Which of the following will not undergo `S_N^1` reaction with `OH^-`

Options

A.
B.
C.
D.

Answer

To determine which of the given compounds will not undergo an \( S_N^1 \) reaction with \( OH^- \), we need to analyze the stability of the carbocations formed during the reaction. The \( S_N^1 \) mechanism involves two steps: the formation of a carbocation intermediate followed by a nucleophilic attack. The stability of the carbocation is crucial, as more stable carbocations will favor the \( S_N^1 \) pathway.

### Step-by-Step Solution:

1. **Understanding \( S_N^1 \) Mechanism**:
- The \( S_N^1 \) reaction is unimolecular, meaning it depends only on the concentration of one reactant (the substrate).
- It proceeds through the formation of a carbocation intermediate.

2. **Formation of Carbocations**:
- When a leaving group (like Cl) departs, it forms a carbocation. The stability of this carbocation is key to determining whether the \( S_N^1 \) reaction will occur.

3. **Analyzing Each Option**:
- **Option 1**: Forms a carbocation that is stabilized by resonance.
- **Option 2**: Forms a tertiary carbocation, which is stabilized by inductive effects from surrounding alkyl groups.
- **Option 3**: Forms a primary carbocation, which is less stable due to fewer electron-donating groups.
- **Option 4**: Forms a carbocation that is also stabilized by resonance.

4. **Stability of Carbocations**:
- Carbocations are stabilized by resonance and inductive effects. Tertiary carbocations are the most stable, followed by secondary, and then primary carbocations.
- The third option, which forms a primary carbocation, is the least stable and therefore will not favor the \( S_N^1 \) pathway.

5. **Conclusion**:
- The compound that will not undergo an \( S_N^1 \) reaction with \( OH^- \) is the one that forms the least stable carbocation, which is **Option 3**.

### Final Answer:
**Option 3** will not undergo \( S_N^1 \) reaction with \( OH^- \).