Give the decreasing order of reactivity of the following alkyl halides in the Williamson's reaction.
i. `(CH_(3))_(3)C-CH_(2)Br` ii. `CICH_(2)CH=CH_(2)`
iii. `(CICH_(2)CH_(2)CH_(3)` iv. `BrCH_(2)CH_(2)CH_(3)`

Options

A. ii gt iv gt iii gt i
B. i gt iii gt iv gt ii
C. ii gt iii gt iv gt i
D. i gt iv gt iii gt ii

Explanation

To determine the decreasing order of reactivity of the given alkyl halides in the Williamson's reaction, we need to consider the factors that influence the SN2 mechanism, which is the mechanism followed in Williamson's synthesis of ethers. The key factors include steric hindrance, bond strength, and the planarity of the molecule.

### Step-by-Step Solution:

1. **Identify the Alkyl Halides:**
- i. `(CH₃)₃C-CH₂Br` (tert-butyl bromide)
- ii. `ClCH₂CH=CH₂` (allyl chloride)
- iii. `ClCH₂CH₂CH₃` (1-chloropropane)
- iv. `BrCH₂CH₂CH₃` (1-bromopropane)

2. **Understand the SN2 Mechanism:**
- The SN2 mechanism is a bimolecular nucleophilic substitution reaction. It is characterized by a single transition state and involves a backside attack by the nucleophile.
- Reactivity in SN2 reactions is influenced by:
- **Steric hindrance:** Less hindered (primary) alkyl halides react faster than more hindered (tertiary) ones.
- **Bond strength:** Weaker carbon-halogen bonds break more easily, leading to higher reactivity.
- **Planarity:** Planar structures allow for better overlap during the nucleophilic attack.

3. **Analyze Each Compound:**
- **Compound ii (ClCH₂CH=CH₂):** This compound is planar due to the presence of the double bond and has a carbon-chlorine bond, which is relatively weaker than carbon-bromine. Therefore, it is the most reactive.
- **Compound iv (BrCH₂CH₂CH₃):** This is a primary alkyl bromide with a weaker carbon-bromine bond compared to carbon-chlorine. It is less hindered than the others, making it the second most reactive.
- **Compound iii (ClCH₂CH₂CH₃):** This is a primary alkyl chloride. Although it is also primary, the carbon-chlorine bond is stronger than the carbon-bromine bond, making it less reactive than compound iv.
- **Compound i ((CH₃)₃C-CH₂Br):** This is a tertiary alkyl bromide, which has significant steric hindrance. Tertiary alkyl halides are the least reactive in SN2 reactions due to steric hindrance.

4. **Order of Reactivity:**
- Based on the analysis, the decreasing order of reactivity in the Williamson's reaction is:
- ii. `ClCH₂CH=CH₂` (most reactive)
- iv. `BrCH₂CH₂CH₃`
- iii. `ClCH₂CH₂CH₃`
- i. `(CH₃)₃C-CH₂Br` (least reactive)

### Final Answer:
The decreasing order of reactivity of the alkyl halides in the Williamson's reaction is:
**ii > iv > iii > i**