Decreasing order of reactivity in Williamson's ether synthesis of the following is:
I. `Me_(3)C CH_(2)Br`
II. `CH_(3)CH_(2)CH_(2)Br`
III. `CH_(3)=CHCH_(2)Cl`
IV. `CH_(3)CH_(2)CH_(2)Cl`

Options

A. IIIgtIIgtIVgtI
B. IgtIIgtIVgtIII
C. IIgtIIIgtIVgtI
D. IgtIIIgtIIgtIV

Explanation

To determine the decreasing order of reactivity in Williamson's ether synthesis for the given compounds, we need to analyze the alkyl halides based on their structure and the mechanism of the reaction. The reactivity in Williamson's ether synthesis primarily depends on the ability of the leaving group and the steric hindrance around the carbon atom bonded to the leaving group.

### Step-by-Step Solution:

1. **Identify the Compounds:**
- I. `Me_(3)C CH_(2)Br` (tert-butyl bromide)
- II. `CH_(3)CH_(2)CH_(2)Br` (1-bromopropane)
- III. `CH_(3)=CHCH_(2)Cl` (allyl chloride)
- IV. `CH_(3)CH_(2)CH_(2)Cl` (1-chloropropane)

2. **Consider the Mechanism:**
- Williamson's ether synthesis typically follows an SN2 mechanism, which is favored by primary alkyl halides due to less steric hindrance. Secondary alkyl halides can also react but are less favorable than primary. Tertiary alkyl halides are generally not reactive in SN2 due to steric hindrance.

3. **Evaluate the Leaving Groups:**
- The leaving ability of halides follows the trend: Iodine > Bromine > Chlorine. Therefore, bromine is a better leaving group than chlorine.

4. **Analyze Each Compound:**
- **I. `Me_(3)C CH_(2)Br`:** This is a tertiary alkyl halide (due to the tert-butyl group), which is less reactive in SN2 reactions due to steric hindrance.
- **II. `CH_(3)CH_(2)CH_(2)Br`:** This is a primary alkyl halide (1-bromopropane), which is very reactive in SN2 reactions.
- **III. `CH_(3)=CHCH_(2)Cl`:** This is an allylic halide (allyl chloride), which is more reactive than a simple primary halide due to resonance stabilization in the transition state.
- **IV. `CH_(3)CH_(2)CH_(2)Cl`:** This is a primary alkyl halide (1-chloropropane), but since chlorine is a poorer leaving group than bromine, it will be less reactive than II.

5. **Order the Reactivity:**
- Based on the analysis:
- II (1-bromopropane) is the most reactive.
- III (allyl chloride) is next due to resonance stabilization.
- IV (1-chloropropane) follows since it's a primary halide but with a poorer leaving group.
- I (tert-butyl bromide) is the least reactive due to steric hindrance.

### Final Order of Reactivity:
The decreasing order of reactivity is:
II > III > IV > I