For the reaction
`RCH_2 Br+I^(-) overset ("Acetone") to underset "major"RCH_2I+Br^(-)`
The correct statement is

Options

A. `Br^(–)` can act as competing nucleophile.
B. The reaction can occur in acetic acid also
C. The transition state formed in the above reaction is less polar than the localised anion.
D. The solvent used in the reaction solvates the ions formed in rate determining step

Answer

To solve the problem, we need to analyze the given reaction and the statements provided. The reaction involves the substitution of bromine (Br) with iodine (I) in the presence of a nucleophile (I^-) and a solvent (acetone).

### Step-by-Step Solution:

1. **Identify the Reaction Type**:
The reaction given is:
\[ RCH_2Br + I^- \xrightarrow{\text{Acetone}} RCH_2I + Br^- \]
This is a nucleophilic substitution reaction. Since the nucleophile (I^-) is attacking the carbon atom that is bonded to bromine, and the reaction proceeds with the simultaneous departure of bromide ion, this indicates that it is an SN2 reaction.

**Hint**: Look for the mechanism of the reaction to determine if it's SN1 or SN2.

2. **Analyze the Statements**:
- **Statement 1**: "Br^- can act as a competing nucleophile."
- In an SN2 reaction, the nucleophile (I^-) is more reactive than the leaving group (Br^-). Therefore, Br^- does not act as a competing nucleophile in this case.

- **Statement 2**: "The reaction can occur in acetic acid also."
- Acetic acid is a polar protic solvent, which can stabilize ions but is not ideal for SN2 reactions. SN2 reactions are favored in polar aprotic solvents like acetone.

- **Statement 3**: "The transition state formed in the above reaction is less polar than the localized anion."
- In SN2 reactions, the transition state is often less polar than the ions formed because the transition state involves a partial bond formation and bond breaking, leading to a less charged environment compared to fully formed ions.

- **Statement 4**: "The solvent used in the reaction solvates the ions formed in the rate-determining step."
- In SN2 reactions, the rate-determining step involves the transition state, not the fully formed ions. Thus, this statement is misleading.

3. **Conclusion**:
Based on the analysis of the statements, the correct statement is:
- **Statement 3**: "The transition state formed in the above reaction is less polar than the localized anion."

### Final Answer:
The correct statement is: **The transition state formed in the above reaction is less polar than the localized anion.**