# Ionic Mechanism for Wurtz Reaction

Instead of going for the normal free radical mechanism for Wurtz reaction, What reasons prevent us from believing in the following ionic mechanism?

\begin{align} \ce{R-X + 2Na &-> R^--Na+ + NaX}\\ \ce{R-X + R^--Na+ &-> R-R + NaX}\\ \ce{2R-X + 2Na &-> R-R + 2NaX} \end{align}

Related but doesn't answer my question here

• Maybe the fact that Wurtz reaction occurs even in mediums that do not solvate ions properly (like non-polar ones) ? Jan 8 '17 at 10:00
• That is only a case Jan 9 '17 at 18:34

The actual mechanism is believed to be strongly dependent on the nature of halide, metal (not only $$\ce{Na}$$ is used in this reaction), and solvent. The 'ionic' mechanism you have written here is the part of the most accepted mechanism which involves both, an 'ionic' and a radical reaction.
First step. Through the single-electron transfer from $$\ce{Na}$$ to the halogen atom, an alkyl radical $$\ce{R^.}$$ and sodium halide is formed: $$\ce{Na + R-X -> Na+X^- + R^.}$$
Alkyl radical further acts as an electron acceptor from another $$\ce{Na}$$ atom to form highly nucleophilic intermediate (can be isolated in some cases): $$\ce{R^. + Na -> R^-Na+}$$
This alkyl sodium intermediate reacts with another alkyl halide in the aliphatic nucleophilic substitution: $$\ce{R^-Na+ + RX -> R-R + NaX}$$