For the reaction:RCH_2Br + I^{-} stackrel{tex | vedclass

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(A) This is the $Finkelstein$ reaction,which proceeds via an $S_N2$ mechanism.In the $S_N2$ transition state,the negative charge is dispersed over the

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The transition state formed in the above reaction is less polar than the localised anion.

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(A) This is the $Finkelstein$ reaction,which proceeds via an $S_N2$ mechanism.In the $S_N2$ transition state,the negative charge is dispersed over the incoming nucleophile $(I^-)$ and the leaving group $(Br^-)$. Therefore,the transition state is less polar than the localized anion $(I^-)$,which facilitates the reaction in a polar aprotic solvent like acetone.Acetic acid is a polar protic solvent,which solvates the nucleophile $(I^-)$ through hydrogen bonding,thereby decreasing its nucleophilicity and hindering the $S_N2$ reaction.Acetone is a polar aprotic solvent; it does not solvate the anion effectively,keeping the nucleophile reactive.$NaBr$ is insoluble in acetone and precipitates out,which drives the reaction forward according to $Le$ $Chatelier's$ principle,preventing $Br^-$ from acting as a competing nucleophile.Thus,the correct statement is that the transition state is less polar than the localized anion.

For the reaction:
$RCH_2Br + I^{-} \stackrel{\text{Acetone}}{\longrightarrow} RCH_2I + Br^{-}$
The correct statement is:

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For the reaction:$RCH_2Br + I^{-} \stackrel{\text{Acetone}}{\longrightarrow} RCH_2I + Br^{-}$The correct statement is: