Aldol Cyclization Cascade Reaction between Two Ketones in Basic Alcoholic Solution

My professor gave a hint that the reaction scheme is a multiple carbonyl reaction which involves cyclization. He emphasizes that the reactivity of carbonyl compound in terms of C-H acidity and carbonyl reactivity will not be obeyed if cyclization is involved.


![Reaction Scheme](https://i.stack.imgur.com/CaPj9.png)


My initial guess is the formation of a vinylogous carbonyl compound for **B** by the elimination of one methoxy group. Subsequently, it react with the β-dicarbonyl compound, **C**. My another guess is that **C** reacts as methylene component either at the α-carbon or at the methyl group and attacks **B** by vinylogous substitution with the elimination of second methoxy group.


Next, intramolecular aldol reaction and cyclization take place. However, **A** (C12H16O4) and **C** (C10H10O2) react and give a compound with 20 carbon atoms by the elimination of two methoxy groups.


The carbonyl and methylene components are played around but I am very inclined that **B** is a vinylogous carbonyl compound and sp2-carbon will not undergo further deprotonation. However, I could not get a compound with 19 carbon atoms. My desperate guess is the formation of ester group with the oxygen atom acting as the O-anion instead of C-anion from the enolate during cyclization. Basic hydrolysis of ester will then give COOH group, yet methyl ester does not exist. Another guess is retro-aldol reaction which seems not possible...


Any suggestion for the reaction scheme is really appreciated. Thank you.
(No reaction condition is specified other than basic alcoholic solution)


