I need to solve this problem for homework. In an article, I saw a quinone formation, but I don't understand the mechanism. In these reactions, the authors used KI (potassium iodide) and AcOH (acetic acid) to generate an alcohol, but I do not understand why an alcohol is generated. Also, they used toluene and p-TsOH·H2O for the opening of the epoxide ring and the formation of a double bond, but I am not sure about this. Here I leave the name of the article (Maruo et al., 2013, "Biomimetic Synthesis of Zeylanone and Zeylanone Epoxide by Dimerization of 2‑Methyl-1,4-naphthoquinone"). I appreciate your help.

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1 Answer 1


Although I'm not 100% positive, I assume the first step is for both conversion of the epoxide and removal of the MOM protecting groups. Acidic conditions are generally sufficient for removing MOM protecting groups, so I would expect the acidic conditions in the first step to remove the MOM groups; however, the iodide also plays a critical role in converting the epoxide to the alkene.

In this case, conversion of an epoxide to an alkene is a net reduction. Therefore, iodide will act as a reducing agent. I assume the first step (A) to be nucleophilic attack by the iodine on the electrophilic epoxide, facilitated by the acidic conditions. We can then perform a net protodehalogentation by attacking the iodo subsitutent with another nucleophilic iodide (B). This is the key redox step. Iodide is oxidized to iodine and the substrate is reduced. Here, the only step left is to eliminate a molecule of water (C) to produce the alkene product. I assume this is the purpose of the second step of reflux in toluene with TsOH. Elimination of water generally goes better in nonpolar solvents like toluene as it can be removed directly with something like a Dean-Stark trap, shifting equilibrium towards the alkene.



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