The given question tests your understanding of the Wurtz reaction in organic chemistry. Let's analyze the problem step by step:
- **Understanding the Reaction:**
The Wurtz reaction involves the coupling of two alkyl halides in the presence of sodium and dry ether to form a higher alkane. In this question, ethyl iodide (C\(_2\)H\(_5\)I) is heated with sodium in dry ether. - **Reaction Equation:**
The general reaction formula is:
\(2\,\text{R--X} + 2\,\text{Na} \rightarrow \text{R--R} + 2\,\text{NaX}\)
Where R is an alkyl group and X is a halogen.
Applying this to ethyl iodide:
\(2\,\text{C}_2\text{H}_5\text{I} + 2\,\text{Na} \rightarrow \text{C}_2\text{H}_5\text{--C}_2\text{H}_5 + 2\,\text{NaI}\) - **Determine the Product:**
The reaction results in the formation of butane (C\(_4\)H\(_{10}\)), which is comprised of two ethyl groups joined together. - **Evaluate the Options:**
- Option 1: C\(_4\)H\(_{10}\): Correct, as it is the expected product of the Wurtz reaction with ethyl iodide. - Option 2: C\(_2\)H\(_6\): Incorrect, as this is ethane, which is not formed in this reaction. - Option 3: C\(_3\)H\(_8\): Incorrect, as this is propane, and not a possible product from the reaction. - Option 4: C\(_2\)H\(_5\)OH: Incorrect, as this is ethanol, an alcohol, and cannot be formed under these reaction conditions.
Thus, the correct answer is C\(_4\)H\(_{10}\) (butane).
To remember, in Wurtz reaction, always look for the coupling of two identical alkyl halides leading to the formation of a new carbon-carbon bond, forming a higher alkane.