The reaction `CH_3 Br + OH^(-) → CH_3 OH + Br^(-)` follows

Options

A. `S_(N^1)` mechanism
B. `S_(N^2)`mechanism
C. either of the above two
D. none of the above two

Explanation

To determine the mechanism of the reaction \( \text{CH}_3 \text{Br} + \text{OH}^- \rightarrow \text{CH}_3 \text{OH} + \text{Br}^- \), we need to analyze the nature of the reactants and the mechanism of nucleophilic substitution.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are methyl bromide (\( \text{CH}_3 \text{Br} \)) and hydroxide ion (\( \text{OH}^- \)).

2. **Determine the Type of Alkyl Halide**:
Methyl bromide (\( \text{CH}_3 \text{Br} \)) is a primary alkyl halide. Primary alkyl halides are those where the carbon atom bonded to the halogen is attached to only one other carbon atom.

3. **Understand the Nucleophile**:
The hydroxide ion (\( \text{OH}^- \)) is a strong nucleophile. Nucleophiles are species that donate an electron pair to form a chemical bond.

4. **Consider the Mechanisms**:
There are two main mechanisms for nucleophilic substitution:
- **SN1 Mechanism**: This involves the formation of a carbocation intermediate and is favored by tertiary alkyl halides.
- **SN2 Mechanism**: This involves a direct attack by the nucleophile on the carbon atom, leading to a transition state and is favored by primary alkyl halides.

5. **Analyze the Mechanism for \( \text{CH}_3 \text{Br} \)**:
Since \( \text{CH}_3 \text{Br} \) is a primary alkyl halide, it will undergo the SN2 mechanism. In this mechanism, the nucleophile (\( \text{OH}^- \)) attacks the carbon atom from the opposite side of the leaving group (bromide ion, \( \text{Br}^- \)).

6. **Transition State Formation**:
As the nucleophile attacks, a transition state is formed where the bond between the carbon and bromine is partially broken while the bond between the carbon and oxygen is partially formed.

7. **Final Product Formation**:
The reaction results in the formation of methanol (\( \text{CH}_3 \text{OH} \)) and bromide ion (\( \text{Br}^- \)). The configuration of the product is inverted due to the backside attack characteristic of the SN2 mechanism.

### Conclusion:
The reaction \( \text{CH}_3 \text{Br} + \text{OH}^- \rightarrow \text{CH}_3 \text{OH} + \text{Br}^- \) follows the **SN2 mechanism**.

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