Diethyl ether on heating with conc. HI gives two moles of:

Options

A. Ethanol
B. Iodoform
C. Ethyl iodide
D. Methyl iodide

Explanation

To solve the question "Diethyl ether on heating with conc. HI gives two moles of:", we will analyze the reaction step by step.

### Step-by-Step Solution:

1. **Understanding the Reactants**:
- The reactant is diethyl ether, which has the chemical formula \( \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 \).
- We are reacting it with concentrated hydrogen iodide (HI).

2. **Ionization of Hydrogen Iodide**:
- When concentrated HI is used, it ionizes into \( \text{H}^+ \) and \( \text{I}^- \) ions. This is important as the iodide ion will act as a nucleophile in the reaction.

3. **Nucleophilic Attack**:
- The iodide ion \( \text{I}^- \) will attack the carbon atom that is bonded to the oxygen in diethyl ether. This results in the breaking of the carbon-oxygen bond.

4. **Formation of Ethanol**:
- When the iodide ion attacks, it forms ethyl iodide \( \text{CH}_3\text{CH}_2\text{I} \) and generates a protonated alcohol (ethanol) \( \text{CH}_3\text{CH}_2\text{OH} \):
\[ \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{I} + \text{CH}_3\text{CH}_2\text{OH} \]

5. **Further Reaction of Ethanol**:
- Since we have excess HI, the ethanol produced can also react with another molecule of HI. The iodide ion will again attack the carbon-oxygen bond in ethanol, leading to the formation of another ethyl iodide and water:
\[ \text{CH}_3\text{CH}_2\text{OH} + \text{HI} \rightarrow \text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O} \]

6. **Overall Reaction**:
- Combining both steps, the overall reaction can be summarized as:
\[ \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 + 2 \text{HI} \rightarrow 2 \text{CH}_3\text{CH}_2\text{I} + \text{H}_2\text{O} \]

7. **Conclusion**:
- Therefore, diethyl ether on heating with concentrated HI gives two moles of ethyl iodide \( \text{CH}_3\text{CH}_2\text{I} \).

### Final Answer:
Diethyl ether on heating with concentrated HI gives two moles of **ethyl iodide** \( \text{CH}_3\text{CH}_2\text{I} \).

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