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What will be the major product formed in the given reaction: enter image description here

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.

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This is the Benzilic Acid rearrangement - diagram sourced from here

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

enter image description here

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In addition to the answer by @Waylander, I would add something since his answer is about symmetrical benzil.

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.

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    $\begingroup$ You are correct $\endgroup$
    – Waylander
    Jul 16 at 14:20
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    $\begingroup$ Although your suggestion is reasonable based upon accumulated knowledge, how does one demonstrate the selectivity in the p-methoxy example? One might argue that although the anisyl carbonyl is less reactive toward base, the anisyl ring is favored in a bridged migration. Certainly nothing in the structure of the benzylic acid proves which carbonyl is attacked by hydroxide. $\endgroup$
    – user55119
    Jul 16 at 16:03
  • $\begingroup$ @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... $\endgroup$
    – TRC
    Jul 17 at 3:32
  • $\begingroup$ Imagine that the benzyl carbonyl is ten times more reactive than the anisyl AND that this step is reversible. If the migratory aptitude (look up the term) is 100X more likely for anisyl than phenyl and that this step is irreversible, then which step determines which group is migrating? For a point of comparison, there is more equatorial bromocyclohexane than axial but it is the axial conformation that has the greater rate constant for E2 elimination than the equatorial. You may wish to construct an energy diagram to illustrate these reactions. Read about the Curtin-Hammett principle. $\endgroup$
    – user55119
    Jul 17 at 15:22
  • $\begingroup$ Ran out of characters.... The anisyl group may be expected to have a greater migratory aptitude than the phenyl group owing to the electron donating ability of the p-methoxy group. $\endgroup$
    – user55119
    Jul 17 at 15:33

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