CH_3-CH_2-COOH xrightarrow{Cl_2/Fe} X xrighta | vedclass

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(C) Step $1$: $CH_3-CH_2-COOH$ reacts with $Cl_2$ in the presence of $Fe$ (or $P$) to undergo $\alpha$-halogenation,known as the Hell-Volhard-Zelinsky

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$CH_2=CH-COOH$

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(C) Step $1$: $CH_3-CH_2-COOH$ reacts with $Cl_2$ in the presence of $Fe$ (or $P$) to undergo $\alpha$-halogenation,known as the Hell-Volhard-Zelinsky reaction,to form $X$ $(CH_3-CHCl-COOH)$.Step $2$: $X$ $(CH_3-CHCl-COOH)$ reacts with alcoholic $KOH$ to undergo dehydrohalogenation (elimination of $HCl$),resulting in the formation of $Y$ $(CH_2=CH-COOH)$,which is acrylic acid.Reaction sequence: $CH_3-CH_2-COOH$ $\xrightarrow{Cl_2/Fe} CH_3-CHCl-COOH (X)$ $\xrightarrow{	ext{Alcoholic } KOH} CH_2=CH-COOH (Y)$.

$CH_3-CH_2-COOH$ $\xrightarrow{Cl_2/Fe} X$ $\xrightarrow{\text{Alcoholic } KOH} Y$. Compound $Y$ is:

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