Benzilic acid rearrangement

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


I'm in need for the exact mechanism of the above reaction. I attempted some tries(mentioned in the mechanism above) and I felt the yield quite unsatisfactory. 
- the last step remains stuck in an equilibrium. Is the energy of the product sufficiently high to surpass the equilibrium created??
- the probability of the product formation (according to the mechanism I assumed) is too low as the pair of electron would hardly prefer an attack on the 5th carbon (as it doesn't possess any partial positive charge as the 6th carboxy carbon does)
[![enter image description here](https://i.stack.imgur.com/diXCS.jpg)](https://i.stack.imgur.com/diXCS.jpg)




Czakó/Kürti$^{[1]}$ suggest a [1,2] shift, which in your case would mean a shift of the 5-6-bond to the 1-carbon; in concert with a free electron pair of the anionic oxygen neutralizing the 6-carbon. The carbonyl oxygen becomes anionic and accepts a proton in the next step. So instead of the 1-6-bond breaking, the 5-6-bond breaks.


Responding to a comment regarding yield: Czakó/Kürti give several references and cite one example yielding 63% over the two steps (addition of base and acid).$^{[2]}$ Other examples are given, but without yield.


1) L Kürti, B Czakó: Strategic Applications of Named Reactions in Organic Synthesis, Elsevier (2005).


2) P A Grieco, J L Collins, J C Huffman, *J. Org. Chem.*, 1998, 63 (25), pp 9576–9579, DOI: [10.1021/jo981419z](http://dx.doi.org/10.1021/jo981419z) (as given by Czakó/Kürti).


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