Which of the following alkyl halide will undergo `S_N 1` reaction is

Options

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

Answer

To determine which alkyl halide will undergo an \( S_N 1 \) reaction, we need to consider the characteristics of the alkyl halides and the mechanism of the \( S_N 1 \) reaction.

### Step-by-Step Solution:

1. **Understanding \( S_N 1 \) Mechanism**:
- The \( S_N 1 \) (Substitution Nucleophilic Unimolecular) reaction involves two main steps:
1. Formation of a carbocation by the departure of the leaving group (halogen).
2. Nucleophilic attack on the carbocation.

2. **Stability of Carbocation**:
- The stability of the carbocation formed is crucial for the \( S_N 1 \) reaction. More stable carbocations are favored. The order of stability is:
- Tertiary > Secondary > Primary > Methyl
- Therefore, alkyl halides that can form more stable carbocations will undergo \( S_N 1 \) reactions more readily.

3. **Leaving Group Ability**:
- The leaving group (halogen) should be able to leave easily. The ability of halogens to act as leaving groups decreases in the order:
- Iodine (I) > Bromine (Br) > Chlorine (Cl) > Fluorine (F)
- This means that iodine is the best leaving group, followed by bromine and then chlorine.

4. **Analyzing the Given Alkyl Halides**:
- We need to evaluate the given alkyl halides based on the above points. The one that can form a stable carbocation and has a good leaving group will be the one that undergoes \( S_N 1 \).

5. **Conclusion**:
- Based on the analysis, the alkyl halide that is likely to undergo \( S_N 1 \) reaction is the one that is tertiary or has a good leaving group (like iodine).
- For example, if we have \( CH_3CH_2C(Br)(CH_3) \) (a tertiary bromide) and \( CH_3CH_2Cl \) (a primary chloride), the tertiary bromide would undergo \( S_N 1 \) due to the stability of the carbocation and the good leaving ability of bromine.

### Final Answer:
The alkyl halide that will undergo \( S_N 1 \) reaction is the one that forms a stable carbocation and has a good leaving group. For instance, \( CH_3CH_2C(I)(CH_3) \) would be a suitable candidate.