In the following reaction
`C_(2)H_(5)OC_(2)H_(5)+4Hoverset(Red P+HI)(rarr)2X+H_(2)O` `X` is

Options

A. Ethane
B. Ethylene
C. Butane
D. Propane

Answer

To solve the given reaction, we need to analyze the reduction of diethyl ether (C2H5OC2H5) in the presence of red phosphorus and hydroiodic acid (HI). The reaction can be summarized as follows:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactant is diethyl ether, which has the molecular formula C2H5OC2H5.

2. **Understand the Reaction Conditions**:
The reaction is carried out in the presence of red phosphorus and hydroiodic acid (HI). Red phosphorus acts as a reducing agent.

3. **Reduction of Diethyl Ether**:
When diethyl ether is treated with HI, it undergoes reduction. The ether bond (C-O) is broken, and the ether is converted into alkane products.

4. **Formation of Ethyl Carbocation**:
The proton from HI attacks the ether oxygen, leading to the formation of an ethyl carbocation (C2H5+). The ether bond breaks, resulting in the formation of ethanol (C2H5OH) and an ethyl carbocation.

5. **Formation of Ethane**:
The ethyl carbocation can further lose a proton to form ethylene (C2H4). The presence of red phosphorus provides hydrogen atoms to the alkene (C2H4), resulting in the formation of ethane (C2H6).

6. **Final Products**:
Since the reaction starts with one molecule of diethyl ether and produces two molecules of ethane, we can summarize that the products of the reaction are 2 moles of ethane (C2H6) and water (H2O).

7. **Conclusion**:
Therefore, the value of X in the reaction is ethane (C2H6).

### Final Answer:
X = C2H6 (Ethane)

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