The product of the reaction of ethyl iodide with Zn-Cu couple/alcohol can also be prepared by the following reaction.

Options

A. By the hydrolysis of the product obtained from the reaction of ethyl iodide with Mg in dry ether.
B. Reaction of iodomethane with Na in dry ether
C. By the reduction of ethyl alcohol with red P + HI
D. All of the above

Answer

To solve the problem, we need to identify the product of the reaction of ethyl iodide (C2H5I) with a zinc-copper couple in alcohol and see how it can also be prepared through other reactions.

### Step-by-Step Solution:

1. **Identify the Reaction with Zinc-Copper Couple:**
- Ethyl iodide (C2H5I) reacts with a zinc-copper couple in the presence of alcohol to produce ethane (C2H6).
- The reaction can be represented as:
\[ \text{C2H5I} + \text{Zn-Cu} \xrightarrow{\text{alcohol}} \text{C2H6} + \text{ZnI2} \]

2. **Option A - Hydrolysis of Product from Reaction with Magnesium:**
- Ethyl iodide can also react with magnesium in dry ether to form a Grignard reagent (C2H5MgI).
- Hydrolysis of this Grignard reagent with water or dilute acid will yield ethane:
\[ \text{C2H5MgI} + \text{H2O} \rightarrow \text{C2H6} + \text{Mg(OH)I} \]
- Thus, option A is valid.

3. **Option B - Reaction of Iodomethane with Sodium in Dry Ether:**
- Iodomethane (CH3I) can react with sodium in dry ether to form a Grignard reagent (CH3Na).
- When this Grignard reagent reacts with ethyl iodide (C2H5I), it forms ethane:
\[ \text{CH3Na} + \text{C2H5I} \rightarrow \text{C2H6} + \text{NaI} \]
- Therefore, option B is also valid.

4. **Option C - Reduction of Ethyl Alcohol with Red Phosphorus and HI:**
- Ethyl alcohol (C2H5OH) can be reduced using red phosphorus and hydroiodic acid (HI) to produce ethane:
\[ \text{C2H5OH} + \text{HI} \xrightarrow{\text{red P}} \text{C2H6} + \text{H2O} + \text{I2} \]
- Hence, option C is valid as well.

5. **Conclusion:**
- Since all options A, B, and C lead to the formation of ethane (C2H6), the correct answer is:
\[ \text{D: All of the above} \]