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)CC1`

Explanation

To determine which compound will react most readily with NaOH to form methanol, we need to analyze the possible candidates and their reactivity.

### Step-by-Step Solution:

1. **Identify the Compounds**: We need to consider compounds that can react with NaOH to produce methanol. The most common candidates are alkyl halides, particularly those that can undergo nucleophilic substitution.

2. **Understand the Reaction Mechanism**: The reaction involves a nucleophilic substitution where NaOH (which provides OH-) attacks the carbon atom bonded to the leaving group (like I- in this case). The leaving group must be able to depart easily for the reaction to proceed.

3. **Evaluate Leaving Groups**: Among the common leaving groups, iodide (I-) is considered a very good leaving group due to its size and ability to stabilize the negative charge after leaving. Therefore, compounds with I- as a leaving group will react more readily.

4. **Consider the Structure of the Compound**: The compound CH3N+(I-) is a quaternary ammonium salt. In this case, the methyl group attached to the nitrogen is likely to be more reactive due to the positive charge on the nitrogen, which makes the carbon attached to the iodide more electrophilic.

5. **Predict the Product**: When CH3N+(I-) reacts with NaOH, the iodide ion (I-) leaves, and the hydroxide ion (OH-) from NaOH attacks the methyl carbon, resulting in the formation of methanol (CH3OH).

6. **Conclusion**: Therefore, the compound that will react most readily with NaOH to form methanol is CH3N+(I-).

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