QJEEM21Chae41: JEE Main 2021 Chemistry - Class 12 JEEM21Chae41 Solution: Match List - I and List - II. .tg…

Question

Verified Solution • 2025–26 Session moderate

Match List - I and List - II.

		List-II
(a)	(i)	$\mathrm{Br}_{2} / \mathrm{NaOH}$
(b)	(ii)	$\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{BaSO}_{4}$
(c)	(iii)	$\mathrm{Zn}(\mathrm{Hg}) / \mathrm{H}$ Conc. $\mathrm{HCl}$
(d)	(iv)	$\mathrm{Cl}_{2}$ Red $\mathrm{P}, \mathrm{H}_{2} \mathrm{O}$

Choose the correct answer from the options given below :

Accepted Answer

(A) Alkyl chloride reacts with H₂/Pd-BaSO₄ and reduced to alkyl aldehyde. This is known as Rosenmund reduction.

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 147 English Explanation 1

(B) Carboxylic acid reacts with Cl₂/P in aqueous medium to form 2-chlorocarboxylic acid. This reaction is known as HVZ reaction.

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 147 English Explanation 2

(C) Alkyl amide reacts with Br₂ in presence of NaOH to give alkyl amine. This reaction is known as Hofmann bromide reaction,

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 147 English Explanation 3

(D) Ketone reacts with Zn-Hg in presence of conc.HCl to give alkane. This reaction is known as Clemmensen reduction.

JEE Main 2021 (Online) 24th February Evening Shift Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 147 English Explanation 4

Hence, correct match is (A)-2, (B)-4, (C)-1, (D)-3.

Step-by-Step Verified Solution