Question

When a mixture of two different alkyl halides reacts with metallic sodium in dry ether, the formation of a possible number of alkanes are-

• A. Two
• B. Three
• C. Four
• D. Five

Hint:

Count the pairs AA, BB and AB.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Wurtz reaction: two alkyl halides couple with sodium in dry ether to form an alkane with the two alkyl groups joined, \(2R-X + 2Na \rightarrow R-R + 2NaX\).

Step 2: Apply to a mixture:
Let the two halides be \(R-X\) and \(R'-X\). Three couplings are possible.

Step 3: List the products:
1. \(R\) with \(R\) gives \(R-R\).
2. \(R'\) with \(R'\) gives \(R'-R'\).
3. \(R\) with \(R'\) gives \(R-R'\).

Step 4: Conclusion:
Three different alkanes can form, so the answer is option (B). Two would be the count only if the cross product were missing, and four or five would need more distinct alkyl groups than we have.

Final Answer:
Two different halides give R-R, R'-R' and R-R', so three alkanes. \[ \boxed{B:\ \text{Three}} \]