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

Options

A. (R)-2-iodobutane
B. (S)-2-iodobutane
C. `(+-)`-2-iodobutane
D. `(+-)`-1,2-diiodobutane

Answer

C

Explanation

(R) -2-iodobutane [(+)-2-iodobutane] reacts with NaI in acetone to form `(+-)`-2-iodobutane, which leads to reacemisation.

Each `S_(N)2` attack on (R)-2-iodobutane or (+)-2-iodobutane by `I^(Theta)` occurs with inversion of configuration resulting in the formation of (S)-2-iodobutane or (-)-2-iodobutane. As the concentration of the (-)-enantiomer in the reaction mixture increases, its tendency to reacts with `I^(Theta)` forming a (+)-enantiomer increases. At a cetain time, the mixture contains equimolar amounts of (+) and (-)-2-iodobutane. Since both the emamtimers reacts with `I^(Theta)` at the same rate (because the activation energy is the same), their composition remains unchanged.