Propose a mechanism to account for the following reaction of an enamine with an alkyl halide:

Name: Problem 66, Chapter 23: Organic Chemistry
Uploaded: 2020-07-13T08:30:00-08:00
Duration: 5 min 15 s
Channel: Tom Rutherford
Description: Propose a mechanism to account for the following reaction of an enamine with an alkyl halide:

Propose a mechanism to account for the following reaction of...

Key Concepts

Enamine Formation

Enamines are formed by the condensation of secondary amines with carbonyl compounds, resulting in a compound that contains a double bond adjacent to an amino group. This process involves the elimination of water and creates an electron-rich alkene that is primed for further reactions. The structure of the enamine, with its conjugated system, plays a key role in its reactivity as a nucleophile.

Nucleophilic Reactivity of Enamines

Enamines are known for their nucleophilic character because the electron density from the adjacent nitrogen atom is delocalized onto the double bond, making it electron-rich. This allows enamines to attack electrophilic centers effectively, particularly in alkylation reactions, where the enamine serves as a soft nucleophile that can add to electrophilic alkyl halides.

Alkylation Reaction Mechanism

The mechanism of alkylation with an enamine involves the nucleophilic attack on an electrophilic alkyl halide, typically via an SN2 process if the halide is primary. The enamine double bond donates electrons to the electrophilic carbon of the alkyl halide, leading to the formation of a new carbon–carbon bond. This process highlights the reactivity difference between nucleophiles and electrophiles and is a classic example of carbon–carbon bond formation in synthetic organic chemistry.

Iminium Ion Formation and Hydrolysis

After the alkylation step, the enamine intermediate is converted into an iminium ion. This intermediate can then undergo hydrolysis where water attacks the iminium carbon, eventually leading back to the carbonyl compound, now substituted with the new alkyl group. This sequence of enamine alkylation followed by iminium ion formation and hydrolysis is a common strategy to achieve ?-alkylation of carbonyl compounds.

Transcript

00:01 Okay, this problem is having us react to this anamine with this alkaliobromide to eventually form this product right there.

00:05 Okay, so this whole problem is asking us to show the mechanism.

00:08 So let's start.

00:09 Okay, so i'm going to move the electrons from this nitrogen onto this single bond to create a double bond, and then i'm going to move the electrons from this double bond onto that carbon.

00:17 Okay, so this is going to end up with a resonance form, and i'm just doing that to show the movement of electrons more smoothly.

00:24 Okay, so i still have my benzene there.

00:26 I have my seven -membered rain still, and then i have my five.

00:30 Carbon compound or five atom molecule and then i have my set of loan pairs there.

00:35 Okay and then of course i made my double bond on that nitrogen carbon to form this product.

00:39 Okay so this is my resonance form and again the reason i drew it is to just show the electron movement okay because otherwise we'd have electrons flowing from double bonds which isn't necessarily bad it's just i like to see lone pairs i like to see actual loan pairs moving around.

00:54 Okay so what i'm going to do is i need to hypothesize what are these loan pairs gonna do? i've seen them in other similar problems where those electrons act as a nucleophile to attack a carbonyl.

01:06 I've seen them act as a base.

01:07 We've seen them attack multiple different things.

01:10 And this particular problem, i have two options.

01:12 I can either take off an acidic proton or i can behave as a nucleophile.

01:19 So if it were to behave as a nucleophile, i have a few options, actually.

01:23 I can attack a carbonyl, i can attack this carbonyl, or i could attack this carbon.

01:28 So why would that carbon be considered? electrophilic.

01:31 The reason why is because we have this bromine.

01:34 That bromine, if it were to leave, it would become a bromide ion.

01:37 We know that bromide ions are super stable as leaving groups because they're relatively stable.

01:41 They're not going to be very reactive if they're to be left by themselves.

01:45 So that's exactly what i'm going to do.

01:46 I'm going to move these electrons, behave them as a nucleophile, and attack that carbon.

01:51 And in the process of doing that, i'm going to have to move that bromine off to become a bromide ion.

01:56 Okay, so after that, i should end up with the following.

02:01 And then i have my seven number briterine.

02:07 I think that is one too many carbons.

02:09 Let me see.

02:10 Okay, so here's this.

02:13 Okay, that's a seven member ring.

02:15 Okay, and then attached to here, i have my product.

02:19 Okay, so this is one i just formed.

02:22 So this carbon right here, that corresponds to this carbon right there.

02:27 So we had a bromine, it used to be attached to this, but then we just made it leave.

02:31 So now it's just like that.

02:32 It's only attached to this carbonyl now...

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