Major product formed when 4-Methoxybenzil is reacted with a base

What will be the major product formed in the given reaction:

I am not able to apply any of the reactions or mechanisms I know of in this case. The only thing I am able to predict is that the $$\ce{OH-}$$ will probably do a nucleophilic approach on the carbon atom directly attached to the phenyl group (adjacent to the starred carbon). I am not sure how to proceed after that. It will be of great help if someone could guide me with a proper mechanism as to how this proceeds or whether this is any standard reaction.

This is the Benzilic Acid rearrangement - diagram sourced from here

The Wikipedia entry is as good a place to start as any Wikipedia

In case the substituents on the two phenyl rings are different, it must be noted that the first step occurring in this reaction is attack of $$\ce{OH^-}$$ as a nucleophile on the carbonyl $$\ce{C=O}$$. Nucleophilic addition is preferred at that carbon atom which is more electron deficient. In your case, methoxy group at para position increases electron density on left carbon. So the attack of $$\ce{OH^-}$$ occurs on the right carbon. Proceed with that as the first step, for the final product.
• @user55119 You're right, my answer is completely based on theoretical knowledge, based on what I have read in a textbook. I would say that since the first step favours the existence of intermediate where $\ce{OH^-}$ is attached to the benzyl carbonyl (and not the anisyl one), that intermediate has greater quantity existing after the first step, and hence has a greater chance to react further. I might be completely wrong, of course. My knowledge is extremely limited - I am still yet to enter college. Perhaps someone more qualified can be of more help...