What is the mechanism of the reaction of NaI in acetone with an alkyl halide?

Question

Sarah Angotti · Accepted Answer

This operates by the #S_(N)2# mechanism. The iodide ion is a reasonably good nucleophile, and adds as #I^-#

#R-Cl + Na^+I^-# #rarr# #R-I + NaCldarr#

This reaction results in a glassy precipitate of sodium chloride in the acetone because sodium iodide is much more soluble in acetone than sodium chloride is. I believe this reaction has a name (I should have finished that bottle of wine last night!), but if you have time in an organic laboratory, it would be worthwhile to perform this reaction in a small test tube so you can notice the turbidity of the sodium chloride when it precipitates.

Santiago Alfonso · Answer

undefined The reaction of sodium iodide (NaI) in acetone with an alkyl halide follows an SN2 (substitution nucleophilic bimolecular) mechanism. In this mechanism, the nucleophile (iodide ion) attacks the electrophilic carbon atom of the alkyl halide, leading to the formation of a new bond and the displacement of the halogen atom. This process occurs in a single step, with the incoming nucleophile and the leaving group being involved in the transition state simultaneously. The SN2 mechanism is characterized by a concerted reaction in which the stereochemistry of the reactant and product are inverted, and it is favored in primary and methyl halides due to less steric hindrance.