The order of decreasing `S_(N)1` reactivities of the halides
(I )`CH_(3)CH_(2)CH_(2)Cl` (II) `CH_(2) = CHCHCICH_(3)` (III) `CH_(3)CH_(2)CHCICH_(3)`

Options

A. `I gt II gt III`
B. `II gt I gt III`
C. `II gt III gt I`
D. `III gt II gt I`

Answer

To determine the order of decreasing SN1 reactivities of the given halides, we need to analyze the stability of the carbocations that would form when the halides undergo an SN1 reaction. The stability of carbocations is crucial because the SN1 reaction rate is directly proportional to the stability of the carbocation formed.

### Step-by-Step Solution:

1. **Identify the Halides:**
- (I) CH₃CH₂CH₂Cl (1-chloropropane)
- (II) CH₂=CHCHClCH₃ (3-chlorobutene)
- (III) CH₃CH₂CHClCH₃ (2-chlorobutane)

2. **Determine the Carbocations Formed:**
- For (I) CH₃CH₂CH₂Cl: Removing the Cl⁻ ion forms a primary carbocation (CH₃CH₂CH₂⁺).
- For (II) CH₂=CHCHClCH₃: Removing the Cl⁻ ion forms a secondary carbocation (CH₂=CHCH⁺CH₃) which is also conjugated due to the adjacent double bond.
- For (III) CH₃CH₂CHClCH₃: Removing the Cl⁻ ion forms a secondary carbocation (CH₃CH₂C⁺HCH₃) which is not conjugated.

3. **Assess Carbocation Stability:**
- **Primary Carbocation (I)**: Least stable due to being a primary carbocation.
- **Secondary Carbocation (III)**: More stable than primary but less stable than conjugated secondary.
- **Conjugated Secondary Carbocation (II)**: Most stable due to resonance stabilization from the adjacent double bond.

4. **Order of Stability:**
- The order of stability of the carbocations is:
- (II) > (III) > (I)
- Therefore, the order of decreasing SN1 reactivity is:
- (II) > (III) > (I)

5. **Final Answer:**
- The order of decreasing SN1 reactivities of the halides is:
- (II) > (III) > (I)