JEE Main Chemistry question

Salicylic acid is treated with bromine under two different conditions. [Y] ←Br₂/Water– Salicylic acid –Br₂ in CH₃COOH→ [X]. Predict the nature of [X] and [Y] in the above reactions.

(A)[X] = 3,5-dibromosalicylic acid; [Y] = 2,4,6-tribromophenol (with loss of CO₂)Correct
(B)[X] = 2,4,6-tribromophenol; [Y] = 3,5-dibromosalicylic acid
(C)Both are 3,5-dibromosalicylic acid
(D)Both are 2,4,6-tribromophenol

Answer: (A)

Solution

In a non-aqueous, weaker-electrophile medium (bromine in acetic acid, milder conditions), simple ring bromination occurs at the positions activated by the phenolic –OH (ortho/para to it, avoiding the –COOH-occupied position), giving the dibromo-substituted salicylic acid, X. In aqueous bromine water (a more vigorous,

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Question ID VK-149485