 Good morning. Good morning. Hey. All right. Well, we now start chapter 24 and we really come to what I think is the pinnacle of organic chemistry in terms of reactivity, mechanism, synthesis, retro synthetic analysis. And this is the carbonyl chemistry of various types of carbonyl condensation reactions. The aldol reaction, the Michael addition, the Claisen condensation, the Robinson annulation reaction, the intramolecular Claisen called the Dieckmann cyclization, the intramolecular aldol reaction. It's putting together the reactivity of carbonyls that we learned about as electrophiles and the reactivity on the alpha carbon as nucleophiles and being able to assemble some very complex and important and beautiful molecules as a result of this. There's also a lot to think about. And so what I'd like to do is to start talking about the aldol reaction, which really lays the groundwork for all of these carbonyl reactions and indeed even some of the sugar chemistry you'll be learning about later on in chapter, I think, 28 or 26, I guess. All right, so in the aldol reaction in general, what happens is an enolate reacts with another carbonyl compound. So you have a carbonyl compound and I'm going to write this very, very generically. The carbonyl compound with an alpha hydrogen and honestly we're talking about just about any of them, both the ketone or aldehyde family but also the ester family honestly, even though I've written this as carbonyl X, I said in this broad ester family in this broad carboxylic acid family, you have carboxylic acids but also nitrials. And so the chemistry I'll talk about today even is applicable in certain cases to nitrials. But the main thing is we have a carbonyl compound with an alpha hydrogen and we have some sort of base and we'll be talking a lot about bases right now. I'm just giving this sort of as a generic cartoon. And under some conditions we get an enolate and we allow that carbonyl compound to react with a ketone or aldehyde, sometimes all mixed together, sometimes done in sequence. We'll talk more about specifics in a moment. And the enolate, let me draw my negative charge here, the enolate of course is nucleophilic at the alpha position. The ketone or aldehyde is electrophilic. And so we form a bond between the enolate and the ketone or aldehyde. And in the most generic form then, the initially formed product is going to be an aldolate and under some conditions, sometimes as part of the reaction condition, sometimes in a sequential separate step, proton source can become available or we can add a proton source. And I'm writing this very generically as H plus. Could be added acid, it could even be alcohol or water in the reaction mixture. We'll go through details in a moment. But whatever your proton source is here, again, very generics, I'll write proton source. Whatever your proton source is, we protonate the oxyanion of the aldolate. And the overall result is an aldol product. And you can recognize an aldol product because it's a beta hydroxy carbonyl compound. Later on, I'll also show you aldol products that are alpha, beta unsaturated carbonyl compounds. All right, so that's sort of the overview at 5,000 feet. I now want to go to a very specific example. I love the aldol reaction. I love it intellectually because there's so much to think about mechanistically, synthetic. I also love it because it's beautiful to do. Synthetically in the laboratory. And so I'll give you an example that I use in my Chem 160 laboratory class. I wanted a simple sort of cookbook aldol recipe for my students. It's the first day of laboratory. This is an upper division laboratory course. I urge you to take it. It's a lot of fun. And the first day, all in one day, they do an aldol reaction and purify the product by one of my favorite recipes, which is basically you take some LDA and THF. We talked before about how often you will make your own LDA by reacting butyl lithium, a very, very strong base with diisopropyl amine, somewhat less basic but still very strong and sterically hindered base. Organic chemists often cool things with dry ice. Dry ice is negative 78 degrees Celsius, which is why you'll often see that otherwise odd sounding temperature. Then in a separate step, one adds the carbonyl compound so you generate the enolate at negative 78 degrees. The carbonyl compound that I chose for my Chem 160 students was benzophenone. PH is just a benzene ring, a C6H5 unit. Benzophenone is just a simple carbonyl compound. And then the source of aqueous acid that I like to use is aqueous ammonium chloride. Just a mild acid, right? NH4 plus is acidic. So I'll write this in parenthesis as H3O plus. You could write either underneath the arrow. But this is sort of my favorite cookbook, what's called a directed aldol reaction. Directed meaning we're first generating the enolate. We're then in a separate step adding a carbonyl compound and isolating the product. So I'll write directed aldol reaction. I've capitalized aldol here as a title, but aldol is a word. It's not a proper noun. So if you're just writing it, you wouldn't capitalize it. Michael, which we'll hear about in at the end of today's lecture, is actually somebody's name. All right. So the aldol product that you get in this specific example that my students in Chem 160 get and then recrystallize, beautiful, beautiful white crystals, great yield. It's just a fun laboratory experiment to do. And it is this beta hydroxy ester. All right. I want to give you another example. This particular recipe for directed aldol really just works and works and works. I'll give you