(R)-2-Bromobutane is treated with $NaI$ in dry acetone to give X by $S_{N} 2$ . X is then treated with aqueous $NaOH$ under conditions where only $S_{N} 2$ occurs to give Y. The product Y is

Answer: A

First

S_{N} 2

inverts: X is (S)-2-iodobutane (priorities

I > C X_{2} H X_{5} > CH X_{3} > H

as in the bromide). Second

S_{N} 2

inverts again: Y has the original spatial arrangement. With

OH > C X_{2} H X_{5} > CH X_{3} > H

the label is (R).