Which of the following amines cannot be prepared by Gabriel phthalimide synthesis ?
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Aniline
The Gabriel phthalimide synthesis gives primary amines from alkyl halides. The three steps are:
\( \displaystyle \ce{C6H4(CO)2NH + KOH} \xrightarrow{\text{ethanol}} \ce{C6H4(CO)2N^-K^+ + H2O} \)
\( \displaystyle \ce{C6H4(CO)2N^-K^+ + R-X -> C6H4(CO)2N-R + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-R + 2NaOH -> C6H4(COONa)2 + R-NH2} \)
The second step is a nucleophilic substitution: the phthalimide anion replaces the halogen. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, because the C–X bond of an aryl halide has partial double bond character. So aromatic primary amines cannot be made this way.
Option (1) Ethylamine — Can be prepared by Gabriel synthesis. It is a primary aliphatic amine, made from an ethyl halide.
Option (2) Aniline — Cannot be prepared by Gabriel synthesis. It would need bromobenzene or chlorobenzene in the second step, and an aryl halide does not react with the phthalimide anion.
Option (3) Methylamine — Can be prepared by Gabriel synthesis. It is a primary aliphatic amine, made from a methyl halide.
Option (4) Isopropylamine — Can be prepared by Gabriel synthesis. It is a primary amine (the –\( \ce{NH2} \) group is on a secondary carbon), made from an isopropyl halide.
NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis
