Gabriel phthalimide synthesis is used for the preparation of:

The correct answer is Option (1) → Primary aliphatic amines

In the Gabriel phthalimide synthesis, the N–H hydrogen of phthalimide is acidic because the nitrogen is flanked by two C=O groups. Ethanolic KOH removes it, and the phthalimide anion is alkylated by an alkyl halide. Since nitrogen then carries no hydrogen, it cannot be alkylated a second time. Alkaline hydrolysis finally releases the amine:

    \( \displaystyle \ce{C6H4(CO)2NH + KOH} \xrightarrow{\text{ethanol}} \ce{C6H4(CO)2N^-K+ + H2O} \)

    \( \displaystyle \ce{C6H4(CO)2N^-K+ + R-X -> C6H4(CO)2N-R + KX} \)

    \( \displaystyle \ce{C6H4(CO)2N-R + 2NaOH(aq) -> R-NH2 + C6H4(COONa)2} \)

So the method gives a pure primary aliphatic amine, free from secondary and tertiary amines. Aryl halides do not react with the phthalimide anion, so aromatic primary amines such as aniline cannot be made this way.

Why the other options are incorrect

• Option (2) Secondary aliphatic amines — Incorrect. A secondary amine needs a second alkyl group on nitrogen, but the N-alkylphthalimide has no N–H left to react.

• Option (3) Primary aromatic amines — Incorrect. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, so primary aromatic amines such as aniline cannot be made by this method.

• Option (4) Secondary aromatic amines — Incorrect. Both steps it would need fail: an aryl halide does not react with the phthalimide anion, and the N-substituted phthalimide has no N–H left for a second group.

NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis

Note: The same/similar question has been asked in different years by NTA as per the details given below: