So today I'd like to continue our discussion of carbonyl chemistry and remember we've now started on the second aspect of carbonyl chemistry, reactivity of the alpha carbon and so for this lecture and then the following two lectures we're going to continue to be developing this idea. So today we're going to discuss a few more reactions of enols and enolates and what we do today is going to set the stage for chapter 24. So we'll be finishing up chapter 23 and setting the stage for chapter 24 which really gets into some absolutely beautiful and profound carbonyl chemistry with the aldol reaction and the Claisen reaction. So I wanted to start out just by coming back to enols and enolates and so here's sort of a generic structure of an enol maybe I'll draw in a couple of lone pairs of electrons just so we get a nice resonance structure and here's a generic structure of an enolate and of course enols we've already talked about in the sense that we've said they're not stable species that you can isolate in general that while some compounds are largely anolic most carbonyl compounds have only a tiny percentage of enol that forms at any given moment and it's transiently formed and it's acid or base catalyzed. Enolates on the other hand we've already seen with say LDA chemistry that you can go ahead and quantitatively convert a ketone to an enolate by treatment with LDA. So if I wanted to make some generalizations because we're going to be talking about principles of reactivity both today and then in the next two lectures if I wanted to talk about some generalizations I'd say that enols are more reactive than alkenes. In other words we've seen acid promoted reactions of enols we were seeing to atomization and these reactions occur with very mild acid. You've seen reactions of alkenes. So for example you've seen treatment of an alkene like isobutylene with aqueous sulfuric acid to form an alcohol, the hydration, the addition of water across an alkene by way of a carbocation. That requires very strong acid typically. Typically you're talking about taking an alkene with a concentrated sulfuric acid solution whereas in the chemistry of enols with electrophiles because you're getting an oxocarbenium ion intermediate not a regular carbocation that really is suffering from a lack of a complete octet you have a much greater propensity to react with electrophiles. So if I wanted to sort of characterize an enol I'd say it reacts with very reactive electrophiles but again that's sort of a relative term in the sense that alkenes react with even or require even more reactive electrophiles to react. So we're going to see reaction with halogen. I'll just write this as X2 to indicate really chlorine, bromine or iodine. Fluorine's always in a sort of special category. It'll react but it'll react with absolutely everything because the fluorine you know every position in the molecule because the fluorine hydrogen bond is so strong. So alkenes also react with halogens but now where you start to see enhanced reactivity is for example protonated carbonyls typically if you had a ketone in acid sure you'd generate a protonated carbonyl transiently but it's not going to react with a typical alkene under typical conditions. Again I can give you exceptions to that whereas this is going to be the basis for what we'll see in the next lecture on the acid catalyzed aldol reaction. Although I won't show examples we're not going to talk in detail about the chemistry. I'll give you one sort of comparison if we talk about in a mini-mion so the sort of a mini-mion intermediate we talked about before these two react with enols in what's called the Manic reaction. Remember how I've talked about the aldol? I've talked about carbon oxygen double bonds and said in many ways carbon nitrogen double bonds are very similar in their reactivity so an emine is very similar to a ketone in its reactivity just a little bit less so because nitrogen's a little bit less electrophilic, a little bit less electronegative and a mini-mion is very similar to a protonated ketone and so there's this manifold of reactivity whereas a typical alkene wouldn't under usual conditions react within a mini-mion or a protonated ketone. Now coming to enolates since we now hear why is this more reactive you've got the oxygen pushing electron density into the enol double bonds so you're donating in electron density it's more electron rich than a regular alkene and along the same lines as I said when it reacts with electrophiles the carbocation that forms gets extra stabilization from that oxygen. Well enolates are even more nucleophilic. In other words they're even more reactive toward electrophiles so in addition to things that are only moderately electrophilic and very elect or in addition to things that are very reactive electrophiles enolates can react with moderately reactive electrophiles. So sure an enolate will react with a halogen because halogens are good electrophiles but you can also react with things like alcohol halides in SN2 reactions. We saw this before when we talked about the direct alkylation of enolates taking a ketone converting it to the enolate with LDA and then alkylating it say with methyl iodide as we saw in the last lecture. Ketones and aldehydes are a lot less reactive than protonated ketones and aldehydes and so these two react with an enolate because an enolate is so much more electron rich, so much more nucleophilic. And again this is chemistry we'll be picking up on in chapter 24 when we start to talk about the aldol reaction. And then similarly under the right condition esters which are even a little less reactive than ketones and a good bit less reactive than aldehydes also under the right conditions can react with enolates. All right where I'd like to begin today's discussion then is to talk about reactions with halogens. So specifically the alpha-halogenation of ketones and aldehydes and so in sort of a very generic equation if I have some sort of ketone or aldehyde obviously with alpha protons and we treat it with some halogen and as I said before we're really talking chlorine, bromine or iodine with fluorine sort of being in a special