A compound ' $X$ ' on treatment with $\mathrm{Br}_{2} / \mathrm{NaOH}$, provided $\mathrm{C}_{3} \mathrm{H}_{9} \mathrm{~N}$, which gives positive carbylamine test. Compound ' X ' is:

(C) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CONH}_{2}$

(A) $\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{NHCH}_{3}$

(B) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{2} \mathrm{NH}_{2}$

(D) $\mathrm{CH}_{3} \mathrm{CON}\left(\mathrm{CH}_{3}\right)_{2}$

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Compounds Containing Nitrogen

A compound '

X

' on treatment with

Br_{2} / NaOH

, provided

C_{3} H_{9} N

, which gives positive carbylamine test. Compound ' X ' is:

Answer: (C)

The Hofmann bromamide degradation (

Br X_{2} / NaOH

) turns a primary amide into a primary amine with one fewer carbon. The amine formed,

C X_{3} H X_{9} N

, gives a positive carbylamine test, so it is the primary amine propylamine,

CH X_{3} CH X_{2} CH X_{2} NH X_{2}

. Working back, the starting amide must have had four carbons,

CH X_{3} CH X_{2} CH X_{2} CONH X_{2}

(butanamide), which on degradation loses one carbon to give propylamine.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option