A compound X on treatment with Br_2/NaOH prov | vedclass

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Question

(C) The reaction of a compound with $Br_2/NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine

Vedclass · Accepted Answer

$CH_3CH_2CH_2CONH_2$

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(C) The reaction of a compound with $Br_2/NaOH$ is the Hofmann bromamide degradation reaction,which converts an amide $(RCONH_2)$ into a primary amine $(RNH_2)$ with one carbon atom less.Given that the product $C_3H_9N$ gives a positive carbylamine test,it must be a primary aliphatic amine $(CH_3CH_2CH_2NH_2)$.Since the product has $3$ carbon atoms,the starting amide must have $4$ carbon atoms.Among the options,$CH_3CH_2CH_2CONH_2$ (butanamide) is the only primary amide with $4$ carbon atoms that would yield $CH_3CH_2CH_2NH_2$ upon reaction with $Br_2/NaOH$.

$A$ compound $X$ on treatment with $Br_2/NaOH$ provided $C_3H_9N$,which gives a positive carbylamine test. Compound $X$ is

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