The back-side attack on 2-bromobutane by meth | vedclass

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(D) $S_N2$ reactions proceed with inversion of configuration (Walden inversion). In the given product,the methoxy group $(OCH_3)$ is attached to the c

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(D) $S_N2$ reactions proceed with inversion of configuration (Walden inversion). In the given product,the methoxy group $(OCH_3)$ is attached to the chiral center. To obtain this product via an $S_N2$ mechanism,the nucleophile $(CH_3O^{-})$ must attack from the side opposite to the leaving group $(Br^{-})$. Looking at the provided product structure,the reactant $2$-bromobutane must have the bromine atom positioned such that the incoming methoxide group results in the observed stereochemistry. Based on the $S_N2$ inversion,the reactant $2$-bromobutane is represented by the Fischer projection where $Br$ is at the top,$Et$ is at the bottom,$Me$ is on the left,and $H$ is on the right. This corresponds to option $D$.

The back-side attack on $2$-bromobutane by methoxide $(CH_3O^{-})$ gives the product shown below. Which Fischer projection represents $2$-bromobutane used as the reactant in this reaction?

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