Which will undergo fastest `S_(N)2` substitution reaction when treated with NaOH ?

Options

A. `H_(5)C_(2)-overset(CH_(3))overset(|)underset(H)underset(|)C-Br`
B. `H_(3)C-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-Br`
C. `H-overset(CH_(3))overset(|)underset(C_(2)H_(5))underset(|)C-Br`
D. `H-overset(H)overset(|)underset(Br)underset(|)C-CH_(2)-CH_(2)-CH_(3)`

Answer

To determine which compound will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH, we need to analyze the steric hindrance around the carbon atom that is bonded to the leaving group (in this case, bromine).

### Step-by-Step Solution:

1. **Identify the Compounds**:
We need to look at the structures of the compounds provided in the question. Each compound will have a halogen (bromine) attached to a carbon atom that is also connected to other groups.

2. **Understand \( S_N2 \) Mechanism**:
The \( S_N2 \) reaction mechanism involves a nucleophile (in this case, hydroxide ion, \( OH^- \)) attacking the carbon atom bonded to the leaving group (bromine) from the opposite side. This results in the inversion of configuration at that carbon center.

3. **Evaluate Steric Hindrance**:
The rate of \( S_N2 \) reactions is significantly affected by steric hindrance. The more bulky groups attached to the carbon atom, the slower the reaction will be.
- **Primary Alkyl Halides**: These have the least steric hindrance and will react the fastest.
- **Secondary Alkyl Halides**: These have moderate steric hindrance and will react slower than primary.
- **Tertiary Alkyl Halides**: These have the most steric hindrance and will react the slowest, often not undergoing \( S_N2 \) reactions at all.

4. **Analyze Each Compound**:
- If a compound has a primary carbon (attached to one alkyl group), it will undergo the \( S_N2 \) reaction faster than a secondary or tertiary carbon.
- If a compound has a tertiary carbon (attached to three alkyl groups), it will be the slowest or may not react via \( S_N2 \).

5. **Conclusion**:
After analyzing the steric hindrance for each compound, the one with the least steric hindrance (usually a primary alkyl halide) will undergo the fastest \( S_N2 \) substitution reaction.

### Final Answer:
The compound with the least steric hindrance, typically a primary alkyl halide, will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH.