[johno91]

How would this reaction mechanistically proceed?



My first idea is to take the alpha hydrogen between the ketone and hydroxy off, but there isn't one. I assume the hydroxy hydrogen is the most acidic so it would be deprotonated. So that gives a negatively charged oxygen with a quaternary carbon on one side and a phenol on the other and I don't know how to go on from there.

Does something intramolecular happen? I can't figure out how the methyl seems to move to the other side.

Thanks for any help.

[Babcock_Hall]

Try taking off the hydrogen on oxygen with a base and see if anything comes to mind.

[johno91]

Like this?



I don't know what to do after that. I would think that makes the carbon electrophilic and the alcohol could attack, but there is not a good leaving group.

[Babcock_Hall]

OK, now try breaking the carbon-carbon bond by making a double bond between oxygen and carbon.  The carbon-carbon bond that I am suggesting you break is between the carbon with the charged oxygen atom and the carbon that is part of the 5-membered ring.

[johno91]

So that should lead to this



But when I try to react those I just end up condensing them back into the original reactant.

[orgopete]

Hint, the pKa of a ketone is about 20 while hydroxide is 18. Also, just think about what must happen to get to the product.

[johno91]

I figured out that if I get to this



I can finish up the reaction just fine. Does your hint mean that if I protonate, it will be more stable as a ketone instead of a hydroxy? So does that mean I can protonate the carbon here with water?

[Babcock_Hall]

If you deprotonate the -CH2- group next to the ketone, you should be on your way.

[swmmr1928]

Hi

My professor presented us with this very reaction in lecture only a few days ago. 
He said it is the same difficulty as his exams. 

You also?

The way we were showed it was
1)use O-R^- to attack carbonyl
2)ring opening
3)proton transfer
4)Diekmann intermolecular ring formation
5)proton transfer

[orgopete]

Hi

My professor presented us with this very reaction in lecture only a few days ago. 
He said it is the same difficulty as his exams. 

You also?

The way we were showed it was
1)use O-R^- to attack carbonyl
2)ring opening
3)proton transfer
4)Diekmann intermolecular ring formation
5)proton transfer

The current reaction is a retro-aldol and then an aldol from the opposite enolate.

The numbered sequence is a retro-Claisen or retro-Dieckmann and Dieckmann from a different enolate. These are related. The retro-Claisen can occur on a beta-keto ester in which it no alpha-hydrogens are present.