H_3CCONH_2 + Br_2 + 4NaOH longrightarrow Y + | vedclass

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(B) The given reaction is the Hofmann bromamide degradation reaction.In this reaction,a primary amide $(RCONH_2)$ reacts with bromine $(Br_2)$ and a s

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$H_3CNH_2$

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(B) The given reaction is the Hofmann bromamide degradation reaction.In this reaction,a primary amide $(RCONH_2)$ reacts with bromine $(Br_2)$ and a strong base ($NaOH$ or $KOH$) to form a primary amine $(RNH_2)$ with one carbon atom less than the original amide.The balanced chemical equation is:$H_3CCONH_2 + Br_2 + 4NaOH \longrightarrow CH_3NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$Comparing this with the given reaction,$Y$ is $CH_3NH_2$ (methylamine),which corresponds to option $B$.

$H_3CCONH_2 + Br_2 + 4NaOH \longrightarrow Y + Na_2CO_3 + 2NaBr + 2H_2O$
What is $Y$ in the above reaction?

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$H_3CCONH_2 + Br_2 + 4NaOH \longrightarrow Y + Na_2CO_3 + 2NaBr + 2H_2O$What is $Y$ in the above reaction?