Intramolecular Aldol Condensation - Online Tutor, Practice Problems & Exam Prep

Intramolecular condensation involves the self-condensation of dicarbonyl compounds to form stable 5 or 6-membered rings. For example, diketones can generate a beta hydroxycarbonyl, which readily dehydrates to yield a cyclic enone. Similarly, diesters undergo intramolecular Claisen condensation, producing cyclic beta ketoesters. The formation of enolates is crucial, as it determines the size of the resulting ring. Understanding these mechanisms is essential for mastering aldol and Claisen reactions in organic synthesis.

Diketones

Hey everyone. In this next page, we're going to be focusing on a special type of condensation called an intramolecular condensation. All right, guys. It gets a little bit more complicated. Dicarbonyl compounds have the ability to self-condense through cyclization if it will generate a 5 or 6 membered ring. Here's the deal. Any ring below 5 members, 4, 3, way too unstable to cyclize on its own. You're going to need to put a lot of energy into that system to make it happen. In order for these reactions to be favorable, we're going to try to make 5 and 6 membered rings. That's actually going to be a theme later on when we talk about other reactions that condensation products can undergo. Here, I've got this diketone. This could happen to diesters, diketones, whatever, dialdehydes. The biggest point here is that when you make your enolate, you want to make your enolate in a place where if you attack the carbonyl, you'll make either a 5 or a 6 membered ring. Let me ask you a question. I've actually got 4 different positions that could make enolates here. I've got position 1, position 2, position 3, and position 4. You guys understand that positions 3 and 4 are the same thing as 1 and 2, so I'm going to erase them because that's just distracting me. Literally, I just have a choice between do I put the enolate on carbon 1 or on carbon 2. They both have alpha hydrogens, so I could definitely get enolates there. The question is the real way that you decide this is you decide by the ring that you could make after you cyclize. Let's look, for example, at enolate 2. If I were to put my enolate on carbon 2, then it would attack this carbonyl here. How big would that ensuing ring be? This being my first, my nucleophile always gets the 1st carbon in the count, so this would be 1, 2, 3, 4. You always count from nucleophile to electrophile. Is that big enough? No way. Let's talk about 1. If I were to make it on 1, it can't attack that one. That's way too close. It would attack this carbonyl. Then let's count. Let's see if it's the right size. 1, 2, 3, 4, 5. That is a lot better. That's a 6 carbon chain that is going to connect between the 1 and the 6, making a 6 membered ring. The NaOAT is going to specifically make an enolate here so that it can cyclize on its own. The mechanism is just going to be what you expect for any aldol reaction. You would kick up the electrons to the O and then it would protonate. What we could do here is what I like to do, if we already know it's going to be a 6 membered ring, let's just draw a 6 membered ring and then let's just add the groups. From lots of years of experience, I know that it's actually going to be easiest to number this ring if I start at the top left corner and I work my way around counterclockwise. 1, 2, 3, 4, 5, 6. Literally, guys, this is just me trying to make it look as pretty as possible. You could have numbered it however you want. As long as your numbers are in order, it doesn't matter. The only thing you can't do is like 1, 3, 5, 7, or whatever. What am I missing? What groups do I need? It looks like I'm missing a ketone on 2. I'm going to put that there. What else am I missing? It looks like I'm missing some stuff on 6. What am I missing on 6? Could you guys tell me exactly? Think about it. Think about exactly what's not on this chain, on this ring. I'm missing an OH because we said that it's going to protonate. Am I missing anything else? I'm also missing a methyl because this methyl was not part of the ring. It's just extra. There we go. Now we have it drawn correctly. This is called what? What's the name of this product? This is our beta-hydroxycarbonyl. You're getting professional at this at this point. But what do we know about beta-hydroxycarbonyls? They love to dehydrate. We're going to dehydrate this bad boy and we're going to dehydrate between the alpha and the beta. What this is going to make as a final answer is I'm going to get what we call a cyclic enone. This is going to be highly favored because that's really stable. You're going to get a cyclic enone and that's what you get when you have basically an intramolecular reaction that's an aldol. This would be specifically an intramolecular aldol. You might say, Johnny, how do you know it was aldol? Because it's ketones or aldehydes. You're going to get a cyclic enol. In the next video, I'm going to show you what happens for intramolecular condensations of esters.

