Question

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

• A. $(\pm )-2-\mathrm{iodobutane}$
• B. $\left(\mathrm{R}\right)-2-\mathrm{iodobutane}$
• C. $(\pm )-1,2-\mathrm{diiodobutane}$
• D. $\left(\mathrm{S}\right)-2-\mathrm{iodobutane}$

Answer 1

The correct option is A

$(\pm )-2-\mathrm{iodobutane}$

Explanation for the Correct Option:

Option (A): $(\pm )-2-\mathrm{iodobutane}$

• The ${\mathrm{SN}}_2$ reaction is known as a nucleophilic substitution in which two components are included in the rate-determining stage.
• For this type of reaction, the nucleophile should be a good nucleophile.
• We know that ${\mathrm{I}}^{-}$can act as a good nucleophile.
• Hence, When $\left(\mathrm{R}\right)-2-\mathrm{iodobutane}$ is treated with $\mathrm{NaI}$, it leads to the occurrence of the ${\mathrm{SN}}_2$ reactions.
• Thus a racemic mixture of $(\pm )-2\mathrm{iodobutane}$will be obtained as the product.

Explanation for the incorrect Option:

Option (B): $\left(\mathrm{R}\right)-2-\mathrm{iodobutane}$

• When $\left(\mathrm{R}\right)-2-\mathrm{Iodobutane}$ is treated with $\mathrm{NaI}$ in acetone and allowed to stand for a long time, $\left(\mathrm{R}\right)-2-\mathrm{Iodobutane}$ does not form as a product.
• Hence, the given Option is an incorrect one

Option (C): $(\pm )-1,2-\mathrm{diiodobutane}$

• $(\pm )-1,2-\mathrm{diiodobutane}$ is not formed as a product, When $\left(\mathrm{R}\right)-2-\mathrm{Iodobutane}$ is treated with $\mathrm{NaI}$ in acetone.
• Hence, this Option is given here does not represent the correct concept.

Option (D): $-2-\mathrm{iodobutane}$

• When $\left(\mathrm{R}\right)-2-\mathrm{Iodobutane}$ is treated with $\mathrm{NaI}$ in acetone, $-2-\mathrm{iodobutane}$ is not formed as the product.
• Thus the given Option doesn’t coincide with the correct concept.

Hence, When $\left(\mathrm{R}\right)-2-\mathrm{Iodobutane}$ is treated with $\mathrm{NaI}$ in acetone and allowed to stand for a long time. The product eventually formed is: $(\pm )-2-\mathrm{iodobutane}$