[ginkgobilbo]

The given IR peaks are throwing me off, since they're so close.

To get to product A, NaOH will strip the alpha-hydrogen from the reagent, yielding a carbanion that will attack the carbonyl on the benzaldehyde, giving me a beta hydroxy carbonyl. An IR peak at 1690 indicates carbonyl stretching, but my structure was marked incorrect (the beta hydroxy carbonyl: C6H5-CO-CH2-C(OH)-C6H5).

So I'm stuck at A. Should it be an alpha, beta unsaturated carbonyl instead? Is this the 1690 peak? If so, how will the amine attack?... Will it attack the carbonyl or the double bond? Does the 1710 peak indicate a carbonyl? If it attacks the carbonyl, will there be a loss of a water molecule? That would lose the oxygen though... 

Thanks for any help.

[orgopete]

For A, it may be useful to think that the reaction is done in the presence of strong base and/or heat. (Hint, what might happen if the ketone were enolized again?)

[ultima]

You were querying correctly. What you said does happen. Enloate formation - attack of aldehde - dehydration (loss of water, formation of double bond) - 1,4 addtion (attack at double bond).

[orgopete]

While the mechanism posted by ultima contains the correct intermediates and products, it does not explain the formation of missing intermediate, the unsaturated ketone. I am also doubtful that the dehydration is acid catalyzed. In the formation of dibenzalacetone, the product from the elimination occurs without acidification. I personally favor using likely proton sources in a reaction, so I would have used water in the first protonation and in the conjugate addition, I would use an anilinium anion.

[medist]

obviously product A would be an alpha beta unsaturated ketone. Thats why u have an IR strchng at 1690. In product A, ß-carbon is more electrophilic than the carbonyl carbon, so the amine will attack at the ß-carbon (like conjugative addition) to ketone not at the carbonyl carbon. This is the reason for IR strchng at 1710 which is characteristic for carbonyl group in ketones.
I hope my assumption is correct.