Passage
Preparation & Chemical reactions of amines
Preparation:
Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium, or platinum as catalyst. Nitriles on reduction with lithium aluminum hydride or catalytic hydrogenation produce primary amines. The Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide.
Chemical reactions:
Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides, and esters by nucleophilic substitution reaction. This reaction is known as acylation. Both aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide give corresponding isocyanide. Benzenesulphonyl chloride also known as Hinsberg’s reagent reacts with primary and secondary amines to form sulphonamide. Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. Aniline reacts with HNO3 in the presence of H2SO4 to give nitro derivatives. Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid or sulphanilic acid.
Identify the compound which can be obtained from Gabriel phthalimide synthesis.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → benzylamine (phenylmethanamine)
The Gabriel phthalimide synthesis is used to prepare primary amines only:
Phthalimide is treated with ethanolic KOH to form potassium phthalimide.
The phthalimide anion attacks an alkyl halide (SN2) to give an N-alkylphthalimide.
Alkaline hydrolysis then releases the primary amine.
\( \displaystyle \ce{C6H4(CO)2NH + KOH} \xrightarrow{\text{ethanol}} \ce{C6H4(CO)2N^-K^+ + H2O} \)
\( \displaystyle \ce{C6H4(CO)2N^-K^+ + C6H5CH2Cl -> C6H4(CO)2N-CH2C6H5 + KCl} \)
\( \displaystyle \ce{C6H4(CO)2N-CH2C6H5 + 2NaOH} \xrightarrow{\text{heat}} \ce{C6H5CH2NH2 + C6H4(COONa)2} \)
Only one group can be placed on the nitrogen, so the product is always a primary amine. Aromatic primary amines cannot be made, because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. Benzylamine, \( \ce{C6H5CH2NH2} \), is a primary amine whose \( \ce{NH2} \) is on an sp3 \( \ce{CH2} \) carbon, made from the reactive benzyl chloride. So it can be obtained by this method.
Why the other options are incorrect
Option (1) N-methylbenzylamine — Incorrect. It is a secondary amine; the Gabriel synthesis gives only primary amines.
Option (2) aniline — Incorrect. It is an aromatic amine; chlorobenzene does not react with potassium phthalimide.
Option (4) 2-aminopyridine — Incorrect. Here too the –\( \ce{NH2} \) is on an aromatic ring carbon, which the phthalimide anion cannot attack.
NCERT (2026-27 edition):
Gabriel Phthalimide Synthesis