Diesters (Dieckmann Condensation)

When a diester cyclizes and self-condenses, the product is going to be a cyclic beta-ketoester. This just makes sense because a beta-ketoester is always the product of a Claisen. But now it's going to be a ring because it's intramolecular. Chemists just had to make this overly complicated and, apparently, the person who discovered this specific cyclization was called Deichmann. This is also called a condensation. But don't get too hung up on the name because really all it is is an intramolecular Claisen. Intramolecular Claisen and Deichmann are the same thing. Once again, we're going to determine where to form enolates based on whether we should, based on if we can make a 5- or 6-membered ring. Esters are a little bit more straightforward because I don't have possibilities on both sides. I only have really 1. Would that enolate make sense? Would that give me a 5- or 6-membered ring if it attacked the other ester like this? Let's see. That's perfect. A 6-membered ring, we know that that's going to be favored.

What I'm going to do is draw the rest of the mechanism which would be O facing up. Once again, since I have a 6-membered ring, I'm going to draw my six numbers. I'm going to start at this corner and I'm going to move my way around counterclockwise. Really, you can start wherever you want but this is just the way that I like to do it. Now I'm wondering what's missing in the different places. Let's even just do the whole mechanism. At this point of the mechanism, what's missing? It looks like ₁ is missing an ester. Let's add that. ₁ is missing this whole ester thing. What else? ₆ is missing two things. ₆ is missing an O-negative and it's missing an O₁₈. Perfect. Now that looks like it's correctly drawn. I'm going to erase the numbers because we know we did it right.

Now, the next step of this mechanism is? What's the next step? The next step is that I have to kick out the leaving group. I would reform the ketone and kick out the OET. What that's going to give me is a compound that now looks like OET and then ketone. Now, I'm wondering, guys, did I get the right compound? Because remember that whenever you get a Dieckmann, a Dieckmann should always yield a cyclic beta-ketoester. Is this a cyclic compound? Yes. Is it a beta-ketoester? Well, I still have an ester and I have a ketone at the beta. So yes, this is correct. Cool, right? So intramolecular. Got to get used to it in this chapter. This isn't the last time it's going to come up. It's going to come up again. I'm warning you. Get good at it now. Let's move on to the next topic.

Consider the following reaction. Provide a stepwise mechanism to explain the given transformation.

