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Question: \(A\xrightarrow{Ph-S{{O}_{2}}Cl}B\xrightarrow{KOH}C\xrightarrow{{{C}_{2}}{{H}_{5}}-I}D\) 'C' is w...

A→Ph−SO2ClB→KOHC→C2H5−IDA\xrightarrow{Ph-S{{O}_{2}}Cl}B\xrightarrow{KOH}C\xrightarrow{{{C}_{2}}{{H}_{5}}-I}D
'C' is water-soluble.
The correct structure of A and D are?
(A)- R−NH2,Ph−SO2−N(R)−(C2H5)2+I−R-N{{H}_{2}},Ph-S{{O}_{2}}-N(R)-({{C}_{2}}{{H}_{5}})_{2}^{+}{{I}^{-}}
(B)- R−NH−R,Ph−SO2−NR2−(C2H5)R-NH-R,Ph-S{{O}_{2}}-N{{R}_{2}}-({{C}_{2}}{{H}_{5}})
(C)- R−NH2,Ph−SO2−N(R)−IR-N{{H}_{2}},Ph-S{{O}_{2}}-N(R)-I
(D)- R2NH,Ph−SO2−NR2−(C2H5)+I−{{R}_{2}}NH,Ph-S{{O}_{2}}-N{{R}_{2}}-{{({{C}_{2}}{{H}_{5}})}^{+}}{{I}^{-}}

Explanation

Solution

When amine reacts with benzenesulfonyl chloride there is a formation N-Alkyl Benzene sulfonamide. Now after this it reacts with potassium hydroxide, to form a product, if this product further reacts with other reagents, then the first amine taken must be primary.

Complete step by step solution:
A→Ph−SO2ClB→KOHC→C2H5−IDA\xrightarrow{Ph-S{{O}_{2}}Cl}B\xrightarrow{KOH}C\xrightarrow{{{C}_{2}}{{H}_{5}}-I}D
This is a complete step by step reaction of Hinsberg’s test. And it is used to detect the primary amine, secondary amine, and tertiary amine.
It can be detected as, if the reaction ends at the formation of the D compound then the amine is primary. If the reaction ends at C, the amine is secondary. If the reaction doesn't take place, the amine is tertiary.
So, the above reaction takes place till D, so the amine must be primary. So when the primary amine reacts with benzenesulfonyl chloride there is a formation N-Alkyl Benzene sulfonamide. This compound is B. Now this compound N-Alkyl Benzene sulfonamide (B) reacts with potassium hydroxide, there is a formation of potassium salt. This compound is (C). This potassium salt is water-soluble. Now this compound (C) reacts with ethyl iodide, the potassium cation is replaced with ethyl cation and iodide anion. This compound is (D ). So this series of reaction is given below: R−NH2→Ph−SO2ClPh−SO2−NHR→KOHK+[PhSO2NR]−→C2H5−IPh−SO2−NR−(C2H5)2+I−R-N{{H}_{2}}\xrightarrow{Ph-S{{O}_{2}}Cl}Ph-S{{O}_{2}}-NHR\xrightarrow{KOH}{{K}^{+}}{{[PhS{{O}_{2}}NR]}^{-}}\xrightarrow{{{C}_{2}}{{H}_{5}}-I}Ph-S{{O}_{2}}-NR-({{C}_{2}}{{H}_{5}})_{2}^{+}{{I}^{-}}
So, the compound (A) is R−NH2R-N{{H}_{2}} and the compound (D) is Ph−SO2−N(R)−(C2H5)2+I−Ph-S{{O}_{2}}-N(R)-({{C}_{2}}{{H}_{5}})_{2}^{+}{{I}^{-}}.

Therefore, the correct answer is an option (A)- R−NH2,Ph−SO2−N(R)−(C2H5)2+I−R-N{{H}_{2}},Ph-S{{O}_{2}}-N(R)-({{C}_{2}}{{H}_{5}})_{2}^{+}{{I}^{-}}.

Note: The benzenesulfonyl chloride is known as Hisnberg’s reagent, and this Hisnberg’s test is based on the number of hydrogen atoms present in the amine, the more the hydrogen atoms present the more the number of reactions.