetic step and say, oh, this had to come or could come from this ester and this enone. That's a big bite. That's a big bite to take in one mouthful. So I want to show you how to break this apart retrosynthetically in two smaller bite size pieces. Right, so breaking this apart in two smaller bite size pieces. We've seen that alpha beta unsaturated carbonyl compounds can come from an aldol reaction and a dehydration. So I can look at this enone. I can look at this alpha beta unsaturated ketone and say this part of the chemistry has to come together. Remember, mad reaction, right? So this part of the chemistry can come from an aldol and a dehydration. So when I take that backwards retrosynthetically, I say, oh, the immediate precursor to that could have been this. In other words, I will write alpha beta, so I'm going to write here, cyclohexenone. This is my clue when Johnny's clue and Kim's clue that the retrosynthetic chemistry involves a Robinson annulation, but if you can't do that, go and look at this and say, oh, well, this is my clue. If I can't yet be at that point, you say, oh, this is my clue that this could have come from an aldol reaction. So I say, oh, this is an alpha beta unsaturated ketone. And now you look at this next component and you say, oh, this is a 1,5-dicarbonyl compound. I can number to either carbonyl. It doesn't really matter. This is a 1,5-dicarbonyl compound. That could have come from a Michael addition. So there's the M in our MAD. And so we, again, go backwards and now we're at this particular precursor and our enone. Take yours in a sec. I want to see if other people have questions. I hear a lot of mumbling and this is good. I want to hear noise. You should be talking about this right now. All right. So does a Michael addition always add from an alkene question? So the Michael, so this is the same question you asked before in a different form. So basically in our Michael addition, whether you think of your enolate like this or whether you think of it like this, it's the same thing. So yes, if you're thinking about like this, yeah, it's from an alkene. But remember, it's both at the same time. So I think that, and this should be a, at this point in everyone's development here, this should be something you're wrestling with. You actually have to actively wrestle with this chemistry to make it make sense to you. You have to be thinking and saying, I'm not sure on this. I don't understand. So this is good. These are exactly the questions you want to ask. All right. Now I'm going to give you one of the questions that I like to ask, synthesize the following from simpler compounds. And again, I'm taking real examples that have been done in the laboratory and some of the molecules have some complexity and some personality to them. So here's an example of a real compound that can really be synthesized in the laboratory. And I want to show you my thought processes on this. It's the same thought process that I outlined on the upper part of that blackboard. I said, okay, I look at this molecule and I say, I see a cyclohexenone and I'm going to think backwards. And I know in the Robinson annulation immediately, cyclohexenone says to me, Robinson annulation, Robinson annulation. And so I look at this and I say, okay, in the Robinson annulation we form the alphabeta unsaturated bond and we form this bond over here. And so I look and I say, all right, thinking backwards, where does that get me? Well, that's going to get me to this ketone here. Yeah, this is hard the first time around. That's going to get me to this ketone over here. That's going to be this half of the molecule. And it's going to get me to this enone like so. Happens to be the same enone we just used over in that example. And I want to work this through now in the forward sense. And by the way, there's a little secret here. So I sort of was a little hedging to you and I said, you know, cyclohexenone, ketones in general can, without being part of a 1,3-dicarbonyl, can do Robinson annulation, plain old cyclohexenone, plain old acetone, can do a Robinson annulation. But often you're going to get a lot of aldol reaction and other stuff going on. There's something special about this ketone. It's even a little bit more acidic than a regular ketone. Remember, I said regular ketone, pKa20. What's special about the alpha position of this ketone? Close to the benzene, it's more specifically that proton, when you pull it off, you're going to get an anion that's both an enolate and benzylic. So it's a little extra acidic. Instead of being pKa20, it's going to be a little less, maybe pKa17. So a little bit better in the Michael reaction. All right. So to run the reaction in a forward sense, now, really, this ends up being kind of a one note at this point, once you see the secret. To run the reaction in the forward sense, we take our ketone plus our enone, and we mix them together with some base. I'm taking a real example from a real experiment. The authors of this experiment use sodium hydroxide and ethanol. If that bugs you, just in your own mind, you're going to write or NaOET over here. It's not going to matter. The reason they used ethanol is organic compounds generally with lots and lots of