class of its own in reactivity. You react to get an alpha-halo ketone or aldehyde and organic chemists are terribly bad at balancing equations but also hydrogen halide and this is important. The reaction in the acid manifold and the enol manifold is catalyzed by acid. So I'll write reaction proceeds by the enol. Now if a reaction produces its own catalyst that means the reaction is also autocatalytic, self-catalytic it catalyzes itself. No catalytic reactions can be a lot of fun because often that means you mix the reactants and they seem to just be sitting there and then they start to react because you have a little bit of the acid around or because it can go by an uncatalyzed manifold or you always have some enol present so it generates more catalysts which goes faster and faster and faster and if it's exothermic it can take off because with more heat that it's producing it'll go even faster and produce more catalysts which is one of the reasons why when you're in the laboratory you're so often running reactions in solvent rather than mixing chemicals directly together. The solvent dissipates the heat. It can boil with a reflux condenser and be condensed and come back and take away heat. Anyway, so for example I'll write a real example of a reaction if you take cyclohexanone and treat it with chlorine in water so often when we're writing synthetic reactions we'll just write reagent over the arrow and sometimes solvent below the arrow. Then you get the alpha-chlorocyclohexanone, 2-chlorocyclohexanone and I'll write plus HCl although it'll remind us that when we're in water of course you don't have HCl in water when you're in methanol you don't have HCl in methanol because you protonate your solvent, you protonate your water to give hydronium ion. HCl is a much stronger acid, it's negative PK of negative 7 or thereabouts, it's a much stronger acid than the conjugate acid of water, the hydronium ion and ditto for methanol. So if I really want to write things out here I'd say plus H3O plus and Cl minus. So let's take a look at how this reaction occurs. We talked about enolization last time so I'm only going to write out a sort of abbreviated mechanism. I will not write every last lone pair of electrons in writing my mechanism, I will not write every step in gory detail so for example our ketone plus H3O plus is going to be an equilibrium with the protonated ketone in water, water can pull off the alpha proton, you've seen all of this mechanistically and so you have this equilibrium to form the enol. Remember that equilibrium lies way, way to the left so you only have a teeny tiny percentage of enol, only a fraction of a fraction of a percent of enol at equilibrium but that's okay because it's going to keep getting regenerated under the acid catalysis as soon as you use it up. So our enol is nucleophilic on the alpha carbon. Here's our chlorine, I won't draw all the lone pairs of electrons on the chlorine but electrons flow from the oxygen down, they flow from the double bond to the chlorine, we push electron, break the chlorine-chlorine bond, push electrons on to chlorine and now you get the alpha chloro compound and I'm going to remind us here that we have a hydrogen at the alpha position just to help us keep track so I suppose if you wanted to you could go ahead and put in a hydrogen here, I just don't want to lose track of what's connected to what carbon. The other thing that's produced is chloride anion like so and again I'm not going to write in every lone pair of electrons. Then in the next step we're going to just lose our proton and remember technically although chloride's a base, if you're in water reaction like this would be done literally bubbling chlorine into water bubbling chlorine from a gas tank into water to make chlorine water and so if you're in water the strongest base that you'll have is water under acidic conditions and so we can envision water coming along with its lone pair of electrons pulling off a proton, putting them back, putting electrons back onto the protonated oxygen and I guess here if I want to be a good person actually I am going to say this one is essentially irreversible because basically once you form that chlorine bond you are for all intents and purposes not breaking it again and so our overall outcome is now you have the alpha chloro ketone plus H3O plus. Question, okay good question. So the question that's being asked is we have chlorine here and we have alkene and we've learned that with chlorine you can add across a double bond like so. Don't worry about your stereochemistry right now. So you've learned that you can do that and indeed it's possible that chlorine because you think of it as stepwise you have that oxacarbenium ion you can write the other resonance structure just like you do with an alkene you say well chloride could come in and intercept and indeed it could and indeed it can. Do we have any ideas about what this compound would do? Kick out the good leaving group and we're actually going to come back so indeed so sure chlorine could add as an equilibrium and you're never going to have a stable adduct of chlorine into the carbonyl whether it's as an oxyanion or as a hydroxychloro compound. We're going to come back to this idea when we talk about the heliform reaction which is a really weird reaction. We'll talk about that in a moment but talk about it in a little bit but you never have a stable adduct. There's like one exception iodide can add into acetone and certain ketones and form a stable oxyanion adduct. Your book doesn't mention it. It's sort of the exception. They say it never occurs. I can give you the one exception question. These two carbons, the ones next to it, alpha on either side in this case are equivalent and so you basically chlorinate once. If I keep pushing it and letting it go longer, yeah it'll chlorinate again but this reaction pretty much because the chlorine is electron withdrawing can stop very nicely after addition of one chlorine. I'll show you an example in a moment where we have two different ones and we'll talk about the regiochemistry. How does the double bond form in the first step over here? How does that