Which of the following alkyl halides will undergo `S_N1` reaction most readily ?

Options

A. `(CH)_3C-F`
B. `(CH_3)_3C-CI`
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 analyze the stability of the carbocation that will form during the reaction. The `S_N1` mechanism involves two main steps: the formation of a carbocation intermediate followed by the nucleophile attacking the carbocation.

### Step-by-Step Solution:

1. **Understand the `S_N1` Mechanism**:
- The `S_N1` reaction proceeds through the formation of a carbocation. The more stable the carbocation, the more readily the `S_N1` reaction will occur.

2. **Identify the Alkyl Halides**:
- Consider the alkyl halides in question. Common examples include primary, secondary, and tertiary alkyl halides.

3. **Evaluate Carbocation Stability**:
- Carbocation stability increases in the following order: primary < secondary < tertiary. Tertiary carbocations are the most stable due to hyperconjugation and inductive effects from surrounding alkyl groups.

4. **Consider the Leaving Group**:
- The leaving group ability is crucial in `S_N1` reactions. The order of leaving group ability is: Iodine (I) > Bromine (Br) > Chlorine (Cl) > Fluorine (F). Iodine is the best leaving group due to its larger size and lower bond dissociation energy.

5. **Determine the Most Readily Undergoing `S_N1` Reaction**:
- Among the alkyl halides, the one that can form the most stable carbocation and has the best leaving group will undergo the `S_N1` reaction most readily.
- For example, if we have a tertiary alkyl iodide, it will undergo the `S_N1` reaction more readily than a primary alkyl chloride.

6. **Conclusion**:
- Therefore, the alkyl halide that will undergo the `S_N1` reaction most readily is the one that is tertiary and has iodine as the leaving group.

### Final Answer:
The alkyl halide that will undergo the `S_N1` reaction most readily is the tertiary alkyl iodide.