COMEDK 2025 Morning Solved Paper

Concept:Hofmann bromamide degradation converts an amide to a primary amine with one fewer carbon.Explanation:Compound $[\mathrm{X}]$ has formula $\mathrm{C}_5\mathrm{H}_{11}\mathrm{NO}$.Reaction with $\mathrm{Br}_2$/aq. NaOH is the Hofmann rearrangement, giving a primary amine $[\mathrm{Y}]$ with one less carbon atom.$[\mathrm{Y}]$ reacts with $\mathrm{CHCl}_3$ and ethanolic KOH to produce a foul‑smelling isocyanide — this is the carbylamine test, confirming $[\mathrm{Y}]$ is a primary amine.$[\mathrm{Y}]$ also reacts with $\mathrm{HONO}$ to give butan‑1‑ol and $\mathrm{N}_2$ gas.This diazotisation shows $[\mathrm{Y}]$ is a primary amine that yields a 4‑carbon alcohol (butan‑1‑ol).Therefore, $[\mathrm{Y}]$ must be $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{NH}_2$ (n‑butylamine).Since Hoffmann degradation gives an amine with one less carbon, $[\mathrm{X}]$ is the amide with 5 carbons: $\mathrm{CH}_3(\mathrm{CH}_2)_3\mathrm{CONH}_2$ (pentanamide).Answer:Option A: $\mathrm{CH}_3-\left(\mathrm{CH}_2\right)_3-\mathrm{CONH}_2$