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.
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2 Answers
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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.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$
answered May 4, 2014 at 12:19
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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.