carbon and lots and lots of grease don't generally dissolve in pure water unless they're very small. And so it's generally good to have an organic solvent and ethanol is a good solvent for organic compounds. The reason they use sodium hydroxide is you can go into the lab and open a bottle of sodium hydroxide and it doesn't hurt to have some hydroxide around. But if that bugs you, because it's not the one I always use, just go ahead and write sodium ethoxide and ethanol should work just as well. Anyway, the product here, same thing now I've been writing over and over, if you're feeling too lazy, just write PDT and there's your product of your Robinson annulation. As I said, with the Robinson annulation, there's a lot of chemistry occurring and often the details end up being condition dependent. In fact, when I look in a graduate level textbook, I can find you a dozen different conditions that have been used for Robinson annulation. I don't want you to know all of them, but I want to show this to you for one set of principles here. And this is an acid-catalyzed example and I wouldn't expect you to be able to use these conditions. In fact, they're kind of specialized. But what I do want to do is just show you something in mechanism that ties into something that we talked about early on. Again, this is a real example. So this particular example is a cyclohexanone. It's 2-methylcyclohexanone plus methylvinyl ketone plus 3-propane plus 3-butene 2-own MVK. This particular Robinson annulation, if you wanted to write it on paper and you wrote it the same way I did there, you'd be fine. In practice, you actually get a pretty crappy yield that way, if any product. And so in practice, this one experimentally was done with sulfuric acid and benzene at reflux. Again, as I said, you wouldn't be expected to know that. But I want us to look at the mechanism because it ties into something that I talked about before. And the point that I really wanted to make is that we're getting acid-catalyzed chemistry here. So you're getting an acid-catalyzed Michael reaction. I want you to think about that for a moment. Remember early on when we started introducing the chemistry of enols and enolates, I said enols are more nucleophilic than regular alkenes, but they're not super nucleophilic. They only react with really good electrophiles. And so under acid conditions, we enolize at all different places. And so let me show you acid-catalyzed. I just want to show you the gist of the acid-catalyzed Michael reaction. So in sulfuric acid, protons come on, protons go off. You generate an enol, just a little bit of it. Protons come on, protons come off. You generate a protonated ketone. Again, transiently and in small amounts. Now here's my point. Enols aren't super good nucleophiles. They're only moderately nucleophilic. Ketones aren't super good electrophiles. And alpha, beta, and saturated ketones aren't super good Michael acceptors. But by the time you protonate the carbonyl, now you've really souped things up. And so at this point, you go ahead and now you're really nucleophilic. And so you go ahead like so, and I'm not going to write out the whole mechanism in detail. But at this point, you've carried out an acid-catalyzed Michael addition reaction here. And then you've tautomerized over to the ketone. Now all of the reactions we've talked about can happen in this acid-catalyzed manifold. So that's our acid-catalyzed Michael. You can have an aldol in an acid-catalyzed manifold. So just imagine tautomerizing the product that I had written over there before. So now imagine tautomerizing, imagine enolizing, and imagine protonating the carbonyl. And again, you should be able to work this out on your own. In fact, I think there may be a problem along these lines in the textbook. Okay, so you can see just as before, you can protonate, you can generate enol. And again, you're set up for an intramolecular reaction. So again, we're going to push electrons like so in the acid-catalyzed manifold. And again, I'm not writing out a super detailed mechanism. I'm just sort of sketching things out for you. All right, so that's our, and then of course, you pull off a proton, right? You're in acid, you're in sulfuric acid and benzene protons. Come on, protons go off. So now that brings us over here. So if you want, just for the hell of it, I will write out my product over here. And if you want, just for the hell of it, I will write out my product over here. But again, you should be able to go ahead and work through this on your own. All right, so the last step, and this one, I know you can do because you've been doing it back. I think it was in Chapter 8 where we were doing acid-catalyzed dehydration of alcohols to generate alkenes. And I think it's also problem 24.4, if I remember correctly. All right, so under these conditions, again, strong sulfuric acid, protons go on. Protons go off everywhere in the molecule. You can protonate your OH group. You can protonate your hydroxyl group. It becomes a leaving group, leaves with its electrons kicking out water. And so now, again, in this sort of sketch of a mechanism, that sets us up for carbocation formation. That's a positive charge over here. And then that sets us up