Consider the following alkyl halides
1. `(CH_3)_3 "CCH"_2Br`
2. `ClCH_2CH=CH_2`
3. `ClCH_2 CH_2 CH_3`
4. `BrCH_2CH_2CH_3`
Arrange these alkyl halides in decreasing order of reactivity in Williamson reaction.

Options

A. `2 gt 1 gt 3 gt 4`
B. `2 gt 1 gt 4 gt 3`
C. `1 gt 2 gt 3 gt 4`
D. `2 gt 4 gt 3 gt 1`

Answer

To solve the problem of arranging the given alkyl halides in decreasing order of reactivity in the Williamson reaction, we will analyze the structures and their corresponding reactivities step by step.

### Step 1: Identify the Alkyl Halides
The alkyl halides provided are:
1. `(CH_3)_3CCH_2Br` (Tertiary alkyl halide)
2. `ClCH_2CH=CH_2` (Vinylic alkyl halide)
3. `ClCH_2CH_2CH_3` (Primary alkyl halide)
4. `BrCH_2CH_2CH_3` (Primary alkyl halide)

### Step 2: Understand the Williamson Reaction
The Williamson reaction involves the nucleophilic substitution of an alkyl halide by an alkoxide ion to form an ether. The reaction typically follows an SN2 mechanism, where the nucleophile attacks the carbon atom bonded to the leaving group (halide).

### Step 3: Determine the Reactivity Order
1. **Vinylic Halide (2)**: `ClCH_2CH=CH_2` is a vinylic halide. Vinylic halides are generally less reactive in SN2 reactions due to steric hindrance and the stability of the vinylic carbocation formed during the reaction. However, they can still react, making this the most reactive in this context.

2. **Primary Alkyl Halides (3 and 4)**:
- `ClCH_2CH_2CH_3` (3) is a primary alkyl halide with chlorine as the leaving group.
- `BrCH_2CH_2CH_3` (4) is also a primary alkyl halide but with bromine as the leaving group. Bromine is a better leaving group than chlorine, making (4) more reactive than (3).

3. **Tertiary Alkyl Halide (1)**: `(CH_3)_3CCH_2Br` is a tertiary alkyl halide. Tertiary halides are less reactive in SN2 reactions due to steric hindrance, making this the least reactive among the given compounds.

### Step 4: Final Order of Reactivity
Based on the analysis, the decreasing order of reactivity in the Williamson reaction is:
1. `ClCH_2CH=CH_2` (2) - Most reactive
2. `BrCH_2CH_2CH_3` (4)
3. `ClCH_2CH_2CH_3` (3)
4. `(CH_3)_3CCH_2Br` (1) - Least reactive

Thus, the final order is:
**2 > 4 > 3 > 1**

### Summary of Reactivity Order:
- **Most Reactive**: `ClCH_2CH=CH_2` (2)
- **Next**: `BrCH_2CH_2CH_3` (4)
- **Next**: `ClCH_2CH_2CH_3` (3)
- **Least Reactive**: `(CH_3)_3CCH_2Br` (1)