Q.Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg's reagent ?
(The source's own item numbering skips 'vi' at this point -- the printed sequence runs v., then vii., with no vi. anywhere in the chapter's exercises; transcribed exactly as printed.)

Step 1. React each amine with benzenesulfonyl chloride (Hinsberg's reagent, section 13.8). Ethylamine (a primary amine) gives N-ethylbenzenesulfonamide + HCl. Diethylamine (a secondary amine) gives N,N-diethylbenzenesulfonamide + HCl. Triethylamine (a tertiary amine), having no N-H hydrogen to react with the reagent's mechanism, does not react at all.

Step 2. Test each product's solubility in aqueous NaOH. N-Ethylbenzenesulfonamide retains ONE hydrogen on nitrogen, made strongly acidic by the adjacent electron-withdrawing sulfonyl group -- this acidic N-H is deprotonated by NaOH, so the sulfonamide DISSOLVES in alkali. N,N-Diethylbenzenesulfonamide has NO hydrogen left on nitrogen at all (both original N-H hydrogens are now substituted, one by sulfonyl, one by the second ethyl group) -- with no acidic proton to remove, this sulfonamide does NOT dissolve in alkali. Triethylamine gives no product with the reagent in the first place, so there is nothing to test for alkali solubility -- the reaction mixture simply shows unreacted starting materials. …

Applying section 13.8's Hinsberg-test three-way outcome (soluble sulfonamide / insoluble sulfonamide / no reaction) speci …

• Treating 'reacts with the reagent' and 'dissolves in alkali' as the same test outcome, rather than recognising diethylamine DOES react (forms a solid) but that solid then fails the SEPARATE alkali-solubility test, while triethylamine fails at the very first s …