The Sandmeyer reaction is an extremely useful reaction for the functionalization of aromatic rings through a diazonium intermediate. For conversion of a diazonium intermediate to the corresponding chloride, bromide, or cyanide, the copper(I) salt is used. Surprisingly (to me at least) for iodination, potassium iodide is the reagent of choice, as described in Maitland Jones's Organic Chemistry (and elsewhere I'm sure), but no rationale is given. SciFinder searches also show that potassium iodide is most frequently used.

Is there something special about copper(I) iodide that makes it unsuitable for use in the Sandmeyer reaction?


Sandmeyer reactions using iodides, water, and thiols do not require the copper(I) ion as a catalyst for the reaction to proceed whereas your other nucleophiles do (though not necessarily stoichiometrically).

Sandmeyer Mechanism Sandmeyer without copper

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As you can see the copper(I) ion is important in reducing the diazonium to nitrogen and forming a radical. Iodine, oxygen, and sulfur must be able to be oxidized by diazonium and therefore do not require copper.

  • $\begingroup$ Is there any evidence that thiols, water (and presumably alcohols), and iodide are oxidized? Oxidation of thiols and iodide are (relatively) facile, but I don't consider oxidation of water to be in the same class. If it has something to do with nucleophilic strength, surely cyanide is a stronger nucleophile than water, yet cyanide requires presence of copper(I). It's probably not appropriate to call copper a catalyst, since it's present in stoichiometric amounts. $\endgroup$ – jerepierre Dec 15 '15 at 1:09
  • $\begingroup$ The copper is not always added in a stoichiometric amount. The reaction with just water requires an input of heat to effect the reaction. $\endgroup$ – A.K. Dec 15 '15 at 1:13
  • $\begingroup$ The answer says that oxygen must be oxidized by diazonium in order to avoid the need for copper. Is there evidence that water is oxidized in this reaction? $\endgroup$ – jerepierre Dec 15 '15 at 20:33
  • $\begingroup$ This is incorrect conclusion - only catalytic reaction is with redox steps. It's not like it won't work at all with Cl and Br, but Cu makes an improvement. $\endgroup$ – Mithoron Jan 22 at 22:53

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