Alylation of Acetic Acids
King, Susan
This is the third (and final) quarter of the organic chemistry series. Topics covered include: Fundamental concepts relating to carbon compounds with emphasis on structural theory and the nature of chemical bonding, stereochemistry, reaction mechanisms, and spectroscopic, physical, and chemical properties of the principal classes of carbon compounds. 
 We left off last time talking about alternatives to using LDA to deprotonate a ketone followed by adding an alkyl halide because we have problems with thermodynamic kinetic enolate. We want some better alternatives that aren't going to give us mixtures. If you get products from both kinetic and thermodynamic enolates, it's really, really hard to separate the two by chromatography or distillation or recrystallization. They're very hard to separate. We want to be able to do cleanly just form one or the other. An alternative for thermodynamic enolate is to make an enamine. Yeah, it's not showing. Thank you. All right. So an alternative is to make an enamine. So let's label that. This is an alternative to making a kinetic enolate using LDA. So LDA gives you a lot of the kinetic product. It's very selective. The product that you get, remember, it was like 45, 40 something percent, not very good. So what you do is you make an enamine. And so here we saw an example here. You make an enamine and then you treat it with the alkyl halide. And what's going to happen is the electrons on nitrogen come down, boom, you attack the carbon that's bonded to the leaving group and you kick off the leaving group. So notice the ethyl that we just incorporated is on the least substituted side. So this is an alternative to making a kinetic enolate using LDA. All right. So why is it so selective for the least substituted side? And it turns out that all of these bonds here that you see here, I'm going to be making those in red, are all in the same plane. All those bonds are in the same plane. And so you can see that it's much better, including this hydrogen right here and that bond right there. Stereically it wants to be on the least substituted side. If it was over on this side, there's a problem with these two hydrogens right here, getting in each other's way. So if there's really any substitution on that one side, then it's going to be a problem making the enamine. And that's why it's so selective. Okay. So it's not hysterically not good. So this is the one that you form. All right. So here's another example of using this. We can take, we can make the enamine. So again, this is the most common enamine to use. It's not the only one. You could just use dimethylamine, diethylamine, so there's not a problem here. So you can see H3O plus catalyst. Or I might write pH 5.0 to show you that you're going to make it the enamine. That's the best conditions for making an enamine or an emine. All right. So we are going to selectively form this on the least substituted side. The double bond will be here. And we can do things like we can attack acid chlorides just like this, just like we've been used to seeing. Let's see what you get from that. Tetrahedral intermediate that has a leaving group. So acyl substitution, eminium ion here, positively charged nitrogen. Electrons on oxygen come down, kick off chloride. So we've seen that so many times. That's the reactivity that we're used to with carbonyl compounds. I'm going to have to move that arrow a little bit here. So then HCl, H2O, we'll convert that eminium ion to a ketone. So notice you turn that into an active methylene compound. We've just synthesized an active methylene compound. All right. Questions? Anybody? Okay. A couple more reactions that we want to talk about. The first one, alkylation of acetoacetic ester. This is an acetoacetic ester. So the aceto for the acetyl group, yeah? But the mechanism for hydrolysis of the emine to the ketone? I'm missing a methyl group. Let's throw that on there. Now I'm going to hear everybody turning the page. And the one above it? Also, yes, I am. All right. So the aceto is for the acetyl group, the ketone group, and the acetic ester is for the ester portion. That has a pKa of about 11. So I'm completely deprotonated using a thoxide ion. We went through that equilibrium earlier. I don't remember what page that was, but we went through that equilibrium earlier. pKa17 here. So you can see that we get essentially complete conversion. At equilibrium, we have one acetoacetic ester that has not been deprotonated to 10 to the six that has been deprotonated. We call that essentially to completion. It's always going back and forth, but it is essentially to completion. So it turns out the enolates of active methylene compounds are nucleophilic and react with alkyl halides and tosylates in the typical S and 2 reactions to introduce alkyl groups into the alpha position. All right. So this is another alternative to making and using LDA, due to deprotonated ketones. So we'll see how we do that. Coming up here. So this is what that looks like. Nucleophilic carbon. You get a methyl incorporated into the active methylene position. Pretty straightforward. It's just a typical enolate reacting with an alkyl halide. This process can be repeated by adding additional base and methyl bromide. So we can do the same thing again. Methoxide comes in and deprotonates, removes that second alpha hydrogen. So we've formed the enolate again. Now we can add a second equivalent of methyl bromide. Do another nucleophilic substitution. Repeat twice. All right so pretty straightforward. This is an S and 2 reactions, best with methylprimary and primary allylic and primary benzylic substrates. By that I mean I'm referring to the alkyl halides. Primary substrates E2 competes. So this is standard S and 2 considerations. E2 competes but these are not as basic as unstabilized enolates. So when I say unstabilized enolates I'm really kind of saying single-least stabilized enolates versus doubly stabilized enolates. So for example if we make the enolate of a ketone pKa of the conjugate acid is 20. If I make the enolate of an ester pKa of the conjugate acid is 25. I'm referring to these as unstabilized. They are stabilized, they're just less stabilized than active methylene enolates. And if you compare that with something that I didn't leave room for so we'll put it over here and that would be the doubly stabilized. So this would be doubly stabilized and so that would be pKa of the conjugate acid is 11. So you see the big difference in pKa. So like I said it should be doubly stabilized versus singly stabilized but these guys here are referred to as unstabilized enolates even though they have resonance stabilization. And this is considered to be a stabilized enolate. Alright so less elimination than we get if we make the enolate using LDA of either the carboxylic either the ester or the ketone. We're going to get a lot more elimination with a secondary substrate because these are stronger bases. This is less basic and so a lot less elimination. So just something to think about. And what we're going to do is we're going to use this as an alternative to using LDA. But first we need to talk about decarboxylation before we do that. And that's on the next page. Any questions on this? Stop me right now. Up at the top. Excuse me? Oh did I leave off tertiary? Tertiary no reaction at all right? Did I not write that? E2 only. With the exclamation point. Yeah thank you. Alright let's look at decarboxylation. Alright loss of carbon dioxide from a carboxylic acid is called decarboxylation. I'm going to circle the atoms that are missing. This does not happen very readily by the way. Just in certain cases CO2 gas which I'm putting an arrow up meaning that just bubbles away. That's a decarboxylation. Certain carboxylic acids readily decarboxylate. The first type is beta keto acids. This is a beta keto acid. So see what that name comes from. We've got a carboxylic acid. Here's our carboxylic acid. Into the beta position we have a ketone. So here's alpha, here's beta. So beta keto acid those guys readily decarboxylate. So this is acetoacetic acid. In the presence of acid this happens even at room temperature. You get a ketone. So in this case it would be acetone plus CO2. Plus CO2 gas which bubbles off. Alright how does that happen? Let's draw the mechanism. We're going to orient the acetoacetic esters so that we're forming an intramolecular hydrogen bond here between the carbonyl of the ketone with the hydroxyl of the carboxylic acid. And what we're going to do is we're going to push electrons around in a circle. And so you can go clockwise, counterclockwise. You want to start with the double bond of this carbonyl and that's going to move over here to make a new bond. So we're going to move electrons around in a circle kind of like Diels Alder. That part of the reaction is a concerted reaction. And then I'm going to attempt to leave things where they are so you can see what's happening here. So I'll draw the products and then we'll take a look at how that happens. If you follow the arrows here we have carbonyl. I mean we have the carbonyl here, these arrows are going to go over here to make a new bond between oxygen and hydrogen. That's right here. This was a double bond. Now it's a single bond because we need to move those two electrons over so that's a single bond. This arrow here says we're breaking this bond here, the sigma bond, and we're moving over here to make a double bond so that's the double bond here. And then we're taking these electrons from the oxygen-hydrogen's breaking and we're making a double bond here so that's where we get the CO2. So what is the name of the carb, this product right here? Enol right? Not stable. It's going to tautomerize. So this