Amines Part 3

Concept:The reaction sequence involves Hoffmann bromamide degradation, diazotization, Sandmeyer reaction, hydrolysis, and oxidation. The final compound $F$ has exactly two different types of hydrogen atoms, indicating a highly symmetric para‑disubstituted benzene.Explanation:Compound $A$ with molecular formula $\mathrm{C_8H_9ON}$ is a neutral primary amide.On treatment with $\mathrm{Br_2/HO^-}$, it undergoes Hoffmann degradation to give amine $B$ (one carbon less, $\mathrm{C_7H_9N}$).$B$ is soluble in dilute acid, confirming it is an aromatic amine.$B$ with $\mathrm{NaNO_2/HCl}$ at $0-5\,{}^\circ\mathrm{C}$ forms diazonium salt $C$.$C$ reacts with $\mathrm{CuCN/NaCN}$ (Sandmeyer reaction) to give nitrile $D$.Hydrolysis of $D$ yields carboxylic acid $E$ ($\mathrm{C_7H_7O_2}$). The same $E$ is obtained by direct hydrolysis of $A$, so $E$ corresponds to the acyl part of $A$.$E$ on oxidation with acidified $\mathrm{KMnO_4}$ gives $F$.$F$ has only two types of hydrogen atoms. This occurs when $F$ is a para‑dicarboxylic acid (e.g., terephthalic acid).Therefore, $E$ must be a methylbenzoic acid, and oxidation of the methyl group gives the symmetric acid. The para isomer is the only one that yields $F$ with two equivalent ring hydrogens and two equivalent carboxyl hydrogens.Thus $A$ is 4‑methylbenzamide ($\mathrm{CH_3-C_6H_4-CONH_2}$).Shortcut:If the final oxidation product has exactly two types of $\mathrm{H}$, it must be a para‑disubstituted benzene with identical groups (like terephthalic acid). Hence the starting amide must have a methyl group para to the amide group.Answer:The correct option is the structure of 4‑methylbenzamide (para ‑ $\mathrm{CH_3}$ and $\mathrm{-CONH_2}$ on benzene ring).(Since the options are images, select the one with the amide group at position 1 and methyl group at position 4.)