JEE Main 2023 · Chemistry · Carbonyl Compounds, Carboxylic Acids and Amines · medium

An amide C4H9NO undergoes Hofmann bromamide degradation to give amine X. Treatment of X with NaNO2/HCl at 273-278 K gives alcohol Y (assume no…

Single correct option

An amide

\mathrm{C_4H_9NO}

undergoes Hofmann bromamide degradation to give amine X. Treatment of X with

\mathrm{NaNO_2/HCl}

273

278\ \mathrm K

gives alcohol Y (assume no rearrangement), which answers the iodoform test. The amide is

A. 2-Methylpropanamide
B. Butanamide
C. $N$ -Methylpropanamide
D. $N,N$ -Dimethylacetamide

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Answer: (A) 2-Methylpropanamide

Hofmann degradation requires a primary amide

\mathrm{R\!-\!CONH_2}

and gives

\mathrm{R\!-\!NH_2}

(one carbon less).

N

-substituted amides (C, D) do not undergo it. For Y to give the iodoform test, it must contain

\mathrm{CH_3CH(OH)\!-\!}

. 2-Methylpropanamide gives isopropylamine, then propan-2-ol, which is iodoform-positive. Butanamide gives propylamine, then propan-1-ol, which is negative.

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