Gabriel phthalimide synthesis method is used for the preparation of:

The correct answer is Option (2) → Primary Amine

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

• Option (3) Tertiary Amine — Incorrect. For the same reason, three alkyl groups cannot be put on nitrogen.

• Option (4) Quaternary amine — Incorrect. Quaternary salts come from exhaustive alkylation of amines, as in ammonolysis with excess alkyl halide, not from this method.

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: