What enone product would you expect from aldol condensation of each of the following compounds?

Name: Problem 3, Chapter 23: Organic Chemistry
Uploaded: 2020-07-07T01:03:23-08:00
Duration: 11 min 8 s
Channel: Tom Rutherford
Description: What enone product would you expect from aldol condensation of each of the following compounds?

What enone product would you expect from aldol condensation...

Key Concepts

Aldol Condensation

Aldol condensation is a reaction where aldehydes or ketones combine through a mechanism that initially forms a ?-hydroxy carbonyl intermediate. This intermediate can then lose water to generate an ?,?-unsaturated carbonyl compound. The process typically involves the formation of an enolate ion, which acts as a nucleophile and attacks a second carbonyl group.

Enolate Chemistry

A key step in aldol reactions is enolate formation, where a base deprotonates the ?-carbon of a carbonyl compound. The resulting enolate ion is nucleophilic and can attack another carbonyl carbon, forming a carbon–carbon bond and setting the stage for further transformation into the final product.

Enone Formation

Enones are ?,?-unsaturated carbonyl compounds that result from the dehydration of the aldol addition product. The conjugation between the carbonyl group and the alkene stabilizes the enone and also imparts distinct reactivity patterns, making them useful in various organic transformations.

Dehydration Mechanism

Following the initial aldol addition, the dehydration step involves the elimination of water from the ?-hydroxy carbonyl intermediate. This elimination leads to the formation of a double bond between the ? and ? carbons, creating the conjugated system characteristic of enones.

Transcript

00:01 Okay, this problem is asking us to show the products of an aldol condensation reaction.

00:05 So aldo condensation reactions deal with the net loss of water.

00:08 So we don't have to necessarily get rid of a complete water molecule.

00:11 All we have to do is get rid of perhaps a hydrogen and then a hydroxide ion.

00:15 All we have to do is know that we are getting rid of net water.

00:18 So as for my first problem, i have a starting material that looks like this.

00:22 I have a cyclopentane connected to a carbonyl.

00:25 Okay, so that cyclopentane with carbonyl is going to react with sodium hydroxide.

00:30 So i'm just going to write down the hydroxide ion.

00:33 Okay, so that hydroxide ion, we know it's to be very basic.

00:36 It's going to deprotonate the most acidic hydrogen on my molecule.

00:39 And the most acidic hydrogen on my molecule is this alpha hydrogen that is connected to my alpha carbon.

00:44 Okay, so what's going to happen is my hydroxide ion is going to depotinate that, and i'm going to move the electrons from my carbon hydrogen bond onto my carbon.

00:53 Okay, so i should end up with this intermediate in which i have my carbonyl unaffected, and then i have lone pairs on that carbon.

01:00 Okay, so those lone pairs are going to be representative of my nucleophile.

01:05 And by nucleophile, i mean that this carbon is going to attack an electrophile.

01:09 And my electrophile, because it's stated in this problem that we don't have any other electrophiles besides the starting material, we're going to attack that same starting material.

01:19 So actually, i'll do this part in red.

01:22 Okay, so here's my other equivalent of starting material.

01:25 This is going to behave as my electrophile in this reaction.

01:27 So here's my nucleophile, designated by the lone pair, by the lone pair, and that is going to attack the carbonyl portion of my carbonyl.

01:36 So the electrons are going to move up to my oxygen, creating this new product, this intermediate.

01:43 Okay, so i have my carbonale unaffected.

01:46 In fact, i'll represent this part in blue so we can see what's what.

01:51 Okay, so here's my carboneal.

01:53 That's unaffected.

01:54 So this one in blue is this one.

01:57 Okay, so this carbon is this carbon.

02:00 Okay, so connected to that carbon, is this.

02:04 Okay, and it connected to this carbon, so this carbon right here, which i'm dotting, is this carbon.

02:08 And connected to that carbon is an oxygen.

02:13 So i'll do this part in red.

02:17 Okay, so connected to that carbon is an oxygen.

02:19 So this auction right here is this one, just so we're keeping track of things.

02:23 And then connected to that carbon as well is the remainder of my cycle of pentane.

02:27 Okay, so that's where we are.

02:28 Okay, so if we zoom back out, remember that i said that condensation reactions are the net loss of water.

02:34 So right now we've lost one hydrogen.

02:37 Okay, so we have a basically hydroxide ion to go.

02:40 But as we can see here, we don't have any hydroxide ions.

02:42 In fact, we have a negative charge on my oxygen.

02:45 Okay, so within organic chemistry, every time you see a negative charge on the compounding question, you kind of hesitate because we don't like negative charges if they have the ability to be neutral and vice versa if it's a positive charge.

02:58 So what i'm going to do is utilizing the water that i created by the protonation of hydroxide, i'm going to protonate my o -minus.

03:06 Okay, by doing that, i'm going to make a hydroxide ion again.

03:09 So i should form this new intermediate in my total reaction, where i have a carbonyl, and then now i have an alcohol instead of an o -minus, and then my cyclopentae.

03:22 Okay, so that's where i'm at right now.

03:25 Okay, so the next step, if we look back to what a aldo condensation eventually produces, we produce an alkyen.

03:31 And every time we make alkenes, we're going to get rid of a leaving group and get rid of a hydrogen.

03:36 So that is just standard elimination mechanisms.

03:40 Okay, so what i'm going to do is figure out what is going to be my leaving group.

03:44 Okay, so i'm going to use my base, which i just created, my hydroxide ion, and i'm going to deprotonate a hydrogen and get rid of a leaving group.

03:53 So what is the most acidic hydrogen on this molecule? well, if i look closely, usually we have alcohols.

04:01 So alcohols usually have a pca of around 17.

04:04 Whereas this hydrogen, this one right here, is attached to the alpha carbon.

04:09 Usually alpha carbons with their hydrogens have a pca of roughly 20 -ish.

04:17 That's for ketones.

04:19 So when we're comparing 20 to 17, obviously the one with 17 would have the lower pca, sorry, lower pca meaning higher acidity.

04:27 So normally we would think that we would depotinate the hydrogen on the alcohol, but in reality, because this hydrogen right here is sandwiched between a carbonyl, so that carbonyl can pull electrons through resonance and also by the inductive effect, and this alcohol is pulling electrons through the inductive effect, we are in reality, this hydrogen is more acidic than my alcohol.

04:48 And in fact, it is actually facilitated, at least the elimination process is facilitated by the presence of this alcohol, because that alcohol can leave.

04:56 So it would be very similar to an elimination reaction.

04:58 So this hydrogen is going to be deprotonated, and i'm going to move the electrons from that carbon hydrogen bond onto this single bond to create a double bond.

05:08 And when i'm doing that, i'm going to kick off my alcohol in the process.

05:12 So normally we don't usually see alcohols as leaving groups because an alcohol by itself is not very stable.

05:19 It can't stand by itself very well.

05:21 But if we look back to where we started, we started out with a hydroxide ion.

05:25 So if we start out with a hydroxide ion, we're in basic condition.

05:28 So we can end up in basic conditions.

05:31 Okay, so it would be representative of a catalyst.

05:35 So this would be my final product in which i have my cyclopentane, my carbonyl unaffected, and then i have my new alkyne and my cyclopentane.

05:45 That is not a cyclohexane, so i need to drop cyclopentane.

05:49 Okay, so that would be my final product...