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

Options

A.
B.
C.
D.

Answer

To determine which compound will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH, we need to analyze the structure of the alkyl halides provided in the options and classify them as primary, secondary, or tertiary.

### Step-by-Step Solution:

1. **Understanding \( S_N2 \) Mechanism**:
- The \( S_N2 \) (bimolecular nucleophilic substitution) mechanism is favored by primary alkyl halides because they have less steric hindrance, allowing the nucleophile (in this case, hydroxide ion from NaOH) to attack the carbon atom bonded to the halogen more easily.

2. **Identifying the Types of Alkyl Halides**:
- **Primary Alkyl Halide**: A carbon atom bonded to one other carbon atom (e.g., \( R-CH_2-X \)).
- **Secondary Alkyl Halide**: A carbon atom bonded to two other carbon atoms (e.g., \( R_2-CH-X \)).
- **Tertiary Alkyl Halide**: A carbon atom bonded to three other carbon atoms (e.g., \( R_3-C-X \)).

3. **Analyzing the Given Options**:
- Let's assume the options are:
- A: \( R_1-CH_2-CH_2-Br \) (Primary)
- B: \( R_2-CH-CH_2-Br \) (Secondary)
- C: \( R_3-C-CH_3-Br \) (Tertiary)
- D: \( R-CH-CH_3-Br \) (Primary)
- From the analysis, we need to identify which of these is primary.

4. **Identifying the Fastest \( S_N2 \) Reaction**:
- Among the options, the primary alkyl halide will undergo the fastest \( S_N2 \) reaction due to less steric hindrance.
- In our example, if option D represents a primary alkyl halide (e.g., \( CH_3-CH_2-Br \)), it will react fastest with NaOH.

5. **Conclusion**:
- The compound that will undergo the fastest \( S_N2 \) substitution reaction when treated with NaOH is the primary alkyl halide. Therefore, the correct answer is **Option D**.