JEE Advanced 2022 · Paper 1 · Chemistry Q13

ChemistryMultiple correct+4 / -2Question 49 of 54

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

P can be reduced to a primary alcohol using

\ce{NaBH4}

Treating P with conc.

\ce{NH4OH}

solution followed by acidification gives Q.

Treating Q with a solution of

\ce{NaNO2}

in aq. HCl liberates

\ce{N2}

P is more acidic than

\ce{CH3CH2COOH}

Answer and worked solution

Official answer: B, C, D

P. The Hell-Volhard-Zelinsky reaction gives 2-bromopropanoic acid, $\ce{CH3CHBrCOOH}$ .

Q. The Gabriel synthesis (potassium phthalimide, then hydrolysis) gives alanine, $\ce{CH3CH(NH2)COOH}$ , along with phthalic acid.

The options.

(A) False: $\ce{NaBH4}$ does not reduce carboxylic acids.
(B) True: ammonia displaces bromide, giving alanine.
(C) True: a primary aliphatic amine and $\ce{HNO2}$ give $\ce{N2}$ .
(D) True: the $\alpha$ -Br withdraws electrons, so P is more acidic than propanoic acid.

JEE Advanced 2022 · Paper 1, Chemistry question 13