for loss of a proton. So I'll just write minus H plus. All right, so the big picture here is not those reagents. Those reagents happen to be for this particular reaction. The big picture here is that this manifold of enolate with a Michael acceptor or enol with a protonated Michael acceptor. Enolate with a carbonyl or enol with a protonated carbonyl. Dehydration by an E1CB mechanism. Remember, that's what I called it when we talked about the aldol dehydration under base conditions. It's elimination unibolecular conjugate base. It means you pull off a proton and then do an elimination or elimination by way of forming a carbocation. That's what I want to say about the Robinson annulation. Any thoughts or questions at this point? The double bond is where the oxygen on the bond is. Yeah. Okay, somebody needs to answer this question. Why in this reaction does the double bond go here and not let's say over here? Someone needs to answer that question. What? They're both unsaturated. So in one case, you imagine dehydrating conjugated. Yes. So all of these Robinson annulation conditions we're learning about are under thermodynamic control. And even if you go ahead and have a reaction occur and say give the beta-gamma unsaturated compound transiently, thermodynamically the thermodynamic product, the deepest energy well is the alpha beta unsaturated compound. Protons go on, protons come off, you reprotonate, you get a carbocation, you deprotonate to the more thermodynamically stable alpha beta unsaturated carbonyl compound. And of course really this Robinson annulation chemistry because it is under thermodynamic control is much more subject to sort of the whims of thermodynamics and the whims of side reactions than say where I started us out in the directed aldol chemistry where I said dammit we organic chemists want to take control ourselves, we do that by generating the enolate that we want reacting it with the carbonyl compound we want in the place we want. It may or may not be the kinetic product, it may or may not be the thermodynamic product but we've controlled that reactivity. All right, I want to take us now to some more very, very old chemistry, the Claisen reaction or Claisen condensation. So we've been seeing all of these 1,3-dicarbonyl compounds. We saw malonic ester, we saw acetoacetic ester, right, diethyl malonate, ethyl thrioxotubutonate, we've seen acetyl acetone. So we've seen that these 1,3-dicarbonyl compounds and of course we saw cyclohexane dione and we saw a carboethoxycyclohexanone. These 1,3-dicarbonyl compounds come from somewhere and it turns out they come from some other chemistry that's also really beautiful and really simple once you understand it. So if we take ethyl acetate and we treat it with sodium ethoxide and ethanol first, so I'll write 1 to indicate we're doing that in a first synthetic step and then we carry out an aqueous workup with some aqueous acid that means some dilute HCl, you're not cooking things together, you're just throwing it in before shaking things in a separatory funnel. You get ethyl acetoacetate, you get ethyl thrioxobutonate. Okay, so this reaction involves more carbonyl chemistry of the precise sort that we've seen before. Under these conditions you can form an emylate, you don't form a lot of it. Remember, ethanol pKa conjugate acid, pKa of the conjugate acid of sodium ethoxide conjugate acid is ethanol pKa about 16. Ethyl acetate is an ester pKa 25. So you have this equilibrium and it's sort of right at the edge of where I said equilibria kind of count. You have this acid base equilibrium with the enolate plus ethanol. That equilibrium lies way, way, way, way, way to the left. Again, I'll write our pKa here, pKa 25, pKa 16 or 17. Your textbook uses 16, so I'll use 16. It's nine orders of magnitude, equilibrium constant, 10 to the negative 9th to the left and it's just right on the cusp. I said 10 to the 10th, 10 pKa units difference is sort of right where these equilibria can become significant. So you're set up to generate an itsy bitsy, itsy little bit of the enolate and you have plenty of ethyl acetate sitting around so that ethyl acetate can react with that itsy bitsy, ency bit of enolate. And of course, what do we have? We have our enolate is a nucleophile. Our ethyl acetate has carbonyl in it. It's an electrophile. Electrons flow from the nucleophile to the electrophile. By now, we should be used to pushing our arrows this way. If you need a little crutch, write the other resonance structure of the enolate. Electrons flow from the enolate to the ethyl acetate to give rise to a tetrahedral intermediate. Remember, tetrahedral intermediates are unstable. They break down. Our oxygen kicks back in. The carbonyl carbon-oxygen double bond is very strong. We can expel a leaving group. Ethoxide leaves. We get our product. What kind of sort of? See, our ethyl oxide is good base here, PKA of about 16 for the conjugate acid. And our 1,3-dicarbonyl compound is reasonably acidic. PKA of ethyl acetate is 11. And this is really the kicker in the chemistry. This is what drives this chemistry. I'm not going to write in lone pairs here. This is what drives the chemistry. Our equilibrium lies now way to the right to give rise