another example just so we can see some of the diversity. Here we generated an enolate of an ester. We reacted it with a ketone. I'll just for variation, say let's take cyclohexanone 1LDA, THF, I know you stress over what to write on exams. Here I'm not writing the temperature. Probably doesn't matter whether I do or don't. I'm writing THF. Probably doesn't matter whether I do or don't. Don't stress too much. But very important, we're going ahead first. Number one, generating the enolate. Two, and this time I think I'm going to use an aldehyde and this generic in this specific example. Let's use isobuter aldehyde as our carbonyl compound, 2-methylpropanol, and we'll do an aqueous workup. And here I'll be a little more generic and just write H3O plus also okay to do. H3O plus could be aqueous ammonium chloride. It could be dilute aqueous HCl or dilute sulfuric acid. And in a very, very typical situation you would then extract your product into an organic solvent and dry it over Pimag sulfate and filter it and concentrate it. And here's our aldol product. Another beta hydroxy carbonyl compound. In this case, our enolate was generated from a ketone in the previous case from an ester. In this case, our electrophile was an aldehyde. In the previous case, a ketone. All right. What I want us to do at this point is to learn to recognize the aldol synthesis and so I want us to get in the habit of learning to think backwards. We've talked about retrosynthetic analysis before. It's a very important skill to be able to see backwards and so I'll write out a sample problem showing the type of thought process. So this is exactly the sort of thing that students in organic chemistry and research encounter all the time. They're making a complex molecule like a natural product and they need to make it from stuff you can buy. And so I'm going to phrase this as synthesize and I'll draw out the compound here rather than naming it but you can figure the name of it if you like. See the one I've chosen for this particular example is this molecule and I'm going to synthesize this molecule from simpler compounds. Why? Because generally as you build up bigger molecules with more functional groups there's more complexity. Chances are they're not available but simple stuff is available. You can buy it. On the exam I gave you a sort of a restricted chemical catalog of four carbon atoms or fewer. That's sort of an artificial example but it's basically what people deal with. There's a bunch of stuff you can buy. Aldehydes, ketones, alkynes, alcohol halides, alcohols, alkenes, alkanes, go ahead use anything you like to make this. Now let's focus on the thought process and remember the big thought process that we use, this real height of thinking, Nobel Prize winning thinking, E.J. Corey type of thinking is the idea of thinking backwards. Retrosynthetic analysis, use this type of open arrow to show retrosynthetic analysis and the beauty of being able to think backwards is it liberates us from conditions, from reagents, from details. What oxidizing agent do I use? What base do I use to see the entire big picture? And the big picture here is we have a beta hydroxy carbonyl compound, right? This is the alpha carbon. This is the beta carbon. So this is a beta hydroxy carbonyl and when we see a beta hydroxy carbonyl compound immediately that should clue us in. The bond between the alpha and the beta carbon can be made by an aldol reaction and so I look at this, maybe I draw a little squiggly line here to remind myself and then I look and I say oh wait a second, we've got this ketone on the left, 3-methyl-2-butanone and we remember on the right here that this alcohol was part of carbonyl compound before we formed the bond and added in a nucleophile. In other words, that came from a ketone acetone and so that's the retrosynthetic analysis. That's the act of thinking our way through the problem. Now we just go on and say all right, what do we need to do? We need to generate the enolate of 3-methyl-2-butanone and then we need to allow it to react with acetone. It's a directed aldol reaction. Now we can worry about details and we think back, wait a second, wait a second, wait a second. All right, is this going to work? There are two enolates that can be generated when a base pulls off an alpha proton. We can pull off an enolated proton at the 3 position or we can pull off a proton at the 1 position and we think back to regiochemistry and kinetic enolates and say ah, we're in good shape here. We remember that when we treat a compound where you've got a less sterically hindered position like a methyl group and a more sterically hindered position like an isopropyl group, the LDA, this bulky base, preferentially pulls off the more accessible proton to give you the kinetic enolate, to give you the enolate on this side. We then add our acetone and we carry out an aqueous workup. In practice, typically you might add the acetone at negative 78 degrees and then allow the reaction mixture to warm up to 0 degrees before quenching it with some mild aqueous acid like ammonium chloride or acetic acid or dilute HCl. And if you use something like HCl, you're not going to let the reaction get hot. You're not going to stir it with it a long time. You're just going to add a little dilute HCl and immediately do your liquid-liquid extraction. We'll come back to why that's important in a moment and later on. All right, I like to start by teaching about the aldol this way and I like to do the aldol reaction by the directed aldol procedure because it is so intuitive. It's this mindset of taking