JEE Main 2022 · 29 June · morning · Single correct · +4 / −1
Given below are two statements :
Statement I : The esterification of carboxylic acid with an alcohol is a nucleophilic acyl substitution.
Statement II : Electron withdrawing groups in the carboxylic acid will increase the rate of esterification reaction.
Choose the most appropriate option :
- (a) Both Statement I and Statement II are correct.
- (b) Both Statement I and Statement II are incorrect.
- (c) Statement I is correct but Statement II is incorrect.
- (d) Statement I is incorrect but Statement II is correct.
Answer and solution
Answer (a) Both Statement I and Statement II are correct.
Solution
Chemistry ma’am · 3 steps
Esterification is attack on the carbonyl carbon, so making that carbon more positive speeds it up.
- Statement I. In Fischer esterification the alcohol oxygen attacks the carbonyl carbon of the protonated acid. The -OH of the acid then leaves as water, so one group on the acyl carbon is swapped for another. That is nucleophilic acyl substitution, so Statement I is correct.
RCOOH + R'OH ⇌(H⁺) RCOOR' + H₂O
- Statement II. The slow step is the alcohol attacking the carbonyl carbon. An electron withdrawing group on R pulls electron density away from that carbon and makes it more electrophilic. The alcohol attacks it more readily, so the rate goes up and Statement II is correct.
- So. Both statements are correct, which is option (a).
The idea to keep. In any nucleophilic acyl substitution, ask how positive the carbonyl carbon is. Electron withdrawing groups speed up attack. Electron donating and bulky groups slow it down.
Answer from the NTA final answer key. Solution by PrepWiser.