Reaction of methyl bromide with aqueous sodium hydroxide involves

Options

A. Racemization
B. `S_(N)1` mechanism
C. Retention of configuration
D. `S_(N)2` mechanism

Answer

To solve the question regarding the reaction of methyl bromide (CH3Br) with aqueous sodium hydroxide (NaOH), we need to analyze the mechanism of the reaction.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are methyl bromide (CH3Br) and sodium hydroxide (NaOH). In aqueous solution, NaOH dissociates into Na⁺ and OH⁻ ions.

**Hint**: Always start by identifying the reactants involved in the reaction.

2. **Determine the Type of Alkyl Halide**:
Methyl bromide is a primary alkyl halide. In the context of nucleophilic substitution reactions, primary alkyl halides are known to undergo SN2 reactions.

**Hint**: Recall the classification of alkyl halides and their reactivity in different substitution mechanisms (SN1 vs. SN2).

3. **Mechanism of Reaction**:
In an SN2 reaction, the nucleophile (OH⁻) attacks the carbon atom bonded to the bromine atom from the opposite side (backside attack). This leads to the formation of a transition state where the bond between carbon and bromine is partially broken while the bond between carbon and hydroxide is partially formed.

**Hint**: Remember that SN2 reactions involve a backside attack and the formation of a transition state.

4. **Formation of Products**:
After the transition state is formed, the bromine atom (Br⁻) is expelled, resulting in the formation of methanol (CH3OH) as the product. The configuration at the carbon atom is inverted due to the backside attack.

**Hint**: Focus on the final products and the stereochemical outcome of the reaction.

5. **Conclusion**:
The reaction of methyl bromide with aqueous sodium hydroxide proceeds via an SN2 mechanism, resulting in the formation of methanol and the inversion of configuration at the carbon atom.

**Hint**: Conclude by summarizing the mechanism and the product formed.

### Final Answer:
The reaction of methyl bromide with aqueous sodium hydroxide involves an SN2 mechanism, leading to the formation of methanol (CH3OH) and the inversion of configuration at the carbon atom.