The test to distinguish primary, secondary and tertiary amines is:

First, let us recall the need of the question. We must find a single laboratory test that can distinguish among the three classes of amines - primary (1°), secondary (2°) and tertiary (3°).

We have four named possibilities in the options:

1. Sandmeyer’s reaction
2. Carbylamine reaction
3. Iodoform test
4. Use of $$C_6H_5SO_2Cl$$ (benzenesulfonyl chloride, the Hinsberg reagent)

Now we examine each option one by one and note what it actually tests for.

Sandmeyer’s reaction. This is written as, for example, $$ArN_2^+Cl^- \rightarrow ArX$$ in the presence of cuprous halides. It is employed with diazonium salts, which are typically prepared from primary aromatic amines only. It does not give any comparative information when secondary or tertiary amines are present together. Therefore it cannot be a single discriminating test for all three classes.

Carbylamine reaction. The reaction is stated as $$ RNH_2 + CHCl_3 + 3\,KOH \rightarrow RNC + 3\,KCl + 3\,H_2O, $$ and it produces the foul-smelling isocyanide (carbylamine) only from primary amines. Secondary and tertiary amines simply do not respond. Hence, although it identifies primary amines, it fails to say anything about secondary versus tertiary, and therefore cannot be the desired single test.

Iodoform test. The condition for a positive iodoform test is the presence of a $$CH_3CO-$$ (in carbonyl compounds) or $$CH_3CH(OH)-$$ group (in alcohols), nothing to do with $$NH_2$$ substitution patterns. Amines in general do not participate. Consequently this option is irrelevant here.

The reagent $$C_6H_5SO_2Cl$$ (benzenesulfonyl chloride). This reagent is commonly known as Hinsberg’s reagent. Let us state Hinsberg’s test in full because it is the key idea:

For a generic amine $$RNH_2$$ (primary), $$R_2NH$$ (secondary) or $$R_3N$$ (tertiary), the following behaviour is observed when the amine is treated with $$C_6H_5SO_2Cl$$ in the presence of aqueous or alcoholic $$KOH$$ and the product is subsequently subjected to acidification.

• For a primary amine: $$ RNH_2 + C_6H_5SO_2Cl \;\xrightarrow{KOH}\; C_6H_5SO_2NHR \; (\text{sulphonamide}) + HCl $$ The product retains one acidic hydrogen on nitrogen, therefore the sulphonamide is soluble in excess alkali as $$ C_6H_5SO_2NHR + KOH \rightarrow C_6H_5SO_2NR^-K^+ + H_2O. $$ Upon later acidification it precipitates again. Thus the sequence “dissolves in alkali, re-precipitates on acidification” is a sure sign of a primary amine.

• For a secondary amine: $$ R_2NH + C_6H_5SO_2Cl \;\xrightarrow{KOH}\; C_6H_5SO_2NR_2 + HCl $$ There is no acidic hydrogen left on nitrogen in the product, so the sulphonamide remains insoluble in the alkaline medium; it simply precipitates and stays precipitated. That behavior is characteristic of secondary amines.

• For a tertiary amine: $$ R_3N + C_6H_5SO_2Cl \;\text{(no reaction)} $$ The tertiary amine lacks an $$N-H$$ bond, so it does not attack Hinsberg’s reagent at all. The original amine remains unchanged and, being a free base, it dissolves in the acid layer on acidification instead of forming any sulphonamide. This “no reaction” outcome uniquely identifies a tertiary amine.

We therefore see that one single experiment with $$C_6H_5SO_2Cl$$ followed by appropriate treatment lets us distinguish all three classes in a clear, mutually exclusive fashion. None of the other three options provides this triple discrimination.

So, the reagent $$C_6H_5SO_2Cl$$—that is, the Hinsberg test—is the required choice.

Hence, the correct answer is Option 4.