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Aryl diazonium salt on treatment with copper and the corresponding hydrohalic acid, gives the aryl halide. This reaction is called the Gatterman reaction.

What is the mechanism of this reaction? Is it similar to Sandmeyer reaction?

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  • $\begingroup$ That's not the Gatterman reaction. en.wikipedia.org/wiki/Gattermann_reaction $\endgroup$ – Zhe Mar 7 at 19:10
  • $\begingroup$ Have you Googled this? I think you have described the Sandmeyer rxn. $\endgroup$ – user55119 Mar 7 at 19:11
  • $\begingroup$ @Zhe Indeed that is the more famous Gattermann reaction. The Wikipedia link that you have provided also contains a link to the Gatterman reaction I am referring to right at the top of the page. $\endgroup$ – Anubhab Das Mar 7 at 19:13
  • $\begingroup$ @user55119 No, the reagent used in Sandmeyer reaction is cuprous halide. March's Advanced Organic Chemistry specifically mentions Gatterman reaction as an alternative to Sandmeyer reaction. $\endgroup$ – Anubhab Das Mar 7 at 19:19
  • $\begingroup$ March notwithstanding, the Sandmeyer rxn is an improvement on the Gattermann conditions that uses mineral acid (HCl, HBr) and copper salts in the second step. The Sandmeyer rxn uses only CuX. This version of the "Gattermann rxn" is lost to history. The Gattermann and Gattermann-Koch reactions are methods for formylating aromatic rings in the absence of diazonium salt. $\endgroup$ – user55119 Mar 7 at 19:59
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The Gattermann reaction you mentioned in the question uses Copper powder and Hydrohalous acid to substitute diazonium group with a halide in an aromatic diazonium salt. It uses $ArSN_1$ pathway, similar to Sandmeyer Reaction which uses Copper Halide and corresponding Hydrohalous acid instead of copper powder. As diazonium group is a very good leaving group, it departs leading to a phenyl carbocation which is very unstable, this is in fact the slow step of the reaction. The Nucleophile (Halide in this case) attacks the phenyl carbocation leading to the formation of Halobenzene(fast step). Nitrogen gas and Hydrogen chloride are also formed.

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  • $\begingroup$ Can you tell me the source of this information? $\endgroup$ – Anubhab Das Mar 21 at 11:54
  • $\begingroup$ It's what our teacher taught us in the class. @AnubhabDas $\endgroup$ – drake01 Apr 7 at 4:49

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