JEE Main Chemistry question

[IIT 2011] A phthalimide-type ring system is treated with (i) KOH, (ii) 4-bromobenzyl chloride. The major product is

(A)The *N*-(4-bromobenzyl)phthalimide, retaining the aryl bromide untouchedCorrect
(B)A regiochemically altered isomer with the bromine and the new C–N bond on the same ring
(C)The product of aryl C–Br displacement by the phthalimide anion, with the benzylic chloride left untouched
(D)A ring-opened phthalamic acid derivative

Answer: (A)

Solution

KOH deprotonates the phthalimide N–H to the nucleophilic phthalimide anion (a Gabriel-synthesis-type nitrogen nucleophile), which then performs an Sɴ2 displacement at the more reactive benzylic chloride carbon (leaving the unreactive aryl C–Br bond on the other ring

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Question ID VK-140209