control. It's the idea, I'm going to make the nucleophile I want. I'm going to add the electrophile I want and darn it, that nucleophile is going to react with that electrophile the way that I want. Now, historically, that's not the way the aldol reaction was developed and up until just before the 1970s, LDA really wasn't on the scene as a base and a lot of organic chemistry was and continues to be done by mixing together some moderately strong base. LDA is a very strong base and some carbonyl compound or carbonyl compounds. Now, the original aldol reaction was actually discovered at the beginning of the 19th century and I'll show you that reaction. Chemistry, the beautiful thing about chemistry is, of course, it works across all time, all space, and all place. Works on Mars, works in the 15th century. It'll work in the 25th century. So the original aldol was acid aldehyde with a base and your textbook also makes reference to this. I'm going to be a little dicey about the base on the blackboard, I'm going to write hydroxide. Now, the big secret, according to my recollection and I read about this all while back, so I may be fuzzy, I think the base that was used originally was calcium hydroxide. Organic chemists for the most part don't use a lot of calcium salts or calcium metal and synthesis these days. Sodium hydroxide's a lot more common base. So your textbook and probably I am not going to write out COH2 because you probably are never going to see it again but the principle is the same. The reaction happened to be done in water. Acid aldehyde is a small molecule. Water dissolves small organic molecules, right? It's got one oxygen atom for two carbon atoms and it was done at a cold temperature at 5 degrees Celsius for 4 to 5 hours. And the product of this reaction in about 50% yield was the aldol product of bringing together two molecules of acid aldehyde and this compound is called aldol, trivial name. Now you think about it, you think about it and I've already said a lot of the ethyl compounds sort of bear the name of things. When you go and you get a glass of scotch it contains alcohol that's ethyl alcohol, ethanol. When you go ahead and you get a bottle of ether that's diethyl ether. So it kind of makes and we talked about acetoacetic ester, right? It's the ethyl ester. And so it kind of makes sense that the archetypal aldol is the aldol of acid aldehyde, the two carbon unit and the product is called aldol. As I said, not a proper name doesn't get capitalized. All right, so what's going on in this chemistry? Well you should now be on top of things enough to write out all of these mechanisms but I'm going to go through it again just in case you're a little bit rusty. All right, so in this first step and this unlike the directed aldol is a big gamish. It's just acid aldehyde and base sat together in a bottle of water for four to five hours and then worked up. So first step hydroxide anion pulls the alpha proton, one of the three alpha protons off just as we've seen before to generate and enolate anion. All of these steps are equilibria and so I'll write an equilibrium arrow. In the directed aldol we're taking a little more charge and basically because things are occurring at very low temperature and we're using a very strong base, they're irreversible but here for the most part all of our steps are equilibria so we generate our enolate here. I'll even be a good person and write in water. Second step in this big gamish now we have plenty of aldehyde it's just floating around aldehyde and hydroxide and enolate. Our enolate is nucleophilic at the alpha position. Electrons flow from the enolate down from the oxygen to the alpha position attacking the alpha carbon. Remember we have two resonance structures for the enolate. I generally prefer to draw the more the major resonance structure but in the back of your mind is also this resonance structure with negative charge on the alpha carbon and so this flow of electrons is just embodying bringing that charge to the negative carbon. It's not a to the alpha carbon. It's not at one position or the other. It's at both positions at the same time but I can only draw one resonance structure at a time. Electrons flow up onto the carbonyl and this gives rise to the aldolate and now in the last step of the reaction we're in water, in water, we're in alcohol, protons go on, protons go off. We have lots and lots of water present so we have this mass action driving the reaction. We pick up a proton. We regenerate our hydroxide. As you can see, hydroxide is a catalyst for this reaction. It's consumed in the first step but recreated in the last step and so it's catalyzing the reaction bringing together the two components. One of the things that's confusing in learning the aldol reaction is often the exact course of the reaction ends up affecting the products. The products form depend on the exact course of the reaction particularly in the non-directed aldol reaction where we're just mixing things together. Typically we have a series of equilibrations occurring and as the chemistry goes we can drive things further and further downhill thermodynamically. So one of the features of the aldol reaction is that higher temperatures in longer times tend to lead to dehydration and by dehydration I mean the loss of the beta hydroxy group and alpha hydrogen. Let me show you what I mean. So let's take the same aldol reaction and we'll take acid aldehyde and a base. As I said, these days people tend to use sodium hydroxide for the aldol reaction. If you're running it under equilibrating conditions it tends to be run