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Why does rearrangement not take place in Wurtz reaction?
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Muhammad Imran
Why does rearrangement not take place in Wurtz reaction?
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.
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.
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 ...
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.
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 ...
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.
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.
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.
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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.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$
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.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$
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