In the reaction CH_3COOH xrightarrow{PCl_5} ( | vedclass

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Question

(C) The reaction sequence is as follows:$1$. $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$. Product $(A)$ is Acetyl chloride $(CH_3COCl)$.$2$

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Methylamine

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(C) The reaction sequence is as follows:$1$. $CH_3COOH + PCl_5 \rightarrow CH_3COCl + POCl_3 + HCl$. Product $(A)$ is Acetyl chloride $(CH_3COCl)$.$2$. $CH_3COCl + NH_3 \rightarrow CH_3CONH_2 + HCl$. Product $(B)$ is Acetamide $(CH_3CONH_2)$.$3$. $CH_3CONH_2 + NaBrO \rightarrow CH_3NH_2 + Na_2CO_3 + NaBr$. This is the Hofmann bromamide degradation reaction.Product $(C)$ is Methylamine $(CH_3NH_2)$.

In the reaction $CH_3COOH$ $\xrightarrow{PCl_5} (A)$ $\xrightarrow{NH_3} (B)$ $\xrightarrow{NaBrO} (C)$,the final product $(C)$ is

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