What is the reaction mechanism of sodium iodide (NaI) in acetone with an alkyl halide?

The reaction you're referring to is called the nucleophilic substitution reaction, specifically the reaction of sodium iodide (NaI) in acetone with an alkyl halide. This reaction is commonly known as the Finkelstein reaction and is used to convert alkyl chlorides or bromides into alkyl iodides. The mechanism of this reaction involves several steps:

Dissolution: In the first step, NaI is dissolved in acetone to form a solution. Acetone serves as the solvent for the reaction.

Ionization: The alkyl halide (e.g., R-X, where R is the alkyl group and X is the halide) is added to the NaI-acetone solution. The polar nature of acetone helps in the ionization of NaI and the alkyl halide. The alkyl halide molecule is polarized, with the halogen (X) being slightly negatively charged due to its higher electronegativity.

Nucleophilic attack: The iodide ion (I-) from the NaI molecule acts as a nucleophile and attacks the electrophilic carbon atom of the alkyl halide. The nucleophilic attack results in the formation of a carbon-iodine bond, and the leaving group (X) is displaced.

Formation of an intermediate: After the nucleophilic attack, an intermediate called an alkyl iodide ion pair (R-I:Na+) is formed. The positive charge on the sodium ion is balanced by the negative charge on the iodide ion.

Rearrangement: In some cases, rearrangement of the alkyl group occurs at this stage to form a more stable carbocation. However, if the alkyl halide is primary (with no branching), rearrangement is unlikely to occur.

Ion dissociation: The alkyl iodide ion pair dissociates into free alkyl iodide (R-I) and sodium ion (Na+).

The overall reaction can be represented as:

R-X + NaI → R-I + NaX

In summary, the reaction involves the nucleophilic attack of the iodide ion on the alkyl halide, leading to the displacement of the leaving group and the formation of an alkyl iodide.