JEE MAIN - Chemistry (2023 - 6th April Morning Shift - No. 3)
For the reaction
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The correct statement is
The solvent used in the reaction solvates the ions formed in rate determining step.
Br$$^-$$ can act as competing nucleophile.
The reaction can occur in acetic acid also.
The transition state formed in the above reaction is less polar than the localised anion.
Explanation
$$
\mathrm{R}-\mathrm{CH}_2-\mathrm{Br}+\mathrm{I}^{-} \stackrel{\text { Acetone }}{\longrightarrow} \mathrm{R}-\mathrm{CH}_2-\mathrm{I}+\mathrm{Br}^{-}
$$
$\mathrm{NaBr}$ is not soluble in acetone and hence reaction shifts in forward reaction and reaction is called Finkelstein reaction which proceeds through $\mathrm{S}_{\mathrm{N}} 2$ mechanism
For $S_N 2$ reactions, transition state formed is less polar than the localised anion.
$\mathrm{NaBr}$ is not soluble in acetone and hence reaction shifts in forward reaction and reaction is called Finkelstein reaction which proceeds through $\mathrm{S}_{\mathrm{N}} 2$ mechanism
For $S_N 2$ reactions, transition state formed is less polar than the localised anion.
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