get there? In other words, how do we get from here to here? Okay. So here are your protons. Water's your strongest base. You can think of the water whoops as pulling off a proton, electrons flow in, electrons flow onto the oxygen. You're welcome. So that's acid catalyzed enol formation. All right. So let us talk now about some other cases. I'll give you a couple of examples. Now one thing that I've done is broke with your textbook and I know you get bent out of shape when I tell you the way things really work and your textbook gives you slightly simplified answers. So I didn't draw any catalytic acid in there and in reality you really don't need it. Now I'll tell you the secret about chlorine. Chlorine in water. So not only do you have some enol to get things going but chlorine actually disproportionates in water giving HCl, hydrochloric acid and HOCl, hyperchloric acid. So a solution of chlorine water is always acidic. So you don't need to add any acid to get things going. I'll give you an example straight from your textbook if it gets you less bent out of shape. We'll take acetone treated with bromine and as I said this really works for all of the three regular halogens.  So HBr and acetic acid is really strong. You're getting a lot more acidity than that one point, negative one point seven of H3O plus. Anyway, I'll give you another example and come back to the question of regiochemistry. One of the things that's cool about this reaction is the rate determining step is the enolization. As soon as you form enol, it's consumed by your halogen and it's irreversible, hence my writing, my arrow like that. So the enol that forms more quickly is the less substituted enol. We talked before about enolates and we described kinetic enolates, the less substituted enolate and thermodynamic enolate, the one that's more thermodynamically stable, the more substituted enolate. What's kind of the same with enols? So when you do have two different groups like an isopropyl group and a methyl group, then you end up going at the less substituted position. And again, this chemistry works in a variety of solvents so I'll just give you another one, bromine and methanol. You notice all of these are protic solvents, water, acetic acid, methanol, typically those are the solvents that promote enolization and so those are the solvents, typically one would do this chemistry in. So then we go through the more kinetically favorable enol and as soon as that enol forms it's consumed, we get the less substituted compound. Don't get too bent out of shape over these details. There's in so much of the chemistry I can give you, there are many, many layers and it gets deeper and richer and you can go on and on with it. But I wanted to give us a flavor mechanistically and give us a flavor synthetically. Alpha halo carbonyl compounds are really cool. Your textbook talks a little bit about their reactivity. They're really good alkylating agents and SN2 alkylation kind of like allylic and benzylic halides which are more reactive toward SN2 alkylation toward SN2 reaction than a standard alkylating agent has to do with the orbitals on the carbonyl or the orbitals of the benzyl or allyl group. Anyway, your textbook gives you some nice examples including talking about synthesizing LSD. All right. What I'd like to do now is to talk about alpha halogenation promoted by base. I'm not saying catalyzed because you'll see when I balance my equation that we use up a whole molar equivalent of base. Synthetically this is a less useful reaction and remember organic chemists tend to be control freaks. We like to control reactions and make them do what we want to do. Synthetically often enolate chemistry ends up being a little less controlled in the halogen manifold and the halogenation manifold. So let me start with a simple example which is controlled and I'm sort of writing some of these examples around your textbook just to keep things in line with this and to illustrate a few principles of reactivity. So let's imagine that we have this phenyl ketone. So we have one alpha proton and we have a source of base so we can form the enolate in an equilibrium reaction. Remember the pK of a typical ketone is about 20. This will be a little more acidic but generally that means since water, the pKa for water deprotonating is 15.7 you're going to have an equilibrium reaction. Anyway, plus bromine and so very much like the other chemistry we were talking about, you're going to end up alpha brominating and if I write a balanced equation you'll see that the other products of this reaction are plus bromide ion and plus water. So you notice we've actually consumed a molar equivalent of hydroxide in the reaction. So the chemistry is very similar. The enolate is nucleophilic. The enolate is even more nucleophilic and again I'll skip over the curved arrow mechanism for forming the enolate.   Electrons flow down from the oxygen to give us our carbonyl back and over and from the double bond to the bromine and push electrons onto the bromine to give rise to the alpha bromoketone plus bromide. And remember if you're uncomfortable with all of these arrows moving at once, remember that picture in your mind of the two resonance structures of the enolate, the one with the negative charge on the oxygen and the other with the negative charge on the carbon and I said, I'll usually just keep one in my mind and it'll be the more important, the bigger contributing resonance structure, this resonance structure but what I'm really doing with this arrow and moving the electrons from the double bond of course is just the mental process of getting to that other resonance structure. So if you're uncomfortable with that just think about the resonance structure like so and it certainly is more intuitive but Johnny probably for example would never write this as the attacking resonance structure just because we tend to think about the more important resonance structure. Question, would bromine minus deprotonate? Another good question, in general equilibria tend to be active, tend to be significant within about 10 pKa units, HBr very acidic conjugate acid of bromide about negative 8, negative 