in water or in alcohols like ethanol and so I'll say water and let us say heat. In other words now instead of cooking the thing or letting the thing go for four or five hours at five degrees Celsius remember chemistry is very slow at cold temperatures in general. In general reactions go about two or three times faster for every 10 degrees Celsius. So you go up from five degrees up to 50 or 100 degrees and reactions that might not otherwise occur can occur very quickly and the product of this reaction now is 3-buten-al. It is 2-buten-al. All right so what's occurring here? Everything that's occurred before is occurring but now we're getting another reaction in addition occurring as a result of the higher temperature or longer reaction times. And this is what we call an E1CB mechanism. It stands for elimination concerted base. Let me just show you what I'm talking about here. And in this mechanism, let me show you that. So here's our hydroxide, here's our beta hydroxy carbonyl compound and so under these reaction conditions you can again pull off an alpha proton to get an enolate and now in a second step we can spit out hydroxide. So our enolate kicks back. We push out hydroxide. It's a lot easier to think about if you remember the resonance structure in which we have a negative charge on the alpha carbon. In that resonance structure you can just think of it as directly eliminating. As I said I prefer to write my enolate this way but you can think of it just as well like that. Great question. Don't we need to make alcohol a good leaving group? So we see two different themes in organic reactivity. One is the theme of leaving groups in SN2 displacement. In SN2 displacement reaction a good leaving group has to be a very, very weak base. Its conjugate acid has to be strongly acidic. Chloride, bromide, iodide, all are good leaving groups. HCl, HBr, HI all have pKa's in the negative. In general in an SN2 displacement reaction a leaving group is generally going to be a very strong acid for the conjugate base. I draw a cut off right at about positive 5 for the pKa because I can give you examples where you have acetate as a leaving group and that's about it. SN2 reaction you're going through this trigonal bipyramidal transition state. Carbon cannot have 10 electrons around it and that nucleophile is coming in. The leaving group is leaving right at the same time. It's a bad reaction. It's a difficult reaction. It only occurs with a good leaving group. Then you have the other class of reactions with leaving groups and that's the reactions we saw in carbonyl compounds with say ester hydrolysis where now we can have a leaving group that has a pKa of let's say 16 or 17. A nucleophile adds to a carbonyl. We get a tetrahedral intermediate and that tetrahedral intermediate kicks out ethoxide, methoxide, hydroxide in this type of chemistry. The addition elimination chemistry manifold allows much more basic leaving groups, much less good leaving groups, leaving groups with weaker conjugate acids to leave. And this falls into that exact same manifold because you are dead right. Hydroxides are leaving group here. pKa of the conjugate acid 15.7. But it's not going through that manifold where you have to fit 10 electrons only transiently around the carbon. You cannot get a stable species with 10 electrons. But you can get a stable species with a negative charge on an adjacent atom. Everyone's happy it has 8 electrons. You push, everyone still has 8 electrons. Same idea as the breakdown of a tetrahedral intermediate on say an ethyl ester. All right, NU here and we can push out ethoxide happily. Really, really important principle and I'm glad you asked other questions at this point. Great question. Great principle. Couldn't the alkene group form where? That's what we're, oh, could we write the mechanism this way? Yeah. Absolutely. And this and this are just two pictures of the same thing. Remember in your mind's eye you have both resonance structures. All right, let me give another example. So I was playing with acid aldehyde there in that example and all sorts of other aldehydes work. So I'll just take a specific example. This may have been from your textbook. It just shows you the sort of conditions. The acid aldehyde example and in the end because of some reactivity isn't the best example as far as yield and principles of reaction go. But I want to give you a very specific example. Let's take butyraldehyde, sodium hydroxide at 80 to 100 degrees Celsius and this is a real example from the laboratory. You would never be expected to know the exact temperature, the exact time but this reaction happens to run over three hours. And I want us to look at the product for a second because we're going to have to learn how to visualize these sorts of products. And so here's the product of reaction. If you want to name it it's 3 ethyl 2hexenol or 2 ethyl 2hexenol rather. But the trick in recognizing these types of compounds is to be able to say I. In my mind's eye I can see that this double bond of this alpha beta unsaturated carbonyl compound must have come from an aldol reaction. And when you look at that you say oh okay I can see this half had been one molecule of butanol and this half here 1, 2, 3, 4 had been the other half of butanol. Let's try another example and I'm going to phrase this as a problem just as I phrased the previous one. So let's say synthesize and I'll draw this compound. I won't name it. Let's synthesize. I think this one actually may be an example from your textbook but it's a nice example so I figured I'd use it. Let's say synthesize this