to your enolate. I'll write one resonance structure if you prefer to write another one you can plus ethanol. And again, I'll write our PKA is 16. So that equilibrium lies five orders of magnitude to the right. That last reaction drives this chemistry. And you notice in this case our chemistry is actually stoichiometric in base. Because at the end of everything we've used up one mole of sodium ethoxide and generated one mole of our enolate of ethyl acetoacetate. So this chemistry requires a full equivalent. The aldol reaction in theory you could get away with just catalytic base. Absolutely. At the end of the reaction we're not done yet here. I've just given you the mechanism for the conjugate addition. At the end of the reaction you add acid. Literally the way you would do this chemistry would be to take some sodium ethoxide, put it in a flask or take some ethanol, throw in some sodium to generate sodium ethoxide, throw in some ethyl acetate, boil it up for a while. After you've boiled it up for a while until the reaction has gone to completion, you'd let it cool to room temperature, maybe even put it in an ice bath, add a little dilute aqueous acid and do a liquid-liquid extraction or distillation to isolate your product. So some students look and they say, wait a second, we're adding acid, why doesn't the ethyl ester hydrolyze? Well, there's a big difference between cooking between heating and for a long time letting an ethyl ester sit or an ester sit with acid, that'll hydrolyze it. But if you don't heat it or you just let them come in contact for a short time, that doesn't hydrolyze your ester. All right, so one of the things that I always like to do is integrate particularly in this chemistry that is so rich and so full of relationships and synthesis. I like to talk about how do you synthesize molecules, how do you recognize them? So I'll give you a typical problem and show you my thinking on it and show you some of the beauty of this chemistry. So let's go ahead and give us a little diversity here. I'll make the methyl ester. All right, so let's synthesize this methyl ester from simpler compounds. You can even name it, what's it going to be? One, two, three, four, five, six. So it's two ethyl three oxo, methyl two ethyl three oxo hexanoate, all right. So synthesize methyl three ethyl two ethyl three oxo hexanoate from simpler compounds. All right, so I look at this compound and I say, all right, we've now learned a new clue. We've now learned a new retro synthetic clue that we've learned that when we see a beta-dicarbonyl compound, we can cleave the bond between the alpha carbon and one of the carbonyls in a Claisen-type reaction. So I look at this molecule and say, ah, retro synthetically, I'm going to split it back to this. Well, it doesn't matter right now when I'm thinking retro synthetically, some sort of group there that's going to leave. And this, that's methylbutanoate. And then I'll look and say, oh, okay, wait a second, these two are the same thing. They're both four carbon, carboxylic acid family compounds. And so I'll say, all right, I've got a synthesis of this. We've been talking about Claisen chemistry. I'll just take methylbutanoate, treat it with, oh, it's a methyl ester, I better use sodium methoxide and methanol so I don't transesterify if I were to use sodium ethoxide and ethanol, I'd transesterify and get some or all of the ethyl ester. And then I go ahead and after cooking it up, I'll carry out an aqueous workup. And that's going to take me to my desired product. All right, this is where this chemistry really gets beautiful and where this chemistry really gets rich. When different people can look at things the same way in different ways and both come up with right answers, that's profound because ultimately building molecules is a chess game. It's this set of many different right moves that can be used to put together a molecule. And so one person can look at the alpha, beta unsaturated carbonyl compound I've drawn here and say I'm going to break this bond between the alpha carbon and one of the carbonyls and another person can go ahead and look at the same molecule and say I'm going to break that other carbon, I wonder what happens. Now not every move ends up being a right move. Sometimes you get problems and that's where you think through in the forward sense. But this particular example happens to have two right moves to it because I break that bond retro synthetically and now I come back this way and I say where does that lead me? Okay, well I've got this forepinone, I've got this ketone here and it's got two symmetrical alpha carbons and I can imagine another methyl group here. I could do a different sort of Claisen reaction where I take dimethyl carbonate, sometimes I'll write OME, sometimes I'll write OCH3, they're of course interchangeable and I'll take sodium methoxide, they're just different ways of writing the same thing. I'll take sodium methoxide and methanol and aqueous acid, okay, does that make sense? Let's see, I guess well there's only one component that can form an enolate, there's only one type of enolate that can form whether I enolize to the left or the right it's the same thing. That enolate can attack this carbonyl