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

Concept: Practice 1: Intramolecular Aldol Condensation

7m

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!

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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!

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Draw the mechanism of this reaction.

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Predict the product of this reaction. 

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

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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

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What is the product of the reaction shown below?

a A 

b B 

c C 

d D

e none of these

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

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

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

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

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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.

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

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