molecule and again we're going to say this broader thing from simpler compounds. Take a moment for yourself to go through the retrosynthetic analysis. And split the molecule apart just like we envision here splitting this molecule and working backwards with a retrosynthetic arrow to two molecules of butanol. What's the carbonyl compound that generates this? I can hear you saying it. The word, the casual name for this is propyophenone. So we're going to do the same thing. We're going to, if you want the formal name, it's one phenylpropanone. So we'll split it apart. Our clue is that it's an alpha, beta unsaturated carbonyl. We'll split it apart into two halves and recognize that it comes from two molecules of propyophenone. And that's not the answer yet. That's the thought process. That's the showing our work. That's the thinking on this. So now we've learned about one set of conditions for the aldol reaction with dehydration. And so basically we'll just take our propyophenone and our favorite sodium hydroxide. Honestly, you could use calcium, you could use, pardon me, potassium hydroxide interchangeably, water as a solvent or water alcohol mixtures and heat. And it's this art of thinking backwards that is just so incredibly valuable and so incredibly powerful to be able to look at a complex molecule like that and see how it comes together. So we've been having some fun with chemistry that basically works well. And what often means working for organic chemists is giving one thing, it's this notion of control. Either I want the beta hydroxy carbonyl compound or I want the alpha-beta unsaturated carbonyl compound. But I don't want both. Either I want this component to be the nucleophile or I want that component to be the nucleophile but I don't want both because I'll get a mixture. And of course as one starts to have mixtures of things you have problems. So for example, if I mixed propanol and butanol together with sodium hydroxide and water and heat, my goodness. Well, I could have the propanol be the nucleophile. I could have the butanol be the nucleophile. I could have the propanol be the electrophile. I could have the butanol be the electrophile. Even if under these conditions I didn't get any beta hydroxy carbonyl and all of my beta hydroxy carbonyl compound went to alpha-beta unsaturated carbonyl. I'd have four different alpha-beta unsaturated carbonyl compounds. I'd have compounds with propanol and propanol compounds with propanol and butanol compounds with butanol and propanol compounds with propanol with butanol and butanol. So in many circumstances if you just mix together different aldehydes and base you get a mixture of products. But there is one special type of case where you can control mixtures of compounds. And the generalities are if you have a ketone and enolizable ketone obviously it's got to be enolizable just like propiophenone. And you have a nonenolizable aldehyde like benzophenone. See the thing that's special about aldehydes is they're more electrophilic than ketones. So there's only one species in that mixture that can generate an enolate because benzaldehyde can't generate an enolate. It doesn't have any alpha protons. And although there are two electrophiles, a ketone and an aldehyde, an aldehyde is the better electrophile. So if we take these two components under the same conditions sodium hydroxide and water and heat and as I said you could have potassium hydroxide, you could have a little ethanol in that mixture to get things to dissolve. But the main point is under these type of conditions now we get a product and that product has the acetone acting as the nucleophile and the benzaldehyde acting as the electrophile. In other words even though we haven't had that precision we talked about initially with the directed aldol with LDA which you could also do for a reaction like this. Darn it, I'll make that enolate of acetone with LDA then I'll add it into benzaldehyde and then I'll do the elimination separately under some other conditions. But even without that we're able to take charge. We have a special name for this reaction. We call this reaction the crossed aldol reaction. By crossed I just mean that we have two different components, one of them's a ketone that's enolizable, one of them's a nonenolizable aldehyde and honestly there's some variation that one can have. For example, you don't have to have a ketone. Anything that you can make an enolate of reasonably well is okay. In general ketones are good. In general remember the PKA I said if you want to keep two numbers in your mind keep two PKA's in your head, 20 for a ketone and 25 for an ester and you can go ahead and say an aldehyde is just about the same as a ketone. So an enolizable aldehyde if you only want to keep two numbers in your head you can say close enough it's 20, okay so acetaldehyde's more like 17 but that's close to 20. For an ester that same PKA is going to be about the same PKA for a nitrile and other compounds for the alpha proton. But you get a special circumstance if you put two esters together we already saw last time that diethyl malonate for example and ethyl acetoacetate are extra acidic and so you can take a compound like diethyl malonate which now is extra acidic and the same gist of the conditions you can go ahead and take this very enolizable remember PKA 13 for diethyl malonate, acetyl acetone 9, ethyl acetoacetate 11, diethyl malonate 13. You want to keep one number in your head just keep in 10 or 11 for all of these 1, 3 di carbonyl compounds. Extra acidic I'll get to your question in a second. Now let me give us just for the sake of some diversity, a funny looking aldehyde it's called furforal. When you learned about aromatic compounds you learned about benzene but then you probably also read about all the five-membered ring aromatic compounds. Furan, all the fundamental ones, furan, parol, and thiophene for the most part all of those rings like the furan ring behave very much like the benzene ring. So this is just another compound like benzene aldehyde. Imagine mixing this highly enolizable 1, 3 di carbonyl compound PKA 13 and furforal and some base. The only difference is I don't want to use sodium hydroxide. I don't want hydroxide anywhere because I have ester groups. If I use hydroxide I'm going to hydrolyze my esters and get carboxylase. So I'll take sodium ethoxide in ethanol, another favorite solvent base combination. You can just make it by throwing in sodium metal into ethanol. Very easy base to make, heat it up. If you like, I'll write delta here just to show you can heat it up and even though this looks very different, it's all the same principles we just saw. It's an alpha beta unsaturated carbonyl compound. In this case an alpha beta unsaturated 1, 3 di carbonyl compound with a formed from a non-enolizable aldehyde and an enolate of a highly enolizable component. Great question, can the aldol reaction form both E and Z components? Yes, yes and no. So for the most part under these conditions the aldol reaction is thermodynamic. All different products are forming and we're going to the lowest energy product. Particularly when you have something big like benzene, the Z is much higher in energy because of steric interactions and so you get virtually none of it and so in this case we're going to get, you know, so little that you wouldn't detect it. In cases where you had two things that were very close to equal in size, a methyl group and an ethyl group, yeah the answer is yeah, you would get both the E and the Z. So under some circumstance, as you might, but for the most part it's pretty selective. For most of the examples you'll see it's pretty selective for the E geometry, for the trans geometry. Now if your molecule has two carbonyl groups and they're in the right relationship to each other an aldol reaction can occur within the molecule in what we call an intramolecular aldol reaction. In general, five and six-membered rings are particularly stable and particularly easy to form. Three and four-membered rings have substantial ring strain. Seven, eight, nine-membered rings have some other strain associated with them. So in general, if the reaction can occur in a way that gives rise to a five or a six-membered ring, it will. So let's take a look at this one, three, that this di carbonyl compound. It's hexane 2,5-dione, six-carbon chain with two carbonyl groups. Treat it with sodium hydroxide in water and heat or favorite aldol conditions and the product of this reaction is this cyclopentenone. Now, first time that most students see this chemistry, they look and they say, whoa, this looks completely different from everything I've seen before. But think about it, it's an alphabeta unsaturated carbonyl compound. You can look at it and say, oh, wait a second, I learned these things form by way of an enolate and a ketone. Let's see, where does that get me back to? That gets me back to that, oh, that's this. So what's happening? Under the conditions of the reaction, you're forming emulates at every possible position, all four of the emulizable sites. This methyl group, the methylene group here, the methylene group there, and the other methyl group. Some of those enolates just form reversibly others where it's enolized on the methyl group can go on to react. And so in this particular reaction, what's happening is when this enolate forms, it can reach around, attack over here, and that takes us over to this aldolate. And then the same things we saw before, the aldonate, again, we're in equilibrium, protons are going on, protons are coming off, the aldolate can protonate. If we form an enolate over here, it does happen. Reversibly, protons go on, protons come off. You can enolize here, nothing happens. But when you form the enolate on the other side, like so, now we're all set up for that elimination. We spit out hydroxide and we get into this thermodynamic well of the alpha, beta unsaturated compound that has, that's conjugated. Thoughts or questions? Absolutely. So we generate some of this enolate, we generate some of this enolate and it's all reversible. So absolutely this enolate forms, but if it reacts, it forms a three-membered ring that's two strain, 27 kilocalories per mole of ring strain are there about so it never cyclizes, pKa of the ketone is about 20. So think about it, most of our species is the ketone. We get a little bit of this enolate, a little bit of that enolate. The enolate that can react does, we keep generating more and going down this thermodynamic well until we've eventually all made the alpha, beta unsaturated ring compound. All right, let's take a look at a couple of other examples of the intramolecular aldol reaction. We'll just have some fun here. So we did a six-carbon diketone, let's do a seven-carbon diketone, 1, 2, 3, 4, 5, 6, 7. We'll take this seven-carbon diketone, same conditions, our same sort of universal aldol conditions and what do we get here? You go through the same thought processes and you get the six-membered ring. In general, five and six-membered rings can form, let's say they are favored. In other words, in general, we don't form four-membered rings. They're two-strained. In general, we