compound and ultimately then the same chemistry, form a tetrahedral intermediate, kick out methoxide, now we have a 1,3-dicarbonyl, the 1,3-dicarbonyl is more acidic, the methoxide pulls off the proton and I've come up with a reasonable synthetic route, a reasonable alternative synthetic route to the compound that I asked the question about and that's beautiful, that's profound. We call this reaction a cross-Claisen reaction, Claisen's a name so I'm capitalizing it or at least trying to, just like in the crossed aldol reaction where we talked about one enolizable component, generally a ketone reacting with another nonenolizable component, generally an aldehyde, one can have a cross-Claisen reaction. Let me give you another example. It's an example I mentioned earlier, it's the one we used in our Robinson annulation so I thought it would be cool. So, cyclohexanone, simple compound, treat it with soda, this is just a riff on the example above, treat this with sodium ethoxide in ethanol and then treat it with diethyl carbonate. Now diethyl carbonate is the diethyl ester of carbonic acid, dimethyl carbonate is the dimethyl ester of carbonic acid, they're in the same family, carbon's in the plus four oxidation state, one higher than a carboxylic acid, so it's two higher than a ketone, three higher than an aldehyde. Same thing, anqueous acidic workup generates our enolate, enolate attacks the diethyl carbonate, that generates the 1,3-dicarbonyl and ethoxide, ethoxide pulls off the proton, regenerate a stable enolate, after aqueous workup, you protonate the enolate and the reaction gives you your product. All right, this business about generating a stable enolate is really, really important. This is the thermodynamic driving force behind the Claisen reaction and I'll show you how important it is. If I go ahead and take something that can't form a stable enolate, so I'll take ethyl isobuterate, ethyl 2-methyl propanoate and again I'll envision cooking it up with our favorite condition, sodium ethoxide in ethanol followed by an aqueous workup. You might think, oh wait a second, okay, that's just like all these other examples that I've been tossing up on the blackboard and you might think, okay, we envision generating the enolate, enolate adds in, we get a Claisen condensation, we get a 1,3-dicarbonyl compound. But we have that problem, no acidic alpha proton and that thermodynamic driving force is necessary for the reaction. There's five pKa units, remember, every order of magnitude, every factor of 10 in an equilibrium constant at room temperature is worth 1.36 kilocalories per mole. So five times 1.36, we're practically up at close to 10 kilocalories per mole of driving energy for that reaction. Carbon-carbon bond is on the order of 80 or 90 kilocalories per mole. Carbon-hydrogen bond, you know, similar sort of strength, 100 kilocalories per mole thereabouts. Point is we're at an energy difference that's about a 10th sort of forming bond and without that energy, we just can't drive this equilibrating chemistry. So to show you what that means, that means that if I tried to do the reaction, if I mix the product under the reaction conditions, this is an equilibrium, this reaction goes backwards. In other words, if I take this hypothetical product and I somehow manage to make it, and there are other ways to make it, and I treat it with sodium ethoxide in ethanol, we end up breaking this product apart to two molecules of diethyl, of ethyl-ice-abuterate. It's the reverse of a Claisen reaction and we call it the retro-Claisen reaction. I'm going to leave it to you to work out the mechanism. I just got the homework problems up from the textbook on Chapter 24. One of the problems about two-thirds or three-quarters of the way through, I think involves a retro-Claisen. I just looked at it very quickly. So you want to think about the mechanism there. And I'll give you this reminder hint that the mechanism, any mechanism in reverse is any reverse reaction, any reaction in reverse goes through the same mechanism in reverse. Remember, that's what we call microscopic reversibility. So if you can write the mechanism for the Claisen reaction in a forward sense, you can write it in the reverse sense, just start at the last step. Obviously, you don't have that deprotonation, but start at the breakdown of the tetrahedral intermediate and envision adding a nucleophile to that, envision adding ethoxide to that and push some arrows and you'll be right back at your reactants. We saw that the Aldol reaction could occur intramolecularly and we saw that five and six-membered rings were particularly stable. It's the same with the Claisen reaction except it gets a special name, the Dieckmann reaction, D-I-E-C-K-M-A-N-N reaction, also called the Dieckmann condensation because again, we're spitting out a molecule. So I'm going to take the diethyl ester of the 7-carbon di-carboxylic acid. I want to show this to you in the forward sense. We'll see this in the reverse sense in a moment. So 1, 2, 3, 4, 5, 6, 7 and we'll envision treating it with our now favorite conditions of sodium ethoxide in ethanol and aqueous workup with some acid. And the product of the reaction is this ethyl ester. Same ethyl