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 +1 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 |
Hoffmann reaction Gabriel synthesis Kolbe reaction Sandmeyer reaction |
Gabriel synthesis |
The correct answer is Option (2) → Gabriel synthesis Core Concept Gabriel synthesis is used for preparation of primary amines using phthalimide. Reaction Phthalimide → (KOH) → Potassium phthalimide Potassium phthalimide + R–X → N-alkyl phthalimide N-alkyl phthalimide → (Hydrolysis) → R–NH₂ + Phthalic acid Explanation of Each Option 1. Hoffmann reaction converts amides into amines with one carbon less and does not involve phthalimide. R–CONH₂ → (Br₂ / KOH) → R–NH₂ 2. Gabriel synthesis involves alkylation of potassium phthalimide followed by hydrolysis to give primary amine and phthalic acid, matching the given reaction. Phthalimide → (KOH) → Potassium phthalimide Potassium phthalimide + R–X → N-alkyl phthalimide N-alkyl phthalimide → (Hydrolysis) → R–NH₂ + Phthalic acid 3. Kolbe reaction involves electrolysis of carboxylate salts and forms alkanes. 2 RCOO⁻ → (Electrolysis) → R–R + 2 CO₂ 4. Sandmeyer reaction involves diazonium salts converting into aryl halides. C₆H₅N₂⁺Cl⁻ → (CuCl / CuBr / CuCN) → C₆H₅Cl / C₆H₅Br / C₆H₅CN |