[Pranav]

I have attached the problem. B is the answer I marked but it is incorrect.

I approached it this way. Since KOH/I₂ is a good oxidising agent, it oxidises the alcholic group to carbonyl groups, and replaces -CH₂I and -CH₃ with -OH, as it happens in an iodoform reaction. So the answer is B but the key states that its D. I don't even have the slightest idea about how the reaction will yield D.

Any help is appreciated. Thanks!

[Dan]

Basically you were 2/3 of the way there. D arises from from a rearrangement of B in the presence of KOH. Start by attacking one of the ketone carbonyls with hydroxide, can you see what has happened next?

[Pranav]

Basically you were 2/3 of the way there. D arises from from a rearrangement of B in the presence of KOH. Start by attacking one of the ketone carbonyls with hydroxide, can you see what has happened next?

Thanks Dan!

First, the OH^- attacks the lower carbonyl group (I am referring to the structure shown in B). The phenyl group with -COOH and -NO₂ leaves and attacks again at the upper carboyl group forming D. Is this correct?

[Dan]

Yes, it's called a benzilic acid rearrangement.

[Pranav]

Yes, it's called a benzilic acid rearrangement.

Thank you Dan! I think I have this in my notes somewhere. Time to revise them.