Q1(b). One benzyne from two dichlorobenzenes

Introduction. An aryl chloride without a strongly electron-withdrawing group does not undergo addition–elimination. With the very strong base KNH₂ (NH₂⁻ in liquid NH₃) it reacts by elimination–addition: the base removes a ring H ortho to Cl, chloride leaves, and a benzyne (aryne) is formed. Both dichlorobenzenes give the same aryne, 3-chlorobenzyne, so both give the same product.

Formation of 3-chlorobenzyne

Scheme: Both substrates lose HCl from adjacent carbons and reach the same aryne.

1. 1,2-Dichlorobenzene: NH₂⁻ removes H-3 (or the equivalent H-6), a ring H next to Cl (made acidic by the -I effect of Cl). The aryl anion expels the neighbouring chloride (C-2) and gives a benzyne whose triple bond is C2≡C3, with Cl on C-1: 3-chlorobenzyne.
2. 1,3-Dichlorobenzene: H-2 lies between two Cl atoms and is the most acidic ring H (two -I effects). It is removed first; the anion loses one of the flanking chlorides, and the triple bond forms between C-1 and C-2 with the other Cl next to it: again 3-chlorobenzyne.

Addition of amide: why the meta product

Scheme: Amide adds so that the new carbanion sits next to Cl, where -I stabilises it.

The aryne triple bond is a strained in-plane π bond, and NH₂⁻ adds to one of its two carbons. Adding at the carbon remote from Cl puts the negative charge (an sp² in-plane carbanion) on the carbon adjacent to Cl, where the inductive effect of chlorine stabilises it. Addition the other way would place the charge one carbon further away. The anion then takes a proton from NH₃.

Conclusion. 1,2- and 1,3-Dichlorobenzene both lose HCl to give the same intermediate, 3-chlorobenzyne. Amide adds to it regioselectively (carbanion next to Cl), so both give 3-chloroaniline (m-chloroaniline).

Application. The same aryne logic explains the cine substitution found in Roberts' 14C-labelled chlorobenzene experiment and is used to make meta-substituted anilines that ordinary substitution cannot give.