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

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)

  • $\begingroup$ Which is the most acidic site in C? 1,3 dicarbonyl species usually react through the methylene group between the two carbonyls. Draw this product and then consider that you have in the product another site alpha to a carbonyl group that can do a 2nd aldol condensation. $\endgroup$
    – Waylander
    Jan 21, 2022 at 11:19
  • $\begingroup$ I am inclined to think that there is a typo in the molecular formula of E $\endgroup$
    – Waylander
    Jan 21, 2022 at 11:35
  • $\begingroup$ After confirming with my professor, he noted that there is a mistake in the molecular formula. E should be C20H18O4. To whom it may concern, sorry for taking your time in considering the problem. $\endgroup$
    – user4723
    Jan 22, 2022 at 4:31


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.