Which of the following will react with NaOH to form methanol as a major product ?

Options

A. `(CH_3)_4N^(+)I^(-)`
B. `(CH_3)_3S^(+)I^(-)`
C. `(CH_3)_3C Cl`
D. `CH_3-O-CH_3`

Explanation

To determine which compound reacts with NaOH to form methanol as a major product, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Compounds**: Look at the given options and identify the compounds that are likely to react with NaOH. We need to focus on alkyl halides since they can undergo nucleophilic substitution reactions.

2. **Determine the Type of Alkyl Halides**: Check if the alkyl halides are primary, secondary, or tertiary. Tertiary alkyl halides are more likely to undergo nucleophilic substitution reactions due to steric hindrance.

3. **Analyze the Reactivity**: Among the alkyl halides, tertiary alkyl halides will react more readily with nucleophiles like OH⁻ from NaOH.

4. **Consider the Electronegativity**: In the given options, if there is a positively charged nitrogen (N⁺) or sulfur (S⁺) adjacent to the alkyl halide, it can increase the electrophilicity of the carbon attached to the halide. This makes it more susceptible to nucleophilic attack.

5. **Write the Reaction**: For the selected compound (let's say it is a tertiary alkyl halide like CH₃I), the reaction with NaOH can be written as:
\[ CH_3I + NaOH \rightarrow CH_3OH + NaI \]
Here, the hydroxide ion (OH⁻) acts as a nucleophile and substitutes the iodide ion (I⁻), resulting in methanol (CH₃OH) as the major product.

6. **Identify the Major Product**: The major product of this reaction is methanol (CH₃OH).

7. **Conclusion**: Based on the analysis, the correct option that reacts with NaOH to form methanol as a major product is the one that contains a tertiary alkyl halide with an adjacent positively charged atom (like nitrogen).

### Final Answer:
The correct option is **Option A**.

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