is going to tautomerize to a ketone. Let's do some arrow pushing for that. We've looked at that already but let's do it again just for fun. Two ways you can do this. You can use electrons on oxygen to come down and grab the acidic hydrogen like this. Boom. This comes in to grab this hydrogen. We break the hydrogen-oxygen bond. That gives us a protonated ketone. And then water comes in and deprotonates. This is catalyzed by acid so this is acid catalyzed to tautomerization. And then that gives you the ketone reversible reaction. Or let me do this in a different color. If you prefer to do it the way we did it back in 51B or a 51B way to do the tautomerization. We had an enol as a result of the hydration of an alkyne. This is back in chapter 10. If you want to take another look at this this should jog your memory. What we did then is we H3O plus. What we did was we were in the hydration of an alkyne mode right here in that chapter. We did this sort of thing. Then we put the positive charge on the site that would best stabilize the positive charge. Mark Hanukov, right? This wouldn't be Mark Hanukov. This is the most stable carbocation. And then we drew these. The electrons came down. We drew the resonance structure. So some students like to do this extra step here and that is perfectly fine. And then you deprotonate. And that will match exactly what we did back in chapter 10. So your choice. Questions on the tautomerization? Anybody? All right. So we're going to an acetyl acetic acid to a ketone. Melonic acids also readily decarboxylate. You do need a little heat. This from an acetyl acetic acid goes at room temperature. This one you need to heat. So this is more of the dicarboxylic acid. That will give you a carboxylic acid, acetic acid derivative, and plus CO2. The mechanism will look very much the same except you'll have a hydroxyl on one side rather than a methyl. So what we're going to do is convert this. We're going to use this to actually as an alternative to using LDA to make enolate ions. Because we know that reaction doesn't work very well. So here's what we're going to be doing. We're going to be taking acetyl acetic ester and through a series of steps which I'm going to show you right now, we're going to be going to substituted ketone derivatives. We can have two R groups or one. Substituted aceton derivatives. So if we go to the second product with the two R groups, you can see that this is going to give us an alternative for the thermodynamic enolate. So the previous reaction making the enamine gave us an alternative for the connecting enolate. If we introduce two R groups, this gives us an alternative for the thermodynamic enolate. Both of those reactions using LDA are not very clean. So how we're going to do this, we're going to alkylate the acetyl acetic ester. We're going to hydrolyze the ester so we get the carboxylic acid and then we're going to decarboxylate like we just did on the previous page. So let's see how that works. Sodium methoxide. Ethyl bromide. Do we want to use sodium hydroxide for that reaction? You want to use sodium methoxide. Why? What would happen if you use sodium hydroxide? It's going to suponify that ester. This hydroxide is going to attack that ester. You're going to kick off the oxide and you're going to deprotonate. So you're going to get a carboxylate here in the end of reaction. You're not going to get any alkylation here. So you have to very carefully, what you want to do is you want to choose your base to match the ester. If this, what would happen if I use sodium methoxide instead of sodium ethoxide? What would happen? I haven't heard it yet. I haven't heard the magic word. It's going to attack this carbonate. So in other words, if you use some ethoxide here, it's going to attack this carbonyl. Boom. We're going to kick electrons up onto oxygen. Electrons on the oxygen are going to come down. And some of the time it's going to kick off ethoxide. So you're going to start getting a mixture of ethyl and methyl esters here. So I deliberately chose sodium ethoxide here for this reaction. So what we're going to do is we're going to essentially completely deprotonate. Let's go through the process here. Here's ethoxide. We're going to remove the most acidic hydrogen, which is the one here. I'm going to leave electrons on carbon rather than pushing them onto the oxygen. So after I do that, I'm going to get an enolate. We'll leave the other hydrogen off. Then I add ethyl bromide. SN2 reaction. I guess you draw that out. If I'm going to draw arrow pushing, I need to draw that ethyl group out. We want to attack the carbon that's bonded to the leaving group. So we're going to attack the carbon bonded to the leaving group, kick off the leaving group. Product of that step has an ethyl incorporated into the active methylene position. Let's do it again. That was so much fun. We're going to do it again. We can use ethyl. We could use, do we want to change it up here? Let's use methyl this time. And methyl iodide. I'm feeling methyl iodide right now. So we're going to deprotonate again. I'm not going to draw the arrows for that second one, but we're going to deprotonate again. And then we add methyl iodide. So now we put a methyl here. Just like that. So step one is alkylate. We ended up actually doing it in two steps, but the first part is to alkylate. The strategy here, step one, alkylate. We can alkylate once. We can alkylate twice. Whatever you need to synthesize, you're going to decide. Step two is to hydrolyze the ester. We know how to hydrolyze an ester. We can do that acid catalyze or base catalyze. When you look at this reaction in the literature, they mostly do base catalyze using KOH and water. Probably means that it works better than acid catalyze. Do you have any experience with that? Either one of you guys? No, it's KOH. What? It's more straightforward. It's more straightforward, but if you did it all with acid, you could do it all in one step, which I'll show you coming up. All right, so that's going to hydrolyze the ester. That's chapter 22, right? You know how to do that. You get a carboxylate, sub-ponification. Now you add H3O plus in a second step. Ethyl's still there, methyl's still there. Now we have a carboxylic acid. Now that carboxylic acid can readily decarboxylate. Step three is decarboxylate. So we just did that mechanism for that. We're going to decarboxylate. We're going to form an enol, and the enol is going to tautomerize to a ketone. This goes readily at room temperature, so we don't even need to heat it, really. And so that whole entire carboxylic acid is coming off, and it is replaced with a hydrogen right there. So what we've essentially done is we've done the same thing as if we started with a ketone. We used LDA. We deprotonated, and we're alkylating on the most substituted sites on making the thermodynamic enolate. Only this is, although it's more steps, it's much cleaner. Okay, questions on that strategy? Anybody? So the whole entire thing is called an acetoacetic estrous synthesis. And basically what it can be used for is making substituted acetone derivative. So here's retrosynthetic analysis. If I give you a compound and I say, can you make this compound from an acetoacetic estrous synthesis? So this is what you would do. Look for acetone and circle acetone. So that's our, I'm just calling this our acetone unit here. And the two R groups are, we're going to put the two R groups on from the acetoacetic ester. So basically this hydrogen right here comes from the decarboxylation. So I'm going to cross out that hydrogen and I'm going to put a temporary ester group here. So now can you see where the acetoacetic ester is? We have right here, that's the acetone. If that hydrogen is off, that's now our ester right here. So these two R groups are going to come from alkylating acetoacetic, acetoacetic ester. So here we replace hydrogen with a temporary ester group. And so these two R groups are going to come from that and that are going to come from RBRRI, your choice, this is going to be R'BR. Could be a chloride, it could be an iodide, it could be a tosylate. Alright so to make this compound we're going to start with RBR, R'BR and acetoacetic ester. And I will say using acetoacetic ester to synthesis to make this compound. Questions on how I took that apart, we're going to write the synthesis on the next page. That's just how I took that apart. So like I said, the exam would say show how you can make this using an acetoacetic ester synthesis. So that means we're going to start with an acetoacetic ester. It could be OET, it could be O-methyl, OET is probably more common. Step one, I'm going to deprotonate NaOET. I'm going to add one of the two R, let's do RBR first. And then I'm going to deprotonate again, you want to do this stepwise. And then R'BR. That does the alkylation. Do not write excess sodium ethoxide, RBR and R'BR. Then you have no control over what happens. You're going to have some of them with two R's, you're going to have some of them with two R''s, some of them with R and R' different, have fun isolating the correct compound from that mixture. Not very easy. Okay, so we've alkylated twice, so I need to actually draw on the alkyl groups here. You were wondering if I was going to do that. Okay, we've alkylated twice. Now we decarboxylate. KOH, H2O, heat. And then H3O plus H2O. We don't need heat, but you can. Or if you want to make it a little shorter, we can hydrolyze an ester in acid, can't we? We know acid catalyzed hydrolysis of an ester and we know base catalyzed. Or we can do this as one step. We can do H3O plus H2O and heat. And this will hydrolyze the ester, then decarboxylate. So as soon as you hydrolyze that ester, it will decarboxylate very readily at