When propionamide reacts with Br_2 in the pre | vedclass

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(B) The reaction of an amide with $Br_2$ in the presence of an alkali ($KOH$ or $NaOH$) is known as the Hofmann bromamide degradation reaction.In this

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$CH_3CH_2NH_2$

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(B) The reaction of an amide with $Br_2$ in the presence of an alkali ($KOH$ or $NaOH$) is known as the Hofmann bromamide degradation reaction.In this reaction,the amide group $(-CONH_2)$ is converted into a primary amine $(-NH_2)$ with the loss of one carbon atom.Propionamide is $CH_3CH_2CONH_2$.According to the Hofmann bromamide degradation reaction: $CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$.The product formed is ethylamine $(CH_3CH_2NH_2)$.

When propionamide reacts with $Br_2$ in the presence of alkali the product is :

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