don't often form seven-membered rings. Try one more example here. So now we've done a six-carbon chain, seven-carbon chain, let's take an eight-carbon chain. These are real examples taken from real experiments that have been published. This particular example just used potassium hydroxide in water instead of sodium hydroxide in water. No particular difference. Now, I said seven-membered rings don't like to form. So now, remember, we can form all these different enolites. We can enolize on this side, we can enolize on that side. If we start to enolize toward the methyl group here and you imagine cyclizing, that'll form a seven-membered ring. As I said, seven-membered rings have a certain form of strain associated with them with eclipsing and transannular interactions, they don't tend to form. But if we enolize on the more substituted side, all of our enolites form now when we envision cyclizing, now we can form a five-membered ring. And that ends up being the product of this reaction. Now, this process of thinking backwards, again, just apply it over here. Look at this, think backwards. We have an alpha-beta unsaturated carbonyl compound because you'd look and you'd say, okay, everything that he said in class makes sense. And then I get this problem on a test or an exam or a quiz or a homework and I look at it and how do you see it in the other direction? So in other words, if you say, oh, how do you synthesize this molecule from something that's less complicated that doesn't have a ring, you say, oh, wait a second, it's an alpha-beta unsaturated carbonyl. I bet it could come from a dicarbonyl compound. You look and you say, ah, okay, thinking backwards, we have this and you say, wait a second, now you just go ahead and you unwrap that and you say, oh, what's that? That's just one, two, three, four, five, six, seven, eight. Keep track of your carbons. That's just that. Thoughts or questions at this point. You can have diketones. You can have dialdehydes reacting in this manifold. It's really pretty cool. All right, this brings me to the second topic I wanted to talk about, the Michael edition. In textbook calls at the Michael reaction, you'll hear it referred to as both ways. That's just a fancy name for the addition of a nucleophile to the beta position of an alpha-beta unsaturated carbonyl compound. We've already seen that ketones and aldehydes are very fundamental compounds. Now we've seen even alpha-beta unsaturated carbonyl compounds are very fundamental compounds. And it turns out that alpha-beta unsaturated carbonyl compounds have special reactivity of their own. So let me draw sort of a generic alpha-beta unsaturated carbonyl compound. This could be an ester, it could be a ketone, it could be an aldehyde. And there's sort of two general features of the reactivity. The carbonyl, as we've learned with carbonyl compounds, is electrophilic. So it's electrophilic at the carbonyl carbon. But, and you've seen this reactivity already in cuprate chemistry, but it's also reactive at the beta carbon. In other words, it's also electrophilic. Now when we talked about cuprate chemistry, we said we can think about the reactivity in terms of resonance, as did we say when we talked about carbonyl chemistry. So in general, what we've said is, all right, you have some carbonyl compound and we can recognize that this is the most important resonance structure. But we also know that oxygen is more electronegative than carbon, so even though they have a covalent bond between the oxygen and the carbon, the sharing of electrons, you share electrons in the covalent bond, you don't share electrons quite equally. The electrons spend a little more time near oxygen and a little less time near carbon. And so we can write a second resonance structure, a minor resonance structure that reflects that reactivity. It's a minor resonance structure because the carbon doesn't have a complete octet. It's really not a very good resonance structure. It just represents a small fraction, a small fraction of the picture of the molecule. Remember, resonance structures aren't one. They aren't the other. It's not oscillating between them. It's both at the same time. But now with the double bond, we can write yet another resonance structure, also a minor resonance structure. So I will write another minor resonance structure. And collectively, this picture of these three resonance structures really make up a more complete picture of the reactivity of the molecule. So the implications of the second resonance structure there is that the beta carbon can also react with nucleophiles. And if I had to generalize it, I'd say what we could say is that strongly basic nucleophiles, in other words, nucleophiles, let's say, well, I'll tell you in a second. But let's say in general, strongly basic nucleophiles, I'll say tend to react because this is all about tendency. So tend to react at the carbonyl carbon. And I'll say moderately basic nucleophiles tend to react at the beta carbon. Now what do I mean by moderately basic? Probably, let's say, the conjugate acid, let's say, has a pKa of, oh, let's say less than or equal to 15. And these are generalizations and tendencies. What do I mean by strongly basic nucleophiles? Well, let's say the conjugate acid tends to have a pKa of, let's say, greater than or equal to 25. And I'll deliberately leave that region in the middle from about 15 to 25 a little bit vague. That region includes the enolates of ketones and aldehydes. And to some extent, they can react in two different ways. So let me give you then a concrete example of the Michael reaction. We