room temperature. You are heating it up to get the ester to hydrolyze. Yes? Oh, could I say please? Is the second step and the first reaction on the top, the one or is the H2O plus H2O such a step? Instead of a third step, there was an invisible second step here that I didn't tell you about. Yes, that should be a two. Thank you. Thank you so much. All right, okay, more questions on that. You think you could do that on the term two? It's not that bad. It really isn't. Trust me. All right, so we're going to talk about Melonic Ester Synthesis next, which is the same idea except we start with Melonic and Ammonic Ester, so ester ester rather than ester ketone. It's the same idea, so guaranteed on midterm two, you will have to do either an acetoacetic ester synthesis or a Melonic ester synthesis. They're both the same. I'll find some fun, cool thing that you can synthesize and then you can use that. All right, so this is the same idea. So if you get the previous one, you'll get this also. So this is going to give us substituted acetic acids. So here's our acetic acid unit here. If we alkylate the active methylene compound once, we get the R group here. If we alkylate twice, we get two R groups in the alpha position. And this is pretty cool because we really can't deprotonate that alpha position of a carboxylic acid, right? Because if we add base, we deprotonate. That gives us the carboxylate and then we're stuck. So you will not find a pKa for the alpha hydrogen of a carboxylic acid. That's because if you add base, the most acidic hydrogen is going to be removed. So base removes proton on oxygen, not carbon, not on the alpha position. All right, so we could alkylate the corresponding ester and saponify. So we could do LDA and then we could do RX. And then we can alkylate there that way. So now we have RCH2 and then our ester. And then we can hydrolyze the ester, KOH, H2O, heat. Definitely need to heat that. And then step two, H3O plus in heat. Here again, we can bypass that and do acid-catalyzed hydrolysis for one step. Now I guess we don't need heat there because we're not decarboxylating, so let me fix that. We just need H3O plus. Or we can just do acid-catalyzed hydrolysis of the ester. H3O plus H2O and heat. And that will give us the carboxylic acid. All right, so again, we can hydrolyze ester's acid-catalyzed or base-catalyzed. So this is base-catalyzed and this is acid-catalyzed. That's an alternative, but it doesn't work as well for a couple different reasons. It's more expensive. You have to use inert atmospheric techniques. You have to use dry solvents, dry glassware, argon or nitrogen atmosphere. This is better for small-scale reactions. And then also the enolates are stronger base than a stabilized enolate, so you get more E2. So this is really not as good of an alternative. All right, so conceptually we have the same idea here. We alkylate diethylmalony. We hydrolyze both esters and then we decarboxylate. So let's do an example here. All right, so step one, we alkylate. Ethylenacid methylene compound has a little bit higher pKa than the acetoacetic ester, but still works really well. And we're going to get one methyl incorporated. Let's just alkylate once this time. You don't have to alkylate twice. You can just alkylate once. And then we have to hydrolyze both esters. And then in step two, we have to alkylate. Looks like that. And then in step three, we protonate the dicarboxylate. Then we decarboxylate. Only one of the carboxylic acids will come off. So step three, we want H3O plus H2O and heat. And when we do that, we get a substituted acetic acid derivative. Or if we want to save ourselves a little time here, we can do acid catalyzed hydrolysis, H3O plus H2O heat. And so we're hydrolyzing the esters in acid. We get the dicarboxylic acid. As soon as we get the dicarboxylic acid, it decarboxylates. So this step here is acid catalyzed ester hydrolysis. Then decarboxylation. All right, so as you can see, the same idea. We could alkylate twice if we wanted to. You will definitely be asked to do one of the two of those on the test, if not both. Probably not both, just because we have so much to cover on this exam. Retrosynthetic analysis, we want to be able to, if I give you a substituted acetic acid derivative, you want to be able to take it apart and see how to make it. So let's do that right now. All right. Here is our acetic acid right here. Circle that. So we're going to replace the hydrogen with a temporary ester. And then this is also going to become an ester, right? So we'll put the ET there also. Because we need a diester to start with. These two groups right here, this is going to be from RBR and RprimeBR. This is our acetic acid unit. So we replaced the hydrogen with a temporary ester. And then we converted this acid to an ester. All right. So what would we start with? We would start with diethyl malenate right here. RBR and RprimeBR. And if we write out the synthesis in the forward direction, then it's going to look just like the last one. Except we just have a different starting material. So let's show how we would use that. Let's outline a malenic ester synthesis of the following carboxylic acid. All right. So circle acetic acid. Try not to be thrown off by the ring. This is one, our alkyl halide, and this is the other one. We've alkylated twice. Let's put on our temporary ester group. There's a temporary ester group. And then we're going to change this to an ester also. All right. So this is going to be from diethyl malenate. And let's make sure we count carbons here. We want 1, 2, 3, 4, 5. 1, 2, 3, 4, 5. That's going to be our alkylating agent. That's our alkyl halide. That's how we're going to make a ring, because we've got the leaving groups on both sides. So let's write out the synthesis in the forward direction. We've got one minute to do this. Start with diethyl malenate. Step one, NaOET. I've chosen my base to match my ester. Step two, I add this alkyl halide here. I want to do this stepwise. If I use two equivalents of sodium ethoxide, and then I add this, then we're going to have competition. Some of the ethoxides are going to attack the alkyl halide, and that's going to take away our starting material. So you definitely want to do this stepwise. That gives us this compound here. And let's do the one step thing here. H3O plus H2O heat. That gives us the substituted acetic acid compound. And that is the end of chapter 23. We'll stop right there, and we will start chapter 24 next time.
 We left off last time talking about alternatives to using LDA to deprotonate a ketone followed  by adding an alkyl halide because we have problems with thermodynamic kinetic enolate.  We want some better alternatives that aren't going to give us mixtures.  If you get products from both kinetic and thermodynamic enolates, it's really, really  hard to separate the two by chromatography or distillation or recrystallization.  They're very hard to separate.  We want to be able to do cleanly just form one or the other.  An alternative for thermodynamic enolate is to make an enamine.  Yeah, it's not showing.  Thank you.  All right.  So an alternative is to make an enamine.  So let's label that.  This is an alternative to making a kinetic enolate using LDA.  So LDA gives you a lot of the kinetic product.  It's very selective.  The product that you get, remember, it was like 45, 40 something percent, not very good.  So what you do is you make an enamine.  And so here we saw an example here.  You make an enamine and then you treat it with the alkyl halide.  And what's going to happen is the electrons on nitrogen come down, boom, you attack the  carbon that's bonded to the leaving group and you kick off the leaving group.  So notice the ethyl that we just incorporated is on the least substituted side.  So this is an alternative to making a kinetic enolate using LDA.  All right.  So why is it so selective for the least substituted side?  And it turns out that all of these bonds here that you see here, I'm going to be making  those in red, are all in the same plane.  All those bonds are in the same plane.  And so you can see that it's much better, including this hydrogen right here and that  bond right there.  Stereically it wants to be on the least substituted side.  If it was over on this side, there's a problem with these two hydrogens right here, getting  in each other's way.  So if there's really any substitution on that one side, then it's going to be a problem  making the enamine.  And that's why it's so selective.  Okay.  So it's not hysterically not good.  So this is the one that you form.  All right.  So here's another example of using this.  We can take, we can make the enamine.  So again, this is the most common enamine to use.  It's not the only one.  You could just use dimethylamine, diethylamine, so there's not a problem here.  So you can see H3O plus catalyst.  Or I might write pH 5.0 to show you that you're going to make it the enamine.  That's the best conditions for making an enamine or an emine.  All right.  So we are going to selectively form this on the least substituted side.  The double bond will be here.  And we can do things like we can attack acid chlorides just like this, just like we've  been used to seeing.  Let's see what you get from that.  Tetrahedral intermediate that has a leaving group.  So acyl substitution, eminium ion here, positively charged nitrogen.  Electrons on oxygen come down, kick off chloride.  So we've seen that so many times.  That's the reactivity that we're used to with carbonyl compounds.  I'm going to have to move that arrow a little bit here.  So then HCl, H2O, we'll convert that eminium ion to a ketone.  So notice you turn that into an active methylene compound.  