Question

(R)-$2$-iodobutane is treated with $NaI$ in acetone and allow to stand for a long time .the product eventually formed is:
A.(R)- $2$-iodobutane
B.(S)- $2$- iodobutane
C.$ \pm 2$-iodobutane
D.$( \pm ) - 1,2 - $iodobutane

Answer 1

Hint: The $S{N_2}$ reaction is a nucleophilic substitution reaction, in this a bond is broken and another is shaped synchronously. Two reacting species are kept inside the rate determining step of the reaction. The term '$S{N_2}$' stands for – Substitution Nucleophilic Bimolecular.

Complete Step by step solution:
${I^{ - 1}}$ is a good nucleophile as well as a very good leaving group. Therefore, if (R)-$2$ -iodobutane is controlled with $NaI$, then it’s repeated $S{N_2}$ reactions occur. As a final result, later a racemic aggregate of $( \pm ) - 2 - $ iodobutane is acquired.

If we see the above explanation according to it, the correct answer is $ \pm 2$-iodobutane .

So, the correct answer is option C.

Additional information:
The answer proceeds through a bottom attack by the nucleophile on the substrate. The nucleophile methods the given substrate at an altitude of $180$ degree to the carbon-leaving institution bond or to the carbon leaving group. The carbon-nucleophile bond forms and carbon-leaving organization bond breaks simultaneously through a transition nation.
Now, the leaving organization is driven out of the transition state on the other side of the carbon-nucleophile bond, forming the specified product. It is vital to note that the product is shaped with an inversion of the tetrahedral geometry on the atom within the center.

Note: The $S{N_2}$ reaction is a good example of stereospecific reactions. In this one there are in which one of a kind stereoisomers react to provide different stereoisomers of the product. Also, $S{N_2}$ response is the highest common instance of Walden inversion, in which an uneven carbon atom(that is an odd number of carbon) undergoes inversion of configurations.