Which of the following alkyl halides will undergo `S_(N)1` reaction most redily ?

Options

A. `(CH_(3))_(3)C-F`
B. `(CH_(3))_(3)`C-Cl
C. `(CH_(3))_(3)C-Br`
D. `(CH_(3))_(3)C-I`

Explanation

To determine which of the given alkyl halides will undergo an \( S_N1 \) reaction most readily, we need to analyze the structure and the leaving groups of the compounds provided.

### Step 1: Identify the Compounds
We have four compounds:
1. \( \text{(CH}_3)_3\text{CF} \) (tert-butyl fluoride)
2. \( \text{(CH}_3)_3\text{Cl} \) (tert-butyl chloride)
3. \( \text{(CH}_3)_3\text{Br} \) (tert-butyl bromide)
4. \( \text{(CH}_3)_3\text{I} \) (tert-butyl iodide)

### Step 2: Recognize the Type of Alkyl Halides
All four compounds are tertiary alkyl halides. Tertiary alkyl halides are known to undergo \( S_N1 \) reactions due to the stability of the carbocation formed after the leaving group departs.

### Step 3: Analyze the Leaving Groups
The rate of an \( S_N1 \) reaction is largely influenced by the quality of the leaving group. The leaving group ability generally follows the trend:
\[ \text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^- \]
This means that iodide (\( \text{I}^- \)) is the best leaving group, followed by bromide (\( \text{Br}^- \)), chloride (\( \text{Cl}^- \)), and finally fluoride (\( \text{F}^- \)).

### Step 4: Compare the Leaving Groups
- In \( \text{(CH}_3)_3\text{CF} \), the leaving group is \( \text{F}^- \) (least favorable).
- In \( \text{(CH}_3)_3\text{Cl} \), the leaving group is \( \text{Cl}^- \).
- In \( \text{(CH}_3)_3\text{Br} \), the leaving group is \( \text{Br}^- \).
- In \( \text{(CH}_3)_3\text{I} \), the leaving group is \( \text{I}^- \) (most favorable).

### Step 5: Conclusion
Since \( \text{I}^- \) is the best leaving group, the compound \( \text{(CH}_3)_3\text{I} \) will undergo the \( S_N1 \) reaction most readily among the given options.

### Final Answer
The alkyl halide that will undergo \( S_N1 \) reaction most readily is \( \text{(CH}_3)_3\text{I} \).

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