JEE Advanced 2022 — Paper 1 — CHEM-II solution

JEE Advanced2022Paper 1CHEM-II

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.
C
Treating Q with a solution of NaNO $_{2}$ in aq. HCl liberates N $_{2}$ .
D
P is more acidic than CH $_{3}$ CH $_{2}$ COOH.

Solution

Step 1 is the Hell-Volhard-Zelinsky (HVZ) reaction: $α$ -bromination of propanoic acid gives P = 2-bromopropanoic acid (CH $_{3}$ CHBrCOOH).

The Gabriel phthalimide synthesis then converts the alkyl halide centre to a primary amine. After base hydrolysis and acid workup, Q = alanine (2-aminopropanoic acid, CH $_{3}$ CH(NH $_{2}$ )COOH), with phthalic acid as the byproduct.

(A) NaBH $_{4}$ is too weak to reduce carboxylic acids; P retains its –COOH group. Wrong.

(B) Treating an $α$ -haloacid with concentrated ammonia followed by workup also gives the $α$ -amino acid (Q) via SN2 displacement of Br by NH $_{3}$ . Correct.

(C) Alanine has a primary aliphatic amine. With NaNO $_{2}$ /HCl at low temperature, primary aliphatic amines give an unstable diazonium salt that decomposes to liberate N $_{2}$ . Correct.

(D) The electron-withdrawing $-$ I effect of the $α$ -Br stabilises the conjugate base of P, making 2-bromopropanoic acid more acidic than propanoic acid. Correct.

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