Esters of catechol - Fries rearrangement

![enter image description here](https://i.stack.imgur.com/BwMkS.png)


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?




[This Wikipedia article](https://en.wikipedia.org/wiki/Fries_rearrangement) 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](https://i.stack.imgur.com/2g7QZ.png)


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](https://books.google.com/books?id=oYPRsEgL2EUC&pg=PA155&lpg=PA155&dq=dibromination%20of%20phenol&source=bl&ots=oJV6URTq1-&sig=ioYxPBybiYW8c0rvfB5jioDBzQ0&hl=en&sa=X&ei=emguVfrAB4q1sAScrIHQAw&ved=0CE8Q6AEwCQ#v=onepage&q=dibromination%20of%20phenol&f=false)). 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](https://books.google.com/books?id=AcpPyv0HWRsC&pg=SL1-PA172&lpg=SL1-PA172&dq=fries%20rearrangement%20of%20catechol%20esters&source=bl&ots=CXwKMjvnIc&sig=_tibx21YVK4_ghXCX-mUB8x51oM&hl=en&sa=X&ei=NIieVfzgMYPl-QHVjImwBQ&ved=0CFwQ6AEwCQ#v=onepage&q=fries%20rearrangement%20of%20catechol%20esters&f=false), p A-172) - apparently one of the acyl groups is lost in the process.


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