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(C) The structure of $2$-methylbutane is $(CH_3)_2CHCH_2CH_3$.There are four possible monochloro derivatives formed by replacing a hydrogen atom:$1$.

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$2$

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(C) The structure of $2$-methylbutane is $(CH_3)_2CHCH_2CH_3$.There are four possible monochloro derivatives formed by replacing a hydrogen atom:$1$. $1$-chloro-$2$-methylbutane: $ClCH_2-CH(CH_3)-CH_2-CH_3$ (Chiral center at $C_2$,optically active).$2$. $2$-chloro-$2$-methylbutane: $(CH_3)_2C(Cl)-CH_2-CH_3$ (No chiral center,optically inactive).$3$. $2$-chloro-$3$-methylbutane: $(CH_3)_2CH-CH(Cl)-CH_3$ (Chiral center at $C_2$,optically active).$4$. $1$-chloro-$3$-methylbutane: $(CH_3)_2CH-CH_2-CH_2Cl$ (No chiral center,optically inactive).Thus,there are $2$ optically active isomers.

How many of the four monochloro derivatives of $(CH_3)_2CHCH_2CH_3$ are optically active?

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