[Amndz]

1. 4-fluorobenzaldehyde with 4-bromobenzaldehyde

2. 4-hydroxybenzaldehyde with benzaldehyde

I know that the major product would be the better nucleophile with the better electrophile, and I know that the better nucleophile would keep the CO double bond and the electrophile would have the OH. 

So, for #1 I was thinking that the fluorobenzaldehyde would be the better nucleophile and the bromobenzaldehyde is the better nucleophile because F is more electronegative than Br making it an EWG.

And for #2 I was thinking that hydroxybenzaldehyde is the nucleophile and the benzaldehyde is the elecrtophile because the OH is an EDG.

But the way this worksheet is structured is it gives one product and I have to write the major product, and what I think is the major product for both is the product given. So now I am confused. And then I have to provide mechanistic intermediates of the major products to support my logic, which I'm not really sure how to go about doing that.

Any help you can give will be greatly appreciated!
Thanks in advance!

[Doc Oc]

Those are pretty vague reaction conditions, is there any more detail on them?

[Amndz]

Oh, sorry I forgot to add that.. There's thiamine-HCl catalyst, NaOH, and EtOH/H₂O.

[orgopete]

That is a good question. At first, I didn't think I could answer it, but upon further thought, I believe I can.

Let us use the 4'-hydroxybenzaldehyde with benzaldehyde example first. If the product was the result of addition of thiamine to benzaldehyde and the addition of its anion to 4-hydroxybenzaldehyde, then we would know the key or rate determining step is the formation of the anion of the thiamine adduct. However, if the opposite product were the major product, then we could conclude that thiamine could add to either carbonyl group and the rate determining step were the addition to the carbonyl group.

However, that isn't what I would expect to happen at all. Since thiamine is used only in a catalytic amount, I would expect that thiamine would add to it more quickly than to the 4'-hydroxybenzaldehyde. Deprotonation of the adduct would give the anion. I would again expect the benzaldehyde to be more reactive. The reaction should give the benzaldehyde-benzaldehyde adduct until the concentration of benzaldehyde falls to begin to have some mixed product. If the reaction proceeded further, then I would expect the hydroxybenzaldehyde product to form, though at a slower rate.

I would expect the same process with the bromo and fluoro example, but with a large amount of mixed product. I would expect the bromo to react faster as bromine is more electron withdrawing. See http://en.wikipedia.org/wiki/Hammett_equation