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 that would form during the reaction. The \( S_N^1 \) mechanism involves two main steps: the formation of a carbocation intermediate and the attack of the nucleophile (in this case, \( OH^- \)) on the carbocation.

### Step-by-Step Solution:

1. **Identify the Compounds**: List the compounds provided in the question. For this example, let's assume we have the following compounds:
- Compound A: \( CH_3CH_2Cl \) (ethyl chloride)
- Compound B: \( (CH_3)_3CCl \) (tert-butyl chloride)
- Compound C: \( C_6H_5CH_2Cl \) (benzyl chloride)
- Compound D: \( CH_3CHClCH_3 \) (isopropyl chloride)

2. **Determine Carbocation Formation**: For each compound, determine the type of carbocation that would form after the leaving group (Cl) departs:
- **Compound A**: Forms a primary carbocation (\( CH_3CH_2^+ \)), which is unstable.
- **Compound B**: Forms a tertiary carbocation (\( (CH_3)_3C^+ \)), which is very stable.
- **Compound C**: Forms a benzyl carbocation (\( C_6H_5CH_2^+ \)), which is stabilized by resonance.
- **Compound D**: Forms a secondary carbocation (\( CH_3C^+HCH_3 \)), which is relatively stable.

3. **Evaluate Stability of Carbocations**: The stability of carbocations increases with the degree of substitution:
- Tertiary > Secondary > Primary > Methyl
- Benzyl carbocations are also stable due to resonance.

4. **Identify the Compound that Will Not Undergo \( S_N^1 \)**:
- **Compound A** (ethyl chloride) forms a primary carbocation, which is too unstable to undergo an \( S_N^1 \) reaction. Therefore, it will not react with \( OH^- \) via this mechanism.
- The other compounds (B, C, and D) form stable carbocations and can undergo \( S_N^1 \) reactions.

### Conclusion:
The compound that will not undergo an \( S_N^1 \) reaction with \( OH^- \) is **Compound A: \( CH_3CH_2Cl \)**.