Question

A compound ‘X’ with molecular formula \(C_3H_9N\) reacts with Hinsberg reagent to give a product insoluble in alkali. Identify ‘X’.

Hint:

Memory trick: $1^\circ$ amine product = Soluble in base (due to 1 remaining H). $2^\circ$ amine product = Insoluble in base (0 remaining H).

Answer 1

Step 1: Conceptual Overview:
The Hinsberg test is a useful method for distinguishing between primary, secondary, and tertiary amines. This test uses benzene sulphonyl chloride and relies on the solubility of the derivatives formed in alkali to determine the type of amine.
Step 2: Detailed Explanation:
1. Reactivity Rules:
- Primary amines react with benzene sulphonyl chloride to form a sulphonamide with an acidic hydrogen, which makes the derivative soluble in alkali.
- Secondary amines form a sulphonamide with no acidic hydrogen on the nitrogen, making the derivative insoluble in alkali.
- Tertiary amines do not react with the Hinsberg reagent, and thus no derivative is formed.
2. Analysis: Since compound X (\( C_3H_9N \)) forms an alkali-insoluble product, it must be a secondary amine.
3. Structure Identification: The possible structure for the secondary amine with the formula \( C_3H_9N \) is \( N \)-methylethanamine, also known as ethylmethylamine. Its structure is:
- \( N \)-methylethanamine: \( CH_3-NH-CH_2CH_3 \).
Step 3: Final Conclusion:
The compound X is \( N \)-methylethanamine, or ethylmethylamine.