Question:

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

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

Step 1: Understanding the Concept:
The Hinsberg test uses benzene sulphonyl chloride to distinguish between primary, secondary, and tertiary amines based on the solubility of their derivatives in alkali.
Step 2: Detailed Explanation:
1. Reactivity Rules:
- Primary amines give a sulphonamide with an acidic hydrogen, making it soluble in alkali.
- Secondary amines give a sulphonamide with no acidic hydrogen on nitrogen, making it insoluble in alkali.
- Tertiary amines do not react with Hinsberg reagent.
2. Analysis: Since compound X (\(C_3H_9N\)) gives an alkali-insoluble product, it must be a secondary amine.
3. Structure Identification: The secondary amine isomers of \(C_3H_9N\) is:
- \(N\)-methylethanamine: \(CH_3-NH-CH_2CH_3\).
Step 3: Final Answer:
The compound X is \(N\)-methylethanamine or ethylmethylamine.
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