Question

Distinguish between methanamine, dimethanamine and trimethanamine using Hinsberg’s reagent.

Answer 1

1. Methanamine reacts with benzenesulphonyl chloride (Hinsberg’s reagent) to give N-methyl benzenesulphonamide which is soluble in alkali.
   \[\ce{\underset{\text{Methanamine}}{CH3NH2} + C6H5SO2Cl ->[OH-] \underset{\text{N-Methyl benzenesulphonamide (Soluble in alkali)}}{C6H5SO2NHCH3}}\]
2. Dimethanamine reacts with benzenesulphonyl chloride (Hinsberg’s reagent) to give a compound which does not dissolve in alkali.
   \[\ce{\underset{\text{Dimethanamine}}{(CH3)2NH} + C6H5SO2Cl ->[OH-] \underset{\text{N, N-Dimethylbenzene sulphonamide (Insoluble in alkali)}}{C6H5SO2N(CH3)2}}\]
3. Trimethanamine does not react with benzenesulphonyl chloride (Hinsberg’s reagent) due to the absence of H-atom attached N-atom.