When methyl iodide is heated with sodium ethoxide, it forms?

Options

A. `CH_(3)OH`
B. `C_(2)H_(5)OC_(2)H_(5)`
C. `CH_(3)OCH_(3)`
D. `CH_(3)OC_(2)H_(5)`

Answer

To solve the problem of what product is formed when methyl iodide is heated with sodium ethoxide, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- Methyl iodide (CH₃I)
- Sodium ethoxide (C₂H₅O⁻Na⁺)

2. **Draw the Structures**:
- Methyl iodide:
\[ \text{CH}_3\text{I} \]
- Sodium ethoxide:
\[ \text{C}_2\text{H}_5\text{O}^- \text{Na}^+ \]

3. **Understand the Reaction Mechanism**:
- This reaction is a nucleophilic substitution reaction (SN2 mechanism). In this case, the ethoxide ion (C₂H₅O⁻) acts as a nucleophile and attacks the electrophilic carbon in methyl iodide (CH₃I).

4. **Analyze the Electrophilic Site**:
- The carbon in methyl iodide is electron-deficient due to the electronegativity of iodine, which pulls electron density away from the carbon. This makes the carbon susceptible to nucleophilic attack.

5. **Nucleophilic Attack**:
- The oxygen atom in the ethoxide ion, which has a negative charge, attacks the electron-deficient carbon atom of methyl iodide. This results in the displacement of the iodide ion (I⁻).

6. **Formation of the Product**:
- The reaction yields methoxyethane (C₂H₅OCH₃) and sodium iodide (NaI) as a byproduct:
\[ \text{C}_2\text{H}_5\text{O}^- + \text{CH}_3\text{I} \rightarrow \text{C}_2\text{H}_5\text{OCH}_3 + \text{NaI} \]

7. **Final Product**:
- The main product formed is methoxyethane, which can also be represented as:
\[ \text{C}_2\text{H}_5\text{OCH}_3 \text{ or } \text{CH}_3\text{O}\text{C}_2\text{H}_5 \]

8. **Conclusion**:
- The product formed when methyl iodide is heated with sodium ethoxide is methoxyethane (C₂H₅OCH₃).