call the second manifold the Michael reaction. And I think I borrowed this example from your textbook just because it's a nice example and you get lots and lots of chemistry today. And so I don't want to overwhelm you too much. Let's envision cyclohexenone and its reaction with ethyl acetoacetate under basic conditions. Let's say with sodium ethoxide in ethanol. I've chosen sodium ethoxide in ethanol because I want to generate the enolate. Sodium ethoxide is a good enough base to generate the enolate of a ketone or aldehyde, but it's specifically a good enough base to quantitatively deprotonate ethyl acetoacetate. Remember pKa of about 11 for ethyl acetoacetate, pKa of about 16 for ethanol. So that equilibrium between sodium ethoxide and ethyl acetoacetate lies way to the right. The product of this reaction, all right, that is a lot to swallow mechanistically. And so let's think about it, what's happening. So we're generating the enolate. I won't write the mechanism for that because by now you've seen this many times. We generate the enolate. The enolate can add in to the beta position of the alpha-beta unsaturated carbonyl compound. So we push electrons on up onto the oxygen. We have our negative charge on the oxygen. Now we have lots and lots of ethanol present and protons go on, protons go off, so we're just going to protonate our enolate to give our product. Thoughts or questions at this point? Okay. I again want to come back to this idea of seeing it because it is amazingly funny when you're sitting and looking at things in the forward direction and you say everything seems to make sense. Okay. I kind of understand this. We've got protons coming on, protons coming off. We've got sodium ethoxide as the base. We have ethyl acetoacetate. Ethyl acetoacetate is pretty acidic. We generate that enolate. We've got this Michael acceptor, this alpha-beta unsaturated carbonyl compound, sodium ethyl cyclohexenone. I know it's electrophilic at the carbonyl compound. I know it's electrophilic at the beta carbon. And I know that our nucleophile isn't so basic. It's just moderately basic, so we add to the beta carbon. Oh, yeah. We generate an enolate. Okay. We're in ethanol. That enolate can protonate. I get the ketone. And then you look at this in a completely different way. And I go ahead, whether on the blackboard or, God forbid, on a quiz or a test, and I write something like that and I say, how do you make it from simpler compounds? And of course, I don't write that exact compound because organic chemistry isn't about memorizing. It's about thinking your way through stuff. And you say, what the heck? How in the heck can I see that? And then you think, okay, wait a second. We've been learning relationships. Today, we've learned two of them thus far. We've learned about beta hydroxy carbonyl compounds and a light bulb goes on. You say, oh, I see a beta hydroxy carbonyl compound. He can't fool me. Nature can't fool me. I can make it from an aldol reaction. You see an alpha, beta unsaturated carbonyl compound and you say, oh, nature can't fool me there either. I can make that from an aldol reaction with dehydration, even if it's an intramolecular aldol or a cross-aldol. And so now we have a new nemesis here. We have this thing. And you look at the relationship and you say, oh, okay, what's our relationship? It's further apart. It's harder to see. It's a one, two, three, four, five, one, five, di carbonyl compound. Now you just look at that relationship and you say, oh, I already know the trick. In a one, five, di carbonyl compound, I envision a retro synthetic analysis where I break the bond between the four and five carbon, between the three and four carbon. You say, ah, I go ahead. I break that bond and you see backwards and you look and you say, oh, wait a second. That could have come from these two components. And you don't worry about reagents or conditions yet in your retro synthetic analysis. Now the other beauty of retro synthetic analysis is that you can see other things and other relationships and this is the genius of organic synthesis. Because you look at this compound and you might say, oh, okay, I see a one, five, di carbonyl compound, too, but I see it that I'm going to number like that. And you look and you say, oh, I'm going to break the bond between the three and four carbon, like so. And I'm going to apply all my smarts about it being the bond between the two and three carbon that was the alpha, beta unsaturated one and the four carbon that was the nucleophile. And you think about it and you say, oh, I see this as my synthetic precursor. And you look at that and you say, here's another route by which I could possibly generate that product. Now that compound doesn't look so simple. I broke the first one into two parts and here I've only broken it into one and it looks kind of complicated. But then you think again and you say, wait a second, I could see this component coming from a crossed aldol reaction, perhaps. I don't know if this is going to work, so I'm going to write a big question mark, but this is the thought process. I could see this component coming from a big crossed aldol reaction where now I have a ketoaldehyde and ethyl acetoacetate and maybe, I don't know, maybe I could condense ethyl acetoacetate in a crossed aldol reaction with this ketoaldehyde, this 1, 2, 3, 4, 5, 6, this 5-ketohexanal to first make this component and then maybe I could carry out an intramolecular Michael addition and that is the beauty of organic synthesis. There's many ways to put together complex molecules. See you on Thursday.