Question

${{\text{C}}_3}{{\text{H}}_9}{\text{N}}$reacts with Hinsberg reagent and the product is insoluble in alkali but soluble in ether. This nitrogen containing compound is:
A. primary amine
B. secondary amine
C. tertiary amine.
D. nitrite compound

Answer 1

Hint: Primary, secondary, and tertiary amines are traced and discerned using Hinsberg test. Benzene sulfonyl chloride is the Hinsberg reagent. It does not react with tertiary amines properly. Along with the Hinsberg reagent, aqueous ${\text{KOH}}$ is also added. So based on the solubility of the product formed, three of them can be differentiated.

Given data:
Given that a nitrogen containing compound reacts with Hinsberg reagent which gives an alkali insoluble compound.

Complete step by step solution:
Amines are organic compounds having one or more nitrogen groups. Amines are divided into primary, secondary, tertiary, and aromatic amines. Primary, secondary, tertiary amines are detected and distinguished by the Hinsberg test. This was introduced by Oscar Heinrich Daniel Hinsberg.
Hinsberg reagent is the benzene sulfonyl chloride, ${\text{PhS}}{{\text{O}}_2}{\text{Cl}}$. Benzene sulfonic acid or its salt is chlorinated with phosphorus oxychloride which produces benzene sulfonyl chloride.
${\text{PhS}}{{\text{O}}_3}^ - {\text{N}}{{\text{a}}^ + }\xrightarrow[{{\text{POC}}{{\text{l}}_3}}]{{{\text{PC}}{{\text{l}}_5}}}{\text{PhS}}{{\text{O}}_2}{\text{Cl}}$
Benzene sulfonyl chloride.
This benzene sulfonyl chloride is reacted with the amine ${{\text{C}}_3}{{\text{H}}_9}{\text{N}}$in the presence of aqueous KOH. It should be shaken well.
A. When primary amine is reacted with Hinsberg reagent, it forms a sulfonamide which on reaction with base gives sulfonamide salt. It is water-soluble.
${\text{PhS}}{{\text{O}}_2}{\text{N}}\left( {\text{H}} \right){\text{R}} + {\text{NaOH}} \to {\text{N}}{{\text{a}}^ + }\left[ {{\text{PhS}}{{\text{O}}_2}{\text{N}}{{\text{R}}^ - }} \right] + {{\text{H}}_2}{\text{O}}$
Sulfonamide Sulfonamide salt
B. There is a direct formation of an alkali insoluble sulfonamide when secondary amine is treated with Hinsberg reagent.

${\text{PhS}}{{\text{O}}_2}{\text{Cl}} + 2{\text{RR}}'{\text{NH}} \to {\text{PhS}}{{\text{O}}_2}{\text{NRR}}' + \left[ {{\text{RR}}'} \right.{\left. {{\text{N}}{{\text{H}}_2}} \right]^ + }{\text{C}}{{\text{l}}^ - }$
${\text{PhS}}{{\text{O}}_2}{\text{Cl}}$-Benzene sulfonyl chloride
Here ${\text{R}}$is the ${\text{C}}{{\text{H}}_3}$group and ${\text{R}}'$is the ${{\text{C}}_2}{{\text{H}}_5}$group
C. Tertiary amines can react with Hinsberg reagent under certain conditions only. Reaction speed, concentration, solubility and temperature have to be considered.
${\text{PhS}}{{\text{O}}_2}{\text{Cl}} + {{\text{R}}_3}{\text{N}} + {{\text{H}}_2}{\text{O}} \to {{\text{R}}_3}{\text{N}}{{\text{H}}^ + }\left[ {{\text{PhS}}{{\text{O}}_3}^ - } \right] + {\text{HCl}}$
This also gives water-soluble sulfonate salts.

Hence the option B is correct.

Additional information:
Here the nucleophile is amine and the electrophile is sulfonyl chloride. Nucleophile amine is attacking the electrophile sulfonyl chloride. In the case of primary amines, when treated with Hinsberg reagent, the alkali helps to lose the proton and forms a water soluble sulfonamide salt.

Note: Sulfonyl chloride undergoes hydrolysis when reacted with tertiary amines which produces water soluble sulfonate salts. Differentiation of primary, secondary, tertiary amines are based on solubility of the product in alkali.