JEE Main Chemistry question

Salicylic acid (ring–OH, –COOH) --HNO₃--> X; --Br₂, water--> Y. X and Y respectively are

(A)Picric acid; 2,4,6-tribromophenol
(B)5-nitrosalicylic acid; 5-bromosalicylic acidCorrect
(C)o-nitrophenol; o-bromophenol
(D)3,5-dinitrosalicylic acid; 3,5-dibromosalicylic acid

Answer: (B)

Solution

Salicylic acid's phenolic –OH is a strong activating, o/p-director, while the –COOH is deactivating/meta-directing; the phenolic ring's activation dominates, so electrophilic substitution (nitration with dilute HNO₃, or bromination with Br₂ water) occurs preferentially at the position para/ortho to –OH that is not

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Question ID VK-149478