IIT-JEE 2011 · Paper 1 · Chemistry Q7

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ChemistrySingle correct+3 / -1Question 7 of 69

The major product of the following reaction is

Phthalimide treated with (i) KOH, (ii) 4-bromobenzyl chloride (Br-C₆H₄-CH₂Cl).

N-[4-(Chloromethyl)phenyl]phthalimide (N bonded directly to the ring).

O-alkylated isomer: 3-(4-bromobenzyloxy)isoindol-1-one (C=N in the ring, O-CH₂-C₆H₄Br).

O-arylated isomer: 3-[4-(chloromethyl)phenoxy]isoindol-1-one.

Answer and worked solution

Official answer: A

Gabriel synthesis. KOH deprotonates phthalimide; the imide anion attacks the alkyl halide through nitrogen in an SN2 reaction. Only the benzylic C-Cl reacts; the aryl C-Br is inert to SN2.

Product N-(4-bromobenzyl)phthalimide.

(A)

Topics:Amines SN1 and SN2

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IIT-JEE 2011 · Paper 1, Chemistry question 7