Decreasing order of reactivity in Williamson's ether synthesis of the following .
I. `Me_3"CC"H_2Br`
II. `CH_3CH_2CH_2Br`
III. `CH_2=CHCH_2Cl`
IV. `CH_3CH_2CH_2CH_2Cl`

Options

A. `IgtIIgtIV gtIII`
B. `III gtIIgtIVgtI`
C. `IgtIIIgtIIgtIV`
D. `IIgtIIIgtIVgtI`

Answer

To determine the decreasing order of reactivity in Williamson's ether synthesis for the given alkyl halides, we need to analyze each compound based on their structure and the nature of the leaving group. The reactivity in Williamson's ether synthesis generally follows the SN2 mechanism, which is influenced by the type of alkyl halide and the quality of the leaving group.

### Step-by-Step Solution:

1. **Identify the Alkyl Halides**:
- I. `Me_3CCH_2Br` (tert-butyl bromide)
- II. `CH_3CH_2CH_2Br` (n-butyl bromide)
- III. `CH_2=CHCH_2Cl` (allyl chloride)
- IV. `CH_3CH_2CH_2CH_2Cl` (n-butyl chloride)

2. **Evaluate the Leaving Groups**:
- Bromine (Br) is a better leaving group than chlorine (Cl). Therefore, compounds with bromine will generally be more reactive than those with chlorine.

3. **Assess Steric Hindrance**:
- The reactivity in SN2 reactions decreases with increasing steric hindrance. Tertiary alkyl halides are less reactive due to steric hindrance compared to primary and secondary alkyl halides.

4. **Rank the Compounds**:
- **I. `Me_3CCH_2Br`**: This is a tertiary alkyl halide with a bromine leaving group. It will be the least reactive due to steric hindrance.
- **II. `CH_3CH_2CH_2Br`**: This is a primary alkyl halide with a bromine leaving group. It will be more reactive than the tertiary alkyl halide.
- **III. `CH_2=CHCH_2Cl`**: This is an allylic halide with a chlorine leaving group. Allylic halides are generally more reactive due to resonance stabilization, but chlorine is a poorer leaving group than bromine.
- **IV. `CH_3CH_2CH_2CH_2Cl`**: This is a primary alkyl halide with a chlorine leaving group. It will be less reactive than the bromide but more reactive than the tertiary halide.

5. **Final Order of Reactivity**:
- Based on the above analysis, the decreasing order of reactivity is:
- II. `CH_3CH_2CH_2Br` > III. `CH_2=CHCH_2Cl` > IV. `CH_3CH_2CH_2CH_2Cl` > I. `Me_3CCH_2Br`

### Conclusion:
The final decreasing order of reactivity in Williamson's ether synthesis is:
**II > III > IV > I**