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(A) The structure of $2,3-$dimethylbutane is $(CH_3)_2CH-CH(CH_3)_2$.Monochlorination can occur at two types of hydrogen atoms:$1$. At the primary car

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(A) The structure of $2,3-$dimethylbutane is $(CH_3)_2CH-CH(CH_3)_2$.Monochlorination can occur at two types of hydrogen atoms:$1$. At the primary carbon atoms (terminal $CH_3$ groups): There are $12$ equivalent primary hydrogens. Replacing one gives $1-$chloro-$2,3-$dimethylbutane. This molecule has no chiral center,so it is achiral.$2$. At the secondary carbon atoms (the $CH$ groups): There are $2$ equivalent secondary hydrogens. Replacing one gives $2-$chloro-$2,3-$dimethylbutane. This molecule has a chiral center at $C_2$. Thus,it exists as a pair of enantiomers ($R$ and $S$ forms).Therefore,only $1$ pair of enantiomers is obtained.

How many pairs of enantiomers are obtained from the monochlorination of $2,3-$dimethylbutane?

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