(a) Give one chemical test to distinguish between the following pairs of compounds:

(i) Methylamine and dimethylamine.

(ii) Aniline and benzylamine

(b) Write the structures of different isomers corresponding to the molecular formula C3H9N, which will liberate nitrogen gas on treatment with nitrous acid.

a(i)

• Methylamine is an aliphatic primary amine and dimethylamine is a secondary amine, hence they can be distinguish on the basis of their degree and corresponding reaction.

• Carbylamine test might be the most suitable test to distinguish between them in which the only aliphatic primary amine reacts on heating with chloroform in alcoholic KOH medium which produces foul-smelling isocyanides or carbylamines.

• Hence the foul smell is the physical indicator of the reaction and methylamine responds to it while dimethylamine does not which clearly the difference between them.

CH₃-NH₂ + CHCl₃ + 3KOH → CH₃-NC + 3KCl + 3H₂

Methylamine(1°) Methyl isocyanide(foul smell)

(CH₃)₂NH + CHCl₃ + 3KOH → No reaction.

a(ii)

• Aniline and benzylamine this two aromatic compounds can be distinguished by their characteristic reactions towards nitrous acid (diazotization) followed by test of coupling reaction.

• Aniline on reaction with HNO₂ gives a very stable diazonium salt which on coupling/azo-dye test obviously gives azo dye (the colour and texture is the identification)

• whereas same reaction of HNO₂ andbenzylamine gives rise to a very unstable diazonium salt which ultimately produces benzyl alcohol (with water) with the evolution of N₂ gas which is not the case for aniline.

therefore, does not give positive azo-dye test which is the clear distinguishing test for both.

(b) There are 4 types of different structures as follows;

CH₃-CH₂-CH₂_NH₂

Propan-1-amine(Primary)

N-methylethanamine(Secondary)

N, N-Dimethyl methanamine(Tertiary)