Hey guys, let's take a look at the following question. So here I say, consider the following reaction. Provide a stepwise mechanism to explain the given transformation. What we have to really do here is first observe what is going on. Well, we have a dialdehyde, so we have a dicarbonyl. And what happens here is this carbonyl somehow reacts with itself to form a ring with a double bond in it. Here, this is our alpha carbon and this is our beta carbon. So we created a cyclic, alpha, beta, since they're double bonded, we use the word unsaturated aldehyde. So if we have a dicarbonyl and it somehow cyclizes into a ring, then we'd say that this is an example of an intramolecular aldol. Not just an aldol reaction because we made a double bond, it's an aldol condensation reaction. The word condensation means that we form a double bond and unsaturated means double bond. Intramolecular means it's reacting with itself, closing in on itself to make a ring. Now, since we have to show the mechanism here, I'm going to take myself out of the image, guys, so we have more room to work with. What happens first is here is our alpha carbon and here is our alpha carbon. Since it's symmetrical, it doesn't matter which alpha carbon gets attacked. This is a base here; it's ionic, so it's not really together. It's really Na+ OEt-. This base is going to shoot out, rip off this hydrogen, and the alpha carbon holds on to the electrons. So what we're going to get initially is we're going to have this carbon negatively charged because it has a lone pair. It's already connected to 1 carbonyl, but now it's attracted to this carbonyl carbon because it's partially positive. This is going to shoot out and attach to this carbonyl carbon, kicking this bond up. Now, we count 1, 2, 3, 4, 5, 6. There'll be 6 elements that loop around each other to form a ring. That's why our ring is 6 carbons large. Here, we come over here. We have 6 carbons in there. We're going to say that this is 1, 2, 3, 4, 5, and 6. On carbon number 1, there's still this aldehyde. Carbon number 6 has an O negative. Remember, the OET- that basically ripped off an H could come back and do what? Protonate this oxygen here. So what we make initially is this, and this is alpha, this is beta, so initially, we make a beta-hydroxyaldehyde. Now, we need to make a double bond between the alpha and beta carbon. Now, what's going to happen next is here, we could say, ETO- comes back, or maybe OH- comes back because I put water here. Now, here's the thing. This comes back, and what is it going to do? It's going to seek out a hydrogen to take off. It's going to take it from the alpha carbon. Now, here's the thing. It's going to shoot out and rip off this alpha hydrogen here, which falls here to make a double bond, which kicks out the OH group. Now, there should be some bells going off in your heads because we've always heard that weak bases are good leaving groups. This is something we learned in Organic 1. Weak bases are good leaving groups. Now, OH is a strong base, so it should be a bad leaving group. So it should be able to leave. But for some reason, we're allowing it to leave in this case. The reason we're letting it leave is that we create in the process a conjugated aldehyde. Remember the word conjugated means you have, so it's okay that we're letting not such a great, leaving group leave because it's kind of a strong base. It's okay because in the process of its leaving, we make a much more stable conjugated carbonyl group. So in this case, we will allow a strong base to leave. Here, this would be our mechanism to illustrate what's going on here. Remember, we're ripping off a hydrogen from the alpha carbon so that the OH could get kicked out as we make the conjugated system. Now, if that alpha carbon didn't have a hydrogen to rip off, then we wouldn't be able to get rid of that OH, and we would have stopped at the beta hydroxy stage. Remember, this helps to make a conjugated carbonyl group. We call these conjugated carbonyl groups usually enones. Now, E means alkene, but one means ketone. Now, why do we usually call them enones? That's because ketones really love to do this. Ketones hate being in the beta hydroxy form and will react further to help make a double bond between the alpha and beta carbon so that they become a conjugated ketone. Again, conjugation makes you more stable, and that's what's going on here. This whole process, because it's making a cyclic ring, is an intramolecular aldol condensation reaction. So remember that little things about this reaction help guide you to the final answer.

Here’s what students ask on this topic:

What is intramolecular aldol condensation?

Intramolecular aldol condensation is a reaction where a single molecule containing two carbonyl groups (such as aldehydes or ketones) undergoes self-condensation to form a cyclic compound. This reaction typically forms stable 5 or 6-membered rings. The process involves the formation of an enolate ion, which then attacks another carbonyl group within the same molecule, leading to cyclization. The initial product is a beta-hydroxycarbonyl compound, which often dehydrates to form a cyclic enone.

How do you determine the position of enolate formation in intramolecular aldol condensation?

To determine the position of enolate formation in intramolecular aldol condensation, you need to consider the potential ring sizes that can be formed. The enolate should form at a position that allows the resulting ring to be either 5 or 6 members, as these are the most stable. For example, if forming an enolate at carbon 1 leads to a 6-membered ring upon cyclization, while forming it at carbon 2 leads to a 4-membered ring, the enolate should form at carbon 1.

What is the product of intramolecular Claisen condensation?

The product of intramolecular Claisen condensation is a cyclic beta-ketoester. This reaction involves a diester undergoing self-condensation to form a ring structure. The enolate ion formed from one ester group attacks the carbonyl carbon of the other ester group, leading to cyclization. The resulting product is a beta-ketoester, where the keto group is positioned beta to the ester group within the ring.

Why are 5 and 6-membered rings favored in intramolecular condensation reactions?

5 and 6-membered rings are favored in intramolecular condensation reactions because they are the most stable due to minimal ring strain. Smaller rings, such as 3 or 4-membered rings, have significant angle strain and torsional strain, making them less stable and harder to form. Larger rings, while having less angle strain, can suffer from transannular strain and are less entropically favored. Therefore, 5 and 6-membered rings strike a balance, making them the preferred products in these reactions.

What is the mechanism of intramolecular aldol condensation?

The mechanism of intramolecular aldol condensation involves several steps: (1) Formation of the enolate ion from one of the carbonyl groups, (2) Nucleophilic attack of the enolate on the carbonyl carbon of the other group, leading to cyclization, (3) Formation of a beta-hydroxycarbonyl intermediate, and (4) Dehydration of the beta-hydroxycarbonyl to form a cyclic enone. This sequence of steps ensures the formation of a stable 5 or 6-membered ring.