Read the following passage and answer questions.

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like $-CH_3$ group, there is no staric hinderance to H-bonding. In case the alkyl group is bigger than $-CH_3$ group, there will be steric hinderance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from $-CH_3$ to $-C_2H_5$ results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous medium.

Hinsberg's reagent is:

Benzenesulphonyl chloride

The correct answer is Option (4) → Benzenesulphonyl chloride

Hinsberg's reagent is benzenesulphonyl chloride (C₆H₅SO₂Cl). It is used in the Hinsberg test to distinguish between primary, secondary, and tertiary amines.

• Primary amines react with benzenesulphonyl chloride in the presence of NaOH to form sulphonamides that are soluble in alkali.

• Secondary amines form sulphonamides that are insoluble in alkali.

• Tertiary amines do not react with Hinsberg’s reagent.

• OTHER OPTIONS

• OPTION 1 - 2,4-dinitrophenol – not used in the Hinsberg test.

• OPTION  2 - NaNOâ‚‚ + HCl – used in diazotization reactions.

• OPTION 3 - HNO₃ + Hâ‚‚SOâ‚„ mixture – nitrating mixture used in nitration reactions.

Therefore, Hinsberg's reagent is Benzenesulphonyl chloride.