How does benzene sulphonyl chloride react with 3°amines ?
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If this question is being asked with respect to the Hinsberg Test of 1°, 2° and 3° Amines, then we have a simple answer.
Under suitable conditions in the Hinsberg Test, the Hinsberg reagent, that is Benzene Sulphonyl Chloride along with Sodium Hydroxide doesn't react with 3° Amines and they are insoluble in it.
However under some special conditions Benzene Sulphonyl Chloride may react with 3° Amines, but in case of the specific conditions in Hinsberg Test, it doesn't react with it.
Speaking of Hinsberg Test, on reaction with 1° amines the compound formed is soluble and further reacts with an alkali (KOH) and thus a test is performed to show that there is a primary amine present.
In the case of 2° amines, the compound formed is also soluble but it doesn't react with the alkali and thus it can be shown that there's a secondary amine present.
Finally since 3° amines do not react with Benzene Sulphonyl Chloride (again under these conditions) due to lack of acidic hydrogen connected to nitrogen in amine, they are insoluble and hence show that there's a tertiary amine present.
Under suitable conditions in the Hinsberg Test, the Hinsberg reagent, that is Benzene Sulphonyl Chloride along with Sodium Hydroxide doesn't react with 3° Amines and they are insoluble in it.
However under some special conditions Benzene Sulphonyl Chloride may react with 3° Amines, but in case of the specific conditions in Hinsberg Test, it doesn't react with it.
Speaking of Hinsberg Test, on reaction with 1° amines the compound formed is soluble and further reacts with an alkali (KOH) and thus a test is performed to show that there is a primary amine present.
In the case of 2° amines, the compound formed is also soluble but it doesn't react with the alkali and thus it can be shown that there's a secondary amine present.
Finally since 3° amines do not react with Benzene Sulphonyl Chloride (again under these conditions) due to lack of acidic hydrogen connected to nitrogen in amine, they are insoluble and hence show that there's a tertiary amine present.
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