(R)-2-Bromobutane reacts with aqueous $\mathrm{NaOH}$ by a pure SN2 mechanism. The major product is

SN2 proceeds with inversion at the stereocentre. Priorities are Br > $\mathrm{C}{}_2\mathrm{H}{}_5$ > $\mathrm{CH}{}_3$ > H in the halide and OH > $\mathrm{C}{}_2\mathrm{H}{}_5$ > $\mathrm{CH}{}_3$ > H in the alcohol, so the order is unchanged and the label changes from R to S.