For the reaction RCH _ 2 Br + I ^ - Acetone ⟶ RCH _ 2 I _ major +...

$R - {CH}_{2} - Br + {I}^{-}\overset{\text{Acetone}}{⟶}R - {CH}_{2} - I + {Br}^{-}$

As mentioned the reaction follows ${S}_{N}2$ reaction mechanism. The Finkelstein reaction classically involves the conversion of Alkyl bromides into Alkyl iodides by the treatment with a solution of Sodium iodide in Acetone. $NaBr$ is not readily soluble in acetone and hence reaction shifts in forward reaction. For ${S}_{N}2$ reactions, transition state formed is less polar than the localised anion.