Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Altered State on September 15, 2013, 09:53:58 AM
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I need to explain why these 2 compounds can't be prepared using aldol reaction (In case A mixed aldol reaction and in case B intramolecular aldol reaction).
I've made a picture in chemdraw to make it clear.
The only compounds I was given in the exercise are the 2 products, I guessed the reactants, but I think that in case that reaction could be carried out, these 2 would be the appropriated ones.
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fi.imgur.com%2FsldAeIT.png&hash=dcdac5c02525102f4c57753d3c22fe79d9383b4c)
This is what I think:
A), I was almost sure that the problem is that both aldehydes are enolizable, so we would end up with a mixture of 4 compounds that is not worth separating. But since I've studied quite a few reactions where the yield was really low, I think may be another problem with the reaction. Besides, if we prepare in one flask butanal and then we add slowly ethanal to it, we would obtain the product with a reasonably good yield, so now I'm almost sure there is another problem.
B), I only see one possible reason, and it's that the formation of a 4 members cycle is not very favorable (high strain) and in this case molecules would prefer undergoing an intermolecular reaction instead. On the other hand I've been taught that intramolecular reactions like this are entropically favored due to is easier to a molecule to collide with itself that with other molecule, so I'm quite lost in this one.
For both cases, I've drawn and checked mechanism but it doesn't seem to be any problem with that..
Thanks in advance
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... if we prepare in one flask butanal and then we add slowly ethanal to it, we would obtain the product with a reasonably good yield...
How would this keep butanal from self-condensing?
B) check Baldwins rules. I don't know the predictions off-hand, but I also think the equilibrium of the forward and reverse aldol will favor the reverse reaction.
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How would this keep butanal from self-condensing?
That's true, didn't realize, but... In some examples we saw during the lessons there were similar conditions. That make me think that may be something else.
B) check Baldwins rules. I don't know the predictions off-hand, but I also think the equilibrium of the forward and reverse aldol will favor the reverse reaction.
I know what those rules are about, but I've never seen them in my lessons, so I guess there should be any other way to explain why isn't this possible.
What comes to my mind is to say that the H of -CH2- in the cyclobutane are going to be more acidic than the ones in the starting product, but I'm not sure of this...
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It is always difficult to suggest an answer to a class we didn't attend. I had some anxiety in suggesting the Baldwin's rule.
I don't know if this was taught, but cyclopropanol is base labile. The ring opens to form a carbonyl and an anion. The anion picks up a proton. This occurs without any stabilizing groups to assist in the cleavage. In this case, you have a cyclobutanol and a carbonyl group to form an enolate.
Even without knowledge of the cyclopropanol, I suggest you invoke relief of ring strain. It can shift the addition to a carbonyl group to a retro-aldol. However, upon reflection, I think the dicarbonyl compound will undergo a retro-Claisen. Hydroxide will readily add to the aldehyde carbonyl group. This can undergo a retro-Claisen to give formic acid/formate and the enolate of 3-methyl-2-butanone.
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... if we prepare in one flask butanal and then we add slowly ethanal to it, we would obtain the product with a reasonably good yield...
How would this keep butanal from self-condensing?
B) check Baldwins rules. I don't know the predictions off-hand, but I also think the equilibrium of the forward and reverse aldol will favor the reverse reaction.
If I'm correct the 2nd example would be a 4-enolendo-exo-trig cyclization which is not favored by Baldwin's rules
The simple way to explain Baldwin's rules would be that for intramolecular cyclizations, due to geometric constraints the nucleophile (lone pair on enolate) cannot attack the electrophile (Π*C-O) at the proper trajectory to make the reaction work. i.e the orbital overlap isn't there or is non productive.
It should be noted that Baldwin's rules CAN be broken, they are simply used for trends but they are usually a great, reliable way to predict cyclization products.