Question

$\text{CH}_3\text{CHO}$ + $\overset{\text{(iv) C}_2\text{H}_6\text{OH}}{\leftarrow}$ $\text{N}_2^+\text{Cl}^-$

Br

Answer 1

The final product is bromobenzene, produced via the Sandmeyer reaction.

Answer 2

A key step in many aromatic substitution syntheses is the conversion of an arenediazonium salt into a halo‐substituted arene via the Sandmeyer reaction. In the scheme shown the benzene–diazonium chloride (–N₂⁺Cl⁻, prepared earlier from an aniline derivative) is converted to bromobenzene by treatment with a “Br⁻” source (usually in the presence of CuBr as catalyst). (The preceding steps – for example, the use of CH₃CHO and the ethanolic reagent – are part of the overall synthesis but the final substitution step is the Sandmeyer reaction.)

Thus, the reaction depicted at the end:

$$\text{Ar–N}_2^+ \text{Cl}^- \xrightarrow{\text{CuBr}} \text{Ar–Br} + N_2\uparrow$$

identifies the conversion of the diazonium salt to bromobenzene.

Core Explanation:

1. A diazonium salt (Ar–N₂⁺Cl⁻) is prepared (from an aniline derivative, not shown).

2. On treatment with CuBr (or a bromide source), the diazonium group is replaced by Br via the Sandmeyer reaction.

3. Hence, bromobenzene is obtained.