(S)-Butan-2-ol is treated with $SOCl X_{2}$ in the absence of a base to give A. A is then treated with $NaI$ in dry acetone to give B by $S_{N} 2$ . The compound B is

Answer: A

Without base,

SOCl X_{2}

reacts by

S_{N} i

with retention, so A is (S)-2-chlorobutane (same priority order as the alcohol). Iodide then displaces chloride by

S_{N} 2

with inversion; the order

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

is unchanged, so B is (R)-2-iodobutane.