An organic compound A on treatment with benze | vedclass

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Question

(D) The reaction of an organic compound $A$ with benzene sulphonyl chloride (Hinsberg's reagent) to form a product $B$ that is soluble in $NaOH$ indic

Vedclass · Accepted Answer

$C_6H_5-CH(CH_3)-NH_2$

Vedclass · Answer

(D) The reaction of an organic compound $A$ with benzene sulphonyl chloride (Hinsberg's reagent) to form a product $B$ that is soluble in $NaOH$ indicates that $A$ is a primary amine $(R-NH_2)$.Primary amines react with benzene sulphonyl chloride to form $N$-alkylbenzene sulphonamide,which contains an acidic hydrogen atom attached to the nitrogen,making it soluble in $NaOH$.Secondary amines form $N,N$-dialkylbenzene sulphonamides,which lack an acidic hydrogen and are insoluble in $NaOH$.Tertiary amines do not react with benzene sulphonyl chloride.Among the given options:$A$: $C_6H_5-N(CH_3)_2$ is a tertiary amine.$B$: $C_6H_5-NH-CH_2-CH_3$ is a secondary amine.$C$: $C_6H_5-CH_2-NH-CH_3$ is a secondary amine.$D$: $C_6H_5-CH(CH_3)-NH_2$ is a primary amine.Therefore,the correct compound is $C_6H_5-CH(CH_3)-NH_2$.

$Test/Method$	$Reagent$
$(i)$ Lucas Test	$(a)$ $C_6H_5SO_2Cl/aq. KOH$
$(ii)$ Dumas method	$(b)$ $HNO_3/AgNO_3$
$(iii)$ Kjeldahl's method	$(c)$ $CuO/CO_2$
$(iv)$ Hinsberg Test	$(d)$ Conc. $HCl$ and $ZnCl_2$
	$(e)$ $H_2SO_4$

An organic compound $A$ on treatment with benzene sulphonyl chloride gives compound $B$. $B$ is soluble in dil. $NaOH$ solution. Compound $A$ is:

Similar Questions

Which of the following properties is $NOT$ applicable to $CH_3CN$ (Ethanenitrile)?

How can primary amines be distinguished from secondary and tertiary amines?

Match the following:
$Test/Method$ $Reagent$
$(i)$ Lucas Test $(a)$ $C_6H_5SO_2Cl/aq. KOH$
$(ii)$ Dumas method $(b)$ $HNO_3/AgNO_3$
$(iii)$ Kjeldahl's method $(c)$ $CuO/CO_2$
$(iv)$ Hinsberg Test $(d)$ Conc. $HCl$ and $ZnCl_2$
$(e)$ $H_2SO_4$

Which of the following is correctly matched?

Which of the following reactions is exhibited only by primary amines?

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