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This question that I recently encountered asked what the most suitable substrate for the Ullmann reaction would be. The answer given is (b) but I think it has to be (d) as the nitro group acts as EWG and thus facilitates the reaction.

However I do not know the mechanism and thus am not sure if (d) is indeed the correct answer.

Any reason why the answer might be (b) and not (d)?

enter image description here

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I think that the answer given in your textbook is wrong as they are trying to conduct ULLMANN reaction on a chloride a strong EWD needs to be present at the ortho/para positions for the reaction to be feasible.

There Are generally 2 Mechanisms Proposed for the reaction :

  1. Free Radical Mechanism
  2. Ionic Mechanism

FREE RADICAL MECHANISM
The Copper Metal (used as catalyst) Generate a Aryl Radical By transferring a electron, and then 2 Aryl Radical combine to form the final product. FREE RADICAL MECHANISM

IONIC MECHANISM
Acording to this a Organometalic intermideate of Aryl copper halide is fromed. The copper is added to the Aryl halide by Oxidative addition. then a electron is transfered to the intermidiate to for a organocuppric radical which again perform the oxidative addition on another aryl halide which result in formation of a biaryl copper complex which in turn undergo reductive elimination and make the final product. enter image description here

You can easily see that we need a EWG to make the Reagent more reactive thus making the reaction feasible thus your answer must be either wrong or you might have checked it wrong. nevethless I also find this paper affirming mine and your belief.

Source:

  1. https://link.springer.com/article/10.1007/s40828-016-0036-2
  2. www.name-reaction.com/ullmann-reaction
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    $\begingroup$ I may be wrong,but isn't a free radical destabilized by an EWG,considering that it is an electron deficient species? So why would EWGs facilitate the reaction? $\endgroup$ – YUSUF HASAN Feb 11 at 5:39
  • $\begingroup$ @YUSUFHASAN no I don't think so as the free readical is formed on a sp2 carbon $\endgroup$ – Advil Sell Feb 11 at 6:21
  • $\begingroup$ Okaay thankyou! $\endgroup$ – Rutwik Feb 11 at 6:30
  • $\begingroup$ @AdvilSell See this It clearly states that a free radical is an electron deficit species,and it's stability is similar to a carbocation $\endgroup$ – YUSUF HASAN Feb 11 at 7:02
  • $\begingroup$ So even if -M effect is not operative then also -I effect surely is via the sigma bonds $\endgroup$ – YUSUF HASAN Feb 11 at 7:12

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