Looks like this can be formed via TWO aldol condensations. Nice! Let’s see…

Alright, let’s start with these two compounds, then:

The mechanism will involve a base (we’ll use ##”OH”^(-)## here for simplicity in drawing the mechanism, but ##”LDA”## is better for yield) and a solvent for that base (water is fine). I would also expect this to be driven with heat to facilitate the condensation portion of the aldol condensation.

Note that the explicitly shown protons are the most acidic on dibenzyl ketone. There are two, so that’s how there can be TWO aldol condensations.

DISCLAIMER: This mechanism is very advanced and will be long. It is possible that this may be on an exam, but more likely you should expect something similar that takes less time to draw. Still, this is a good practice problem.

Also, before I draw it out, I will abbreviate the phenyl groups as ##”Ph”## as is conventionally done. That way I can fit it onto one image. 😛

1. Typical enolate formation via a base grabbing a proton from the ##alpha##-carbon on a ketone.

2. Enolate nucleophilic attack of the target, the 1,2-diphenylethanedione.

3. Proton transfer to form the ##beta##-hydroxyketone intermediate. This step concludes the aldol addition portion of the overall first aldol condensation.

4. Another base comes in to grab an ##alpha##-proton and form another enolate. At this point this is done under high heat.

5. ##pi## electron conjugation, heat, and the antiperiplanar conformation are enough to make the hydroxide leave without protonating it. This is the “condensation” step in the overall first aldol condensation.

6. Repeat step 1.

7. Repeat step 2, except for the fact that the nucleophilic carbon on the enolate is exactly four carbons away from the other carbonyl carbon on the compound. If a bond forms via the nucleophilic attack, it forms a five-membered ring, which is stable. Thus, intramolecular aldol condensation can occur here.

8. The bond has formed. This technically unnecessary step is to make it easier to visualize.

9. The oxide grabs a proton to form another ##beta##-hydroxyketone intermediate, as in step 3. This step concludes the second aldol addition that occurs during the second aldol condensation.

10. A base comes in yet again to grab a proton, as in step 4. At this point this is done under high heat.

11. The same dehydration-elimination of a hydroxide occurs, as in step 5. This is the “condensation” step in the overall second aldol condensation.

PHEW! ##11## steps. There we go! Depending on how fast you are, this should take you about ##12 ~ 18## minutes to draw out (use the ##”Ph”## abbreviation, and your professor would understand).