$(R)$-2-Iodobutane is treated with $\mathrm{NaI}$ in acetone and allowed to stand for a long time. The product eventually formed is

• A

  (R)-2-iodobutane
• B

  $(S)$-2-iodobutane
• Correct answer
  C

  $( \pm)-2$-iodobutane
• D

  $( \pm)-1,2$-diiodobutane.

Solution

$\mathrm{I}^{-}$is a good nucleophile as well as a good leaving group. Therefore, if $(R)$-2-iodobutane is treated with $\mathrm{NaI}$, repeated $\mathrm{S}_{\mathrm{N}} 2$ reactions occur. As a result, eventually a racemic mixture of $( \pm)$-2-iodobutane is obtained.

Hence, the correct answer is option: 'C'.