Consider the reaction: `C_(2)H_(5)O^(-)`+ Dimethyl sulphate `(Me-O-overset(O)overset(||)underset(O)underset(||)S-O-Me)rarr`
The leaving group is:

Options

A. `MeOSO_(2)O^(-)`
B. `MeO^(-)`
C. `Me(-)`
D. `MeOSO_(2)^(-)`

Explanation

To solve the question regarding the reaction of ethoxide ion (C₂H₅O⁻) with dimethyl sulfate, we need to identify the leaving group in the reaction. Here’s a step-by-step breakdown of the solution:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants in the reaction are ethoxide ion (C₂H₅O⁻) and dimethyl sulfate (Me₂SO₄). The structure of dimethyl sulfate consists of a sulfur atom bonded to two methoxy groups (–OCH₃) and has a sulfate group (–SO₄).

2. **Understanding the Mechanism**: The reaction between a nucleophile (C₂H₅O⁻) and dimethyl sulfate typically proceeds through an SN2 mechanism. In this mechanism, the nucleophile attacks the electrophilic carbon atom of the dimethyl sulfate, leading to the displacement of a leaving group.

3. **Identify the Electrophilic Site**: In dimethyl sulfate, the carbon atoms bonded to the methoxy groups (–OCH₃) are electrophilic due to the presence of the electronegative sulfur atom. This makes them susceptible to nucleophilic attack.

4. **Determine the Leaving Group**: When the ethoxide ion attacks one of the carbon atoms in dimethyl sulfate, one of the methoxy groups (–OCH₃) will be displaced as a leaving group. The leaving group in this case is the methoxy group (–OCH₃), which can be represented as MeO.

5. **Final Product**: The reaction will yield ethyl methyl ether (C₂H₅OCH₃) and the leaving group (MeO). The overall reaction can be summarized as:
\[ C₂H₅O⁻ + Me₂SO₄ \rightarrow C₂H₅OCH₃ + MeO⁻ \]

6. **Conclusion**: Therefore, the leaving group in this reaction is the methoxy group (–OCH₃), which can also be represented as MeO.

### Final Answer:
The leaving group is: **Methoxy group (–OCH₃)**.