1 bromo 3 clorocyclobutane + Metallic sodium = ? ( the medium of reaction is dioxane )?

Ah, this is the Wurtz reaction.

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

#"R"-"Br" + "R"'-"Cl" stackrel("Na"(s)" ")(->) "R"-"R"'#
(with loss of #"NaCl"# and then #"NaBr"#)

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

1. 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.
2. The reactive alkyl radical then reacts with another sodium atom to become a nucleophile. Since sodium has a low electronegativity compared to carbon (#0.93# vs. #2.5#), #"R"-"Na"# is really more like #"R":^((-))""^((+))"Na"#.
3. Finally, the nucleophile acts in an #bb("S"_N2)# backside-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.

   A bond is made between the two halide centers.

   Remember that the more electronegative halogen leaves first, while the better leaving group during the substitution step is the larger halogen (more acidic conjugate acid).

   Since the intramolecular reaction occurs faster than the intermolecular reaction, the major product is made through the nucleophilic center attacking within the same molecule.