For the reaction of 2-methylbutane with Cl_2 | vedclass

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Question

(B) The starting material is $2$-methylbutane,$CH_3-CH(CH_3)-CH_2-CH_3$.Chlorination with $Cl_2/hv$ can occur at four different types of hydrogen atom

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$6, 4$

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(B) The starting material is $2$-methylbutane,$CH_3-CH(CH_3)-CH_2-CH_3$.Chlorination with $Cl_2/hv$ can occur at four different types of hydrogen atoms:$1$. $CH_3-CH(CH_3)-CH_2-CH_2-Cl$ ($1$-chloro$-2-$methylbutane,achiral)$2$. $CH_3-CH(CH_3)-CHCl-CH_3$ ($2$-chloro$-2-$methylbutane,chiral,gives $d$ and $l$ enantiomers)$3$. $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ ($2$-chloro$-2-$methylbutane,achiral)$4$. $ClCH_2-CH(CH_3)-CH_2-CH_3$ ($1$-chloro$-3-$methylbutane,chiral,gives $d$ and $l$ enantiomers)Total number of isomers $N = 1 + 2 + 1 + 2 = 6$.Fractional distillation separates compounds based on boiling points. Enantiomers ($d$ and $l$ pairs) have identical boiling points and cannot be separated by fractional distillation.Thus,the number of fractions $M = 4$ (corresponding to the four constitutional isomers).

For the reaction of $2-$methylbutane with $Cl_2$ in the presence of $hv$,let $N$ be the total number of isomeric monochlorinated products (including stereoisomers). If these products are separated by fractional distillation,let $M$ be the number of fractions obtained. Find the values of $N$ and $M$.

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