Question

An organic compound $“\mathrm{A}”$ on treatment with $\mathrm{benzene}\mathrm{sulphonyl}\mathrm{chloride}$gives compound $\mathrm{B}.\mathrm{B}$ is soluble in $\mathrm{dil}.\mathrm{NaOH}$ solution. Compound $\mathrm{A}$ is:

• A. ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{N}-{\left({\mathrm{CH}}_3\right)}_2$
• B. ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{NHCH}}_2{\mathrm{CH}}_3$
• C.

  

• D. ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_2{\mathrm{NHCH}}_3$

Answer 1

The correct option is C

The correct option is C.

The explanation for the correct option :

Hinsberg reagent (Benzene sulphonyl chloride) is used to distinguish $1°,2°\mathrm{and}3°$amine. In the Hinsberg test, an amine is reacted with a benzene sulfonyl chloride. If a product forms, the amine is either a primary or secondary amine, because tertiary amines do not form stable sulphonamides. If the sulfonamide that forms dissolves in aqueous sodium hydroxide solution, it is a primary amine. If the sulfonamide is insoluble in aqueous sodium hydroxide, it is a secondary amine.

In the reaction scheme given below the amine act as a nucleophile and attacks the electrophilic benzene sulphonyl chloride. As a result of this reaction, the chloride is displaced and N-alkylbenzene sulphonamide (B) is produced. This compound (B) is soluble in an aqueous sodium hydroxide (NaOH) when the primary amine produces a sulphonamide.

The scheme of the reaction of structure in option C is as follows:

Hinsberg reagent (Benzene sulphonyl chloride) gives a reaction product with $1°$amine and it is soluble in dil. NaOH.

Hence, the compound $\mathrm{A}$ is the given structure in option C.

Explanation for incorrect options :

Options A, B, and D are incorrect.

Hinsberg reagent (Benzene sulphonyl chloride) doesn't give a reaction product with $3°$amine and is insoluble in dil. NaOH for $2°$ amine.

Hence, the structures given in options A, B and D are incorrect.