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 alkyl halide will undergo an \( S_N1 \) reaction most readily, we need to consider two main factors: the stability of the carbocation formed during the reaction and the leaving group ability of the halide.

### Step-by-Step Solution:

1. **Understanding the \( S_N1 \) Mechanism**:
- The \( S_N1 \) (Substitution Nucleophilic Unimolecular) reaction involves two steps. The first step is the formation of a carbocation after the leaving group departs, and the second step is the attack of the nucleophile on the carbocation.

2. **Carbocation Stability**:
- The stability of the carbocation is crucial because the rate of the \( S_N1 \) reaction is determined by the formation of this carbocation. The more stable the carbocation, the faster the reaction will proceed.
- Carbocations are stabilized by alkyl groups. Tertiary (3°) carbocations are more stable than secondary (2°) and primary (1°) carbocations due to hyperconjugation and inductive effects.

3. **Evaluating the Alkyl Halides**:
- The question provides several alkyl halides, all of which are tertiary (3°) halides. This means they will form stable carbocations.

4. **Leaving Group Ability**:
- The leaving group ability is also important in \( S_N1 \) reactions. The better the leaving group, the faster the reaction. The order of leaving group ability for halides is:
- Iodine (I) > Bromine (Br) > Chlorine (Cl) > Fluorine (F)
- This means that iodine is the best leaving group, followed by bromine, chlorine, and then fluorine.

5. **Conclusion**:
- Since all the alkyl halides are tertiary and will form stable carbocations, the rate of the \( S_N1 \) reaction will depend on the leaving group. Therefore, the alkyl halide with iodine as the leaving group will undergo the \( S_N1 \) reaction most readily.

### Final Answer:
The alkyl halide that will undergo \( S_N1 \) reaction most readily is the one with iodine as the leaving group.