[swmmr1928]

Hi.  I can't believe how long it took to write this mechanism in paint.  

I am having trouble to get from the product of an aldol (β-hydroxy carbonyl) to turn into enolate so it can attack a ketone intramolecularly.  See below:

[discodermolide]

Just eliminate 2H2O from the aldol product and enolise the ketone to give the product?
Driving force, formation of aromatic ring.

[swmmr1928]

Just eliminate 2H2O from the aldol product and enolise the ketone to give the product?
Driving force, formation of aromatic ring.

Yes I understand that part.  I am having trouble with the mechanism from A+B => C, not C=>D.

[stewie griffin]

Try using A as a enolate two separate times and each carbonyl of B as an electrophile

[swmmr1928]

Try using A as a enolate two separate times and each carbonyl of B as an electrophile

Lovely, thankyou very much. 

[orgopete]

Try using A as a enolate two separate times and each carbonyl of B as an electrophile

There had been several questions semi-related to this problem. Although the suggestion is expedient, you should realize there is good reason for this suggestion. You can expect the pKa of any alcohol to be the the 16-18 category. What you need to look at is the reverse reaction. If you make the anion, you will release a the enolate of a beta-diketone, a much weaker base. This reverse or retro-aldol reaction should also be favorable. This is similar to why you don't find the HCl addition product to a carbonyl group. It isn't that it cannot occur, but the reverse reaction is far more favorable.