Ethanoic acid, on heating with ammonia, forms compound A, which on treatment with bromine and sodium hydroxide, gives compound B. Compound B on treatment with NaNO2dil.HCl, gives compound C. The compounds A, B and C respectively are

A
Ethanamide, methanamine, methanol
B
Propanamide, ethanamine, ethanol
C
N-ethylpropanamide, methaneisonitrile, methanamine
D
Ethanamine, bromoethane, ethanediazonium chloride

Solution:

${CH}_{3} COOH + {NH}_{3} \overset{∆}{\to} \underset{ethanamide (A)}{{CH}_{3} {CONH}_{2}} {CH}_{3} {CONH}_{2} + {Br}_{2} + 3 NaOH \to \underset{methyl amine (B)}{{CH}_{3} {NH}_{2}} + {Na}_{2} {CO}_{3} + NaBr + H_{2} O {CH}_{3} {NH}_{2} \to_{({NaNO}_{2} / dil . HCl)}^{{HNO}_{2}} \underset{methanol (C)}{{CH}_{3} OH}$

Ethanoic acid, on heating with ammonia, forms compound A, which on treatment with bromine and sodium hydroxide, gives compound $B$ . Compound B on treatment with $\frac{{NaNO}_{2}}{dil . HCl}$ , gives compound C. The compounds A, B and $C$ respectively are

Solution:

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