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

Options

A. `(CH_3)_3C-F`
B. `(CH_3)_3C-Cl`
C. `(CH_3)_3C-Br`
D. `(CH_3)_3C-I`

Explanation

To determine which alkyl halide will undergo an \( S_N1 \) reaction most readily, we need to consider the factors that influence the rate of this reaction. The \( S_N1 \) mechanism involves the formation of a carbocation intermediate, and the rate of the reaction is primarily dependent on the stability of this carbocation and the ability of the leaving group to depart.

### Step-by-Step Solution:

1. **Identify the Alkyl Halides**:
- List the alkyl halides provided in the question. For example, let's assume the options are:
- A: \( CH_3Cl \) (methyl chloride)
- B: \( CH_3Br \) (methyl bromide)
- C: \( CH_3I \) (methyl iodide)
- D: \( (CH_3)_3CBr \) (tert-butyl bromide)

2. **Determine Carbocation Stability**:
- The stability of carbocations increases with the degree of substitution. Tertiary carbocations are more stable than secondary, which are more stable than primary.
- In our options, \( (CH_3)_3C^+ \) (from D) is a tertiary carbocation, which is the most stable.

3. **Evaluate the Leaving Groups**:
- The leaving group ability is crucial for \( S_N1 \) reactions. The order of leaving group ability from best to worst is:
- Iodide (I\(^-\)) > Bromide (Br\(^-\)) > Chloride (Cl\(^-\)) > Fluoride (F\(^-\)).
- In our options, \( CH_3I \) has iodide as the leaving group, which is very good, while \( CH_3Br \) has bromide, and \( CH_3Cl \) has chloride, which is less favorable.

4. **Combine Stability and Leaving Group Ability**:
- For \( (CH_3)_3CBr \), we have a tertiary carbocation and bromide as a leaving group.
- For \( CH_3I \), we have a primary carbocation (not favorable) but a very good leaving group.
- For \( CH_3Br \), we have a primary carbocation and a decent leaving group.
- The best combination of a stable carbocation and a good leaving group is found in option D: \( (CH_3)_3CBr \).

5. **Conclusion**:
- Therefore, the alkyl halide that will undergo \( S_N1 \) reaction most readily is option D: \( (CH_3)_3CBr \).