Question

When ethyl bromide and \textit{n}-propyl bromide are allowed to react with Na metal in dry ether, the number of different alkanes formed is:

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

In Wurtz reaction of two different haloalkanes, use combination logic: \textit{n} compounds → up to \(\frac{n(n+1)}{2}\) alkanes.

Answer 1

The Correct Option is C

This is a Wurtz reaction where alkyl halides react with sodium in dry ether to form alkanes: \[ \begin{aligned} &\ce{C2H5Br + C2H5Br→[Na/dry~ether] C2H5-C2H5} →(\text{butane})
&\ce{C3H7Br + C3H7Br →[Na/dry~ether] C3H7-C3H7}→ (\text{hexane})
&\ce{C2H5Br + C3H7Br →[Na/dry~ether] C2H5-C3H7}→(\text{pentane}) \end{aligned} \] Thus, 3 different alkanes are formed.