ester we used in our Robinson annulation reaction. The mechanism, it's just a Claisen reaction. Often when students see rings for the first time and intramolecular reactions, they get freaked out. So let's go ahead and do sort of an abbreviated version of the mechanism. Okay, we're under equilibrating conditions. We generate a teeny-eeny bitsy-bitsy bit of enolate. I'm getting punchy up here. We generate a teeny bit of our enolate. Our enolate can react intramolecularly. We push our arrows, push our arrows up onto the oxygen, forms a six-membered ring, our intermediate, our tetrahedral intermediate is a six-membered ring. Six-membered rings generally are good things. They generally can form easily. A tetrahedral intermediate can break down, kicking out ethoxide. Member, you can generally kick out anything up to about and including 25 pKa as a leaving group from a tetrahedral intermediate. Remember, it's different than an SN2 displacement. That pKa of 25, which I said was kind of the cutoff. We saw that in a heliform reaction. That also plays in the retro-Claisen reaction. So there's one step. When you write out that mechanism, you're going to have to do it and say, I don't know about that step. You're kicking out a leaving group with a pKa of a conjugate acid of 25. Okay, so our tetrahedral intermediate breaks down plus ethoxide. I'm not going to draw in all my lone pairs of electrons. And that final step, that equilibrium that lies way, way, way, way, way to the right pulls off the proton, generates our enolate. If you don't like the way I draw my enolate, draw the negative charge on the alpha carbon. All right, so let's think about what we just saw. Here's our product. We use that product in a Robinson annulation. We use that product to make something very complicated from something a lot simpler. We've now seen how to make that product from simpler, to make that reactant from simpler molecules. We've seen it in two different ways. For that matter, we also saw how to make our alpha beta unsaturated carbonyl compounds. Those are made by an aldol reaction with dehydration. So your MVK, you can make it from an aldol reaction with dehydration. Okay, but what about this one? Well, remember, this is our retro synthetic analysis that we've been learning about. It's a 1,3-dicarbonyl compound. The way we see it over here, you break it apart retro synthetically, like so, to this diester, and if you count, that's just 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7. That's just, and so you can see one way to break it apart. But you can also see a second way to break it apart. We saw that way just a moment before. That way of breaking it apart, breaks it apart to cyclohexanone and diethylcarbonate. And that's really the beauty of being able to recognize these relationships. And so let me end with a very simple problem. Simple, I was being lazy. I figured, all right, I'd end this with a problem that's analogous, because we've seen so much chemistry in these past two lectures and in the past four lectures that your heads are probably spinning right now, and that's okay. When you work the problems, that spinning is going to stop. Things are going to crystallize in your head and you have to work the problems. All right, so here's a sort of noic exam problem. I didn't say it would be an exam problem. I just said it's the sort I always like to ask. You've seen three of them today. I think it's three. Synthesize this compound from simpler molecules. I said compounds. All right, well by this point, I don't think I can fool you. This is just an analogy to what I've written over there. And of course you have it up on the blackboard. So 1,3-dicarbonyl compound. We can split it apart in two ways. We can split it apart at this bond here. Then I just need to be able to count correctly 1, 2, 3, 4, 5, 6. Usually at this point in the class I'll miscount. So I've got to make dimethyl adipoate. The dimethyl ester of adipic acid. You could make that from cyclohexene if you wanted. And we'll treat it with sodium methoxide and methanol. I use sodium methoxide because I have the methyl esters. I don't want to hydrolyze them with sodium hydroxide. Suponify them if you want to be technical. I don't want to go ahead and transesterify them. We carry out N aqueous workup because we've got to protonate that anion and there's our product. And we also know, and I'll draw a real long arrow onto the other blackboard, we also know that we can form that bond. And so we say, oh wait a second, we can do a crust clasin. We take cyclopentanone, we'll take our favorite base sodium methoxide, now I'm making life a little easier for you. I've done these reactions myself and I've often done these reactions with sodium hydride as a base. But there are about a bunch of bases you can use. And I'll make it simple. We'll use sodium methoxide and methanol. I've used sodium hydride and dimethyl carbonate as solvent, aqueous acid workup, same old, same old. We haven't seen a lot of variation there. And bing. Two different ways to the same product to write answers. All right, we will pick up next time with, I think our quiz next time, is that right, Johnny? And talking about Chapter 25. Have a good day and good weekend.