2-chlorobutane + Cl_2 rightarrow C_4H_8Cl_2 ( | vedclass

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(A) The free radical chlorination of $2$-chlorobutane $(CH_3CHClCH_2CH_3)$ can occur at different positions:$1$. Substitution at $C_1$: $ClCH_2CHClCH_

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$6$

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(A) The free radical chlorination of $2$-chlorobutane $(CH_3CHClCH_2CH_3)$ can occur at different positions:$1$. Substitution at $C_1$: $ClCH_2CHClCH_2CH_3$ ($1$-chloro$-2-$chlorobutane). This has one chiral center at $C_2$,so it exists as $2$ enantiomers.$2$. Substitution at $C_2$: $CH_3CCl_2CH_2CH_3$ ($2$,$2$-dichlorobutane). This is achiral (optically inactive).$3$. Substitution at $C_3$: $CH_3CHClCHClCH_3$ ($2$,$3$-dichlorobutane). This has two chiral centers. The meso form is optically inactive,while the $(2R, 3R)$ and $(2S, 3S)$ forms are optically active ($2$ isomers).$4$. Substitution at $C_4$: $CH_3CHClCH_2CH_2Cl$ ($1$,$3$-dichloro$-2-$chlorobutane is incorrect,it is $1,3-$dichlorobutane or $2,4-$dichlorobutane depending on numbering). The product is $CH_3CHClCH_2CH_2Cl$ ($1$,$3$-dichlorobutane). This has a chiral center at $C_2$,so it exists as $2$ enantiomers.Total optically active isomers = $2$ (from $1-$chloro$-2-$chlorobutane) + $2$ (from $2,3-$dichlorobutane) + $2$ (from $1,3-$dichlorobutane) = $6$.

Column $I$	Column $II$
$A$. $CH_3-CHBr-CD_3$ on treatment with alc. $KOH$ gives $CH_2=CH-CD_3$ as a major product.	$P$. $E1$ reaction
$B$. $Ph-CHBr-CH_3$ reacts faster than $Ph-CHBr-CD_3$.	$Q$. $E2$ reaction
$C$. $Ph-CH_2-CH_2Br$ on treatment with $C_2H_5OD / C_2H_5O^-$ gives $Ph-CD=CH_2$ as the major product.	$R$. $E1cb$ reaction
$D$. $PhCH_2CH_2Br$ and $PhCD_2CH_2Br$ react with same rate.	$S$. First order reaction

$2$-chlorobutane $+ Cl_2 \rightarrow C_4H_8Cl_2$ (isomers)
Total number of optically active isomers shown by $C_4H_8Cl_2$,obtained in the above reaction is......................

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$2$-chlorobutane $+ Cl_2 \rightarrow C_4H_8Cl_2$ (isomers)Total number of optically active isomers shown by $C_4H_8Cl_2$,obtained in the above reaction is......................