Question

In the reaction of 1-bromo-3-chlorocyclobutane with two equivalents of sodium in ether, the major product is

Answer 1

Typically, you would react two alkyl halides to form an alkane that consists of their respective Rgroups. So, as an example, you would have:

R−Br+R'−ClNa(s) −−−−−→R−R'
(with loss of NaCl and then NaBr)

An example of the mechanism with methyl chloride and ethyl bromide would be:



Since sodium has a radical 3s electron, it donates it to the more electronegative halogen (Cl), and Cl− donates one bonding electron, to form NaCl. The remaining radical electron goes to form an alkyl radical.

The reactive alkyl radical then reacts with another sodium atom to become a nucleophile. Since sodium has a low electronegativity compared to carbon (0.93vs. 2.5), R−Na is really more like R:(−)(+)Na.

Finally, the nucleophile acts in an SN2backside-attack on the other alkyl halide to form the final product.

So the end result is that the R and R' groups of two alkyl halides combine. Don't forget to count your carbons!

In the case of your reaction, dioxane is a refluxed solvent, and at the temperature this reaction is performed for 1-bromo-3-chlorocyclobutane, sodium is a liquid

Answer 2

When 1-bromo-3-chloro cyclobutane is treated with 2 equivalents of Na in presence of ether, bicyclo[1.1. 0]butane is the product. In this reaction, a molecule of BrCl is eliminated and a C-C bond is formed which gives bicyclo compound.