Which of the following amines cannot be prepared by Gabriel phthalimide synthesis?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → N-Methylbenzylamine
Gabriel phthalimide synthesis puts one alkyl group on the nitrogen of phthalimide and then frees it as \( \ce{R-NH2} \), so it gives only primary amines:
\( \displaystyle \ce{C6H4(CO)2NH + KOH} \xrightarrow{\text{ethanol}} \ce{C6H4(CO)2N^-K^+ + H2O} \)
\( \displaystyle \ce{C6H4(CO)2N^-K^+ + R-X} \xrightarrow{\Delta} \ce{C6H4(CO)2N-R + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-R + 2NaOH -> C6H4(COONa)2 + R-NH2} \)
N-Methylbenzylamine, \( \ce{C6H5CH2-NH-CH3} \), has two carbon groups on nitrogen: it is a secondary amine, so this method cannot make it.
Why the other options are incorrect
Option (1) Butylamine — Incorrect. \( \ce{CH3CH2CH2CH2NH2} \) is a primary amine; it forms from 1-bromobutane as R–X.
Option (2) Isopropylamine — Incorrect. \( \ce{(CH3)2CHNH2} \) has one carbon group on nitrogen, so it is a primary amine; it forms from 2-bromopropane as R–X.
Option (3) 2-Phenylethylamine — Incorrect. \( \ce{C6H5CH2CH2NH2} \) is a primary amine whose –NH2 is not on the ring; it forms from the alkyl halide \( \ce{C6H5CH2CH2Br} \). Only aromatic primary amines, which would need an aryl halide, are out of reach.
NCERT (2026-27 edition):
9.4 Preparation of Amines – “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.”