Answer
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Hint: Hinsberg test differentiates primary, secondary amines, and tertiary amines as the primary amines form products with Hinsberg reagents that are base soluble. The amine having N atom bonded to only one alkyl chain is base soluble.
Complete step by step answer:
Before jumping to the options, first, we need to see what the Hinsberg reagent is and what its function is.
So, the Hinsberg reagent is an alternative name for benzene sulphonyl chloride. The chemical formula for benzene sulphonyl chloride can be ${{C}_{6}}{{H}_{5}}S{{O}_{2}}Cl$.
It is an organosulphur compound. So, now the question arises what is the organosulphur compound.
The organosulphur compounds are the compounds in which Sulphur is attached with organo i.e. hydrocarbon part. It is similar to the organozinc compound in which zinc is attached to the hydrocarbon part. Hinsberg test is commonly used to differentiate between primary, secondary, and tertiary amines. The differences are observed in the solubility of the sulphonamide product in the base. In the case of primary and secondary amines, the sulphonamides formed are precipitates and of these, the product with primary amine is base soluble.
Now, let’s see our options. The option a.) is Neopentyl amine which has the structure as-
The Neopentyl amine is a primary amine. Thus, it gives the base soluble product.
Now, let’s see option b.) Secpropyl amine. It has structure-
This is also a primary amine. This will also give the base soluble product.
Moving to option c.) that is diethylamine. The diethylamine has structure-
This is secondary amine and it will not form a product that is base soluble. So, this option is not correct.
Now the option d.) is ethyl methylamine which has structure-
This is a tertiary amine and even this will not form a product that is base soluble.
So, the correct answer is “Option A and B”.
Note: In the Hinsberg test, the amines act as nucleophiles and attack the Sulphonyl Chloride which acts as electrophile resulting in the displacement of chloride. As a result, sulphonamides are generated.
Complete step by step answer:
Before jumping to the options, first, we need to see what the Hinsberg reagent is and what its function is.
So, the Hinsberg reagent is an alternative name for benzene sulphonyl chloride. The chemical formula for benzene sulphonyl chloride can be ${{C}_{6}}{{H}_{5}}S{{O}_{2}}Cl$.
It is an organosulphur compound. So, now the question arises what is the organosulphur compound.
The organosulphur compounds are the compounds in which Sulphur is attached with organo i.e. hydrocarbon part. It is similar to the organozinc compound in which zinc is attached to the hydrocarbon part. Hinsberg test is commonly used to differentiate between primary, secondary, and tertiary amines. The differences are observed in the solubility of the sulphonamide product in the base. In the case of primary and secondary amines, the sulphonamides formed are precipitates and of these, the product with primary amine is base soluble.
Now, let’s see our options. The option a.) is Neopentyl amine which has the structure as-
The Neopentyl amine is a primary amine. Thus, it gives the base soluble product.
Now, let’s see option b.) Secpropyl amine. It has structure-
This is also a primary amine. This will also give the base soluble product.
Moving to option c.) that is diethylamine. The diethylamine has structure-
This is secondary amine and it will not form a product that is base soluble. So, this option is not correct.
Now the option d.) is ethyl methylamine which has structure-
This is a tertiary amine and even this will not form a product that is base soluble.
So, the correct answer is “Option A and B”.
Note: In the Hinsberg test, the amines act as nucleophiles and attack the Sulphonyl Chloride which acts as electrophile resulting in the displacement of chloride. As a result, sulphonamides are generated.
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