Read the passage carefully and answer the Questions.
Amines are classified as primary (1°), secondary (2°) and tertiary (3°) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecules. Aliphatic amines are a stronger bases due to +I effect of an alkyl groups. Intermolecular hydrogen bonding in primary and secondary amines affects the boiling points of amines. The intermolecular association is more in primary amines than in secondary amines as there are more number of hydrogen atoms available for hydrogen bond formation. Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in an acidic medium. Gabriel synthesis is used for the preparation of primary amines. Primary amines react with nitrous acid to give unstable diazonium-salts which can be used to synthesize a variety of compounds.
The reaction is an example of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Gabriel synthesis
The reaction follows the three steps of the Gabriel phthalimide synthesis, which gives a pure primary amine:
\( \displaystyle \ce{C6H4(CO)2NH + KOH -> C6H4(CO)2N^-K^+ + H2O} \)
\( \displaystyle \ce{C6H4(CO)2N^-K^+ + C2H5X -> C6H4(CO)2N-C2H5 + KX} \)
\( \displaystyle \ce{C6H4(CO)2N-C2H5 + 2H2O -> C2H5NH2 + C6H4(COOH)2} \)
Because nitrogen can take only one alkyl group here, no secondary or tertiary amine is formed. So the reaction is the Gabriel synthesis.
Why the other options are incorrect
Option (1) Hoffmann reaction — Incorrect. In the Hofmann bromamide reaction an amide is heated with bromine and NaOH and gives an amine with one carbon fewer: \( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \). No phthalimide is involved.
Option (3) Kolbe reaction — Incorrect. Sodium phenoxide is heated with carbon dioxide and then acidified to give salicylic acid (2-hydroxybenzoic acid); it makes a carboxylic acid, not an amine:
\( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2},\ \text{400 K, 4-7 atm}} o\text{-}\ce{HOC6H4COOH} \)
Option (4) Sandmeyer reaction — Incorrect. A diazonium salt is treated with \( \ce{Cu2Cl2/HCl} \) or \( \ce{Cu2Br2/HBr} \) to give an aryl halide: \( \ce{ArN2^+X^- -> ArX + N2} \).
NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis
