Give one chemical test to distinguish between the following pairs of compounds:
$(i)$ Methylamine and dimethylamine
$(ii)$ Secondary and tertiary amines
$(iii)$ Ethylamine and aniline
$(iv)$ Aniline and benzylamine
$(v)$ Aniline and $N$-methylaniline.

(N/A) $(i)$ Methylamine and dimethylamine can be distinguished by the carbylamine test. Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. Methylamine (being an aliphatic primary amine) gives a positive carbylamine test,while dimethylamine does not.
$CH_3NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} CH_3NC + 3KCl + 3H_2O$
$(CH_3)_2NH + CHCl_3 + 3KOH \xrightarrow{\Delta} \text{No reaction}$
$(ii)$ Secondary and tertiary amines can be distinguished by Hinsberg's test using benzenesulphonyl chloride $(C_6H_5SO_2Cl)$. Secondary amines react to form a product insoluble in alkali,whereas tertiary amines do not react.
$(iii)$ Ethylamine and aniline can be distinguished using the azo-dye test. Aniline (aromatic primary amine) reacts with $HNO_2$ at $0-5^{\circ}C$ to form a diazonium salt,which couples with $2$-naphthol to form an orange dye. Ethylamine (aliphatic primary amine) does not form a stable diazonium salt and evolves $N_2$ gas.
$(iv)$ Aniline and benzylamine can be distinguished by their reaction with nitrous acid $(HNO_2)$. Benzylamine reacts with $HNO_2$ to form an unstable diazonium salt,which decomposes to give benzyl alcohol and $N_2$ gas. Aniline forms a stable diazonium salt at $0-5^{\circ}C$.
$(v)$ Aniline and $N$-methylaniline can be distinguished using the carbylamine test. Aniline (primary amine) gives a positive carbylamine test,while $N$-methylaniline (secondary amine) does not.