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Orthoboric Acid

A compound Z ...

Question

A compound Z with molecular formula $C_{3} H_{9} N$ reacts with $C_{6} H_{5} S O_{2} C l$ to give a solid, insoluble in alkali. Identify $Z .$

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Solution

Identifying the compound $Z$

Benzene sulphonyl chloride $(C_{6} H_{5} S O_{2} C l),$ which is also known as Hinsberg reagent, reacts with primary and secondary amines to form sulphonamides.

In the reaction with secondary amine, sulphonamide formed is $N, N - dialkyl benzene sulphonamide.$ It does not contain any hydrogen atom attached to nitrogen atom. So it is not acidic and hence insoluble in alkali. Hence, $Z$ must be a secondary amine i.e. $N - Methylethanamine.$

Reaction of Benzene sulphonyl chloride with N-Methylethanamine

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Orthoboric Acid

Standard XII Chemistry

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