Q.Write reaction of p-toluenesulfonyl chloride with diethylamine.

p-Toluenesulphonyl chloride (Hinsberg's reagent) gives a KOH-soluble sulfonamide with a primary amine but a KOH-insoluble one with a secondary amine.

p-Toluenesulphonyl chloride ($p$-$CH_3C_6H_4SO_2Cl$), also called Hinsberg's reagent, reacts with amines to form sulfonamides, and the solubility of the product in alkali distinguishes primary from secondary amines (the Hinsberg test).

With ethylamine (a primary amine):

$C_2H_5NH_2 + ClSO_2C_6H_4CH_3 \rightarrow C_2H_5-NH-SO_2C_6H_4CH_3 + HCl$

The product, N-ethyl-p-toluenesulphonamide, still has one hydrogen on nitrogen. This N–H has become acidic due to the strongly electron-withdrawing sulfonyl group, so it is readily removed by KOH, and the resulting sodium/potassium salt is soluble in the alkaline solution (clear solution forms).

With diethylamine (a secondary amine):

$(C_2H_5)_2NH + ClSO_2C_6H_4CH_3 \rightarrow (C_2H_5)_2N-SO_2C_6H_4CH_3 + HCl$

The product, N,N-diethyl-p-toluenesulphonamide, has no N–H remaining (the single hydrogen the secondary amine had is used up in forming the sulfonamide bond), so there is no acidic proton for KOH to remove — the product remains insoluble in KOH (separates as an oily layer/precipitate).

This differing solubility (soluble for 1° amine products, insoluble for 2° amine products, and 3° amines not reacting at all) is the basis of the Hinsberg test used to distinguish primary, secondary and tertiary amines.