[jones106]

Could someone please help me figure out which molecule forms the enolate ion and attacks the other in the double aldol condensation of cinnamaldehyde and acetone?  I would think that the acetone forms the enolate and attacks the cinnamaldehyde.  I think, however, that my TA told me that the cinnamaldehyde will form the enolate and attack the acetone.  When I write out this reaction, the product just looks a bit outrageous to me, so I think I misunderstood him.  Could someone help me figure out what product is formed in this reaction?

Thanks,
jones106

[Mitch]

is this done under basic conditions?

[jones106]

yes, it is done under basic condition

[Mitch]

Well, which one is the most acidic proton?

[jones106]

I think the most acidic proton would be the alpha hydrogens on the acetone.  Am I right?

[Mitch]

you tell me. http://www.chemicalforums.com/index.php?page=pkavalues

[jones106]

I found the pKa of acetone to be 26.5, but could not find the pKa of an aldehyde.

[jones106]

I found the pKa of aldehydes to be around 17, which would make them more acidic than ketones.  This would mean that the aldehyde would form its enolate ion and attack the ketone.  Is this right?

[Mitch]

sounds better at least.

[movies]

Woah, hold up.  The alpha proton on cinnamaldehyde isn't like a regular aldehyde!  It's an enal, so the alpha proton isn't very acidic at all!

[Mitch]

good catch.  

[Yggdrasil]

Would a Michael Addition be likely in a reaction like this?  Enolates are soft nucleophiles, so I would think that a Michael addition would be more likely than an aldol reaction.

[movies]

Yeah, it's possible, but the Michael addition will be reversable, so will the initial aldol reaction.  The elimination of water from the aldol product (completing the aldol condensation) will be irreversable, so the product would likely all funnel to that under thermodynamic conditions.