`S_(N^1)` reaction is fastest in

Options

A. `CH_3CH_2Br`
B. `CH_3-underset(Cl)underset(|)(CH)-CH_3`
C. `CH_3-underset(CH_3)underset(|)overset(CH_3)underset(|)C-Cl`
D. `CH_3-underset(CH_3)underset(|)underset(CH_2)underset(|)(CH)-Cl`

Answer

To determine which compound undergoes an \( S_N^1 \) reaction the fastest, we need to analyze the stability of the carbocations formed during the reaction. The stability of carbocations is crucial because the rate of the \( S_N^1 \) reaction is dependent on how easily the carbocation can form.

### Step-by-Step Solution:

1. **Understanding \( S_N^1 \) Mechanism**:
- The \( S_N^1 \) (Substitution Nucleophilic Unimolecular) reaction involves two main steps:
1. Formation of a carbocation after the leaving group departs.
2. Nucleophilic attack on the carbocation.

2. **Carbocation Stability**:
- The stability of carbocations follows this order:
- Tertiary (3°) > Secondary (2°) > Primary (1°) > Methyl.
- The more stable the carbocation, the faster the \( S_N^1 \) reaction will proceed.

3. **Analyzing the Given Compounds**:
- **Option 1: CH₃CH₂Br (Ethyl bromide)**:
- When Br leaves, it forms a primary carbocation (CH₃CH₂⁺), which is the least stable.

- **Option 2: CH₃CH(CH₃)CH₂Cl (Isobutyl chloride)**:
- When Cl leaves, it forms a secondary carbocation (CH₃C⁺H(CH₃)₂), which is more stable than a primary carbocation but less stable than a tertiary one.

- **Option 3: CH₃C(CH₃)₂Cl (Tert-butyl chloride)**:
- When Cl leaves, it forms a tertiary carbocation (CH₃C⁺(CH₃)₂), which is the most stable.

- **Option 4: CH₃CH(CH₂CH₃)Cl (Butyl chloride)**:
- When Cl leaves, it forms a secondary carbocation (CH₃C⁺H(CH₂CH₃)₂), which is again less stable than a tertiary carbocation.

4. **Conclusion**:
- Among the options analyzed, the third option (tert-butyl chloride) forms the most stable tertiary carbocation, making it the fastest in undergoing an \( S_N^1 \) reaction.

### Final Answer:
The \( S_N^1 \) reaction is fastest in **Option 3: CH₃C(CH₃)₂Cl (Tert-butyl chloride)**.

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