On monochlorination of 2-methylbutane,the tot | vedclass

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(B) The structure of $2-$methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$. Monochlorination can occur at four different types of hydrogen atoms: $1$. At $C-1

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$4$

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(B) The structure of $2-$methylbutane is $CH_3-CH(CH_3)-CH_2-CH_3$. Monochlorination can occur at four different types of hydrogen atoms: $1$. At $C-1$ (terminal methyl): $CH_2Cl-CH(CH_3)-CH_2-CH_3$ (achiral). $2$. At $C-2$ (tertiary carbon): $CH_3-C(Cl)(CH_3)-CH_2-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers). $3$. At $C-3$ (secondary carbon): $CH_3-CH(CH_3)-CHCl-CH_3$ (chiral,exists as $(+)$ and $(-)$ enantiomers). $4$. At $C-4$ (terminal methyl): $CH_3-CH(CH_3)-CH_2-CH_2Cl$ (achiral). Thus,the chiral products are the enantiomers formed by substitution at $C-2$ and $C-3$. Total number of chiral compounds = $2$ (from $C-2$) + $2$ (from $C-3$) = $4$.

On monochlorination of $2-$methylbutane,the total number of chiral compounds is

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