This reaction is a classic example of the Finkelstein reaction, where alkyl bromides or chlorides are converted to alkyl iodides using NaI in the presence of acetone as a solvent. Let's analyze each of the given options:
Acetic acid solvent can take in above reaction:
Acetic acid is not typically used as a solvent in this reaction because it can hinder the precipitation of NaBr. Acetone is preferred for promoting the precipitation of NaBr, driving the reaction forward.
NaI is soluble in acetone but NaBr is precipitate in acetone:
This statement is true. NaI is soluble in acetone, allowing it to react with the alkyl bromide. NaBr, on the other hand, has low solubility in acetone, so it precipitates out, effectively pushing the reaction towards completion.
NaI is precipitated in acetone but NaBr is soluble in acetone:
This is incorrect, as the solubility trends are reversed; NaI is soluble in acetone while NaBr is not.
When acetone is taken in solvent transition state is highly polar:
This does not accurately describe why acetone is used in the Finkelstein reaction. The use of acetone is primarily for its ability to dissolve NaI and precipitate NaBr, not due to polarity concerns affecting the transition state.
Therefore, the correct answer is: NaI is soluble in acetone but NaBr is precipitate in acetone.
