Question

(b) (i) Describe any one method for the identification of primary, secondary and tertiary amines. Also write the chemical reactions. (ii) Write a short note on Gabriel Phthalimide Synthesis. (3+2=5)
OR
How will you obtain (write chemical equations only): (i) Benzene diazonium chloride from Aniline (ii) Methylamine from Acetamide (iii) Aniline from Nitrobenzene. (2+1½+1½=5)

Hint:

For Option 1 use Hinsberg's reagent (C₆H₅SO₂Cl) and judge solubility in KOH; recall Gabriel synthesis gives only primary aliphatic amines. For Option 2 think diazotisation, Hoffmann bromamide degradation, and reduction of nitrobenzene.

Answer 1

Option 1 (alternative treatment):

(i) Distinguishing the three amines: The standard single test is Hinsberg's test with benzenesulphonyl chloride, C₆H₅SO₂Cl. Everything follows from how many hydrogens sit on nitrogen.
• A 1° amine (R-NH₂) still keeps one N-H after reacting, and that N-H is made acidic by the neighbouring -SO₂- group; the sulphonamide forms a salt and dissolves in KOH: C₆H₅SO₂NHR + KOH → C₆H₅SO₂NRK + H₂O (clear solution).
• A 2° amine (R₂NH) loses its only N-H on substitution, so C₆H₅SO₂NR₂ has no acidic proton and stays as an insoluble solid in KOH.
• A 3° amine (R₃N) has no N-H at all and cannot form a sulphonamide, so nothing happens.
Summary of observations: clear solution means primary; insoluble solid means secondary; no reaction means tertiary. (The carbylamine test, R-NH₂ + CHCl₃ + 3KOH → R-NC + 3KCl + 3H₂O, is a confirmatory test for 1° amines only.)

(ii) Gabriel synthesis: a way to obtain primary amines free from 2°/3° contamination. Potassium phthalimide (from phthalimide + KOH) acts as a nitrogen nucleophile toward an alkyl halide, giving N-alkylphthalimide. Cleaving this imide by aqueous acid or base hydrolysis, or better by hydrazine (hydrazinolysis), releases R-NH₂ and returns the phthalic acid (or its hydrazide). Because the key step is an S_N2 displacement on the halide, it works for alkyl halides but fails for aryl halides, so aromatic primary amines cannot be prepared by it.

Option 2 (alternative wording of the routes):
(i) Aniline is diazotised with nitrous acid (NaNO₂ + HCl formed in situ) kept at 273–278 K: C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl^− + 2H₂O.
(ii) Acetamide is degraded with bromine and alkali (Hoffmann's method), losing one carbon to give methylamine: CH₃CONH₂ + Br₂ + 4KOH → CH₃NH₂ + K₂CO₃ + 2KBr + 2H₂O.
(iii) Nitrobenzene is reduced to aniline in acidic medium: C₆H₅NO₂ + 6[H] →(Sn/HCl) C₆H₅NH₂ + 2H₂O, or catalytically with H₂/Ni.