Solution

Hinsberg's reagent is used to distinguish between primary, secondary and tertiary amine.

$$\beta$$−Naphthol is used for the distinction between aliphatic and aromatic amine. Where aromatic amine undergoes diazotization followed by $$\beta$$−Naphthol to form an azo dye but aliphatic amine do not form an azo dye.

Aniline $$(C_6H_5NH_2)$$, on diazotization forms diazonium salt and with $$\beta$$-naphthol gives azo dye.

But, when ethylamine $$(C_2H_5NH_2)$$ reacts with $$HNO_2$$ (diazotisation), $$-NH_2$$ is replaced by $$-OH$$ group forming alcohol $$(C_2H_5OH)$$, which do not form azo dye with $$\beta$$−Naphthol.

Hence, option $$B$$ is correct.