CH_3CH_2CH_2OH xrightarrow{PCl_5} A xrightarr | vedclass

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(D) Step $1$: The reaction of propan$-1-$ol with $PCl_5$ is a nucleophilic substitution reaction where the hydroxyl group is replaced by a chlorine at

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Propene

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(D) Step $1$: The reaction of propan$-1-$ol with $PCl_5$ is a nucleophilic substitution reaction where the hydroxyl group is replaced by a chlorine atom. $CH_3CH_2CH_2OH + PCl_5 ightarrow CH_3CH_2CH_2Cl (A) + POCl_3 + HCl$. Step $2$: The reaction of $A$ ($1$-chloropropane) with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction),which results in the formation of an alkene. $CH_3CH_2CH_2Cl + KOH (	ext{alc.}) ightarrow CH_3CH=CH_2 (B) + KCl + H_2O$. Thus,the final product $B$ is propene.

$CH_3CH_2CH_2OH$ $\xrightarrow{PCl_5} A$ $\xrightarrow{\text{alcoholic } KOH} B$. The final product $B$ is ...... .

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