When acetamide is treated with NaOBr,the prod | vedclass

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(C) The reaction of acetamide $(CH_3CONH_2)$ with $NaOBr$ is known as the Hofmann bromamide degradation reaction.In this reaction,an amide is converte

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$CH_3NH_2$

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(C) The reaction of acetamide $(CH_3CONH_2)$ with $NaOBr$ is known as the Hofmann bromamide degradation reaction.In this reaction,an amide is converted into a primary amine with one carbon atom less than the original amide.$CH_3CONH_2 + NaOBr + 4NaOH \rightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$.Thus,the product formed is methylamine $(CH_3NH_2)$.

When acetamide is treated with $NaOBr$,the product formed is

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