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 factors that influence the \( S_N2 \) mechanism, particularly focusing on steric hindrance.

### Step-by-Step Solution:

1. **Understanding the \( S_N2 \) Mechanism**:
- The \( S_N2 \) (substitution nucleophilic bimolecular) reaction involves a nucleophile attacking an electrophile (alkyl halide) from the opposite side of the leaving group (halogen). This backside attack leads to the formation of a transition state and ultimately results in the substitution of the halogen with the nucleophile.

2. **Identify the Role of Steric Hindrance**:
- Steric hindrance refers to the crowding around the reactive center (the carbon atom bonded to the leaving group). The more bulky groups attached to the carbon, the slower the \( S_N2 \) reaction will be. Therefore, compounds with less steric hindrance will react faster.

3. **Evaluate Different Alkyl Halides**:
- Consider different alkyl halides (for example, primary, secondary, and tertiary alkyl halides):
- **Primary Alkyl Halide**: Has one alkyl group attached to the carbon with the leaving group. This structure has the least steric hindrance.
- **Secondary Alkyl Halide**: Has two alkyl groups attached. This structure has moderate steric hindrance.
- **Tertiary Alkyl Halide**: Has three alkyl groups attached. This structure has the most steric hindrance and is typically unreactive in \( S_N2 \) reactions.

4. **Conclusion**:
- Since the \( S_N2 \) reaction rate is inversely related to steric hindrance, the primary alkyl halide will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH.

5. **Final Answer**:
- The compound that will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH is the primary alkyl halide.