Question

An organic compound X having molecular formula ${C_4}{H_{11}}N$ reacts with $p - \,$ toluene sulphonyl chloride to form a compound $\,Y\,$ that is soluble in aqueous $\,KOH\,$. Compound $\,X\,$ is optically active and reacts with acetyl chloride to form compound $Z$. Identify the compound $Z$.
A. $C{H_3} - C{H_2} - C{H_2} - C{H_2} - NH - CO - C{H_3}$
B. $C{H_3} - C{H_2} - CH(C{H_3}) - NH - CO - C{H_3}$
C. $C{H_3} - CH(C{H_3}) - C{H_2} - NH - CO - C{H_3}$
D. $C{H_3} - C{(C{H_3})_2} - NH - CO - C{H_3}$

Answer 1

Hint: $\,p - \,$ toluene sulphonyl chloride is known as Hinsberg’s reagent. It is an alternate name for benzene sulfonyl chloride . Here, toluene is the benzene used in the reagent. This reagent undergoes a reaction with reactive compounds comprising $\, - OH - \,$and $\, - NH - \,$ bonds..

Complete step by step answer:
First, let us look into the reaction of $\,p - \,$ toluene sulphonyl chloride with the amines. As we saw, this type of reaction is done under the ‘Hinsberg test’ let us first understand what is Hinsberg’s test;
The Hinsberg test is a chemical reaction used to discriminate between primary, secondary and tertiary amines. The German chemist, Oscar Heinrich Daniel Hinsberg, described this reaction.
First, we need to understand the type of amine involved in this reaction. For this we have to see the reaction of primary amine, secondary amine and tertiary amine with this $\,p - $ toluenesulfonyl chloride;
When the compound reacts with primary amine in the end, they will form a compound soluble in a base. When the compound reacts with the secondary amine the compound formed will not react further with $\,KOH\,$. And for the tertiary amine there will be no reaction as there is no free $\,H\,$.
When ${C_4}{H_{11}}N$ reacts with $\,p - \,$ toluene sulphonyl chloride the compound $\,Y\,$ is formed which is soluble in $\,KOH\,$, so for this $\,X\,$ should be a primary amine,
${C_4}{H_{11}}N$+ $\,p - \,$ toluene sulphonyl chloride $ \to \,{C_6}{H_5}S{O_2}NHCH(C{H_3})C{H_2}C{H_3}$
Now, \[C{H_3}C{H_2}^{}CH(C{H_3}) - N{H_2}\, + \,ClCOC{H_3}\]$ \to $ $C{H_3} - C{H_2} - CH(C{H_3}) - NH - CO - C{H_3}$
So, the compound $\,Z\,$ is $C{H_3} - C{H_2} - CH(C{H_3}) - NH - CO - C{H_3}$.
For a compound to be soluble in $\,KOH\,$ and that reacts with $\,p - \,$ toluene sulphonyl chloride they will react with primary amines only.
$\,p - \,$ toluene sulphonyl chloride is white in colour and mainly used for the organic synthesis and also it is a derivative of toluene.

So, Option B is correct.

Note: Only the primary amines will take part in the Hinsberg reaction and the product will be soluble in base, and the base may be $\,NaOH\,$ or $\,KOH\,$. The amines operate as nucleophiles in the Hinsberg Test and attack the electrophile (sulfonyl chloride). This contributes to the chloride being displaced and the sulfonamides being formed. This sulfonamide compound is not soluble as they are formed from primary and secondary amines, and it precipitates as a solid from the solution.