Identify the compound that will react with Hinsberg’s reagent to give a solid which dissolves in alkali.

Correct Answer: CH₃-CH₂-NH₂ (Ethanamine / Ethylamine)

Explanation:
Hinsberg's reagent is benzene sulfonyl chloride (C₆H₅SO₂Cl). It is used to distinguish between primary (1°), secondary (2°), and tertiary (3°) amines based on their reaction behavior and the solubility of the resulting product in alkali.

1. Reaction with Primary Amines (1° Amines):
A primary amine, such as ethanamine (CH₃-CH₂-NH₂ shown in the correct option), reacts with benzene sulfonyl chloride to form N-ethylbenzenesulfonamide (a solid product):
C₆H₅SO₂Cl + CH₃CH₂NH₂ → C₆H₅SO₂NH-CH₂CH₃ + HCl
Since the sulfonamide product contains an acidic hydrogen atom attached to the electronegative nitrogen atom (which is adjacent to the highly electron-withdrawing sulfonyl group), it reacts with alkali (like KOH or NaOH) to form a water-soluble potassium/sodium salt. Hence, the solid dissolves in alkali.

2. Reaction with Secondary Amines (2° Amines):
A secondary amine, such as ethylmethylamine (CH₃-CH₂-NH-CH₃), reacts with Hinsberg's reagent to form a N,N-disubstituted sulfonamide:
C₆H₅SO₂Cl + (CH₃CH₂)(CH₃)NH → C₆H₅SO₂N(CH₂CH₃)(CH₃) + HCl
This product does not have any acidic hydrogen attached to the nitrogen atom. Therefore, it is insoluble in alkali.

3. Reaction with Tertiary Amines (3° Amines):
Tertiary amines, such as diethylmethylamine ((CH₃CH₂)₂N-CH₃), do not have any hydrogen atoms attached to nitrogen and do not react with Hinsberg's reagent under ordinary conditions.

Comparing the compounds from the options:
• Option 1 represents nitroethane (CH₃-CH₂-NO₂), which is not an amine.
• Option 2 represents a secondary amine (CH₃-CH₂-NH-CH₃).
• Option 3 (Correct Option) represents a primary amine (CH₃-CH₂-NH₂).
• Option 4 represents a tertiary amine ((CH₃-CH₂)₂N-CH₃).

Therefore, the compound that reacts with Hinsberg’s reagent to give a solid which dissolves in alkali is the primary amine, CH₃-CH₂-NH₂.