An aromatic compound 'X' on treatment with aqueous ammonia and heating forms the compound 'Y' which on heating with \( Br_2 \) and KOH forms a compound 'Z' of the molecular formula \( C_6H_7N \). The structures of X, Y and Z can be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → X = benzoic acid, Y = benzamide, Z = aniline
Work backwards from Z. The formula \( \ce{C6H7N} \) is \( \ce{C6H5NH2} \), aniline. The reagents \( \ce{Br2} \) and KOH acting on Y point to the Hofmann bromamide degradation, which turns an amide into a primary amine with one carbon atom less. So Y is the seven-carbon amide benzamide:
\( \displaystyle \ce{C6H5CONH2 + Br2 + 4KOH -> C6H5NH2 + K2CO3 + 2KBr + 2H2O} \)
An amide is obtained when a carboxylic acid is treated with ammonia and the ammonium salt formed is heated, which drives out water. So X is benzoic acid:
\( \displaystyle \ce{C6H5COOH + NH3 -> C6H5COONH4} \)
\( \displaystyle \ce{C6H5COONH4} \xrightarrow{\Delta} \ce{C6H5CONH2 + H2O} \)
So X, Y and Z are benzoic acid, benzamide and aniline, the set in Option (1).
Why the other options are incorrect
Option (2) X = benzonitrile, Y = benzamide, Z = aniline — Incorrect. Y and Z are right, but a nitrile is not converted into an amide by aqueous ammonia; partial hydrolysis of \( \ce{C6H5CN} \) needs an acid or a base.
Option (3) X = aniline, Y = benzoic acid, Z = benzaldehyde — Incorrect. Z must contain nitrogen (\( \ce{C6H7N} \)); benzaldehyde is \( \ce{C7H6O} \). Ammonia also cannot turn aniline into benzoic acid.
Option (4) X = benzamide, Y = acetophenone, Z = aniline — Incorrect. Acetophenone has no nitrogen and is not an amide, so it cannot undergo the Hofmann reaction to give aniline, and benzamide does not give acetophenone with ammonia.