Which of the following alkyl halides is hydrolysed by `S_(N^(1))` mechanism?

Options

A. `CH_(3)Cl`
B. `CH_(3)CH_(2)Cl`
C. `CH_(3)CH_(2)CH_(2)Cl`
D. `(CH_(3))_(3)"CCl"`

Answer

To determine which alkyl halide is hydrolyzed by the \( S_N1 \) mechanism, we need to analyze the stability of the carbocations that would form during the hydrolysis process. The \( S_N1 \) mechanism involves two main steps: the formation of a carbocation intermediate and the nucleophilic attack on that carbocation.

### Step-by-Step Solution:

1. **Identify the Alkyl Halides**: List the given alkyl halides. For this example, let's assume we have the following options:
- A: \( CH_3Cl \) (methyl chloride)
- B: \( CH_3CH_2Cl \) (ethyl chloride)
- C: \( CH_3CH_2CH_2Cl \) (n-propyl chloride)
- D: \( (CH_3)_3CCl \) (tert-butyl chloride)

2. **Determine the Carbocation Formed**:
- **Option A**: \( CH_3Cl \) → \( CH_3^+ \) (methyl carbocation)
- **Option B**: \( CH_3CH_2Cl \) → \( CH_3CH_2^+ \) (ethyl carbocation)
- **Option C**: \( CH_3CH_2CH_2Cl \) → \( CH_3CH_2CH_2^+ \) (n-propyl carbocation)
- **Option D**: \( (CH_3)_3CCl \) → \( (CH_3)_3C^+ \) (tert-butyl carbocation)

3. **Evaluate Carbocation Stability**:
- Carbocation stability increases with the degree of substitution:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl (0°)
- **Option A**: Methyl carbocation is very unstable.
- **Option B**: Ethyl carbocation is less stable than a secondary carbocation.
- **Option C**: n-Propyl carbocation is also primary and unstable.
- **Option D**: Tertiary carbocation is the most stable.

4. **Conclusion**: Since the \( S_N1 \) mechanism favors the formation of stable carbocations, the alkyl halide that will undergo hydrolysis via the \( S_N1 \) mechanism is the one that forms the most stable carbocation, which is option D: \( (CH_3)_3CCl \) (tert-butyl chloride).

### Final Answer:
The alkyl halide that is hydrolyzed by the \( S_N1 \) mechanism is **Option D: \( (CH_3)_3CCl \)**.