Question:
Which of the following reaction is correctly matched with their product ?
Which of the following reaction is correctly matched with their product ?
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Gabriel Phthalimide = Aliphatic Primary Amines only. Hoffmann Bromamide = Primary Amine with one less carbon.
Updated On: Feb 3, 2026
- Phthalimide reaction forming Aniline
- $\text{CH}_3\text{CH}_2\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{CH}_3\text{CH}_2\text{NH}_2$
- Carbylamine reaction forming nitrile
- Reduction of amide with $\text{LiAlH}_4$ forming nitrile
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The Correct Option is B
Solution and Explanation
(1) Gabriel Phthalimide Synthesis prepares aliphatic primary amines. It fails for aniline because aryl halides do not undergo nucleophilic substitution with the phthalimide anion. Incorrect.
(2) Hoffmann Bromamide Degradation: $\text{R-CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{R-NH}_2 + 2\text{KBr} + \text{K}_2\text{CO}_3 + 2\text{H}_2\text{O}$.
Here $\text{R} = \text{CH}_3\text{CH}_2$. Product is $\text{CH}_3\text{CH}_2\text{NH}_2$ (Ethylamine). This is correct.
(3) Carbylamine reaction forms Isocyanides ($\text{R-NC}$), characterized by a foul smell. Incorrect.
(4) Reduction of amides ($\text{RCONH}_2$) with $\text{LiAlH}_4$ yields Primary Amines ($\text{RCH}_2\text{NH}_2$). Dehydration with $\text{P}_2\text{O}_5$ would yield nitriles. Incorrect.
(2) Hoffmann Bromamide Degradation: $\text{R-CONH}_2 + \text{Br}_2 + 4\text{KOH} \to \text{R-NH}_2 + 2\text{KBr} + \text{K}_2\text{CO}_3 + 2\text{H}_2\text{O}$.
Here $\text{R} = \text{CH}_3\text{CH}_2$. Product is $\text{CH}_3\text{CH}_2\text{NH}_2$ (Ethylamine). This is correct.
(3) Carbylamine reaction forms Isocyanides ($\text{R-NC}$), characterized by a foul smell. Incorrect.
(4) Reduction of amides ($\text{RCONH}_2$) with $\text{LiAlH}_4$ yields Primary Amines ($\text{RCH}_2\text{NH}_2$). Dehydration with $\text{P}_2\text{O}_5$ would yield nitriles. Incorrect.
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