`CH_(3)-CH_(2)-CI+NaI overset("Dry acetone")rarr [X]+NaCI` Product `[X]` is -

Options

A. a.`CH_(3)-CH_(3)`
B. b.`CH_(3)-CH_(2)-I`
C. c.`CH_(3)-I`
D. d.`CH_(3)-CH_(2)-CH_(2)-CH_(3)`

Explanation

To solve the question, we need to analyze the reaction between the alkyl chloride (CH₃-CH₂-Cl) and sodium iodide (NaI) in the presence of dry acetone.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants given are CH₃-CH₂-Cl (ethyl chloride) and NaI (sodium iodide).

2. **Understand the Reaction Conditions**:
The reaction occurs in dry acetone, which is a polar aprotic solvent. This type of solvent is known to favor nucleophilic substitution reactions.

3. **Recognize the Type of Reaction**:
This reaction is a Finkelstein reaction, which involves the substitution of one halogen for another. In this case, chlorine (Cl) is being replaced by iodine (I).

4. **Write the Reaction Mechanism**:
In the presence of NaI, the iodide ion (I⁻) acts as a nucleophile and attacks the carbon atom bonded to the chlorine atom. The chlorine atom is a good leaving group, and it will leave as chloride ion (Cl⁻).

5. **Determine the Product**:
The product of the reaction will be CH₃-CH₂-I (ethyl iodide) after the substitution of Cl with I. The byproduct of this reaction will be NaCl (sodium chloride).

6. **Final Product Representation**:
Therefore, the product [X] is CH₃-CH₂-I.

### Conclusion:
The final answer for product [X] is **ethyl iodide (CH₃-CH₂-I)**.