Benzilic acid rearrangement 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)
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 (as given by Czakó/Kürti).