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)I`

Answer

To determine which compound reacts most readily with NaOH to form methanol, we will analyze the given options step by step.

### Step 1: Identify the Structure of the Given Compound
Let's consider the first option provided in the video, which is \( \text{CH}_3^4\text{N}^+ \text{I}^- \). This compound can be represented as a quaternary ammonium salt, where a nitrogen atom is positively charged and is bonded to four methyl groups.

### Step 2: Reaction with NaOH
When this compound reacts with NaOH, the sodium ion (Na\(^+\)) will not participate in the reaction, but the hydroxide ion (OH\(^-\)) will act as a nucleophile. The reaction can be represented as follows:

\[ \text{CH}_3^4\text{N}^+ \text{I}^- + \text{NaOH} \rightarrow \text{CH}_3^3\text{N}^+ \text{OH} + \text{NaI} \]

### Step 3: Formation of Methanol
In the above reaction, the hydroxide ion (OH\(^-\)) replaces the iodide ion (I\(^-\)) in the quaternary ammonium salt. After this substitution, we need to heat the compound to facilitate the formation of methanol. The nitrogen atom, being positively charged, makes the adjacent methyl group more reactive. Thus, upon heating, one of the methyl groups will release and form methanol.

\[ \text{CH}_3^3\text{N}^+ \text{OH} \xrightarrow{\text{heat}} \text{CH}_3\text{OH} + \text{CH}_3^3\text{N}^+ \]

### Step 4: Conclusion
From the above steps, we can conclude that the compound \( \text{CH}_3^4\text{N}^+ \text{I}^- \) reacts most readily with NaOH to form methanol due to the presence of a good leaving group (I\(^-\)) and the electron-deficient nitrogen atom, which enhances the reactivity towards the nucleophile (OH\(^-\)).

### Final Answer
The compound that will react most readily with NaOH to form methanol is \( \text{CH}_3^4\text{N}^+ \text{I}^- \).

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