Question

Which of the following alkanes is NOT obtained when a mixture of ethyl iodide and n-propyl iodide is subjected to Wurtz reaction?

• A. n-Butane
• B. Propane
• C. n-Pentane
• D. n-Hexane

Hint:

Wurtz couples two alkyl groups: C2+C2, C3+C3 and C2+C3 are the only products.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In the Wurtz reaction, two alkyl halides react with sodium in dry ether to join their alkyl groups: \(\text{R-X} + \text{R'-X} + 2\text{Na} \rightarrow \text{R-R'} + 2\text{NaX}\).

Step 2: Detailed Explanation:
With \(\text{C}_2\text{H}_5\text{I}\) (ethyl) and \(\text{n-C}_3\text{H}_7\text{I}\) (propyl), three couplings are possible:
Ethyl + ethyl gives butane (4 carbons).
Propyl + propyl gives n-hexane (6 carbons).
Ethyl + propyl gives n-pentane (5 carbons).
Propane has only 3 carbons, and no combination of 2 + 3 or 2 + 2 or 3 + 3 carbons gives 3. So propane is not formed.

Step 3: Final Answer:
Option (B), propane.

Final Answer:
Coupling gives C4, C5 and C6 alkanes, never C3. \[ \boxed{\text{(B) }\text{Propane}} \]