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

(i) Identification of 1°, 2° and 3° amines – Hinsberg's test: The reagent is benzenesulphonyl chloride (C₆H₅SO₂Cl, 'Hinsberg's reagent'). The amine is shaken with it and the mixture is then treated with aqueous KOH.

Primary amine: gives an N-alkylbenzenesulphonamide. The H left on nitrogen is acidic (activated by the -SO₂- group), so the product dissolves in KOH.
C₆H₅SO₂Cl + H₂N-R → C₆H₅SO₂NH-R (soluble in alkali).

Secondary amine: gives an N,N-dialkylbenzenesulphonamide. No N-H is left, so it is insoluble in KOH (a solid separates).
C₆H₅SO₂Cl + R₂NH → C₆H₅SO₂NR₂ (insoluble in alkali).

Tertiary amine: has no replaceable N-H, so it does not react and stays as an insoluble layer.
So: soluble product means 1°, insoluble product means 2°, no reaction means 3°.

(ii) Gabriel phthalimide synthesis: a route to pure primary aliphatic amines. Steps: (1) Phthalimide is treated with KOH to give potassium phthalimide. (2) This is heated with an alkyl halide (R-X) to give N-alkylphthalimide. (3) Alkaline or acidic hydrolysis (or hydrazinolysis) then sets the primary amine free together with phthalic acid.
Phthalimide →(KOH) potassium phthalimide →(R-X) N-alkylphthalimide →(hydrolysis) R-NH₂ + phthalic acid.
Aromatic primary amines cannot be made this way because aryl halides do not undergo the needed nucleophilic substitution with potassium phthalimide.

Option 2: Preparations (equations)

(i) Benzene diazonium chloride from Aniline (diazotisation): C₆H₅NH₂ + NaNO₂ + 2HCl →(273–278 K) C₆H₅N₂⁺Cl^− + NaCl + 2H₂O.
(ii) Methylamine from Acetamide (Hoffmann bromamide degradation): CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O. The amine has one carbon fewer than the amide.
(iii) Aniline from Nitrobenzene (reduction): C₆H₅NO₂ + 3H₂ →(Ni/Pt/Pd) C₆H₅NH₂ + 2H₂O. Sn + HCl or Fe + HCl also reduces nitrobenzene to aniline.