JEE Main Chemistry question

A deuterium-labelled phenol derivative (ring–D, ortho to OD) --NaOH--> [Intermediate] --(i) CO₂ (ii) D⁺--> A. Here, A is

(A)A salicylic-acid-type product retaining both ring-D and the newly installed carboxyl group, with the acidic proton replaced by D (Kolbe–Schmitt-type carboxylation under D-labelled conditions)Correct
(B)A salicylic-acid-type product with the ring-D lost
(C)A salicylic-acid-type product, alternate regiochemistry, ring-D retained
(D)The reaction does not proceed under these labelled conditions

Answer: (A)

Solution

This is the Kolbe–Schmitt reaction: sodium phenoxide (formed from the phenol with NaOH) reacts with CO₂ under pressure/heat to install a carboxylate ortho to the original oxygen (via a six-membered cyclic transition state), and acidic

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Question ID VK-149482