Question

An organic compound A (C\(_5\)H\(_9\)N) upon reaction with Na/Hg/C\(_2\)H\(_5\)OH gives compound B. B reacts with NaNO\(_2\)/HCl at 274 K to form C with quantitative liberation of N\(_2\) gas. B also reacts with Hinsberg's reagent to form a compound which is soluble in alkali. Identify compound B.

• A. \([ \text{(CH}_3\text{)}_3 \text{N} - \text{CH}_3 ]^+\)
• B. CH\(_3\) - (CH\(_2\))\(_3\) - NH - CH\(_3\)
• C. CH\(_3\) - NH\(_2\)
• D. CH\(_3\) - (CH\(_2\))\(_2\) - NH\(_2\)

Hint:

Hinsberg's reagent is used to distinguish between primary, secondary, and tertiary amines. Primary amines form soluble products in alkali, secondary amines form insoluble products, and tertiary amines do not react.

Answer 1

The Correct Option is C

Step 1: Analyze compound A.
Compound A is an amine (C\(_5\)H\(_9\)N). The reaction with Na/Hg/C\(_2\)H\(_5\)OH is characteristic of a reduction reaction, which suggests that A is an amide or a nitrile group, reduced to an amine group. The product formed in this step, B, is most likely an amine.

Step 2: Analyze the reaction of B with NaNO\(_2\)/HCl.
When B reacts with NaNO\(_2\) and HCl at 274 K, it liberates N\(_2\) gas. This reaction is typical of a primary aromatic amine undergoing diazotization, leading to the formation of a diazonium salt.

Step 3: Analyze the reaction of B with Hinsberg's reagent.
B also reacts with Hinsberg's reagent, which is used to test for primary, secondary, and tertiary amines. A primary amine reacts with Hinsberg’s reagent to form a soluble product in alkali. This confirms that compound B is a primary amine.

Step 4: Identify compound B.
Given the reactions described, compound B must be a primary amine. The correct structure matching all the reactions is CH\(_3\) - NH\(_2\), which is methylamine.

Step 5: Conclusion.
Thus, compound B is CH\(_3\) - NH\(_2\), and the correct answer is option (C).