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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...

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

The acidity of phenol is not great enough to dissolve zinc (at least, not easily), but it could adsorb on the zinc/zinc oxide surface. First step: probably a phenolation of an oxygen to phenolate and hydroxide:

$\ce{Zn - Zn - ZnO - Zn - Zn- (surface) -> Zn - Zn - Zn (OH, OC6H5) - Zn - Zn}$.

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.

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