Compound A, C4H9NO, gives amine B with Br2 + NaOH. B on treatment with NaNO2/HCl gives propan-2-ol with evolution of N2. A is:

2-methylpropanamide, (CH3)2CHCONH2

N,N-dimethylacetamide, CH3CON(CH3)2

2-methylpropanamide, (CH3)2CHCONH2

N-methylpropanamide, CH3CH2CONHCH3

JEE Advanced levelSingle correct optionDifficulty 3 of 3nc018

Compound A, $C X_{4} H X_{9} NO$ , gives amine B with $Br X_{2} + NaOH$ . B on treatment with $NaNO X_{2} / HCl$ gives propan-2-ol with evolution of $N X_{2}$ . A is:

An aliphatic primary amine with nitrous acid gives the alcohol with the same carbon skeleton, so B is propan-2-amine

(CH X_{3}) X_{2} CHNH X_{2}

. Hofmann degradation adds one carbon back: the amide is

(CH X_{3}) X_{2} CH - CONH X_{2}

, 2-methylpropanamide, of formula

C X_{4} H X_{9} NO

. Butanamide would give propan-1-amine, and N-substituted amides do not undergo the reaction.