Why does rearrangement not take place in Wurtz reaction?
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 $\ce{R-X}$:
$$\ce{Na + R-X -&gt; Na+ +[R-X].^{-}}$$
The alkyl halide radical anion fragments to a carbon radical and a halide anion:
$$\ce{[R-X].^{-}-&gt;R. + X-}$$
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 ...
$$\ce{R-CH2-CH2. +.CH2-CH2-R-&gt;R-CH=CH2 + CH3-CH2-R}$$
... or coupling:
$$\ce{R-CH2-CH2. + .CH2-CH2-R -&gt; R-CH2-CH2-CH2-CH2-R}$$
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:
$$\ce{R. + Na -&gt; R:^{-} + Na+}$$
The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:
$$\ce{R:^{-} + X-R -&gt; R-R + X-}$$
The formation of the anion is actually problematic, especially if $\ce{R-X}$ is secondary or tertiary, since E2 elimination will compete.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -&gt; (CH3)3C-H + (CH3)C=CH2 + X-}$$