Which carbonyl gets attacked in aldol???

[navneetdh]

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Ok I know this may sound like a stupid question, but I have been doing aldol reactions and so far i have been gettin them just fine. But I just realized that if I have 2 aldehydes (could be 2 ketones too or even 1 adehyde 1 ketone), and one of them had a longer chain while the other has a shorter chair. Then which of the two carbonyl carbons will be attacked and get the OH group and which one will retain its C=O group. Long chain or the shorter chain one? Or will there be 2 possible products?

also between aldehyde and ketone which one would retain the C=O group an which one will have OH?

 

[solstice]

Ok I know this may sound like a stupid question, but I have been doing aldol reactions and so far i have been gettin them just fine. But I just realized that if I have 2 aldehydes (could be 2 ketones too or even 1 adehyde 1 ketone), and one of them had a longer chain while the other has a shorter chair. Then which of the two carbonyl carbons will be attacked and get the OH group and which one will retain its C=O group. Long chain or the shorter chain one? Or will there be 2 possible products?

also between aldehyde and ketone which one would retain the C=O group an which one will have OH?

You will have a mixture of products in real experiments and that is the reason it is best to either use two similar Aldehyde/ketone or have one of the reactants without an Alpha H.
Now if you have two different reactants, the one who has the most acidic H will lose one to OH and the Carbon that lost the H will become your Nuc and attack the carbonyl on the other Ald/Ketone. The one that attacks gets to keep its carbonyl and the other one will have OH. Remember that your double bonds should be in conjugation which is why the product whether it is an Aldol or Enal(actual condensated product and hence the name) is very stable.

Hope this helps.

 

[navneetdh]

You will have a mixture of products in real experiments and that is the reason it is best to either use two similar Aldehyde/ketone or have one of the reactants without an Alpha H.
Now if you have two different reactants, the one who has the most acidic H will lose one to OH and the Carbon that lost the H will become your Nuc and attack the carbonyl on the other Ald/Ketone. The one that attacks gets to keep its carbonyl and the other one will have OH. Remember that your double bonds should be in conjugation which is why the product whether it is an Aldol or Enal(actual condensated product and hence the name) is very stable.

Hope this helps.

yes thanks you that helps...but i guess i dont understand which alpha hydrogen will be more acidic? the ones next to C=O with longer chain or the ones next to C=O on shorter chain?

 

[solstice]

In that case think of electron donating and withdrawing groups. If it is all alkyl, then the longer chain reduces acidity of a-Hydrogen (due to more e-donating groups) and the shorter chain will loose its H first.
However; remember for all practical purposes there would not be any difference in acidity between methyl and decyl and there will not be any questions to ask you distinguish to this minute detail. You need to know which of the 5 answer choices is actually a right end product. 4 out of 5 will have a wrong connection somewhere or missing something.

 

[navneetdh]

cool! thanks a lot 🙂