Question

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

(a) (R)-2-iodobutane 

(b) (S)-2-iodobutane 

(c) (+-)-2-iodobutane 

(d) (+-)-1.2-diiodobutane

Answer 1

The correct option is (c) (+-)-2-iodobutane.

Explanation:

I^- is a good nucleophile as well as a good leaving group. Therefore, if (R)-2-iodobutane is treated with NaI, repeated S_N2 reactions occur. As a result, eventually a racemic mixture of (+-)-2-iodobutane is obtained.