Propanenitrile is reduced with LiAlH4 to give A. Propanamide is treated with Br2 + NaOH to give B. The correct relation between A and B is:

A is propan-1-amine and B is ethanamine, so A is the next higher homologue of B

A is a secondary amine and B is a primary amine

JEE Advanced levelSingle correct optionDifficulty 3 of 3nc014

Propanenitrile is reduced with $LiAlH X_{4}$ to give A. Propanamide is treated with $Br X_{2} + NaOH$ to give B. The correct relation between A and B is:

AA and B are the same compound
BA is propan-1-amine and B is ethanamine, so A is the next higher homologue of B
CA and B are positional isomers
DA is a secondary amine and B is a primary amine

Reduction of

CH X_{3} CH X_{2} CN

keeps all carbons and adds one

CH X_{2}

to the chain end, giving

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

(propan-1-amine). Hofmann degradation of

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

loses one carbon and gives

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

(ethanamine). Both are primary amines, and A has one more

CH X_{2}

than B.