In the following reaction sequence,X and Z,re | vedclass

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(B) $1$. The reaction of propan$-1-$ol $(CH_{3}CH_{2}CH_{2}OH)$ with phosphorus pentachloride $(PCl_{5})$ is given by:$CH_{3}CH_{2}CH_{2}OH + PCl_{5}

Vedclass · Accepted Answer

$X = POCl_{3}$; $Z = CH_{3}CH_{2}CH_{2}-Br$

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(B) $1$. The reaction of propan$-1-$ol $(CH_{3}CH_{2}CH_{2}OH)$ with phosphorus pentachloride $(PCl_{5})$ is given by:$CH_{3}CH_{2}CH_{2}OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + POCl_{3} + HCl$Thus,$X = POCl_{3}$.$2$. The product $CH_{3}CH_{2}CH_{2}Cl$ reacts with alcoholic $KOH$ upon heating to undergo dehydrohalogenation,forming propene $(Y = CH_{3}CH=CH_{2})$.$3$. Propene $(CH_{3}CH=CH_{2})$ reacts with $HBr$ in the presence of peroxide $((C_{6}H_{5}CO)_{2}O_{2})$ via the anti-Markownikoff addition mechanism to yield $1-$bromopropane $(Z = CH_{3}CH_{2}CH_{2}Br)$.Therefore,$X = POCl_{3}$ and $Z = CH_{3}CH_{2}CH_{2}Br$.

In the following reaction sequence,$X$ and $Z$,respectively,are:
$CH_{3}CH_{2}CH_{2}-OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + X + HCl$
$CH_{3}CH_{2}CH_{2}Cl \xrightarrow{alc. KOH, \Delta} Y$
$Y \xrightarrow{HBr, (C_{6}H_{5}CO)_{2}O_{2}} Z$

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In the following reaction sequence,$X$ and $Z$,respectively,are:$CH_{3}CH_{2}CH_{2}-OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + X + HCl$$CH_{3}CH_{2}CH_{2}Cl \xrightarrow{alc. KOH, \Delta} Y$$Y \xrightarrow{HBr, (C_{6}H_{5}CO)_{2}O_{2}} Z$