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(A) The reaction of $(R)-2-bromo-1-methoxypropane$ with $NaCN$ in acetone follows an $S_N2$ mechanism. In an $S_N2$ reaction,there is an inversion of

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$(R)$

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(A) The reaction of $(R)-2-bromo-1-methoxypropane$ with $NaCN$ in acetone follows an $S_N2$ mechanism. In an $S_N2$ reaction,there is an inversion of configuration at the chiral center. However,the $(R/S)$ designation depends on the priority of the groups according to Cahn-Ingold-Prelog $(CIP)$ rules. In the reactant,the priorities are: $-Br (1) > -CH_2OCH_3 (2) > -CH_3 (3) > -H (4)$. In the product,the priorities are: $-CH_2OCH_3 (1) > -CN (2) > -CH_3 (3) > -H (4)$. Because the priority of the groups changes (the nucleophile $-CN$ is priority $2$,while the leaving group $-Br$ was priority $1$),the spatial inversion results in the product still having the $(R)$ configuration.

The correct description of the product in the following reaction is:
$CH_3-CH(Br)-CH_2OCH_3 \text{ } ((R)) \xrightarrow{NaCN, \text{acetone}} CH_3-CH(CN)-CH_2OCH_3$

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The correct description of the product in the following reaction is: $CH_3-CH(Br)-CH_2OCH_3 \text{ } ((R)) \xrightarrow{NaCN, \text{acetone}} CH_3-CH(CN)-CH_2OCH_3$