# Mechanism of oxidation of phenol to quinone by chromium(VI)

In the supplementary information for Org. Lett. 2005, 7 (19), 4297–4300, the following experimental method for the synthesis of 2-bromo-6-methylcyclohexa-2,5-diene-1,4-dione is given:

To a suspension containing 20.0 g (75.8 mmol) of 2,4-dibromo-6-methylphenol in 50 mL of 80% acetic acid in water and 25 mL of acetonitrile, was added 8.34 g of $$\ce{CrO3}$$ in 25 mL of water. The reaction mixture was heated to 60℃ for 1.5 h, cooled to room temperature, diluted with 400 mL of water and extracted with three 200 mL portions of chloroform.

I've summarised the reaction here:

How does this work, mechanistically? My first guess would be a nucleophilic substitution of $$\ce{Br}$$, followed by an oxidation by $$\ce{CrO3}$$; but I can't imagine there are many free hydroxide ions floating around in a solution of acetic acid.

• A possibility: the phenol OH attacks CrO3, which creates a Cr leaving group on the phenol oxygen. Then some O-based nucleophile attacks the para position. You can push the double bonds around the ring onto the phenolic O and expel the Cr leaving group. Now the para position has a Br and a OR group, which can collapse to a carbonyl. Not sure about the finer details, but some phenol -> quinone oxidations (e.g. with hypervalent iodine) work like this mechanistically. – orthocresol Oct 30 '18 at 18:32