We've just synthesized an active methylene compound.  All right.  Questions?  Anybody?  Okay.  A couple more reactions that we want to talk about.  The first one, alkylation of acetoacetic ester.  This is an acetoacetic ester.  So the aceto for the acetyl group, yeah?  But the mechanism for hydrolysis of the emine to the ketone?  I'm missing a methyl group.  Let's throw that on there.  Now I'm going to hear everybody turning the page.  And the one above it?  Also, yes, I am.  All right.  So the aceto is for the acetyl group, the ketone group, and the acetic ester is for  the ester portion.  That has a pKa of about 11.  So I'm completely deprotonated using a thoxide ion.  We went through that equilibrium earlier.  I don't remember what page that was, but we went through that equilibrium earlier.  pKa17 here.  So you can see that we get essentially complete conversion.  At equilibrium, we have one acetoacetic ester that has not been deprotonated to 10 to the  six that has been deprotonated.  We call that essentially to completion.  It's always going back and forth, but it is essentially to completion.  So it turns out the enolates of active methylene compounds are nucleophilic and react with  alkyl halides and tosylates in the typical S and 2 reactions to introduce alkyl groups  into the alpha position.  All right.  So this is another alternative to making and using LDA, due to deprotonated ketones.  So we'll see how we do that.  Coming up here.  So this is what that looks like.  Nucleophilic carbon.  You get a methyl incorporated into the active methylene position.  Pretty straightforward.  It's just a typical enolate reacting with an alkyl halide.  This process can be repeated by adding additional base and methyl bromide.  So we can do the same thing again.  Methoxide comes in and deprotonates, removes that second alpha hydrogen.  So we've formed the enolate again.  Now we can add a second equivalent of methyl bromide.  Do another nucleophilic substitution.  Repeat twice.  All right so pretty straightforward.  This is an S and 2 reactions, best with methylprimary and primary allylic and primary benzylic  substrates.  By that I mean I'm referring to the alkyl halides.  Primary substrates E2 competes.  So this is standard S and 2 considerations.  E2 competes but these are not as basic as unstabilized enolates.  So when I say unstabilized enolates I'm really kind of saying single-least stabilized enolates  versus doubly stabilized enolates.  So for example if we make the enolate of a ketone pKa of the conjugate acid is 20.  If I make the enolate of an ester pKa of the conjugate acid is 25.  I'm referring to these as unstabilized.  They are stabilized, they're just less stabilized than active methylene enolates.  And if you compare that with something that I didn't leave room for so we'll put it over  here and that would be the doubly stabilized.  So this would be doubly stabilized and so that would be pKa of the conjugate acid is  11.  So you see the big difference in pKa.  So like I said it should be doubly stabilized versus singly stabilized but these guys here  are referred to as unstabilized enolates even though they have resonance stabilization.  And this is considered to be a stabilized enolate.  Alright so less elimination than we get if we make the enolate using LDA of either the  carboxylic either the ester or the ketone.  We're going to get a lot more elimination with a secondary substrate because these are  stronger bases.  This is less basic and so a lot less elimination.  So just something to think about.  And what we're going to do is we're going to use this as an alternative to using LDA.  But first we need to talk about decarboxylation before we do that.  And that's on the next page.  Any questions on this?  Stop me right now.  Up at the top.  Excuse me?  Oh did I leave off tertiary?  Tertiary no reaction at all right?  Did I not write that?  E2 only.  With the exclamation point.  Yeah thank you.  Alright let's look at decarboxylation.  Alright loss of carbon dioxide from a carboxylic acid is called decarboxylation.  I'm going to circle the atoms that are missing.  This does not happen very readily by the way.  Just in certain cases CO2 gas which I'm putting an arrow up meaning that just bubbles away.  That's a decarboxylation.  Certain carboxylic acids readily decarboxylate.  The first type is beta keto acids.  This is a beta keto acid.  So see what that name comes from.  We've got a carboxylic acid.  Here's our carboxylic acid.  Into the beta position we have a ketone.  So here's alpha, here's beta.  So beta keto acid those guys readily decarboxylate.  So this is acetoacetic acid.  In the presence of acid this happens even at room temperature.  You get a ketone.  So in this case it would be acetone plus CO2.  Plus CO2 gas which bubbles off.  Alright how does that happen?  Let's draw the mechanism.  We're going to orient the acetoacetic esters so that we're forming an intramolecular hydrogen  bond here between the carbonyl of the ketone with the hydroxyl of the carboxylic acid.  And what we're going to do is we're going to push electrons around in a circle.  And so you can go clockwise, counterclockwise.  You want to start with the double bond of this carbonyl and that's going to move over  here to make a new bond.  So we're going to move electrons around in a circle kind of like Diels Alder.  That part of the reaction is a concerted reaction.  And then I'm going to attempt to leave things where they are so you can see what's happening  here.  So I'll draw the products and then we'll take a look at how that happens.  If you follow the arrows here we have carbonyl.  I mean we have the carbonyl here, these arrows are going to go over here to make a new bond  between oxygen and hydrogen.  That's right here.  This was a double bond.  Now it's a single bond because we need to move those two electrons over so that's a  single bond.  This arrow here says we're breaking this bond here, the sigma bond, and we're moving over  here to make a double bond so that's the double bond here.  And then we're taking these electrons from the oxygen-hydrogen's breaking and we're making  a double bond here so that's where we get the CO2.  So what is the name of the carb, this product right here?  Enol right?  Not stable.  It's going to tautomerize.  So this is going to tautomerize to a ketone.  Let's do some arrow pushing for that.  We've looked at that already but let's do it again just for fun.  Two ways you can do this.  You can use electrons on oxygen to come down and grab the acidic hydrogen like this.  Boom.  This comes in to grab this hydrogen.  We break the hydrogen-oxygen bond.  That gives us a protonated ketone.  And then water comes in and deprotonates.  This is catalyzed by acid so this is acid catalyzed to tautomerization.  And then that gives you the ketone reversible reaction.  Or let me do this in a different color.  If you prefer to do it the way we did it back in 51B or a 51B way to do the tautomerization.  We had an enol as a result of the hydration of an alkyne.  This is back in chapter 10.  If you want to take another look at this this should jog your memory.  What we did then is we H3O plus.  What we did was we were in the hydration of an alkyne mode right here in that chapter.  We did this sort of thing.  Then we put the positive charge on the site that would best stabilize the positive charge.  Mark Hanukov, right?  This wouldn't be Mark Hanukov.  This is the most stable carbocation.  And then we drew these.  The electrons came down.  We drew the resonance structure.  So some students like to do this extra step here and that is perfectly fine.  And then you deprotonate.  And that will match exactly what we did back in chapter 10.  So your choice.  Questions on the tautomerization?  Anybody?  All right.  So we're going to an acetyl acetic acid to a ketone.  Melonic acids also readily decarboxylate.  You do need a little heat.  This from an acetyl acetic acid goes at room temperature.  This one you need to heat.  So this is more of the dicarboxylic acid.  That will give you a carboxylic acid, acetic acid derivative, and plus CO2.  The mechanism will look very much the same except you'll have a hydroxyl on one side  rather than a methyl.  So what we're going to do is convert this.  We're going to use this to actually as an alternative to using LDA to make enolate ions.  Because we know that reaction doesn't work very well.  So here's what we're going to be doing.  We're going to be taking acetyl acetic ester and through a series of steps which I'm going  to show you right now, we're going to be going to substituted ketone derivatives.  We can have two R groups or one.  Substituted aceton derivatives.  So if we go to the second product with the two R groups, you can see that this is going  to give us an alternative for the thermodynamic enolate.  So the previous reaction making the enamine gave us an alternative for the connecting  enolate.  If we introduce two R groups, this gives us an alternative for the thermodynamic enolate.  Both of those reactions using LDA are not very clean.  So how we're going to do this, we're going to alkylate the acetyl acetic ester.  We're going to hydrolyze the ester so we get the carboxylic acid and then we're going  to decarboxylate like we just did on the previous page.  So let's see how that works.  Sodium methoxide.  Ethyl bromide.  Do we want to use sodium hydroxide for that reaction?  You want to use sodium methoxide.  Why?  What would happen if you use sodium hydroxide?  It's going to suponify that ester.  This hydroxide is going to attack that ester.  You're going to kick off the oxide and you're going to deprotonate.  So you're going to get a carboxylate here in the end of reaction.  You're not going to get any alkylation here.  