enter image description here

Esters of catechol undergo the Fries rearrangement

But one product is formed more than the other.

I believe it is due to the electron withdrawing nature of the $\ce{OCOCH3}$ group which prefers the attack on meta position.I was able to write a reaction mechanism for the product which is minor in this case. But I couldn't write one for the major product. I wonder how it is formed?

Does the minor product itself rearrange to give the major product or are both of the products formed directly in this reaction?

  • $\begingroup$ Your products appear to have lost an acyl group compared to the reactants. Is this intentional? $\endgroup$
    – bon
    Jul 9, 2015 at 14:26
  • $\begingroup$ The last step is hydrolysis after which an acyl group is lost. $\endgroup$
    – miyagi_do
    Jul 9, 2015 at 15:20

1 Answer 1


This Wikipedia article provides a nice overview of the Fries rearrangement. They note that the mechanism of the rearrangement is not fully understood, but the following mechanism is generally accepted.

enter image description here

The key step in the reaction is thought to involve the interaction of a lewis acid with a phenyl ester to generate an acylium ion ($\ce{[R-C=O]^+}$). Once the acylium ion is generated the reaction proceeds like any other electrophilic aromatic substituion. The phenolic oxygen is a strong ortho-para director, so both ortho- and para-products are produced by the same mechanism. The $\ce{OCOR}$ group does not play a role in "directing" the substitution pattern.

The ratio of ortho/para isomers in most aromatic electrophilic substitution reactions is very temperature dependent. This is because of kinetic vs. thermodynamic control. Typically, the para product is the kinetic product (first formed), while the ortho product is the thermodynamically more stable product and will form if the reaction is run at higher temperatures and allowed to equilibrate. For example, when phenol is mono-brominated at low temperature, the para-bromo product is formed in >95% yield (reference). So your specific reaction is likely being run under kinetically controlled conditions to produce the para product as the major product. As the Wikipedia link points out, the para product can equilibrate at higher temperatures to produce more of the ortho product.

Finally, just a word to address @bon 's comment. Indeed, Fries rearrangement of catechol diesters does produce mono-acyl catechols (reference, p A-172) - apparently one of the acyl groups is lost in the process.

  • $\begingroup$ So the mechanism is still not clear? Because i have gone through a concerted mechanism where no free acylium ion is generated. $\endgroup$
    – miyagi_do
    Jul 9, 2015 at 15:25
  • 1
    $\begingroup$ Yes, the mechanism is not completely understood, but the reaction is not concerted and appears to involve an acylium ion. $\endgroup$
    – ron
    Jul 9, 2015 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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