Ullmann's Reaction:

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My textbook says with $\ce{X} = \ce{Cl}$, the reaction gives less yield compared to using $\ce{X} = \ce{I}$ since $\ce{Cu}$ is not as reactive as $\ce{Na}$ ($\ce{C - I}$ bond is weak).

But if aryl halides with substituents such as $\ce{NO_2}$ are used, they can help in obtaining good yield.

What I observe from both the mechanisms is that adding deactivating groups in the ortho or para position relative to the halogen should actually give lesser yield since the deactivating group would cause a partial positive charge to exist on the $\alpha$-carbon hence destabilizing the free radical intermediate.

Can anyone clarify my doubt? I suspect my last argument to be wrong regarding the $\alpha$-carbon. The $\alpha$-carbon is already electron-deficient and the partial positive charge should make it worse?

Courtesy: http://www.name-reaction.com/ullmann-reaction


1 Answer 1


Your assumption that electron-withdrawing groups destabilize radicals is wrong. Radicals are in fact stabilized by both electron-withdrawing and electron-donating groups and even more so if both types are attached to the same carbon atom (captodative effect).

I modified an image from Ian Fleming's Molecular Orbitals where you have the proof. In the image you can see that the combination of the carbonyl group's (which is electron-withdrawing) pi bonding/antibonding pair of orbitals with the p orbital of the radical results in an overall decrease in energy:

enter image description here

  • $\begingroup$ What if the first step in the mechanism was the RDS? Then the deactivating group would improve the reaction rate since it enhances the ability to knock out the halide. $\endgroup$
    – Yashas
    Feb 26, 2016 at 15:12
  • $\begingroup$ I guess so. ... $\endgroup$
    – EJC
    Feb 26, 2016 at 16:41

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