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Distinguish between ethylamine, diethylamine and triethylamine by using Hinsberg's reagent?
Ans. This reaction is useful for the distinction of primary, secondary and tertiary amines. (i) Primary amine (like ethyl amine) is treated with Hinsberg's reagent (benzene sulphonyl chloride) forms N-alkyl benzene sulphonamide which dissolve in aqueous KOH solution to form a clear solution of potassium salt and upon acidification gives insoluble N-alkyl benzene sulphonamide.

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(ii) Secondary amine like diethyl amine is treated with benzene sulphonyl chloride forms N,N-diethyl benzene which sulphonyl amide remains insoluble in aqueous KOH and does not dissolve in acid.

So as given in the reaction, when we react N-Ethyl benzene sulphonamide with KOH it form a potassium salt. When the potassium salt is acidified, why is the N-alkyl benzene sulphonamide insoluble when it was earlier soluble?

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  • $\begingroup$ In what solvent? $\endgroup$
    – Waylander
    Commented Feb 18 at 8:16
  • $\begingroup$ Potassium hydroxide solution, presumably $\endgroup$
    – Gaurav Sai Maddipati
    Commented Feb 18 at 8:24
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    $\begingroup$ Please substitute images with typed text. You might keep relevant images of structures but at least add typed text. $\endgroup$
    – Buck Thorn ♦
    Commented Feb 18 at 8:52
  • $\begingroup$ Why would you expect an organic compound like N-ethyl benzenesulfonamide to be aqueous soluble? The potassium salt is soluble because it is an ionic species, the neutral compound will not be. $\endgroup$
    – Waylander
    Commented Feb 18 at 11:07
  • $\begingroup$ @Waylander For the record the schematic reports that the K-salt is insoluble. That might be the source of the confusion. $\endgroup$
    – Buck Thorn ♦
    Commented Feb 23 at 6:40

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