JEE Main 2026 Apr 6 Shift 2, Chemistry Q18: Preparation of Amines

JEE Main2026Apr 6, Shift 2Chemistry

The number of compounds from the following which can undergo reaction with Br $_{2}$ /KOH (alcoholic) to give respective products, and these respective products can also be obtained separately by Gabriel phthalimide reaction, is:

Solution

Hofmann bromamide (Br $_{2}$ /KOH) requires an unsubstituted primary amide (R–CONH $_{2}$ ) and gives the primary amine with one carbon less. Gabriel synthesis gives primary aliphatic amines only (R must be a substrate amenable to S $_{N}$ 2 on phthalimide salt — aryl halides do not work).

Check each amide:

(i) PhCONH $_{2}$ → aniline. Aniline is NOT accessible by Gabriel (S $_{N}$ 2 on Ph–X fails). Excluded.

(ii) PhCH $_{2}$ CONH $_{2}$ → PhCH $_{2}$ NH $_{2}$ (benzylamine). Benzylamine is obtainable by Gabriel (PhCH $_{2}$ X + phthalimide salt). Counts.

(iii) CH $_{3}$ CONH $_{2}$ → CH $_{3}$ NH $_{2}$ . Methylamine is accessible by Gabriel (CH $_{3}$ X). Counts.

(iv) Cyclohexyl–CO–NH–CH $_{2}$ CH $_{3}$ : N is substituted, so Hofmann does not yield a primary amine. Excluded.

(v) (CH $_{3}$ ) $_{3}$ C–CO–NH–CH $_{3}$ : same problem — N is substituted. Excluded.

(vi) C $_{6}$ H $_{11}$ –CONH $_{2}$ (cyclohexanecarboxamide) → cyclohexylamine. Cyclohexylamine is obtainable by Gabriel (cyclohexyl bromide + phthalimide salt — secondary alkyl halide, gives moderate yield). Counts.

Total = 3.

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JEE Main 2026 Apr 6 Shift 2, Chemistry Q18: Preparation of Amines

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