p -Bromotoluene reacts with potassium amide to give a mixture of m - and p -methylaniline. Expla

P -Bromotoluene reacts with potassium amide to give a...

Key Concepts

Benzyne Mechanism

Under extremely strong basic conditions, such as those provided by potassium amide, aromatic compounds can undergo an elimination reaction that expels a leaving group and a hydrogen from an adjacent carbon, resulting in the formation of a highly reactive intermediate known as benzyne. This intermediate possesses a strained triple bond on the aromatic ring, and its formation bypasses the traditional addition?elimination mechanism seen in many aromatic nucleophilic substitutions.

Aromatic Nucleophilic Substitution

Aromatic nucleophilic substitution describes a class of reactions where a nucleophile replaces a leaving group on an aromatic ring. In cases where the aromatic system lacks strongly electron-withdrawing substituents, the reaction proceeds via the elimination-addition (benzyne) pathway rather than the standard addition-elimination mechanism. The formation of a benzyne intermediate in these reactions explains the non-classical regioselectivity observed.

Substituent Effects on Regioselectivity

In reactions proceeding through the benzyne mechanism, the usual directing effects of substituents are diminished or altered because the reactive intermediate is nearly symmetric. As a result, when a substituent like a methyl group is present, the addition of the nucleophile to the benzyne intermediate can occur at positions that lead to mixtures of products. This lack of strong electronic control by the substituent explains the formation of both meta and para isomers in the reaction.

Transcript

0:00 All right.

00:01 So when we react for bromotanuin with potassium amid, which is a very strong base, right? we get a mixture of the para and meta methyl anilans, right? so we're substituting out, well, i shouldn't say we're substituting.

00:22 We are, we are.

00:24 The bromid is being replaced by an amino group.

00:28 And in one of the products it is changing its position on the ring.

00:34 So this was mentioned subtly in the chapter in section 16 .8.

00:45 Right.

00:46 So when you have a ring that has a leaving group on it and you have a strong base like potassium amid, or in this case just a mid.

01:05 So this is the conjugate base of ammonia.

01:07 Amonia has a pca of 36.

01:10 So those protons are not acidic at all.

01:12 So if you manage to pop one of those guys off, you're going to have a pretty strong base.

01:18 Right.

01:18 So what you can do is you can deprotonate hydrogen atoms on the benzene ring adjacent to decent leaving groups.