9, somewhere around there, bromide's not going to deprotonate significantly. Water, pKa of the conjugate acid negative 1.7 not going to deprotonate significantly. So if you really ran out of hydroxide we could slip into the acid catalyzed manifold. If I tried to play a trick on this chemistry and used half an equivalent of hydroxide and a full molar equivalent of bromide I suppose the first half of the reaction would run in the base promoted manifold and then as the base was consumed eventually we'd start going through the enol pathway. I've never tried it. These are good questions. I like this, this is a lot more fun and simply babbling on. It means your mind is working and while we're at it how about next class we all get down more toward the front row? What's, there's no fun in being in the back row. All right, so let's continue with this because I want to take us to some principles and something that's cool, kind of counterintuitive, no longer super important in terms of analytical chemistry but nevertheless very cool. All right, now let's imagine because I'm going to walk us from something where we have one alpha proton to something where we have two alpha protons and then to a compound where we have three alpha protons all on the same carpet. So let's imagine we take this same manifold, a base and you notice I mean remember I make this point.  They'll look at you funny. Well, which hydroxide do you want? Potassium hydroxide, sodium hydroxide. Usually organic chemists will use one of those too. Sodium is common. All right, so I'll take sodium hydroxide and bromine and I'll indicate excess. All right, I'm going to do it. I'll write excess rather than the word excess. All right, and I'll probably keep doing that. I think your textbook writes excess, E-X-C-E-S-S. We also in the laboratory when our gas tanks get empty from nitrogen we run reactions under them write M-T on the gas tank in chalk. All right, so we get the di-bromocompound.  Make some noise, come on. The real thing I want to say is the reaction doesn't stop at the monobromocompound. In other words, what I'm saying is if I wanted to make the monobromocompound by this route I'd be in trouble. If I wanted to make the monobromocompound we could go in the acid-catalyzed manifold, but if I wanted to use the base-catalyzed manifold I'd be in trouble. The problem is bromine's electron withdrawing and so once you put one bromine on there, once that enolate has reacted once and we get a bromine on there, now this proton's more acidic. And so you get this competition. Now you've got this molecule because you know partway through the reaction you've got this molecule with two protons that are less acidic and then the molecule with the bromine next to the proton making it more acidic. So if I wanted to be a troublemaker and say go ahead and use one equivalent, so I'll write NaOH, Br2, one equivalent EQUIV meaning equivalent and see what happened. Then I'd end up with a mixture of products. I'd end up with some of the di-bromo. I haven't done this myself but it would be cool to monitor by NMR because you'd see peaks for all of these. Some of the unreacted starting material you'd probably see. Some of the monobromo in there as well. It's fine, it's cool, it's okay but remember what I said before about organic chemists being control freaks, typically we want to take a reactant and convert it to a single product and so from the point of view of carrying out a synthetic reaction it's not so useful, it's not so helpful to be able to get a mixture of, or to not be able to get a single product better to do it in the acid catalyzed manifold. So where all this is leading is to some weird and funky chemistry. So let's take the example now where we have three alpha protons so instead of propyophenone or isobuterophenone the first one we had we'll take acetophenone and let's make our live simple. We'll treat this with sodium hydroxide and bromine and again I'll write excess. And now what's really wild is we break a carbon-carbon bond. We get the carboxylate and CHBR3 which is called bromiform and this reaction is general, it works with all three of the main halogens, chlorine, bromine and iodine, all the important ones and so this is called the heliform reaction so in other words I can write it, it's general for methyl ketone so I can write this sort of in general terms of a methyl ketone drawn sort of generically with CL2, BR2 or I2, in other words molecular, chlorine, bromine or iodine and some base sodium hydroxide but again potassium hydroxide would do just fine generates RCOO minus or RCO2 minus depending on how you prefer to write it either of them is fine plus CHCl3 chloroform CHBR3 we already said that's bromiform so I'll say plus one or the other or CHI3 iodiform and historically this reaction's been used as a chemical test for methyl ketones nowadays if you wanted to take a methyl ketone well just take an NMR and okay you've got a methyl peak and you've got a ketone in the IR and the methyl peaks at about two parts per million you say I have a methyl ketone but historically this was used as a test and typically it was done with iodine because iodiform is really cool it's a yellow organic compound so you take a methyl ketone like acetone a drop of it in a test tube with some sodium hydroxide and you throw in iodine and this yellow precipitate comes out which is really cool so you've got this chemical test and it only occurs with methyl ketones. Iodiforms also need it's got this really need odor to it it's got this characteristic smell very, very pretty however the thing that I want your textbook presents this this is a reaction that historically I thought do I really want to teach this to sophomores it's counterintuitive and it's actually cool and it's counterintuitiveness and it also is relevant and so I do think this is worth learning about. Alright so what's going on in this chemistry we're breaking a carbon-carbon bond so there's something weird going on here so I'm going to start we'll do this sort of on our generic methyl ketone like so and I'll say as above by as above of course I mean form the enolate alpha halogenate form the enolate