Question:
Which intermediate is formed during the reaction of acetamide with bromine in the presence of excess KOH solution?
Which intermediate is formed during the reaction of acetamide with bromine in the presence of excess KOH solution?
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Think about what species forms when the N-Br bond breaks with loss of bromide ion.
Updated On: Jul 3, 2026
- Carbocation
- Carbanion
- Carbene
- Nitrene
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The Correct Option is D
Solution and Explanation
Step 1: Acetamide reacts with bromine and KOH in the Hofmann bromamide degradation reaction. Bromine first reacts with KOH to form KOBr, which brominates the amide nitrogen.
\[ CH_3CONH_2 + Br_2 + KOH \rightarrow CH_3CONHBr + KBr + H_2O \]
Step 2: The N-bromoamide loses a proton to a second equivalent of KOH, giving an anion.
\[ CH_3CONHBr + KOH \rightarrow CH_3CONBr^- + K^+ + H_2O \]
Step 3: This anion undergoes alpha-elimination, expelling a bromide ion and generating an electron-deficient nitrogen species called an acyl nitrene.
\[ CH_3CONBr^- \rightarrow CH_3CON: + Br^- \]
Step 4: The nitrene is unstable and rearranges immediately, with the methyl group migrating from carbon to nitrogen, forming an isocyanate.
\[ CH_3CON: \rightarrow CH_3N=C=O \]
Step 5: The isocyanate hydrolyzes in the basic medium to give a primary amine with loss of carbon dioxide as carbonate.
\[ CH_3N=C=O + 2KOH \rightarrow CH_3NH_2 + K_2CO_3 \]
The key intermediate formed during this degradation is the nitrene.
\[\boxed{\text{Nitrene}}\]
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