A mixture of ethyl iodide and methyl iodide is subjected to the Wurtz reaction. The product(s) formed is/are

Options

A. ethane
B. butane
C. propane
D. 2-methylpropane

Answer

To solve the problem of determining the products formed when a mixture of ethyl iodide (C2H5I) and methyl iodide (CH3I) is subjected to the Wurtz reaction, we can follow these steps:

### Step 1: Understand the Wurtz Reaction
The Wurtz reaction involves the coupling of alkyl halides in the presence of sodium metal (Na) to form higher alkanes. The reaction proceeds through the formation of free radicals.

### Step 2: Identify the Reactants
The reactants in this case are:
- Methyl iodide (CH3I)
- Ethyl iodide (C2H5I)

### Step 3: Generate Free Radicals
When the alkyl iodides are treated with sodium, they undergo homolytic cleavage to form free radicals:
- From methyl iodide:
\[ CH3I \rightarrow CH3\cdot + I\cdot \]
- From ethyl iodide:
\[ C2H5I \rightarrow C2H5\cdot + I\cdot \]

### Step 4: Possible Coupling Reactions
The free radicals can couple in different ways:
1. **Self-coupling of methyl radicals:**
\[ CH3\cdot + CH3\cdot \rightarrow C2H6 \quad (\text{Ethane}) \]

2. **Self-coupling of ethyl radicals:**
\[ C2H5\cdot + C2H5\cdot \rightarrow C4H10 \quad (\text{Butane}) \]

3. **Cross-coupling between methyl and ethyl radicals:**
\[ CH3\cdot + C2H5\cdot \rightarrow C3H8 \quad (\text{Propane}) \]

### Step 5: Summary of Products
From the above reactions, the products formed are:
- Ethane (C2H6)
- Butane (C4H10)
- Propane (C3H8)

### Step 6: Exclude Other Possibilities
2-Methylpropane (C5H12) cannot be formed in this reaction because it would require a different combination of alkyl groups that are not present in the reactants.

### Final Answer
The products formed from the Wurtz reaction of a mixture of ethyl iodide and methyl iodide are:
- Ethane (C2H6)
- Butane (C4H10)
- Propane (C3H8)