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 determine the leaving group in the reaction between ethoxide ion (C₂H₅O⁻) and dimethyl sulfate (Me₂SO₄), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in the reaction are the ethoxide ion (C₂H₅O⁻) and dimethyl sulfate (Me₂SO₄).

2. **Understand the Structure of Dimethyl Sulfate**:
Dimethyl sulfate can be represented as:
\[ \text{Me-O-S(=O)(=O)-O-Me} \]
Here, the sulfur atom is bonded to two methyl groups and has a sulfonate structure.

3. **Recognize the Nucleophile**:
The ethoxide ion (C₂H₅O⁻) acts as a nucleophile. It has a negative charge and is capable of attacking electrophilic centers.

4. **Mechanism of Reaction**:
When the ethoxide ion attacks one of the methyl groups in dimethyl sulfate, the electrons from the C-O bond will move towards the oxygen, creating a negative charge on the oxygen and causing the methyl group to become a good leaving group.

5. **Identify the Leaving Group**:
The leaving group in this reaction is the group that departs when the nucleophile attacks. In this case, the leaving group is:
\[ \text{MeOSO}_2^- \]
This group is formed when the ethoxide ion attacks the methyl group, resulting in the formation of ethyl methyl ether and the departure of the methyl sulfonate ion.

6. **Conclusion**:
The leaving group in the reaction is:
\[ \text{MeOSO}_2^- \]
which corresponds to option A.

### Final Answer:
The leaving group is: **MeOSO₂⁻** (option A).

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