Acetoacetic Ester Synthesis

Concept: Concept: General Reactions

6m
Video Transcript

In this video, I want to review the general reactions that acetoacetic esters can go through. This is like your cheat sheet for all the different reactions that you can do with acetoacetic ester synthesis. The first step for all of these is always going to be deprotonation and formation of an enolate. That enolate is going to selectively form right here because that the easiest proton to pull off. Remember, it’s the most acidic. We’re just going to go one by one down these reactions. Once you have your enolate, if you react that enolate with an alkyl halide, you’re going to get an alkylation. That just makes sense. We’ve done that plenty.
It turns out that usually you're going to have two hydrogens on this middle part, on this methylene carbon. Usually most of these acetoacetic esters start off with not just one hydrogen but two. If you wanted to, after you get the first alkylation on, you could deprotonate the other one with another equivalent of base and then react with another alkyl halide. That way, by reacting twice, you would remove both of these hydrogens one after the other. You would actually get now two R groups attached to the alpha-carbon. So this would be dialkylation. Just keep in mind, to do a dialkylation, you need to react with the base twice in order to get both protons off. It can't be two equivalents of base at the same time. It needs to be base, then alkyl halide, then base again, then alkyl halide again.
Then finally the last two steps, I’m skipping in all of these. Just so you guys know, I'm always assuming that at the end of this reaction, I’m going to do plus H3O+ and plus heat. Why am I assuming that? Why did I not draw it? I guess I could have drawn it but it happens every single time. It’s a waste of space. What I’m trying to show you is the stuff that changes. The stuff that stays the same that you always hydrolyze and you always decarboxylate. That's going to hold true for all of these.
Next. What if you use, instead of using a halide or an alkyl halide, what if you use a terminal alkyl dihalide? Then what you can do is you can actually cycloalkylate. What you can do is you can react with your base, react with one of your alkyl halide. Then react another cool enough base, then react with the other side of your alkyl halide. That’s going to form a ring around the alpha-carbon. You're going to get a ring with a size of n+1, where n is equal to the number of carbons in between in your chain. If you reacted with a three-carbon dihalide, you’re going to get a four-carbon chain. Why? Because these are the three carbons, and this is the alpha-carbon that you started with. It’s a four-member ring but only three of those new carbons came from the dihalide.
Finally, this one is easy – acylation. You know that you could attack here, kick out the Cl and you would get an acyl group after you do your H3O+ and your heat. I know this is a lot to go over. The only way you’re going to learn is just start applying it. Let's do this practice problem. Noticed it’s five steps. You’re going to have to figure out what's going on. Try to do this five-step product. Don’t worry about mechanisms. Just worry about the product, and then I’ll help out. By the way, I don’t want anyone to get stuck on molecule 2 so I’m going to give it to you. Notice that it says (BrCH2CH2)2O. What is that? That’s an ether. Let me draw it for you. That would look like this. O in the middle with these two groups coming off of both sides. It would be CH2CH2Br, CH2CH2Br. Now you have no excuse to get this one wrong because I'm actually giving hint to you. Go ahead and try to do it yourself and then I'll step in and show you how it's done.

Concept: Example: Predict the products

6m
Video Transcript

Alright guys, let's start off by drawing this ester out because I hate working with condensed formulas. So, that's just not helpful. So, I'm going to draw this out like this, my ketone ch2 COOR, it's going to be O, O Et so that is aceto acidic ester, okay? Now, the first reagent I'm reacting with is OC2H5, is this a good base to use to form an enolate on this dicarbonyl, what do you think? Okay guys so remember that we said that you want to make sure that your R group is the same between your ester and your oxide or you're going to get what? a transesterification, right? So, is this a good choice, yes it is guys, Ethel and C2H5 are the same thing, it's just different ways of writing it, sorry, that was a little tricky, right? So, it's fine, this is a good base to use, this is exactly what we are going to use. So, like the first reaction is going to be that the O, grabs one of the H's and forms an enolate here, okay? Now, that enolate let me see if I can move this down a little bit, I can't. So, this enolate is now going to react with my alkyl halide and my alkyl halide looks like this. So again, it's Br, O, Br, okay? So, this enolate is going to do a backside attack, that's not really a backside attack, let's do a backside attack, on one of the bromines and it's going to give me this, let me move it up a little bit. So, it's going to give me carbonyl, carbonyl OEt, plus I have this huge chain attached. So, that's going to be 1-2, O, 1-2, Br, okay? So, now that whole thing is attached, okay? Now, look at my third reagent, usually we would be thinking, okay, let's decarboxylate, I mean, let's hydrolyzed and decarboxylate but no I'm reacting with a second equivalent of base, what's that going to do? Well, guys I've got one more proton there that can react. So, I'm going to take my base and I'm going to make an enolate there, okay?

Now, what's going to happen guys, that enolate can do another backside attack, okay? So, what we're going to get is now a compound that looks like this, with now, what does it have? Well, it has, it's a ring, right? It's going to be a ring of what size? Well, to count out the size of a ring I always start from nucleophile to electrophile. So, I would say 1, 2, 3, 4, 5, 6, guys, this is going to make a six membered ring coming off of the middle like this, where you guys just have to tell me if I'm missing anything, so this is 1, 2, 3, 4, 5, 6, are there any atoms I should change, anything I should add, should I add a Br, okay? So, I heard loud and clear I need to change the four to an oxygen, right? So, let's do that, you'd be surprised how much I can hear over the interwebs, okay? What else how's the Br, guy's the Br is just a leaving group. So, I'm just going to put plus Br negative, I don't need that, okay? Excellent. So, what's next? guys it says now we're going to use water and acid, what's that going to do, that's going to do a hydrolysis reaction on my ester, that's going to give me this, carboxylic acid instead of ester, okay? So, that's what my H3O plus does, and then heat, what is heat going to do guys? heat is going to decarboxylate, meaning that I take that entire carboxylic acid off, okay? if you're curious about the mechanism for that definitely search decarboxylation, and what you're going to get is that now that whole thing is gone. So, I think the stick for a ring is just coming straight off of this alpha carbon and that's it, it's gonna be that plus CO2, and that's going to give us our final answer. So, once again. So, this was actually what? this was a cyclone alkylation reaction because it reacted with a dihalide and two equivalents of base and I was able to get our ring, cool, right? Awesome guys. So, let's move on to the next set of videos.

Problem: Provide the major product for the following reaction

4m

Concept: Practice 2: Acetoacetic Ester Retrosynthesis

3m
Video Transcript

Hey guys, let's take a look at the following retrosynthesis question. So, here it says, beginning from ethyl acetoacetate and using any other necessary reagents, show the necessary compounds needed to form the following compound. So, remember, what does this type of reaction do? Well, this reaction changes in alkyl halide into a ketone, it adds three additional carbons in order to form a ketone. So, here if we look, this is 1, 2, 3, those portions had to have been part of the ethyl acetoacetate.

So, what does that mean? Well, that means that these two methyl groups they are originally alkyl halides, so the process was this, we used ethyl acetoacetate. So, we use this originally, we used NaOEt to rip off the first acidic hydrogen, then we brought in our methyl halide to add the first methyl group to the alpha carbon in the middle then we used what a second mole of NaOEt to rip off the second alpha hydrogen from our acidic alpha carbon and then what do we do? we brought in a second mole of CH3Br. So, that's how we got two methyl groups onto that alpha carbon, what do we do then? we did, we could have done acid-base hydrolysis or just simply acidic hydrolysis, that would have changed this ester end here, into a carboxylic acid back into the parent, then all we did at the end as we used some heat to lose CO2 and finally get to this answer here. So, that's how you approach these types of questions, we know that this process is going to make the ketone by adding three additional carbons, find those three additional carbons, they were originally part of the ethyl acetoacetate, all of the carbons must have been part of some kind of alkyl group, alkyl halide group. So, make those the alkyl halides since there's two of them here that means we have to do NaOEt strong base twice to rip each one of the Alpha hydrogen's off in order to add a methyl group to the alpha carbon twice.