What is the action of following on ethylamine –(i) Hinsberg reagent(ii) Nitrous acid(b) Write increasing order of basicity for CH3NH2, (CH3)2NH and (CH3)3N

Answer:

(a) (i) Hinsberg reagent (benzenesulfonyl chloride): Ethylamine (a 1° amine) forms N-ethylbenzenesulfonamide, which dissolves in aqueous KOH because it has an acidic H on nitrogen:
C₆H₅SO₂Cl + H₂N–C₂H₅ → C₆H₅SO₂NH–C₂H₅ + HCl

(ii) Nitrous acid (NaNO₂ + HCl): Ethylamine gives ethyl alcohol with brisk effervescence of nitrogen gas:
C₂H₅NH₂ + HONO → C₂H₅OH + N₂↑ + H₂O

(b) In aqueous solution: (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH. The +I effect of methyl groups increases basicity, but in water the cation is also stabilised by hydrogen bonding (solvation); the tertiary ammonium ion has only one N–H for H-bonding and the bulky groups cause steric hindrance, so trimethylamine is the weakest of the three. (In the gas phase, where only the inductive effect operates, the order is CH₃NH₂ < (CH₃)₂NH < (CH₃)₃N.)