Arrange the following halides in the decreasing order of `SN^1` reactivity :
(I) `CH_3CH_2CH_2Cl,`
(II) `CH_2=CHCH(Cl)CH_3`
(III) `CH_3CH_2CH(Cl)CH_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`

Explanation

To arrange the given halides in decreasing order of SN1 reactivity, we need to analyze the stability of the carbocations formed during the SN1 reaction. The more stable the carbocation, the more reactive the halide will be in an SN1 reaction.

### Step-by-Step Solution:

1. **Identify the Structures and Their Carbocations**:
- (I) `CH3CH2CH2Cl` → Forms a primary carbocation: `CH3CH2CH2+`
- (II) `CH2=CHCH(Cl)CH3` → Forms a secondary carbocation: `CH2=CH-CH+CH3` (which can be stabilized by resonance)
- (III) `CH3CH2CH(Cl)CH3` → Forms a secondary carbocation: `CH3-CH+-CH2-CH3` (which is also stabilized by hyperconjugation)

2. **Stability of Carbocations**:
- Primary carbocations (like in (I)) are the least stable due to having only one alkyl group to donate electron density.
- Secondary carbocations (like in (II) and (III)) are more stable because they have two alkyl groups that can donate electron density.
- The carbocation from (II) can be further stabilized by resonance due to the adjacent double bond.

3. **Comparing the Stability**:
- (II) has resonance stabilization, making it more stable than the secondary carbocation formed in (III).
- (III) is a secondary carbocation but lacks the resonance stabilization present in (II).
- (I) is a primary carbocation, which is the least stable.

4. **Order of Reactivity**:
- Therefore, the decreasing order of SN1 reactivity based on the stability of the carbocations formed is:
1. (II) `CH2=CHCH(Cl)CH3` (most stable due to resonance)
2. (III) `CH3CH2CH(Cl)CH3` (secondary, stable due to hyperconjugation)
3. (I) `CH3CH2CH2Cl` (least stable primary carbocation)

### Final Answer:
The decreasing order of SN1 reactivity is:
**(II) > (III) > (I)**