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: Count combinations:
Each coupling joins two alkyl radicals. With two kinds of radicals, A and B, the possible pairs are AA, BB and AB.

Step 2: Check formula:
For $n$ different halides the number of possible alkanes is $n(n+1)/2$. For $n=2$ this is $2\times 3/2 = 3$.

Step 3: Select:
Three alkanes, so option (B).

Final Answer:
With two different halides, three coupling products are possible. \[ \boxed{B} \]