again alpha halogenate again form the enolate once again alpha halogenate once again and so at this point we have the trihalomethyl ketone and this is unstable to base and that's really the key it's unstable under the reaction conditions it breaks down we've now put three electron withdrawing halogens on carbon even though it's an alkyl carbon it's getting to be a very unusual alkyl carbon it's rendering by pulling electron density away from the carbonyl it's rendering the carbonyl more electrophilic and you've got hydroxide present so here's our hydroxide hydroxide can add in I'll put in a few electrons and try to be a good person hydroxide now adds in to this extra electrophilic carbon now here's where things get fun and I'll only draw the lone pairs that we need just to prevent us from going crazy here all right so at this point we formed a tetrahedral intermediate but what's special here is that that trihalomethyl group is electron withdrawing unlike say a methyl group or an alkyl group in general it can stabilize electron density so here's the big surprise electrons come in and they push it out like a leaving group because indeed it is a leaving group in this particular context so electrons come back we go ahead we end up with a carboxylic acid and again I'm drawing abbreviated mechanism I'm not putting in every gosh darn lone pair of electrons our car whoops here's the kicker a carbanion normally see carbanions you see a satellite anion but normally I mean we see a grignyard I said a grignyard's like a carbanion but you really have a covalent bond to the metal you can just think of it as being like a carbanion because the metals elect are positive organolithium reagents same thing an organolithium like butyl lithium is like a butyl anion but it's not really a butyl anion but here we can kick out a carbanion there's something special so I'm going to write exclamation point just to bring home the point and of course the carbanion is still basic so it can pick up a proton the carboxylic acids the most acidic thing around and so we have the carboxylic acid plus our heliform but we're dancing around this very uncomfortable step here we're dancing around this fact that I've just kicked out a carban with a negative charge on this now you've seen this before you've seen this before you've seen this in cyanohydrin chemistry where we know that we can form a cyanohydrin but we also know under basic conditions the cyanohydrin formation is reversible and so in cyanohydrin formation we learned it forward and backward we encountered it on a quiz we became comfortable with the idea of kicking out cyanide anion and getting a carbonyl back and again I'm just drawing in the relevant lone pairs of electrons and the key on this of course was that well cyanide gets a negative charge on carbon but it's special I mean you can go you can buy a bottle of sodium cyanide HCN right we always talk about how stable an anion is by thinking about the pKa of the conjugate acid HCN has a pKa of about 7 up 9 so what's going on here in our heliform reaction well chloroform has a pKa of about 25 the textbook uses that number that's the number I use 2 so when we kick out the trichloromethyl anion it's not like a carb anion remember methane is 50 this is much more like a stabilized carb anion not as good as cyanide I was curious because I didn't have the numbers at my fingertips I wouldn't expect you to either but the pKa for bromiform is about 23 off of at least one source and Iota form iodine is more polarizable than chlorine even though it's less electronegative that means it can better stabilize the I'll use a little till day there to mean approximately can better stabilize the negative charge pKa of 14 so the Iota form reaction the one that you can do in a test tube really really works very easily because basically you're kicking out something that's kind of like an anion and cyanohydrin formation now what's cool to me about this chemistry here we always calibrate ourselves I've emphasized this idea pKa and what are markers to have in your head for various values I said a factor of 10 10 pKa units is sort of my waiting on whether an equilibrium occurs to any significant extent so you can take away that for kicking out an anion out of a carbonyl that 25 sort of is that teetering edge point now this is going to come back in chapter 24 when we discussed the Claisen reaction but more specifically when we discuss its mechanistic reverse the Claisen reaction involves an ester enolate adding to a carbonyl the mechanistic reverse involves kicking out an ester enolate it's called the retro Claisen it involves kicking out an ester enolate remember that number I said keep in your head 25 for an ester enolate so that's sort of right at the same value of chloroform and so I think this I think the heliform reaction provides us with this very nice calibration point for breaking carbon carbon bonds I saw a question there ketone primarily would get one side you I'm not sure how clean it would be I'm pretty actually I'm pretty sure you would get it clean on the less substituted side under basic conditions so yeah I'm pretty sure because the kinetic enol remember this reaction is rate determining in enol formation or enolate formation so you form that kinetic enol or the kinetic enolate it goes now it's even more acidified so it goes at that position again so I think the answer is yeah if we took that other one of the isopropyl methyl example that I was showing you before and tried it under these conditions would you get all isobutyric acid or would you get some bromo isobutyric acid a little bit I don't know for sure we could try it literally in the lab all right so that kind of that kind of brings me to what I wanted to say about halogenation and now I wanted to start talking about some more anion chemistry and really this is going to set the stage for the next chapter but it's also some very cool chemistry in its own and this is older enolate alkylation chemistry and going to present to us two reactions one of them is called the acetoacetic ester synthesis and one of them is called the melonic ester synthesis the acetoacetic ester