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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

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    $\begingroup$ Maybe the fact that Wurtz reaction occurs even in mediums that do not solvate ions properly (like non-polar ones) ? $\endgroup$ Jan 8 '17 at 10:00
  • $\begingroup$ That is only a case $\endgroup$ Jan 9 '17 at 18:34
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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: $$\ce{R^-Na+ + RX -> R-R + NaX}$$

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Actually, nothing prevents us .This mechanism is well accepted but only for 1 degree & 2 degree carboanions . It does not give results in case of 3 degree carboanions because the resulting alkyl sodium would act as a strong base for the R-X and give elimination products

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