# What is the complete mechanism of reaction for sodium benzoate and ascorbic acid?

What are the complete steps of reaction from sodium benzoate and ascorbic acid to benzene? I can't seem to visualize it, and, much less, find anything online regarding the actual steps of the reaction.

Additionally: how does it make use of, say Cu(II) or Fe(III) to complete the reaction? It seems that the reaction can only proceed with these present.

$\ce {Cu^{2+} + H_2Asc \rightarrow Cu^+ + HAsc^.}$

$\ce {Cu^+ + O_2 \rightarrow Cu^{2+} + O_2^-}$

$\ce {2O_2^- + 2H^+ \rightarrow H_2O_2 + O_2}$

$\ce {Cu^+ + H_2O_2 \rightarrow Cu^{2+} + OH^- + OH^.}$

Then the hydroxyl radical reacts with benzonic acid as opposed to benzoate.

The hydroxl radical abstracts hydrogen from the acidic proton of the benzoic acid, producing water and a benzoate radical. This benzoate releases $\ce{CO2}$, producing a $\ce{C6H5}$ phenyl radical, which can abstract a hydrogen from water to regenerate the hydroxyl radical and produce the benzene. https://en.wikipedia.org/wiki/Benzene_in_soft_drinks

• Which hydroxyl radical? Answers should be self-contained. Please edit it to include how to generate the radical or at least reference another answer if it is explained there. – Jan Mar 28 '17 at 18:41