So you have to very carefully, what you want to do is you want to choose your base to match  the ester.  If this, what would happen if I use sodium methoxide instead of sodium ethoxide?  What would happen?  I haven't heard it yet.  I haven't heard the magic word.  It's going to attack this carbonate.  So in other words, if you use some ethoxide here, it's going to attack this carbonyl.  Boom.  We're going to kick electrons up onto oxygen.  Electrons on the oxygen are going to come down.  And some of the time it's going to kick off ethoxide.  So you're going to start getting a mixture of ethyl and methyl esters here.  So I deliberately chose sodium ethoxide here for this reaction.  So what we're going to do is we're going to essentially completely deprotonate.  Let's go through the process here.  Here's ethoxide.  We're going to remove the most acidic hydrogen, which is the one here.  I'm going to leave electrons on carbon rather than pushing them onto the oxygen.  So after I do that, I'm going to get an enolate.  We'll leave the other hydrogen off.  Then I add ethyl bromide.  SN2 reaction.  I guess you draw that out.  If I'm going to draw arrow pushing, I need to draw that ethyl group out.  We want to attack the carbon that's bonded to the leaving group.  So we're going to attack the carbon bonded to the leaving group, kick off the leaving  group.  Product of that step has an ethyl incorporated into the active methylene position.  Let's do it again.  That was so much fun.  We're going to do it again.  We can use ethyl.  We could use, do we want to change it up here?  Let's use methyl this time.  And methyl iodide.  I'm feeling methyl iodide right now.  So we're going to deprotonate again.  I'm not going to draw the arrows for that second one, but we're going to deprotonate  again.  And then we add methyl iodide.  So now we put a methyl here.  Just like that.  So step one is alkylate.  We ended up actually doing it in two steps, but the first part is to alkylate.  The strategy here, step one, alkylate.  We can alkylate once.  We can alkylate twice.  Whatever you need to synthesize, you're going to decide.  Step two is to hydrolyze the ester.  We know how to hydrolyze an ester.  We can do that acid catalyze or base catalyze.  When you look at this reaction in the literature, they mostly do base catalyze using KOH and  water.  Probably means that it works better than acid catalyze.  Do you have any experience with that?  Either one of you guys?  No, it's KOH.  What?  It's more straightforward.  It's more straightforward, but if you did it all with acid, you could do it all in one  step, which I'll show you coming up.  All right, so that's going to hydrolyze the ester.  That's chapter 22, right?  You know how to do that.  You get a carboxylate, sub-ponification.  Now you add H3O plus in a second step.  Ethyl's still there, methyl's still there.  Now we have a carboxylic acid.  Now that carboxylic acid can readily decarboxylate.  Step three is decarboxylate.  So we just did that mechanism for that.  We're going to decarboxylate.  We're going to form an enol, and the enol is going to tautomerize to a ketone.  This goes readily at room temperature, so we don't even need to heat it, really.  And so that whole entire carboxylic acid is coming off, and it is replaced with a hydrogen  right there.  So what we've essentially done is we've done the same thing as if we started with a ketone.  We used LDA.  We deprotonated, and we're alkylating on the most substituted sites on making the thermodynamic  enolate.  Only this is, although it's more steps, it's much cleaner.  Okay, questions on that strategy?  Anybody?  So the whole entire thing is called an acetoacetic estrous synthesis.  And basically what it can be used for is making substituted acetone derivative.  So here's retrosynthetic analysis.  If I give you a compound and I say, can you make this compound from an acetoacetic estrous  synthesis?  So this is what you would do.  Look for acetone and circle acetone.  So that's our, I'm just calling this our acetone unit here.  And the two R groups are, we're going to put the two R groups on from the acetoacetic  ester.  So basically this hydrogen right here comes from the decarboxylation.  So I'm going to cross out that hydrogen and I'm going to put a temporary ester group here.  So now can you see where the acetoacetic ester is?  We have right here, that's the acetone.  If that hydrogen is off, that's now our ester right here.  So these two R groups are going to come from alkylating acetoacetic, acetoacetic ester.  So here we replace hydrogen with a temporary ester group.  And so these two R groups are going to come from that and that are going to come from  RBRRI, your choice, this is going to be R'BR.  Could be a chloride, it could be an iodide, it could be a tosylate.  Alright so to make this compound we're going to start with RBR, R'BR and acetoacetic ester.  And I will say using acetoacetic ester to synthesis to make this compound.  Questions on how I took that apart, we're going to write the synthesis on the next page.  That's just how I took that apart.  So like I said, the exam would say show how you can make this using an acetoacetic ester  synthesis.  So that means we're going to start with an acetoacetic ester.  It could be OET, it could be O-methyl, OET is probably more common.  Step one, I'm going to deprotonate NaOET.  I'm going to add one of the two R, let's do RBR first.  And then I'm going to deprotonate again, you want to do this stepwise.  And then R'BR.  That does the alkylation.  Do not write excess sodium ethoxide, RBR and R'BR.  Then you have no control over what happens.  You're going to have some of them with two R's, you're going to have some of them with  two R''s, some of them with R and R' different, have fun isolating the correct compound from  that mixture.  Not very easy.  Okay, so we've alkylated twice, so I need to actually draw on the alkyl groups here.  You were wondering if I was going to do that.  Okay, we've alkylated twice.  Now we decarboxylate.  KOH, H2O, heat.  And then H3O plus H2O.  We don't need heat, but you can.  Or if you want to make it a little shorter, we can hydrolyze an ester in acid, can't we?  We know acid catalyzed hydrolysis of an ester and we know base catalyzed.  Or we can do this as one step.  We can do H3O plus H2O and heat.  And this will hydrolyze the ester, then decarboxylate.  So as soon as you hydrolyze that ester, it will decarboxylate very readily at room temperature.  You are heating it up to get the ester to hydrolyze.  Yes?  Oh, could I say please?  Is the second step and the first reaction on the top, the one or is the H2O plus H2O  such a step?  Instead of a third step, there was an invisible second step here that I didn't tell you about.  Yes, that should be a two.  Thank you.  Thank you so much.  All right, okay, more questions on that.  You think you could do that on the term two?  It's not that bad.  It really isn't.  Trust me.  All right, so we're going to talk about Melonic Ester Synthesis next, which is the same idea  except we start with Melonic and Ammonic Ester, so ester ester rather than ester ketone.  It's the same idea, so guaranteed on midterm two, you will have to do either an acetoacetic  ester synthesis or a Melonic ester synthesis.  They're both the same.  I'll find some fun, cool thing that you can synthesize and then you can use that.  All right, so this is the same idea.  So if you get the previous one, you'll get this also.  So this is going to give us substituted acetic acids.  So here's our acetic acid unit here.  If we alkylate the active methylene compound once, we get the R group here.  If we alkylate twice, we get two R groups in the alpha position.  And this is pretty cool because we really can't deprotonate that alpha position of a  carboxylic acid, right?  Because if we add base, we deprotonate.  That gives us the carboxylate and then we're stuck.  So you will not find a pKa for the alpha hydrogen of a carboxylic acid.  That's because if you add base, the most acidic hydrogen is going to be removed.  So base removes proton on oxygen, not carbon, not on the alpha position.  All right, so we could alkylate the corresponding ester and saponify.  So we could do LDA and then we could do RX.  And then we can alkylate there that way.  So now we have RCH2 and then our ester.  And then we can hydrolyze the ester, KOH, H2O, heat.  Definitely need to heat that.  And then step two, H3O plus in heat.  Here again, we can bypass that and do acid-catalyzed hydrolysis for one step.  Now I guess we don't need heat there because we're not decarboxylating, so let me fix that.  We just need H3O plus.  Or we can just do acid-catalyzed hydrolysis of the ester.  H3O plus H2O and heat.  And that will give us the carboxylic acid.  All right, so again, we can hydrolyze ester's acid-catalyzed or base-catalyzed.  So this is base-catalyzed and this is acid-catalyzed.  That's an alternative, but it doesn't work as well for a couple different reasons.  It's more expensive.  You have to use inert atmospheric techniques.  You have to use dry solvents, dry glassware, argon or nitrogen atmosphere.  This is better for small-scale reactions.  And then also the enolates are stronger base than a stabilized enolate, so you get more  E2.  So this is really not as good of an alternative.  All right, so conceptually we have the same idea here.  We alkylate diethylmalony.  We hydrolyze both esters and then we decarboxylate.  So let's do an example here.  All right, so step one, we alkylate.  Ethylenacid methylene compound has a little bit higher pKa than the acetoacetic ester,  but still works really well.  And we're going to get one methyl incorporated.  Let's just alkylate once this time.  