Amines Part 1

The amine on reaction with benzene sulphonyl chloride (Heisenberg reagent) produces a compound insoluble in alkali. It indicates the amine is a $2^{\circ}$ amine. i.e. all options are possible except option (b) which is a $1^{\circ}$ amine $(\mathrm{CH_3CH_2NH_2})$. As this $2^{\circ}$ amine can be prepared by ammonolysis of ethyl chloride, the $2^{\circ}$-amine should have at least one ethyl $(\mathrm{C_2H_5})$ group. (a) $\mathrm{Ph-NH-CH_2CH_2CH_3}$ does not have ethyl group. (c) $\mathrm{CH_3CH_2CH_2-NH-CH_3}$ does not have ethyl group. (d) $\mathrm{CH_3CH_2CH_2-NH-CH_2CH_3}$ has one ethyl group. So, option (d) is the correct answer. Preparation $\mathrm{CH_3CH_2CH_2-NH-CH_2CH_3}$ by ammonolysisof $\mathrm{CH_3CH_2Cl}$