Intramolecular Condensation

Concept: Concept: Diketones

6m
Video Transcript

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.

Concept: Concept: Diesters (Dieckmann Condensation)

4m
Video Transcript

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.

Concept: Practice 1: Intramolecular Aldol Condensation

7m
Video Transcript

Hey guys, let's take a look at the following question. So, here I say consider the following reaction, provide a step wise mechanism to explain the given transformation. So, 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 bond and, 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. Now, not just an aldol reaction because we made a double bond it's an aldol condensation reaction, so the word condensation means that we form a double bond and unsaturated means double bond, okay? So, 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. So, 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 is ionic. So, it's not really together it's really NA positive OEt negative. So, this space is going to shoot out, rip off this hydrogen, alpha carbon holds onto the electrons. So, 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 one carbonyl but now it's attracted this carbonyl carbon because it's partially positive so 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 six elements that loop around each other to form a ring. So, that's why our ring it six carbons large. So, here we come over here. So, we have six carbons in there, we're going to say that this is 1, 2, 3, 4, 5 and 6 on carbon number one there's still this aldehyde, carbon number six has an O Negative. Remember, the OH or the OEt minus 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 make up beta hydroxy aldehyde. 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 negative comes back or maybe an OH minus 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 off the OH group. Now, there should be some bells going off in your heads, because we've always heard that good bases, no, weak bases are good leaving groups, this is something you learned in organic one, weak bases are good leaving groups. Now, OH is a strong base, so it should be a bad leaving group. So, should it 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 because we create in the process a conjugated aldehyde and remember the word conjugated, conjugated means you have alternating double and single bonds. So, we go double, single, double, conjugation is a very stabilizing effect because of resonance. 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 it leaving, we make a much more stable conjugated carbonyl group. So, in this case we will allow a strong base to leave. So, here this would be our mechanism to illustrate what's going on here, and remember, we're ripping off a hydrogen from the alpha carbon so that the OH 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. So, remember is this helps to make a conjugated carbonyl group, we call these conjugated carbonyl groups usually enone. Now, en means alkyne but on 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 intramolecular aldol condensation reaction. So, remember that little things about this reaction to help guide you to the final answer.

Intramolecular Condensation Additional Practice Problems

Heating of compound 1 with sodium hydroxide in water led to isolation of the cyclic product 2. Write a detailed mechanism that explains the conversion of 1 to 2. Show all intermediates!

Watch Solution

Heating of compound 1 with sodium hydroxide in water led to isolation of the cyclic product 2. Write a detailed mechanism that explains the conversion of 1 to 2. Show all intermediates!

Watch Solution

Draw the mechanism of this reaction.

Watch Solution

Predict the product of this reaction. 

Watch Solution

Show your inderstanding of the following Dieckmann Condensation reaction by showing the full arrow pushing mechanism, including ALL RESONANCE FORMS and predicting the products. 

Watch Solution

Complete the mechanism for the following Dieckmann reaction. Be sure to show arrows to indicate movement of all electrons, write all lone pairs, all formal charges, and all the products for each step. Remember, I said all the products for each step. IF A NEW CHIRAL CENTER IS CREATED IN AN INTERMEDIATE OR THE PRODUCTS, MARK IT WITH AN ASTERISK AND LABEL AS "RACEMIC" IF RELEVANT. IN THE BOX BY EACH SET OF ARROWS, WRITE WHICH OF THE 4 MECHANISTIC ELEMENTS IS INDICATED IN EACH STEP OF YOUR MECHANISM

Watch Solution

What is the product of the reaction shown below?

a A 

b B 

c C 

d D

e none of these

Watch Solution

Complete the following reaction. Pay careful attention to the stereochemistry of the product.

Watch Solution

What will be the  major product of the following reaction? Pay careful attention to the stereochemistry of the product.

Watch Solution

Predict the product(s) in the following aldol condensation using aqueous sodium hydroxide.

Watch Solution

Predict the product(s) in the following aldol condensation using aqueous sodium hydroxide.

Watch Solution

Aldol reactions can be conducted either under basic or acidic reaction conditions. Show the mechanism for the following aldol cyclization.

Watch Solution

In the presence of a base such as NaOEt, the following reaction proceeds to form the bicyclic product that is shown below. Some of the carbon atoms have been labeled from 1‐9 in the starting material. Number the atoms in the product that correspond to those numbered in the starting material.

Watch Solution

List the reagents A-D in the order necessary to complete the following transformation. 

Watch Solution