You don't have to alkylate twice.  You can just alkylate once.  And then we have to hydrolyze both esters.  And then in step two, we have to alkylate.  Looks like that.  And then in step three, we protonate the dicarboxylate.  Then we decarboxylate.  Only one of the carboxylic acids will come off.  So step three, we want H3O plus H2O and heat.  And when we do that, we get a substituted acetic acid derivative.  Or if we want to save ourselves a little time here, we can do acid catalyzed hydrolysis,  H3O plus H2O heat.  And so we're hydrolyzing the esters in acid.  We get the dicarboxylic acid.  As soon as we get the dicarboxylic acid, it decarboxylates.  So this step here is acid catalyzed ester hydrolysis.  Then decarboxylation.  All right, so as you can see, the same idea.  We could alkylate twice if we wanted to.  You will definitely be asked to do one of the two of those on the test, if not both.  Probably not both, just because we have so much to cover on this exam.  Retrosynthetic analysis, we want to be able to, if I give you a substituted acetic acid  derivative, you want to be able to take it apart and see how to make it.  So let's do that right now.  All right.  Here is our acetic acid right here.  Circle that.  So we're going to replace the hydrogen with a temporary ester.  And then this is also going to become an ester, right?  So we'll put the ET there also.  Because we need a diester to start with.  These two groups right here, this is going to be from RBR and RprimeBR.  This is our acetic acid unit.  So we replaced the hydrogen with a temporary ester.  And then we converted this acid to an ester.  All right.  So what would we start with?  We would start with diethyl malenate right here.  RBR and RprimeBR.  And if we write out the synthesis in the forward direction, then it's going to look just like  the last one.  Except we just have a different starting material.  So let's show how we would use that.  Let's outline a malenic ester synthesis of the following carboxylic acid.  All right.  So circle acetic acid.  Try not to be thrown off by the ring.  This is one, our alkyl halide, and this is the other one.  We've alkylated twice.  Let's put on our temporary ester group.  There's a temporary ester group.  And then we're going to change this to an ester also.  All right.  So this is going to be from diethyl malenate.  And let's make sure we count carbons here.  We want 1, 2, 3, 4, 5.  1, 2, 3, 4, 5.  That's going to be our alkylating agent.  That's our alkyl halide.  That's how we're going to make a ring, because we've got the leaving groups on both sides.  So let's write out the synthesis in the forward direction.  We've got one minute to do this.  Start with diethyl malenate.  Step one, NaOET.  I've chosen my base to match my ester.  Step two, I add this alkyl halide here.  I want to do this stepwise.  If I use two equivalents of sodium ethoxide, and then I add this, then we're going to  have competition.  Some of the ethoxides are going to attack the alkyl halide, and that's going to take  away our starting material.  So you definitely want to do this stepwise.  That gives us this compound here.  And let's do the one step thing here.  H3O plus H2O heat.  That gives us the substituted acetic acid compound.  And that is the end of chapter 23.  We'll stop right there, and we will start chapter 24 next time.
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We left off last time talking about alternatives to using LDA to deprotonate a ketone followed by adding an alkyl halide because we have problems with thermodynamic kinetic enolate. We want some better alternatives that aren't going to give us mixtures. If you get products from both kinetic and thermodynamic enolates, it's really, really hard to separate the two by chromatography or distillation or recrystallization. They're very hard to separate. We want to be able to do cleanly just form one or the other. An alternative for thermodynamic enolate is to make an enamine. Yeah, it's not showing. Thank you. All right. So an alternative is to make an enamine. So let's label that. This is an alternative to making a kinetic enolate using LDA. So LDA gives you a lot of the kinetic product. It's very selective. The product that you get, remember, it was like 45, 40 something percent, not very good. So what you do is you make an enamine. And so here we saw an example here. You make an enamine and then you treat it with the alkyl halide. And what's going to happen is the electrons on nitrogen come down, boom, you attack the carbon that's bonded to the leaving group and you kick off the leaving group. So notice the ethyl that we just incorporated is on the least substituted side. So this is an alternative to making a kinetic enolate using LDA. All right. So why is it so selective for the least substituted side? And it turns out that all of these bonds here that you see here, I'm going to be making those in red, are all in the same plane. All those bonds are in the same plane. And so you can see that it's much better, including this hydrogen right here and that bond right there. Stereically it wants to be on the least substituted side. If it was over on this side, there's a problem with these two hydrogens right here, getting in each other's way. So if there's really any substitution on that one side, then it's going to be a problem making the enamine. And that's why it's so selective. Okay. So it's not hysterically not good. So this is the one that you form. All right. So here's another example of using this. We can take, we can make the enamine. So again, this is the most common enamine to use. It's not the only one. You could just use dimethylamine, diethylamine, so there's not a problem here. So you can see H3O plus catalyst. Or I might write pH 5.0 to show you that you're going to make it the enamine. That's the best conditions for making an enamine or an emine. All right. So we are going to selectively form this on the least substituted side. The double bond will be here. And we can do things like we can attack acid chlorides just like this, just like we've been used to seeing. Let's see what you get from that. Tetrahedral intermediate that has a leaving group. So acyl substitution, eminium ion here, positively charged nitrogen. Electrons on oxygen come down, kick off chloride. So we've seen that so many times. That's the reactivity that we're used to with carbonyl compounds. I'm going to have to move that arrow a little bit here. So then HCl, H2O, we'll convert that eminium ion to a ketone. So notice you turn that into an active methylene compound. We've just synthesized an active methylene compound. All right. Questions? Anybody? Okay. A couple more reactions that we want to talk about. The first one, alkylation of acetoacetic ester. This is an acetoacetic ester. So the aceto for the acetyl group, yeah? But the mechanism for hydrolysis of the emine to the ketone? I'm missing a methyl group. Let's throw that on there. Now I'm going to hear everybody turning the page. And the one above it? Also, yes, I am. All right. So the aceto is for the acetyl group, the ketone group, and the acetic ester is for the ester portion. That has a pKa of about 11. So I'm completely deprotonated using a thoxide ion. We went through that equilibrium earlier. I don't remember what page that was, but we went through that equilibrium earlier. pKa17 here. So you can see that we get essentially complete conversion. At equilibrium, we have one acetoacetic ester that has not been deprotonated to 10 to the six that has been deprotonated. We call that essentially to completion. It's always going back and forth, but it is essentially to completion. So it turns out the enolates of active methylene compounds are nucleophilic and react with alkyl halides and tosylates in the typical S and 2 reactions to introduce alkyl groups into the alpha position. All right. So this is another alternative to making and using LDA, due to deprotonated ketones. So we'll see how we do that. Coming up here. So this is what that looks like. Nucleophilic carbon. You get a methyl incorporated into the active methylene position. Pretty straightforward. It's just a typical enolate reacting with an alkyl halide. This process can be repeated by adding additional base and methyl bromide. So we can do the same thing again. Methoxide comes in and deprotonates, removes that second alpha hydrogen. So we've formed the enolate again. Now we can add a second equivalent of methyl bromide. Do another nucleophilic substitution. Repeat twice. All right so pretty straightforward. This is an S and 2 reactions, best with methylprimary and primary allylic and primary benzylic substrates. By that I mean I'm referring to the alkyl halides. Primary substrates E2 competes. So this is standard S and 2 considerations. E2 competes but these are not as basic as unstabilized enolates. So when I say unstabilized enolates I'm really kind of saying single-least stabilized enolates versus doubly stabilized enolates. So for example if we make the enolate of a ketone pKa of the conjugate acid is 20. If I make the enolate of an ester pKa of the conjugate acid is 25. I'm referring to these as unstabilized. They are stabilized, they're just less stabilized than active methylene enolates. And if you compare that with something that I didn't leave room for so we'll put it over here and that would be the doubly stabilized. So this would be doubly stabilized and so that would be pKa of the conjugate acid is 11. So you see the big difference in pKa. So like I said it should be doubly stabilized versus singly stabilized but these guys here are referred to as unstabilized enolates even though they have resonance stabilization. And this is considered to be a