Which is the most effective ion in an `SN^(2)` displacement on methyl bromide?

Options

A. `C_(6) H_(5)O^(theta)`
B. `overset(theta)(O)H`
C. `C_(6)H_(5)O^(theta)`
D. `CH_(3)COO^(theta)`

Explanation

To determine the most effective ion in an \( \text{S}_\text{N}2 \) displacement on methyl bromide (CH₃Br), we need to analyze the nucleophiles provided and their respective strengths. Here’s a step-by-step solution:

### Step 1: Identify the Nucleophiles
The question provides four anions as potential nucleophiles for the \( \text{S}_\text{N}2 \) reaction with methyl bromide. The anions are:
1. \( \text{C}_2\text{H}_5\text{O}^- \) (Ethoxide ion)
2. \( \text{OH}^- \) (Hydroxide ion)
3. \( \text{C}_6\text{H}_5\text{O}^- \) (Phenoxide ion)
4. \( \text{CH}_3\text{COO}^- \) (Acetate ion)

### Step 2: Understand the Mechanism of \( \text{S}_\text{N}2 \)
The \( \text{S}_\text{N}2 \) mechanism involves a nucleophile attacking the electrophilic carbon (in this case, the carbon in methyl bromide) and displacing the leaving group (bromine). The strength of the nucleophile is crucial for the rate of this reaction.

### Step 3: Compare Basicity of the Nucleophiles
The strength of a nucleophile is often related to its basicity. A stronger base is typically a better nucleophile. To compare the nucleophiles, we can look at their conjugate acids:
- For \( \text{C}_2\text{H}_5\text{O}^- \), the conjugate acid is \( \text{C}_2\text{H}_5\text{OH} \) (Ethanol).
- For \( \text{OH}^- \), the conjugate acid is \( \text{H}_2\text{O} \) (Water).
- For \( \text{C}_6\text{H}_5\text{O}^- \), the conjugate acid is \( \text{C}_6\text{H}_5\text{OH} \) (Phenol).
- For \( \text{CH}_3\text{COO}^- \), the conjugate acid is \( \text{CH}_3\text{COOH} \) (Acetic acid).

### Step 4: Determine the Acidity Order
The acidity of these conjugate acids can be ranked:
1. Acetic acid \( (\text{CH}_3\text{COOH}) \) - Strongest acid
2. Phenol \( (\text{C}_6\text{H}_5\text{OH}) \)
3. Water \( (\text{H}_2\text{O}) \)
4. Ethanol \( (\text{C}_2\text{H}_5\text{OH}) \) - Weakest acid

### Step 5: Infer Basicity from Acidity
From the acidity order, we can deduce the basicity:
- The weaker the acid, the stronger the conjugate base.
- Therefore, the order of basicity (and thus nucleophilicity, since the nucleophilic center is the same) is:
1. \( \text{C}_2\text{H}_5\text{O}^- \) (Strongest base)
2. \( \text{OH}^- \)
3. \( \text{C}_6\text{H}_5\text{O}^- \)
4. \( \text{CH}_3\text{COO}^- \) (Weakest base)

### Step 6: Conclusion
Since \( \text{C}_2\text{H}_5\text{O}^- \) (Ethoxide ion) is the strongest base, it is also the most effective nucleophile for the \( \text{S}_\text{N}2 \) displacement on methyl bromide.

Thus, the answer is:
**Ethoxide ion \( (\text{C}_2\text{H}_5\text{O}^-) \)** is the most effective ion in \( \text{S}_\text{N}2 \) displacement on methyl bromide.

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