What is the source of nitrogen in Gabriel synthesis of primary aliphatic amines?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Potassium phthalimide
Gabriel phthalimide synthesis gives pure primary amines, free from secondary and tertiary amines. Phthalimide is first treated with ethanolic KOH to form potassium phthalimide. The phthalimide anion is a nucleophile; it displaces the halide from an alkyl halide to give an N-alkylphthalimide. Alkaline hydrolysis then sets the primary amine free. The nitrogen of the amine comes from potassium phthalimide, and since this nitrogen carries no H after alkylation, it cannot be alkylated a second time:
\( \displaystyle \ce{C6H4(CO)2NH} \xrightarrow{\text{ethanolic KOH}} \ce{C6H4(CO)2N^-K^+} \)
\( \displaystyle \ce{C6H4(CO)2N^-K^+ + R-X -> C6H4(CO)2N-R + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-R + 2NaOH} \xrightarrow{\text{aq.}} \ce{R-NH2 + C6H4(COONa)2} \)
So the source of nitrogen is potassium phthalimide. Aromatic primary amines cannot be made this way, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion.
Why the other options are incorrect
Option (1) Sodium azide — Incorrect. \( \ce{NaN3} \) gives alkyl azides, \( \ce{R-N3} \), which are reduced to amines in a separate method (azide synthesis); it is not used in the Gabriel synthesis.
Option (2) Sodium nitrile — Incorrect. There is no such reagent. (Sodium nitrite, \( \ce{NaNO2} \), is used with HCl to diazotise amines, not to make them.)
Option (3) Potassium cyanide — Incorrect. KCN converts R–X into a nitrile, \( \ce{R-CN} \), whose reduction gives \( \ce{R-CH2NH2} \), an amine with one carbon more; this is a different route.
NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis
