15
$\begingroup$

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...

Improve this question
$\endgroup$
1
  • 3
    $\begingroup$ 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, 2018 at 18:33

1 Answer 1

Reset to default
2
$\begingroup$

The net result is $\ce{C6H5-OH + Zn -> ZnO + C6H5-H}$, but under the reaction conditions (zinc DUST and strong heating; b.p. of phenol is $\pu{182 ^\circ C}$), 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 ($\ce{C6H5-H}$), 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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ @andselisk: I get multiple SE requests to confirm your rejection of a proposed edit (which I do confirm). The message says the matter has been handled, and yet asks for my agreement. I can't seem to make the matter go away. Am I missing something? $\endgroup$
    – James Gaidis
    Jul 25, 2021 at 12:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.