# Reaction of phenol with Zinc dust [duplicate]

I have seen this reaction many times:

$\text {Phenol}+\text{Zn}\ce {->}\text{Benzene}+\text{ZnO}$

But what is the mechanism?! Tried a lot but couldn't cook up one...

Thanks for any help...

• How about a reference. I see folks using it on SE but I have not found it in the primary literature. – user55119 Jan 25 '18 at 18:33

The net result is $$\ce{C6H5OH + Zn -> ZnO + C6H5H}$$, but under the reaction conditions (zinc DUST and strong heating (b.p. of phenol $$= \pu{182 ^\circ C}$$)), I suspect the reaction mechanism involves the zinc surface rather than individual $$\ce{Zn}$$ atoms or $$\ce{Zn^{2+}}$$ ions. And the zinc surface will be a complex oxide. I have read that the reaction is not clean and produces several byproducts, typical of high temperature reactions.
$$\ce{Zn - Zn - ZnO - Zn - Zn- (surface) -> Zn - Zn - Zn (OH, OC6H5) - Zn - Zn}$$.
At high temperature, the $$\ce{H}$$ on the hydroxyl could reach the phenyl group in a five-membered ring and go off as benzene ($$\ce{C6H5H}$$), leaving $$\ce{Zn - Zn - ZnO - ZnO - Zn}$$.
Magnesium and zinc have some similarities (ionic radius, oxide coat), but magnesium is significantly more reactive. I suspect that magnesium in phenol would react at lower temperatures and give magnesium phenolate and hydrogen: $$\ce{Mg + 2 C6H5OH -> Mg(OC6H5)2 + H2}$$, no benzene. In this case, the acidity of phenol would be enough to dissolve the magnesium; fine powder (pyrophoric) would not be needed.