C_4H_9Cl (Optically Active Compound A) xright | vedclass

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(A) The reaction of an alkyl halide with $NaCN$ in $DMSO$ proceeds via an $S_N2$ mechanism. In an $S_N2$ reaction,there is an inversion of configurati

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(A) The reaction of an alkyl halide with $NaCN$ in $DMSO$ proceeds via an $S_N2$ mechanism. In an $S_N2$ reaction,there is an inversion of configuration (Walden inversion) at the chiral center. The product formed is $(R)-2-$methylbutanenitrile. Since the reaction involves inversion of configuration,the starting material (compound $A$) must have the $(S)$ configuration. The structure of $2-$methylbutanenitrile is $CH_3-CH(CH_3)-CH_2-CN$. Therefore,the starting material $A$ is $(S)-2-$chlorobutane,which is $CH_3-CH(Cl)-CH_2-CH_3$. Among the given options,the structure corresponding to $(S)-2-$chlorobutane is represented by option $A$.

$C_4H_9Cl$ (Optically Active Compound $A$) $\xrightarrow[DMSO]{NaCN}$ $(R)-2-$methylbutanenitrile; Identify compound $A$?

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