JEE MainChemistry

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Finkelstein Reaction Mechanism and Transition State Polarity

Q: Why is the transition state considered less polar than the reactant anion?

A free anion has a full localized -1 charge on a single atom, yielding high charge density and strong electrostatic interactions. In the extS_()extN2 transition state, this -1 charge is delocalized over both entering and leaving halogen atoms, lowering the net local charge density and polarity.

For the reaction: RCH_2Br + I^(-) xrightarrowAcetone undersetmajorRCH_2I + Br^(-) The correct statement is:

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AThe transition state formed in the above reaction is less polar than the localised anion.BThe reaction can occur in acetic acid also.CThe solvent used in the reaction solvates the ions formed in rate determining step.DBr^(-) can act as competing nucleophile.

Step-by-step solutionView

Correct answer

In an $S_{N} 2$ substitution involving a neutral substrate and an anionic nucleophile, the transition state disperses the negative charge across two atoms, making it less polar than the localised reactant anion.

Option analysis

Why each option works or fails

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

This is the correct statement. In the $S_{N} 2$ transition state $[I^{δ -} \dots {RCH}_{2} \dots {Br}^{δ -}]^{‡}$ , the full negative unit charge of the incoming iodide ion is delocalised over two halogen centers, reducing local charge density and overall solvation demand relative to the reactant anion $I^{-}$ .

B · The reaction can occur in acetic acid also.

Assuming that the Finkelstein reaction operates identically in all polar solvents, ignoring the role of differential precipitation. The Finkelstein reaction is driven to completion in acetone because $NaI$ is soluble while $NaBr$ precipitates. In polar protic solvents like acetic acid, $NaBr$ is soluble and $I^{-}$ is strongly hydrogen-bonded, suppressing nucleophilicity and preventing the forward shift via precipitation.

C · The solvent used in the reaction solvates the ions formed in rate determining step.

Believing acetone strongly solvates anions or that an intermediate ionic species forms during the rate-determining step. Acetone is a polar aprotic solvent; it solvates cations well via its carbonyl oxygen, but poorly solvates anions. Furthermore, $S_{N} 2$ is a concerted single-step mechanism that does not generate free ionic intermediates in the rate-determining step.

D · Br^(-) can act as competing nucleophile.

Assuming the displaced ${Br}^{-}$ remains in solution and competes effectively in the reverse reaction. In acetone, the displaced bromide precipitates out as an insoluble salt (such as $NaBr$ ), removing ${Br}^{-}$ from the solution phase and preventing reverse nucleophilic attack.

Reviewed route

Solution

StepWorking

01Concept

This reaction is the classic Finkelstein reaction ( $S_{N} 2$ substitution of alkyl bromide by iodide in dry acetone). The reaction proceeds via a single bimolecular transition state $[I^{δ -} \dots {RCH}_{2} \dots {Br}^{δ -}]^{‡}$ where the full negative charge is dispersed across two atoms, making it less polar/more charge-delocalized than the localized free reactant/product anions ( $I^{-}$ , ${Br}^{-}$ ).

02Option verdict

In the $S_{N} 2$ transition state $[I^{δ -} \dots C \dots {Br}^{δ -}]^{‡}$ , the unit negative charge is distributed over both halogens ( $δ -$ on each), so charge density is lower and the transition state is less polar than a compact localized anion like $I^{-}$ or ${Br}^{-}$ . Hence, this statement is correct.

03Option verdict

Acetic acid is a polar protic solvent. It solvates nucleophiles like $I^{-}$ through hydrogen bonding and drastically slows down the $S_{N} 2$ reaction rate. Also, $NaBr$ does not precipitate effectively to drive the equilibrium forward as it does in acetone.

04Option verdict

Acetone is a polar aprotic solvent; it readily solvates cations ( ${Na}^{+}$ ) via its oxygen dipole but poorly solvates anions, keeping the nucleophile $I^{-}$ 'naked' and reactive.

05Option verdict

$NaBr$ is insoluble in acetone and precipitates out of the reaction mixture. By Le Chatelier's principle, its removal prevents the reverse reaction, so ${Br}^{-}$ cannot act as a competing nucleophile.

✓Discriminator

In $S_{N} 2$ , the transition state disperses a single $- 1$ charge across two halogen centers, making it less localized and less polar than free $I^{-}$ / ${Br}^{-}$ .

Hints that build this answer step by step

What mechanism does the conversion ${RCH}_{2} Br + I^{-} \overset{acetone}{\to} {RCH}_{2} I + {Br}^{-}$ follow?
Concerted bimolecular nucleophilic substitution ( $S_{N} 2$ )
How does charge distribution in the $S_{N} 2$ transition state compare to the reactant state?
A full negative charge is dispersed across two centers ( $[I^{δ -} \dots {RCH}_{2} \dots {Br}^{δ -}]^{‡}$ ), lowering charge density relative to free $I^{-}$ .
Why does the reverse reaction by displaced ${Br}^{-}$ not compete effectively in acetone?
The bromide salt (e.g., $NaBr$ ) is insoluble in acetone and precipitates out, shifting the equilibrium forward.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

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Students also ask

Why is the transition state considered less polar than the reactant anion?