synthesis is a synthesis of methyl ketones the melonic ester synthesis is a synthesis of carboxylic acids they're both really useful reactions they can be done on a bucket scale literally to make kilograms of compound most of the LDA chemistry we were talking about sort of fussy you do it in a little flask with a little bit of butyl lithium making you know 10 millimoles or 100 millimoles of compound maybe a mole if you're lucky but this is bucket chemistry all right I want to lay the groundwork of this and I want to come back to the two numbers I told you before so I said if you want to keep two numbers in your head for pKa's the numbers you can keep are pKa of 20 and this is sort of a generic ketone so remember I'm writing a generic structure here say 24 generic ketone acetone the specific example here happens to be 19 but that's not so important basically figure 20 and remember I said for an aldehyde yeah it's a little less than 20 if you want to keep another number in your head keep 17 if you can only fit a certain number of numbers in your head just keep 20 for all of them all right Esther about 25 so we'll use a little till day here to mean about all right what I want to show you now is what happens if we start to have two carbonyl groups and it shouldn't surprise us that if we have two carbonyl groups the corresponding enolate is going to be more stable in other words the ketone with two ketone groups is going to be more acidic the Esther with two Esther groups flanking a hydrogen is going to be more acidic so this compound here often called acetyl acetone yeah if you want to get the IUPAC name you can call it 24 pentane dione but it's nickname it's common name is acetyl acetone has a pka of about 9 if you have one ketone and one Esther probably think it's going to be a little less acidic pka is about 11 for this one this one's often referred to as ethyl aceto acetate again if you want to get all IUPAC you can say ethyl thrioxobutano weight but a nickname a common name is ethyl aceto acetate now give you an even more common common name in just a moment two Esther groups even yet a little bit less acidic now we're talking about pk of about 13 this is called diethomalinate and it kind of makes sense that diethomalinate should be a little less acidic all of these should be more acidic than the monocarbonial compound if you want to keep one number in your head you know put in 11 if you don't want to 10 if you don't want to keep track of all these numbers but the main point is they're way more acidic than a regular carbonyl compound all right what's the reason for this greater acidity well we already talked about how you can write a carbonyl like resonance form for an enolate but you've got lots and lots of extra stabilization by sticking the negative charge onto oxygen and so you can write an enolate like resonance structure I suppose I'll be a good person here for once put in all my lone pairs of electrons you can write an enolate like resonance structure or resonance structure with a negative charge on oxygen but the cool thing about all three of these one three di carbonyl compounds is the extra delocalization just like a carboxylic acid spreads that negative charge out over two oxygen atoms and so it's more acidic than an alcohol here we can spread out the negative charge not just on this oxygen but on the other oxygen and so it's more acidic than a regular ketone hence our pka of nine now what does nine mean nine means that unlike acetone if I go ahead and treat acetyl acetone if I treat two four pentane dione with sodium hydroxide I'm going to have an equilibrium that's going to lie to the right to give the corresponding anion the corresponding enolate sodium will be the counter ion and if I want to write a balance equation I'll remind us that water balances our equation so we can quantitatively we can stoichiometrically make this corresponding enolate because it gets extra resonance stabilization if I take ethyl acetoacetate and I treat it with sodium ethoxide again I have an equilibrium that lies to the right for generation of the enolate I'll write the structure where I've enolized the ketone that's probably the slightly more contributing resonance structure over the one of enolizing the ester but remember there are all three resonance structures contributing together negative charge on this oxygen negative charge on that oxygen negative charge on carbon plus ethanol and think about it in terms of pKa's pKa of water 15.7 pKa of the acetyl acetone seven that equilibrium lies to the right by seven orders of magnitude pKa of ethanol 16 or 17 depending on which number you want to keep in your head pKa of ethyl acetoacetate 11 that equilibrium lies to the right by five orders of magnitude I want to use sodium ethoxide for the second one the first one doesn't matter so much why don't I want to use sodium hydroxide for the second one what does sodium hydroxide do with an ester makes it leave a tax the ester carbonyl kicks out ethoxide gives the carboxylic acid the carboxylic acid swaps a proton we saponify the ester now if you're thinking you'll say well he keeps saying ethoxide is just like hydroxide in many of its reactions so can't ethoxide attack here we had transesterification on the midterm sure it can be a tag it can attack form a tetrahedral intermediate kick out ethoxide you're back where you started it's this degenerate equilibrium sure it happens lots of things happen in carbonyl chemistry but it doesn't matter it doesn't make any particular consequence this is also why I'm choosing to pair sodium ethoxide with ethyl acetoacetate rather than say sodium methoxide sodium methoxide of course unlike hydroxide won't give you a carboxylic acid but now we're going to get a mixture of the methyl ester and the ethyl ester as you transesterify all right so where all of this is going is to a synthetically useful reaction which I was hinting at before the aceto acetic ester synthesis and I'll put parenthetically of methyl ketones because this is a reaction that's useful for making them by the bucket now I hinted that there's another name in addition to ethyl acetoacetate that name for this compound in addition to calling it ethyl threoxobutanate butanowate in addition to calling it ethyl acetoacetate also gets a called acetoacetic ester it's a historical thing when you go to the lab and you do an extraction you ask for ether and it means diethyl ether when you go to a bar and you ask for a glass of scotch you say I'm drinking alcohol because you're drinking ethyl alcohol not methyl alcohol and here acetoacetic ester means the ethyl ester of acetoacetic acid from historical terms let me introduce the reaction and then we'll talk about the mechanism because it's cool as an in addition to being useful all right so here's our acetoacetic ester we'll treat it with sodium ethoxide in ethanol very easy to make you throw a chunk of sodium into ethanol it fizzes it doesn't catch fire and ethanol fortunately if you throw it into water it catches fire you take an alkyl halide and we're talking about one that's good at SN2 alkyl halides in other words not not tert butyl bromide but something better like ethyl iodide you can probably get away with isopropyl but you can't go with tert butyl because it doesn't do SN2 we finally we throw in some acid and we heat this gamish and out pops a methyl ketone where we now have the R group on it this sounds kind of abstract so I want to give us a concrete example so let's say we take and then we'll walk through how it occurs so let's say we take ethyl acetoacetate we treat it with sodium ethoxide in ethanol we'll take an alkyl halide I'll choose propyl bromide I could use propyl iodide bromides are cheaper than iodides propyl chloride is I think a gas or a teeters on the border being gas a gas propyl bromide is a liquid I can buy it alkyl bromides are good at SN2 displacement reactions and we do an aqueous work up with acid acid unspecified but as I keep saying you can't go and buy a bottle of H3O plus we're talking about either HCl or sulfuric acid and water and the methyl ketone will get out of here we've got a lot of carbon so I want to help us keep track that's why I've given us a concrete example the methyl ketone we get out of here is two hexanone alright what's going on here first part and I'm going to give us an abbreviated mechanism because we've seen a lot of these steps alright first part's easy you have your anion drawing one of the resonance structures of it make our lives easier you want to draw a different one that's okay too electrons flow from our nucleophile from the alpha carbon to the electrophile we kick electrons onto bromine in an SN2 displacement so now we've alkylated with a propyl group no I'm not going to balance my equation and show the bromide we're just getting a flavor of the mechanism okay next step we treat with acid do this aqueous acid and what is this it's acid catalyzed ester hydrolysis we've snatched before you protonate the carbonyl water attacks you kick out and you swap some protons around you kick out ethanol so at this point we have a beta keto acid what the heck I'll be a good person right ethanol here but this is the guy that we're going to focus on the beta keto acid beta keto acids are special one three dye acids are special they can undergo decarboxylation and the mechanism is really cool going to write a different confirmation of the molecule this is probably the preferred confirmation where the OH group is hydrogen bonded to the carbonyl all I've done is spin around this bond here here's our propyl group we can move six electrons in a ring like so and that occurs upon heating when I push our arrows let's think about it what does this mean we're forming a bond between this oxygen and hydrogen we're breaking this bond we're forming a bond between this carbon we're forming a double bond over here we formed carbon dioxide that's a right a little plus sign call this a decarboxylation that's an OH up there if you're having trouble seeing call this a decarboxylation because we're losing carbon dioxide so that's kind of kind of intuitive but this also fits into this broad class of reactions you learned about in 51b it's a pericyclic reaction what's the archetypal pericyclic reaction you learned about in 51b deals alder reaction this is just like the deals alder reaction six electrons moving in a ring if you want to get technical I can call it a retroene reaction try that one in one of your lab TAs they may not have seen that before but however you want to talk about it pericyclic reaction decarboxylation reaction retroene reaction something like the deals alder reaction we broke in a carbon carbon bond and gotten carbon dioxide our ketone undergoes enolization or well keto enol I will write tautomerization and that gives rise to the two methyl ketone because of course enols aren't stable right so let's just put all of this together and I'll show you a cool synthetic example something a little bit different we're going to have some fun with this this really is useful chemistry it's chemistry by the bucket you can prepare methyl ketones so we'll take our acetoacetic ester we'll treat it with our proverbial sodium methoxide in ethanol we're going to do a double alkylation we're going to alkylate once and alkylate again so I'll take ethyl iodide now I'm going to continue so we've alkylated once but in your mind's I imagine we're now going to treat with some more sodium methoxide not going to write it out for you picture this in your mind's I we treat with some more sodium methoxide we again generate the enolate we repeat and now just for fun let's use some methyl iodide CH3I now we carry out an aqueous workup and heat it and decarboxylate the same way we've done up there but we've done a double alkylation we've put on a methyl group and an ethyl group and so we use this as a reaction to synthesize three methyl two pentanone so that's cool that's useful you can synthesize a bunch of methyl ketones by way of this chemistry slays the groundwork for the last section that I want to talk about the melonic ester synthesis of carboxylic acids I already showed you diethyl malonate that was the one with the two carboxylic acids the one three diacid again the old nickname for it is melonic ester because it's the diethyl ester