# Ullmann's Reaction (similar to Wurtz) - aryl chloride with deactivating groups - stability of free radical

## Question:

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?