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) = $6$.

$2$-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......................