of melonic acid we already said well ethyl is sort of the default so we take our diethyl malonate also available cheap like aceto aceto acetate I'll talk more about how all of these are formed as we get into the Claisen reaction in chapter 24 so we take our melonic ester and now we're going to do the same type of chemistry we're going to first treat with sodium ethoxide in ethanol we'll then treat with an alkyl halide again we're talking about an alkyl halide that's good at SN2 displacement so not tert-butyl bromide but a primary alkyl halide or maybe even a secondary alkyl halide and finally we'll carry out an aqueous workup and heat with acid and the product of this reaction is the corresponding carboxylic acid chemistry is very very similar make the enolate alkylate hydrolyze the ester groups to the corresponding carboxylic acid the diacid here's our diacid I guess I'll write it once sort of normally so you can see it and then I'll write it once in a more realistic confirmation with an internal hydrogen bond just like we did in the other case again we're going to push our electrons in a ring like so we lose carbon dioxide just as we did before we get the enol form of the carboxylic acid and we to atomize alright where I want to conclude is coming back to the idea of seeing backwards to retro synthetic analysis we've seen the aceto-acetic ester in the forward synthesis in the forward sense and that's fine because you'll figure it out for the backward sense I want to show you this in the backward sense we already saw in the midterm exam how you can look at an emine and take it apart to the corresponding ketone and amine to an enamine and we can do the same thing to various types of alcohols and we can do the same thing so let me give you one more retro synthetic disconnection let me write out a little problem here let's say we can want to prepare three phenyl propanoic acid by the aceto-acetic by the melanin by the melonic ester synthesis so we look at this compound and we look at the chemistry we've seen in the melonic ester synthesis we look at the same chemistry we've seen in the aceto-acetic ester synthesis and we say all right what are we really doing we are really forming this bond the bond between the alpha and beta carbon by an SN2 reaction and so in our mind's eye when we look at three phenyl propanoic acid we look and we say ah okay that's the bond that we're going to make by an SN2 reaction in the melonic ester synthesis and so we see something in a big piece now this is what retro synthetic analysis is all about this is what seeing that acetone emine and saying oh that's the emine from acetone and ethylamine that's the big site that you get so we look at this and now we say oh yeah we're going to form that bond we're going to use the melonic ester synthesis with a benzene halide and you'll just go to your chemical shelves and pick your favorite benzene halide we'll go with the bromide it's nice and reactive we could go with the chloride but bromides are a little bit more reactive so we'll start with our melonic ester we'll start with our diethyl malonate we'll treat it with sodium ethoxide and ethanol see and this is all pre LDA chemistry LDA really came about in the late 1960s and to me it was transformative in chemistry because it really meant that you're orchestrating a reaction no tricks I have a ketone I'm going to make an enolate I'm going to alkylate it I have an ester I'm going to make an enolate where I want it I'm going to alkylate it here we have a trick because so ethyl acetyl because sodium ethoxide isn't basic enough to quantitatively pull off the proton on ethyl acetate you have an equilibrium that lies eight orders of magnitude nine orders of magnitude to the left on that reaction but by going ahead and using the one three dicharmial compound with this trick we're going and saying yeah we're going to quantitatively get that enolate then we're going to alkylate it so sodium ethoxide and ethanol benzyl bromide aqueous acid that's like sulfuric acid and water or hydrochloric acid heated up and we have our phenyl propanoic acid alright one last example from your textbook and I think it's a good one so I want to show it to you and really really think our way through it for beginning students in organic chemistry rings can really mess with your mind and so often when a beginning student sees a cyclic acetal it's very very different in their mind than a regular acetal and of course being able to break out of that and see beyond that's really valuable so here's an example I'm going to say prepare cyclopentane carboxylic acid and presumably since we're talking about the malonic ester synthesis we should be thinking about that and you look at cyclopentane carboxylic acid and you say well wait a second we got this ring here do with this we learn that we're breaking the alpha beta bond well here's an alpha beta bond here's an alpha beta bond we know we can form two we saw in the aceto-acetic ester synthesis where I used ethyl iodide and methyl iodide that you can do it twice and you can do it twice here and if you think about it what that means thinking backwards is it means that we're going to this but we know we put this in quotes because that's the piece coming from the aceto-acetic ester synthesis and this and so that becomes the crux recognizing that we have a dihalobutate and so now all we have to do is put this into action in the forward manifold we start with diethyl malonate now the nice thing the reactions for giving you can throw stuff together I've been writing step one step two step three you can throw but you can't throw the acid together with the base you'll quench the base but I can go ahead and throw this bromide with sodium methoxide in ethanol and in this commission sodium methoxide is going to react deprotonate alkylate deprotonate again alkylate so that's step one at that point we have this stuff you don't even need to isolate this cyclopentane diester now we just take this I'll just write this with an arrow H3O plus delta we hydrolyze we decarboxylate and we get our cyclopentane carboxylic acid all in one fell swoop and that ends chapter 23 see you on