The reagent used for the distinction of 1^{ci | vedclass

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Question

(C) The reagent used to distinguish between primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines is benzenesulfonyl chlorid

Vedclass · Accepted Answer

Benzenesulfonyl chloride ($C_6H_5SO_2Cl$,Hinsberg reagent)

Vedclass · Answer

(C) The reagent used to distinguish between primary $(1^{\circ})$,secondary $(2^{\circ})$,and tertiary $(3^{\circ})$ amines is benzenesulfonyl chloride $(C_6H_5SO_2Cl)$,commonly known as the Hinsberg reagent.$1$. Primary $(1^{\circ})$ amines react with benzenesulfonyl chloride to form $N$-alkylbenzenesulfonamide,which is soluble in alkali due to the presence of an acidic hydrogen atom on the nitrogen.$2$. Secondary $(2^{\circ})$ amines react to form $N$,$N$-dialkylbenzenesulfonamide,which is insoluble in alkali because it lacks an acidic hydrogen atom on the nitrogen.$3$. Tertiary $(3^{\circ})$ amines do not react with benzenesulfonyl chloride in the absence of an acidic hydrogen atom on the nitrogen,and thus remain insoluble in the reaction mixture.

List-$I$ (Reagents used)	List-$II$ (Compound with functional group detected)
$A.$ Alkaline solution of copper sulphate and sodium citrate	$I.$ $C_6H_5CH(OH)CH_3$
$B.$ Neutral $FeCl_3$ solution	$II.$ $CH_3CH(NH_2)CH_3$
$C.$ Alkaline chloroform solution	$III.$ $C_6H_5CH_2CHO$
$D.$ Potassium iodide and sodium hypochlorite	$IV.$ $m-C_2H_5C_6H_4OH$

The reagent used for the distinction of $1^{\circ}$,$2^{\circ}$ and $3^{\circ}$ amines is

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