Amines Part 3

Concept:The Hofmann bromamide degradation converts an amide ($\mathrm{RCONH_2}$) to a primary amine with one less carbon atom ($\mathrm{RNH_2}$) using $\mathrm{Br_2}$ and $\mathrm{KOH}$.Explanation:The final compound $\mathrm{R}$ has formula $\mathrm{C_6H_7N}$.Its degree of unsaturation is $4$, indicating a benzene ring.Thus, $\mathrm{R}$ must be aniline ($\mathrm{C_6H_5NH_2}$).Since $\mathrm{R}$ is formed from $\mathrm{Q}$ by Hofmann degradation, $\mathrm{Q}$ must be benzamide ($\mathrm{C_6H_5CONH_2}$).Benzamide is prepared by treating benzoic acid ($\mathrm{C_6H_5COOH}$, compound $\mathrm{P}$) with hot aqueous ammonia.Thus, $\mathrm{P}$ = benzoic acid, $\mathrm{Q}$ = benzamide, $\mathrm{R}$ = aniline.Answer:Option B: Benzoic acid, benzamide, aniline.