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

The correct answer is:

Understanding Hinsberg's Test:
Hinsberg’s reagent is benzenesulfonyl chloride (${\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{Cl}$). This reagent is used to distinguish between primary (1°), secondary (2°), and tertiary (3°) amines based on their reactivity and the solubility of the resulting products in alkali (such as aqueous $\text{KOH}$ or $\text{NaOH}$).

Reactivity of Different Types of Amines:

1. Primary Amine (1°):
A primary amine reacts with benzenesulfonyl chloride to form an N-alkylbenzenesulfonamide. The hydrogen atom directly attached to the nitrogen in this sulfonamide is strongly acidic due to the electron-withdrawing nature of the sulfonyl group ($-{\text{SO}}_2-$). Therefore, it readily dissolves in aqueous alkali to form a soluble salt.
${\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{Cl}+\text{R}-{\text{NH}}_2\to {\text{C}}_6{\text{H}}_5{\text{SO}}_2\text{NHR}\text{(solid insoluble in water)}\stackrel{\text{KOH}}{\to }\text{Soluble salt}$

2. Secondary Amine (2°):
A secondary amine reacts to form an N,N-dialkylbenzenesulfonamide. Since there is no acidic hydrogen attached to the nitrogen atom in the resulting product, the solid formed does not dissolve in alkali.

3. Tertiary Amine (3°):
A tertiary amine does not have any hydrogen atom on the nitrogen atom and therefore does not react with benzenesulfonyl chloride at room temperature.

Analysis of Options:

• The compound shown in the correct option is ethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$), which is a primary (1°) amine. Hence, it reacts with Hinsberg's reagent to produce a solid (${\text{C}}_6{\text{H}}_5{\text{SO}}_2{\text{NHCH}}_2{\text{CH}}_3$) that contains an acidic hydrogen atom and dissolves in aqueous alkali.

• The compound is ethylmethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NHCH}}_3$), a secondary amine, so its sulfonamide product is insoluble in alkali.

• The compound is diethylmethylamine ($({\text{CH}}_3{\text{CH}}_2{)}_2{\text{NCH}}_3$), a tertiary amine, so it does not react with Hinsberg's reagent.

• The compound is nitroethane (${\text{CH}}_3{\text{CH}}_2{\text{NO}}_2$), which is not an amine.

Thus, ethylamine (${\text{CH}}_3{\text{CH}}_2{\text{NH}}_2$) is the primary amine that reacts with Hinsberg's reagent to give a solid which dissolves in alkali.