stabilized enolate. Alright so less elimination than we get if we make the enolate using LDA of either the carboxylic either the ester or the ketone. We're going to get a lot more elimination with a secondary substrate because these are stronger bases. This is less basic and so a lot less elimination. So just something to think about. And what we're going to do is we're going to use this as an alternative to using LDA. But first we need to talk about decarboxylation before we do that. And that's on the next page. Any questions on this? Stop me right now. Up at the top. Excuse me? Oh did I leave off tertiary? Tertiary no reaction at all right? Did I not write that? E2 only. With the exclamation point. Yeah thank you. Alright let's look at decarboxylation. Alright loss of carbon dioxide from a carboxylic acid is called decarboxylation. I'm going to circle the atoms that are missing. This does not happen very readily by the way. Just in certain cases CO2 gas which I'm putting an arrow up meaning that just bubbles away. That's a decarboxylation. Certain carboxylic acids readily decarboxylate. The first type is beta keto acids. This is a beta keto acid. So see what that name comes from. We've got a carboxylic acid. Here's our carboxylic acid. Into the beta position we have a ketone. So here's alpha, here's beta. So beta keto acid those guys readily decarboxylate. So this is acetoacetic acid. In the presence of acid this happens even at room temperature. You get a ketone. So in this case it would be acetone plus CO2. Plus CO2 gas which bubbles off. Alright how does that happen? Let's draw the mechanism. We're going to orient the acetoacetic esters so that we're forming an intramolecular hydrogen bond here between the carbonyl of the ketone with the hydroxyl of the carboxylic acid. And what we're going to do is we're going to push electrons around in a circle. And so you can go clockwise, counterclockwise. You want to start with the double bond of this carbonyl and that's going to move over here to make a new bond. So we're going to move electrons around in a circle kind of like Diels Alder. That part of the reaction is a concerted reaction. And then I'm going to attempt to leave things where they are so you can see what's happening here. So I'll draw the products and then we'll take a look at how that happens. If you follow the arrows here we have carbonyl. I mean we have the carbonyl here, these arrows are going to go over here to make a new bond between oxygen and hydrogen. That's right here. This was a double bond. Now it's a single bond because we need to move those two electrons over so that's a single bond. This arrow here says we're breaking this bond here, the sigma bond, and we're moving over here to make a double bond so that's the double bond here. And then we're taking these electrons from the oxygen-hydrogen's breaking and we're making a double bond here so that's where we get the CO2. So what is the name of the carb, this product right here? Enol right? Not stable. It's going to tautomerize. So this is going to tautomerize to a ketone. Let's do some arrow pushing for that. We've looked at that already but let's do it again just for fun. Two ways you can do this. You can use electrons on oxygen to come down and grab the acidic hydrogen like this. Boom. This comes in to grab this hydrogen. We break the hydrogen-oxygen bond. That gives us a protonated ketone. And then water comes in and deprotonates. This is catalyzed by acid so this is acid catalyzed to tautomerization. And then that gives you the ketone reversible reaction. Or let me do this in a different color. If you prefer to do it the way we did it back in 51B or a 51B way to do the tautomerization. We had an enol as a result of the hydration of an alkyne. This is back in chapter 10. If you want to take another look at this this should jog your memory. What we did then is we H3O plus. What we did was we were in the hydration of an alkyne mode right here in that chapter. We did this sort of thing. Then we put the positive charge on the site that would best stabilize the positive charge. Mark Hanukov, right? This wouldn't be Mark Hanukov. This is the most stable carbocation. And then we drew these. The electrons came down. We drew the resonance structure. So some students like to do this extra step here and that is perfectly fine. And then you deprotonate. And that will match exactly what we did back in chapter 10. So your choice. Questions on the tautomerization? Anybody? All right. So we're going to an acetyl acetic acid to a ketone. Melonic acids also readily decarboxylate. You do need a little heat. This from an acetyl acetic acid goes at room temperature. This one you need to heat. So this is more of the dicarboxylic acid. That will give you a carboxylic acid, acetic acid derivative, and plus CO2. The mechanism will look very much the same except you'll have a hydroxyl on one side rather than a methyl. So what we're going to do is convert this. We're going to use this to actually as an alternative to using LDA to make enolate ions. Because we know that reaction doesn't work very well. So here's what we're going to be doing. We're going to be taking acetyl acetic ester and through a series of steps which I'm going to show you right now, we're going to be going to substituted ketone derivatives. We can have two R groups or one. Substituted aceton derivatives. So if we go to the second product with the two R groups, you can see that this is going to give us an alternative for the thermodynamic enolate. So the previous reaction making the enamine gave us an alternative for the connecting enolate. If we introduce two R groups, this gives us an alternative for the thermodynamic enolate. Both of those reactions using LDA are not very clean. So how we're going to do this, we're going to alkylate the acetyl acetic ester. We're going to hydrolyze the ester so we get the carboxylic acid and then we're going to decarboxylate like we just did on the previous page. So let's see how that works. Sodium methoxide. Ethyl bromide. Do we want to use sodium hydroxide for that reaction? You want to use sodium methoxide. Why? What would happen if you use sodium hydroxide? It's going to suponify that ester. This hydroxide is going to attack that ester. You're going to kick off the oxide and you're going to deprotonate. So you're going to get a carboxylate here in the end of reaction. You're not going to get any alkylation here. So you have to very carefully, what you want to do is you want to choose your base to match the ester. If this, what would happen if I use sodium methoxide instead of sodium ethoxide? What would happen? I haven't heard it yet. I haven't heard the magic word. It's going to attack this carbonate. So in other words, if you use some ethoxide here, it's going to attack this carbonyl. Boom. We're going to kick electrons up onto oxygen. Electrons on the oxygen are going to come down. And some of the time it's going to kick off ethoxide. So you're going to start getting a mixture of ethyl and methyl esters here. So I deliberately chose sodium ethoxide here for this reaction. So what we're going to do is we're going to essentially completely deprotonate. Let's go through the process here. Here's ethoxide. We're going to remove the most acidic hydrogen, which is the one here. I'm going to leave electrons on carbon rather than pushing them onto the oxygen. So after I do that, I'm going to get an enolate. We'll leave the other hydrogen off. Then I add ethyl bromide. SN2 reaction. I guess you draw that out. If I'm going to draw arrow pushing, I need to draw that ethyl group out. We want to attack the carbon that's bonded to the leaving group. So we're going to attack the carbon bonded to the leaving group, kick off the leaving group. Product of that step has an ethyl incorporated into the active methylene position. Let's do it again. That was so much fun. We're going to do it again. We can use ethyl. We could use, do we want to change it up here? Let's use methyl this time. And methyl iodide. I'm feeling methyl iodide right now. So we're going to deprotonate again. I'm not going to draw the arrows for that second one, but we're going to deprotonate again. And then we add methyl iodide. So now we put a methyl here. Just like that. So step one is alkylate. We ended up actually doing it in two steps, but the first part is to alkylate. The strategy here, step one, alkylate. We can alkylate once. We can alkylate twice. Whatever you need to synthesize, you're going to decide. Step two is to hydrolyze the ester. We know how to hydrolyze an ester. We can do that acid catalyze or base catalyze. When you look at this reaction in the literature, they mostly do base catalyze using KOH and water. Probably means that it works better than acid catalyze. Do you have any experience with that? Either one of you guys? No, it's KOH. What? It's more straightforward. It's more straightforward, but if you did it all with acid, you could do it all in one step, which I'll show you coming up. All right, so that's going to hydrolyze the ester. That's chapter 22, right? You know how to do that. You get a carboxylate, sub-ponification. Now you add H3O plus in a second step. Ethyl's still there, methyl's still there. Now we have a carboxylic acid. Now that carboxylic acid can readily decarboxylate. Step three is decarboxylate. So we just did that mechanism for that. We're going to decarboxylate. We're going to form an enol, and the enol is going to tautomerize to a ketone. This goes readily at room temperature, so we don't even need to heat it, really. And so that whole entire carboxylic acid is coming off, and it is replaced with a hydrogen right there. So what we've essentially done is we've done the same thing as if