Which of the following amines can be prepared by Gabriel synthesis?
(A) Isobutyl amine
(B) 2-Phenyl ethylamine
(C) N-Methyl benzyl amine
(D) Aniline
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) only
Gabriel phthalimide synthesis gives only primary aliphatic amines: potassium phthalimide is alkylated by an alkyl halide (an SN2 step), and alkaline hydrolysis then frees \( \ce{R-NH2} \).
(A) Isobutylamine, \( \ce{(CH3)2CHCH2NH2} \) — Can be prepared. Primary amine from isobutyl bromide.
(B) 2-Phenylethylamine, \( \ce{C6H5CH2CH2NH2} \) — Can be prepared. Primary amine; the halide \( \ce{C6H5CH2CH2Br} \) reacts readily at its \( \ce{CH2-Br} \) carbon.
(C) N-Methylbenzylamine, \( \ce{C6H5CH2NHCH3} \) — Cannot be prepared. It is a secondary amine; Gabriel synthesis gives only primary amines.
(D) Aniline — Cannot be prepared. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, so aromatic primary amines cannot be made this way.
NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis – “Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.”