Statement-I: Optically active 2-idoibutane on treatment with `NaI` in acetone undergoes racemisation.
Because Statement-II: Repeated Walden inversions on the reactant and its product evantually gives a racemic mixure.

Options

A. Statement-I is True, Statement-II is True, Statement-II is a correct explanation for Statement-I
B. Statement-I is True, Statement-II is True, Statement-II is NOT a correct explanation for Statement-I
C. Statement-I is True, Statement-II is False.
D. Statement-I is False, Statement-II is True.

Answer

To solve the problem, we need to analyze both statements provided in the question.

### Step 1: Understand the structure of 2-Iodobutane
2-Iodobutane is an organic compound with the following structure:
- It has four carbon atoms (butane) and an iodine atom attached to the second carbon.
- The molecular formula is C4H9I.

### Step 2: Identify the optical activity
- The compound is optically active, meaning it has a chiral center at the second carbon (C-2) where the iodine is attached.
- This chirality allows the compound to exist in two enantiomeric forms (R and S configurations).

### Step 3: Reaction with NaI in acetone
- When 2-Iodobutane reacts with sodium iodide (NaI) in acetone, it undergoes an SN2 reaction mechanism.
- In SN2 reactions, the nucleophile attacks the electrophile from the opposite side of the leaving group (iodine in this case).
- This results in the inversion of configuration at the chiral center.

### Step 4: Formation of racemic mixture
- As a result of the SN2 reaction, the product formed is the inverted form of 2-Iodobutane.
- Since both the original (R or S) and the inverted product (S or R) coexist in the reaction mixture, this leads to the formation of a racemic mixture (equal amounts of both enantiomers).

### Step 5: Analyze Statement II
- Statement II discusses repeated Walden inversions, which refer to the process of inversion occurring at the chiral center during the reaction.
- The repeated inversion leads to a situation where both the original and inverted forms are present, resulting in a racemic mixture.

### Conclusion
- Both Statement I and Statement II are true.
- Statement II correctly explains Statement I because the repeated Walden inversion results in the racemic mixture observed in the reaction.

### Final Answer
Both statements are true, and Statement II is the correct explanation for Statement I.

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