we started with a ketone. We used LDA. We deprotonated, and we're alkylating on the most substituted sites on making the thermodynamic enolate. Only this is, although it's more steps, it's much cleaner. Okay, questions on that strategy? Anybody? So the whole entire thing is called an acetoacetic estrous synthesis. And basically what it can be used for is making substituted acetone derivative. So here's retrosynthetic analysis. If I give you a compound and I say, can you make this compound from an acetoacetic estrous synthesis? So this is what you would do. Look for acetone and circle acetone. So that's our, I'm just calling this our acetone unit here. And the two R groups are, we're going to put the two R groups on from the acetoacetic ester. So basically this hydrogen right here comes from the decarboxylation. So I'm going to cross out that hydrogen and I'm going to put a temporary ester group here. So now can you see where the acetoacetic ester is? We have right here, that's the acetone. If that hydrogen is off, that's now our ester right here. So these two R groups are going to come from alkylating acetoacetic, acetoacetic ester. So here we replace hydrogen with a temporary ester group. And so these two R groups are going to come from that and that are going to come from RBRRI, your choice, this is going to be R'BR. Could be a chloride, it could be an iodide, it could be a tosylate. Alright so to make this compound we're going to start with RBR, R'BR and acetoacetic ester. And I will say using acetoacetic ester to synthesis to make this compound. Questions on how I took that apart, we're going to write the synthesis on the next page. That's just how I took that apart. So like I said, the exam would say show how you can make this using an acetoacetic ester synthesis. So that means we're going to start with an acetoacetic ester. It could be OET, it could be O-methyl, OET is probably more common. Step one, I'm going to deprotonate NaOET. I'm going to add one of the two R, let's do RBR first. And then I'm going to deprotonate again, you want to do this stepwise. And then R'BR. That does the alkylation. Do not write excess sodium ethoxide, RBR and R'BR. Then you have no control over what happens. You're going to have some of them with two R's, you're going to have some of them with two R''s, some of them with R and R' different, have fun isolating the correct compound from that mixture. Not very easy. Okay, so we've alkylated twice, so I need to actually draw on the alkyl groups here. You were wondering if I was going to do that. Okay, we've alkylated twice. Now we decarboxylate. KOH, H2O, heat. And then H3O plus H2O. We don't need heat, but you can. Or if you want to make it a little shorter, we can hydrolyze an ester in acid, can't we? We know acid catalyzed hydrolysis of an ester and we know base catalyzed. Or we can do this as one step. We can do H3O plus H2O and heat. And this will hydrolyze the ester, then decarboxylate. So as soon as you hydrolyze that ester, it will decarboxylate very readily at room temperature. You are heating it up to get the ester to hydrolyze. Yes? Oh, could I say please? Is the second step and the first reaction on the top, the one or is the H2O plus H2O such a step? Instead of a third step, there was an invisible second step here that I didn't tell you about. Yes, that should be a two. Thank you. Thank you so much. All right, okay, more questions on that. You think you could do that on the term two? It's not that bad. It really isn't. Trust me. All right, so we're going to talk about Melonic Ester Synthesis next, which is the same idea except we start with Melonic and Ammonic Ester, so ester ester rather than ester ketone. It's the same idea, so guaranteed on midterm two, you will have to do either an acetoacetic ester synthesis or a Melonic ester synthesis. They're both the same. I'll find some fun, cool thing that you can synthesize and then you can use that. All right, so this is the same idea. So if you get the previous one, you'll get this also. So this is going to give us substituted acetic acids. So here's our acetic acid unit here. If we alkylate the active methylene compound once, we get the R group here. If we alkylate twice, we get two R groups in the alpha position. And this is pretty cool because we really can't deprotonate that alpha position of a carboxylic acid, right? Because if we add base, we deprotonate. That gives us the carboxylate and then we're stuck. So you will not find a pKa for the alpha hydrogen of a carboxylic acid. That's because if you add base, the most acidic hydrogen is going to be removed. So base removes proton on oxygen, not carbon, not on the alpha position. All right, so we could alkylate the corresponding ester and saponify. So we could do LDA and then we could do RX. And then we can alkylate there that way. So now we have RCH2 and then our ester. And then we can hydrolyze the ester, KOH, H2O, heat. Definitely need to heat that. And then step two, H3O plus in heat. Here again, we can bypass that and do acid-catalyzed hydrolysis for one step. Now I guess we don't need heat there because we're not decarboxylating, so let me fix that. We just need H3O plus. Or we can just do acid-catalyzed hydrolysis of the ester. H3O plus H2O and heat. And that will give us the carboxylic acid. All right, so again, we can hydrolyze ester's acid-catalyzed or base-catalyzed. So this is base-catalyzed and this is acid-catalyzed. That's an alternative, but it doesn't work as well for a couple different reasons. It's more expensive. You have to use inert atmospheric techniques. You have to use dry solvents, dry glassware, argon or nitrogen atmosphere. This is better for small-scale reactions. And then also the enolates are stronger base than a stabilized enolate, so you get more E2. So this is really not as good of an alternative. All right, so conceptually we have the same idea here. We alkylate diethylmalony. We hydrolyze both esters and then we decarboxylate. So let's do an example here. All right, so step one, we alkylate. Ethylenacid methylene compound has a little bit higher pKa than the acetoacetic ester, but still works really well. And we're going to get one methyl incorporated. Let's just alkylate once this time. You don't have to alkylate twice. You can just alkylate once. And then we have to hydrolyze both esters. And then in step two, we have to alkylate. Looks like that. And then in step three, we protonate the dicarboxylate. Then we decarboxylate. Only one of the carboxylic acids will come off. So step three, we want H3O plus H2O and heat. And when we do that, we get a substituted acetic acid derivative. Or if we want to save ourselves a little time here, we can do acid catalyzed hydrolysis, H3O plus H2O heat. And so we're hydrolyzing the esters in acid. We get the dicarboxylic acid. As soon as we get the dicarboxylic acid, it decarboxylates. So this step here is acid catalyzed ester hydrolysis. Then decarboxylation. All right, so as you can see, the same idea. We could alkylate twice if we wanted to. You will definitely be asked to do one of the two of those on the test, if not both. Probably not both, just because we have so much to cover on this exam. Retrosynthetic analysis, we want to be able to, if I give you a substituted acetic acid derivative, you want to be able to take it apart and see how to make it. So let's do that right now. All right. Here is our acetic acid right here. Circle that. So we're going to replace the hydrogen with a temporary ester. And then this is also going to become an ester, right? So we'll put the ET there also. Because we need a diester to start with. These two groups right here, this is going to be from RBR and RprimeBR. This is our acetic acid unit. So we replaced the hydrogen with a temporary ester. And then we converted this acid to an ester. All right. So what would we start with? We would start with diethyl malenate right here. RBR and RprimeBR. And if we write out the synthesis in the forward direction, then it's going to look just like the last one. Except we just have a different starting material. So let's show how we would use that. Let's outline a malenic ester synthesis of the following carboxylic acid. All right. So circle acetic acid. Try not to be thrown off by the ring. This is one, our alkyl halide, and this is the other one. We've alkylated twice. Let's put on our temporary ester group. There's a temporary ester group. And then we're going to change this to an ester also. All right. So this is going to be from diethyl malenate. And let's make sure we count carbons here. We want 1, 2, 3, 4, 5. 1, 2, 3, 4, 5. That's going to be our alkylating agent. That's our alkyl halide. That's how we're going to make a ring, because we've got the leaving groups on both sides. So let's write out the synthesis in the forward direction. We've got one minute to do this. Start with diethyl malenate. Step one, NaOET. I've chosen my base to match my ester. Step two, I add this alkyl halide here. I want to do this stepwise. If I use two equivalents of sodium ethoxide, and then I add this, then we're going to have competition. Some of the ethoxides are going to attack the alkyl halide, and that's going to take away our starting material. So you definitely want to do this stepwise. That gives us this compound here. And let's do the one step thing here. H3O plus H2O heat. That gives us the substituted acetic acid compound. And that is the end of chapter 23. We'll stop right there, and we will start chapter 24 next time.
hydrolysis of the emine to the ketone? I'm compounds are nucleophilic and react with alkyl decarboxylation. Alright loss of carbon dioxide This is the most stable carbocation. And then we Is the second step and the first reaction on the