# Why does rearrangement not take place in Wurtz reaction?

The Wurtz reaction is $$\ce{R-X + Na ->[ether] R-R}$$ where $\ce{R}$ is alkyl. But if we take butylchloride with $\ce{Na}$, the product is n-octane and not 3,4-dimethylhexane. I am unable to understand the fact that rearrangement does not happen even when it's a free radical mechanism.

The Wurtz Reaction mechanism involves radical anions. Neither radicals nor carbanions rearrange the same way that carbocations do.

The first step of the mechanism is a single electron transfer from $$\ce{Na}$$ to $$\ce{R-X}$$:

$$\ce{Na + R-X -> Na+ +[R-X].^{-}}$$

The alkyl halide radical anion fragments to a carbon radical and a halide anion:

$$\ce{[R-X].^{-}->R. + X-}$$

The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...

$$\ce{R-CH2-CH2. +.CH2-CH2-R->R-CH=CH2 + CH3-CH2-R}$$

... or coupling:

$$\ce{R-CH2-CH2. + .CH2-CH2-R -> R-CH2-CH2-CH2-CH2-R}$$

The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. $$\ce{Na}$$ is a good reducing agent, and that primary radical is readily reduced to an anion:

$$\ce{R. + Na -> R:^{-} + Na+}$$

The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:

$$\ce{R:^{-} + X-R -> R-R + X-}$$

The formation of the anion is actually problematic, especially if $$\ce{R-X}$$ is secondary or tertiary, since E2 elimination will compete.

$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$

You are making it too complicated, but it's really simple: $$\ce{Bu-Cl + 2Na + Cl-Bu -> Bu-Bu + 2NaCl}$$ where $\ce{Bu-{}}$ is the n-butyl group and $\ce{Bu-Bu}$ is n-octane.

You can understand wurtz's reaction much easier this way.