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(C) The reaction sequence is as follows:$1$. The first step is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,where propanoic acid reacts with $Cl_2$ in t

Vedclass · Accepted Answer

$CH_2=CH-COOH$

Vedclass · Answer

(C) The reaction sequence is as follows:$1$. The first step is the Hell-Volhard-Zelinsky $(HVZ)$ reaction,where propanoic acid reacts with $Cl_2$ in the presence of red phosphorus to form $\alpha$-chloropropanoic acid $(X)$:$CH_3-CH_2-COOH \xrightarrow[	ext{red } P]{Cl_2} CH_3-CHCl-COOH 	ext{ (} X 	ext{)}$$2$. The second step is a dehydrohalogenation reaction using alcoholic $KOH$,which removes $HCl$ from the $\alpha$-carbon and $\beta$-carbon to form acrylic acid $(Y)$:$CH_3-CHCl-COOH \xrightarrow{	ext{Alc. KOH}} CH_2=CH-COOH 	ext{ (} Y 	ext{)}$Therefore,the end product $Y$ is acrylic acid $(CH_2=CH-COOH)$.

The end product of the following reaction is: $CH_3-CH_2-COOH$ $\xrightarrow[\text{red } P]{Cl_2} X$ $\xrightarrow{\text{Alc. KOH}} Y$

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