In `S_N2` reaction the order of reactivity of the halides. `CH_3X, C_2H_5X, n-C_3H_7X,n-C_4H_9X` is

Options

A. `CH_3X gt C_2H_5X gt n - C_3H_7X gt n - C_4H_9X`
B. `C_2H_5X gt n - C_3H_7X gt n - C_4H_9X gt CH_3X`
C. `C_2H_5X gt n - C_3H_7X gt n - C_4H_9X lt CH_3X`
D. `n-C_4H_9X gt n-C_3H_7X gt C_2H_5X gt CH_3X`

Explanation

To determine the order of reactivity of the halides in an `S_N2` reaction, we need to consider the steric hindrance around the carbon atom that is being attacked by the nucleophile. The less steric hindrance there is, the more reactive the halide will be.

### Step-by-Step Solution:

1. **Identify the Halides**: The halides given are:
- CH₃X (methyl halide)
- C₂H₅X (ethyl halide)
- n-C₃H₇X (n-propyl halide)
- n-C₄H₉X (n-butyl halide)

2. **Understand the `S_N2` Mechanism**: In an `S_N2` reaction, the nucleophile attacks the carbon atom from the opposite side of the leaving group (X). This backside attack is hindered by bulky groups around the carbon.

3. **Analyze Steric Hindrance**:
- **CH₃X**: This has no alkyl groups attached to the carbon, so it has minimal steric hindrance. It is the most reactive.
- **C₂H₅X**: This has one ethyl group attached, which introduces some steric hindrance but is still relatively low. It is less reactive than CH₃X but more reactive than the others.
- **n-C₃H₇X**: This has two ethyl groups, which increases steric hindrance further, making it less reactive than C₂H₅X.
- **n-C₄H₉X**: This has three ethyl groups, leading to the highest steric hindrance among the four halides. Therefore, it is the least reactive.

4. **Order of Reactivity**: Based on the analysis of steric hindrance, the order of reactivity in `S_N2` reactions is:
- CH₃X > C₂H₅X > n-C₃H₇X > n-C₄H₉X

### Final Answer:
The order of reactivity of the halides in `S_N2` reaction is:
**CH₃X > C₂H₅X > n-C₃H₇X > n-C₄H₉X**