Under identical conditions, the `S_(N^(1))` reaction will occur most efficiently with

Options

A. Tert-butyl chloride
B. 1-chlorobutane
C. 2-methyl-1-chloropropane
D. 2-chlorobutane

Explanation

To determine which compound will undergo the \( S_N1 \) reaction most efficiently, we need to analyze the stability of the carbocations formed during the reaction. The \( S_N1 \) mechanism involves two main steps: the formation of a carbocation and then the nucleophilic attack. The rate of the \( S_N1 \) reaction is primarily dependent on the stability of the carbocation formed in the rate-determining step (RDS).

### Step-by-Step Solution:

1. **Identify the Compounds**: We have four compounds to consider:
- Tertiary butyl chloride
- 1-chlorobutane
- 2-methyl-1-chloropropane
- 2-chlorobutane

2. **Determine the Carbocation Formed**:
- **Tertiary butyl chloride**: When the chloride ion leaves, it forms a tertiary carbocation, which is stabilized by three alkyl groups.
- **1-chlorobutane**: This will form a primary carbocation, which is less stable as it has only one alkyl group.
- **2-methyl-1-chloropropane**: This forms a secondary carbocation, which is more stable than a primary but less stable than a tertiary.
- **2-chlorobutane**: This will also form a secondary carbocation.

3. **Analyze Carbocation Stability**:
- **Tertiary Carbocation**: Most stable due to hyperconjugation and inductive effects from three alkyl groups.
- **Secondary Carbocation**: More stable than primary but less stable than tertiary.
- **Primary Carbocation**: Least stable due to having no alkyl groups to stabilize the positive charge.

4. **Conclusion**: The compound that will undergo the \( S_N1 \) reaction most efficiently is the one that forms the most stable carbocation. In this case, tertiary butyl chloride forms a tertiary carbocation, which is the most stable among the options provided.

### Final Answer:
**Tertiary butyl chloride** will undergo the \( S_N1 \) reaction most efficiently.

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