The compound that will react most readily with `NaOH` to from methanol is

Options

A. `(CH_(3))_(4) N^(+)I^(-)`
B. `CH_(3) OCH_(3)`
C. `(CH_(3))_(3)S^(+)I^(-)`
D. `(CH_(3))_(3)Cl`

Answer

To determine which compound reacts most readily with sodium hydroxide (NaOH) to form methanol (CH3OH), we need to analyze the given options.

### Step-by-Step Solution:

1. **Identify the Compounds**: The question provides several compounds, including CH3I (methyl iodide), CH3Cl (methyl chloride), and CH3N(CH3)3I (quaternary ammonium iodide).

2. **Understand the Reaction Mechanism**: The reaction we are considering is a nucleophilic substitution reaction (SN2 mechanism). In this mechanism, a nucleophile (in this case, OH-) attacks the carbon atom bonded to the leaving group (I-, Cl-, etc.), resulting in the formation of methanol and the corresponding ammonium or alkyl halide.

3. **Evaluate the Leaving Groups**: The effectiveness of the leaving group is crucial for the reaction:
- Iodide (I-) is a better leaving group than chloride (Cl-) due to its larger size and lower bond strength with carbon.
- Quaternary ammonium iodide (CH3N(CH3)3I) has no beta hydrogen, which means it cannot undergo elimination reactions and will react via the SN2 mechanism.

4. **Analyze Each Option**:
- **Option 1: CH3N(CH3)3I**: This compound can react with NaOH, where the iodide ion (I-) leaves, allowing the hydroxide ion (OH-) to attach to the carbon, forming methanol.
- **Option 2: CH3Cl**: Chloride is a poorer leaving group than iodide, making this compound less reactive in the SN2 mechanism.
- **Option 3: CH3I**: Methyl iodide is also a good candidate because I- is a good leaving group. However, it is not as readily reactive as the quaternary ammonium iodide due to steric hindrance.

5. **Conclusion**: The compound that will react most readily with NaOH to form methanol is **CH3N(CH3)3I** (quaternary ammonium iodide) because it has the best leaving group and can undergo the SN2 reaction effectively.

### Final Answer:
The compound that will react most readily with NaOH to form methanol is **CH3N(CH3)3I**.