Question

What is the mechanism of the reaction of $\text{NaI}$ in acetone with an alkyl halide?

Answer 1

Hint: Alkyl halides (also known as haloalkanes) are alkanes that have had one or more hydrogen atoms substituted by halogen atoms (fluorine, chlorine, bromine or iodine). The compounds with the general formula $\text{RX}$ where $\text{R}$is an alkyl or substituted alkyl group and $\text{X}$ is a halogen, are known as alkyl halides.

Complete answer:
$$\text{R}\text{ - }\text{Cl}\text{ + }\text{N}{\text{a}}^{\text{ + }}{\text{I}}^{\text{ - }}\to \text{R}\text{ - }\text{I}\text{ + }\text{NaCl}\downarrow$$
The mechanism of the reaction is the ${\text{S}}_{\text{N}}\text{2}$mechanism. The ion iodide is a strong nucleophile, and it adds as ${\text{I}}^{\text{ - }}$. Since sodium iodide is much more soluble in acetone than sodium chloride, we will notice a glassy precipitate of sodium chloride in the acetone when we perform this reaction.
The reagent used in the Finkelstien reaction is $\text{NaI}$ in acetone. To make alkyl iodides, alkyl halides are treated with the reagent. Even though ${\text{I}}^{-}$ is a weak nucleophile, the reaction is accelerated by the lower solubility of the products $\text{NaCl}$ and $\text{NaBr}$ in acetone.

Additional Information:
The nucleophilic substitution reaction of the leaving group (which usually consists of halide groups or other electron-withdrawing groups) with a nucleophile in a given organic compound is the $S_{N}2$reaction mechanism.

Note:
The number two in the term ${\text{S}}_{\text{N}}\text{2}$ stands for bimolecular, implying that two molecules are involved in the rate-determining step. The rate of bimolecular nucleophilic substitution reactions is affected by both the haloalkane and the nucleophile concentrations.
${\text{S}}_{\text{N}}\text{2}$ occurs at a higher rate as it is a one-step procedure. $S_{N}1$is a two-step process in which the first step, carbocation formation, is slow and the second step, nucleophilic attack, is quick.