In the standard advanced thallation-iodination sequence, an arene reacts with thallium(III) trifluor — Haloalkanes and Haloarenes Chemistry Question
Question
In the standard advanced thallation-iodination sequence, an arene reacts with thallium(III) trifluoroacetate, and the bulky intermediate is then cleanly cleaved by aqueous $KI$ to yield the desired aryl iodide. What is the formal oxidation state of the thallium atom in the inorganic precipitate definitively formed as the final reaction byproduct?
💡 Solution & Explanation
The initial reagent is thallium(III) trifluoroacetate, where thallium is in the $+3$ oxidation state. The intermediate is $Ar-Tl(OCOCF_3)_2$, where thallium remains $+3$. When treated with $KI$, an oxidation-reduction reaction occurs at the metal center. The iodide ion acts as a reducing agent and nucleophile, substituting the thallium while transferring electrons. The thallium atom is reduced from $Tl(III)$ to $Tl(I)$, precipitating out of the aqueous mixture as the highly insoluble salt, Thallium(I) iodide ($TlI$). Thus, the oxidation state is $+1$.