Intramolecular Aldol Condensation

In this next page, we're going to be focusing on a special type of condensation called an intramolecular condensation.
It gets a little bit more complicated. Dicarbonyl compounds have the ability to self-condensate through cyclization if it will generate a five or six-membered ring. Here's the deal. Any ring below five members – four, three – way so unstable the 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 five and six-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 five or a six-membered ring. Let me ask you a question. I’ve actually got four different positions that could make enolates here. I’ve got position one, position two, position three, and position four. Now, you guys understand that three and four are the same thing as one and two so I’m going to erase them because that's just distracting me. I literally just have a choice between “Do I put the enolate on carbon one or on carbon two?” They both have alpha-hydrogens so I could definitely get enolates there. The real way that you decided this is you decide by the ring that you could make after you cyclize. Let’s look at for example 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 first carbon into account. 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, so then let’s count. Let’s see if it’s the right size. Guys, that is a lot better. That’s a six-carbon chain that is going to connect between the one and the six, making a six-membered ring. The NaOEt 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 I’d like to do is if we already know it’s going to be a six-membered ring, let’s just draw a six-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. 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.
What am I missing? What groups do I need? It looks like I'm missing a ketone on 2, so 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 ring. I’m missing an OH that it’s going to protonate. Am I missing anything else? I’m also missing a methyl because look, 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 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 did you know it was aldol?” Because it’s a ketone to our aldehydes, and you’re going to get a cyclic enone.
In the next video, I’m going to show you what happens for intramolecular condensations of esters.

So guys when a diester cyclizes and self condensates the product is going to be a cyclic beta keto Ester and this just makes sense because beta keto Ester is always the product of a claisen but now it's going to be a ring because it's intermolecular. Now, chemists just had to make this overly complicated and apparently the guy that discovers this specific cyclization is called Dyckman, so this is also called a Dyckman condensation but don't get too hung up on the name because really all it is, is an intramolecular claisen, okay? So, intermolecular claisen and Dyckman are the same thing, alright? Awesome. So, once again, we're going to determine where to form enolates based on or if we should based on if we can make a five or six membered ring. So, Esters are a little bit more simple because I don't have like possibilities on both sides I only have really one, would that enolate make sense? would that give me a five or six membered ring if it attacks the other Ester, like this, let's see 1, 2, 3, 4, 5, Oh that's perfect, six membered ring, we know that's going to be favored, okay? So, what I'm going to do is draw the rest of the mechanism, which would be O facing up, okay? And once again since I was six 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 and guys really you can start wherever you want but this is just the way that I like to do it, okay? So, now I'm wondering what's missing in the different places and let's even just do the whole mechanism. So, at this point of the mechanism what's missing? Well, looks like one is missing an ester. So, let's add that, one is missing, this whole ester thing, what else? six is missing two things, right? Six is missing and O negative and it's missing an OEt, perfect. So, now that looks is correctly drawn, I'm going to erase the numbers because we know we did it, right? So, guys do you know what 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. So, I would reform the ketone and kick out the OEt and what that's going to give me is a compound that now looks like this, and oops, yeah, and then ketone, okay? So, now what I'm wondering is guys, did I get the right compound? because remember, that whenever you go to Dyckman, a Dyckman should always do a cyclic beta keto Ester. So, is this a cyclic compound? yes, is it a beta keto Ester? Well, I still have an ester and I still have a ketone of the beta. So yes, this is correct, cool, right? So, intermolecular, got to get used to it in this chapter, okay? This isn't the last one is going to come up, it's going to come up again, I'm warning you. So, get good at it now. Alright, so let's move on to the next topic.