Discussion
When ethanamide is treated with Br2 and NaOH in ethanolic solution we get
Answer: D
The product is methanamine.
This is the Hofmann bromamide degradation: an amide treated with bromine and sodium hydroxide gives a primary amine with one carbon fewer.
CH3CONH2 + Br2 + 4NaOH → CH3NH2 + Na2CO3 + 2NaBr + 2H2O
The carbonyl carbon is lost as carbonate, so ethanamide (two carbons) gives methanamine (one carbon), not ethanamine. No brominated compound is the final product.
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