A free anion has a full localized $- 1$ charge on a single atom, yielding high charge density and strong electrostatic interactions. In the $e x t S_{e} x t N 2$ transition state, this $- 1$ charge is delocalized over both entering and leaving halogen atoms, lowering the net local charge density and polarity.

Answer

In an $S_{N} 2$ substitution involving a neutral substrate and an anionic nucleophile, the transition state disperses the negative charge across two atoms, making it less polar than the localised reactant anion.

Why each option works or fails

A · correct: The transition state formed in the above reaction is less polar than the localised anion. - This is the correct statement. In the $S_{N} 2$ transition state $[I^{δ -} \dots {RCH}_{2} \dots {Br}^{δ -}]^{‡}$ , the full negative unit charge of the incoming iodide ion is delocalised over two halogen centers, reducing local charge density and overall solvation demand relative to the reactant anion $I^{-}$ .
B: The reaction can occur in acetic acid also. - Assuming that the Finkelstein reaction operates identically in all polar solvents, ignoring the role of differential precipitation. The Finkelstein reaction is driven to completion in acetone because $NaI$ is soluble while $NaBr$ precipitates. In polar protic solvents like acetic acid, $NaBr$ is soluble and $I^{-}$ is strongly hydrogen-bonded, suppressing nucleophilicity and preventing the forward shift via precipitation.
C: The solvent used in the reaction solvates the ions formed in rate determining step. - Believing acetone strongly solvates anions or that an intermediate ionic species forms during the rate-determining step. Acetone is a polar aprotic solvent; it solvates cations well via its carbonyl oxygen, but poorly solvates anions. Furthermore, $S_{N} 2$ is a concerted single-step mechanism that does not generate free ionic intermediates in the rate-determining step.
D: Br^(-) can act as competing nucleophile. - Assuming the displaced ${Br}^{-}$ remains in solution and competes effectively in the reverse reaction. In acetone, the displaced bromide precipitates out as an insoluble salt (such as $NaBr$ ), removing ${Br}^{-}$ from the solution phase and preventing reverse nucleophilic attack.

Step-by-step solution

Concept: This reaction is the classic Finkelstein reaction ( $S_{N} 2$ substitution of alkyl bromide by iodide in dry acetone). The reaction proceeds via a single bimolecular transition state $[I^{δ -} \dots {RCH}_{2} \dots {Br}^{δ -}]^{‡}$ where the full negative charge is dispersed across two atoms, making it less polar/more charge-delocalized than the localized free reactant/product anions ( $I^{-}$ , ${Br}^{-}$ ).
Option verdict: In the $S_{N} 2$ transition state $[I^{δ -} \dots C \dots {Br}^{δ -}]^{‡}$ , the unit negative charge is distributed over both halogens ( $δ -$ on each), so charge density is lower and the transition state is less polar than a compact localized anion like $I^{-}$ or ${Br}^{-}$ . Hence, this statement is correct.
Option verdict: Acetic acid is a polar protic solvent. It solvates nucleophiles like $I^{-}$ through hydrogen bonding and drastically slows down the $S_{N} 2$ reaction rate. Also, $NaBr$ does not precipitate effectively to drive the equilibrium forward as it does in acetone.
Option verdict: Acetone is a polar aprotic solvent; it readily solvates cations ( ${Na}^{+}$ ) via its oxygen dipole but poorly solvates anions, keeping the nucleophile $I^{-}$ 'naked' and reactive.
Option verdict: $NaBr$ is insoluble in acetone and precipitates out of the reaction mixture. By Le Chatelier's principle, its removal prevents the reverse reaction, so ${Br}^{-}$ cannot act as a competing nucleophile.
Discriminator: In $S_{N} 2$ , the transition state disperses a single $- 1$ charge across two halogen centers, making it less localized and less polar than free $I^{-}$ / ${Br}^{-}$ .

Finkelstein Reaction Mechanism and Transition State Polarity

What feels right?

In an $S_{N} 2$ substitution involving a neutral substrate and an anionic nucleophile, the transition state disperses the negative charge across two atoms, making it less polar than the localised reactant anion.

Why each option works or fails

Solution

01Concept

02Option verdict

03Option verdict

04Option verdict

05Option verdict

✓Discriminator

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In an $S_{N} 2$ substitution involving a neutral substrate and an anionic nucleophile, the transition state disperses the negative charge across two atoms, making it less polar than the localised reactant anion.