JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · Aldehydes Ketones and Carboxylic Acids

Considering the reaction sequence given below, the correct statement(s) is(are)

(A)P can be reduced to a primary alcohol using $NaBH_{4}$ .
(B)Treating P with conc. $NH_{4} OH$ solution followed by acidification gives Q.✓ Correct
(C)Treating $Q$ with a solution of $NaNO_{2}$ in aq. HCl liberates $N_{2}$ .✓ Correct
(D) $P$ is more acidic than $CH_{3} CH_{2} COOH$ .✓ Correct

Answer: (B), (C), (D)

Solution

From the reaction sequence,

P

is benzoic acid and

Q

is aniline.

(A) Benzoic acid (

P

) cannot be reduced to a primary alcohol using

NaBH_{4}

NaBH_{4}

does not reduce carboxylic acids. Incorrect.

(B) Treating benzoic acid with conc.

NH_{4} OH

followed by heating gives benzamide, which on Hofmann degradation gives aniline (

Q

). Correct.

(C) Treating aniline (

Q

) with

NaNO_{2}

in aq. HCl forms benzenediazonium chloride, which can decompose to liberate

N_{2}

. Correct.

(D) Benzoic acid is more acidic than propanoic acid due to the electron-withdrawing effect of the benzene ring. Correct.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 1 · One or more correct options