In the reactionCH_3-CH=CH_2 xrightarrow{Cl_2/ | vedclass

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(C) $1$. The reaction of propene with $Cl_2$ at $500\,^{\circ}C$ is an allylic substitution reaction (Wurtz-like conditions for radical substitution).

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$CH_2=CH-CH_2-CH_2-CH=CH_2$

Vedclass · Answer

(C) $1$. The reaction of propene with $Cl_2$ at $500\,^{\circ}C$ is an allylic substitution reaction (Wurtz-like conditions for radical substitution).$2$. $CH_3-CH=CH_2 + Cl_2 \xrightarrow{500\,^{\circ}C} CH_2Cl-CH=CH_2 + HCl$. Thus,$A$ is allyl chloride $(CH_2Cl-CH=CH_2)$.$3$. The reaction of allyl chloride with $Na$ in dry ether is a Wurtz coupling reaction.$4$. $2 CH_2=CH-CH_2Cl + 2Na \xrightarrow{	ext{dry ether}} CH_2=CH-CH_2-CH_2-CH=CH_2 + 2NaCl$.$5$. The product $B$ is $1,5$-hexadiene.

In the reaction
$CH_3-CH=CH_2$ $\xrightarrow{Cl_2/500\,^{\circ}C} A$ $\xrightarrow{Na/\text{dry ether}} B$
$B$ is

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In the reaction$CH_3-CH=CH_2$ $\xrightarrow{Cl_2/500\,^{\circ}C} A$ $\xrightarrow{Na/\text{dry ether}} B$$B$ is