Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: AlphaScent on April 19, 2015, 05:22:13 PM
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Do you think I can get this reaction to go to completion? Maybe heat and Lewis acid?? Is the diene reactive enough?
(I know cyclopentadiene is very reactive and will be more so with a activating methyl group)
The homoketone (unconjugated) bothers me a bit but it is the only way to get the synthesis I want. I can show the rest if someone would like.
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You probably don't need the catalyst. That will tie up the 2 carbonyls.
But you need to think about the stereochemistry of the product. That chiral auxiliary will have some effect.
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Hmm... I was going off of notes from class and worked out the TS and believe I have the right stereochemistry.
I have been known to be wrong though haha.
Attached is what is in the notes given by the professor.
Cheers!
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Wait....The diene will approach from the opposite face of the auxillary....The notes and I are wrong!!!
The auxillary should be the other way, correct??
The fact the double bond is trans also makes me think. Is that the Endo product...seems to me it is.
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which C=C are you talking about, the one in the first scheme or the second one?
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The C=C from the second scheme is straight from my notes given by the professor.
The first is my method for a reaction in a synthesis.
I have to think if the C=C with the methyl (methylcyclobutadiene) will be on the correct side (regiochemistry).
I am referring to the first scheme. I may have my regiochemistry wrong.
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OK, well is the top face of the C=C not blocked by the auxiliary?
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I have little faith in my ability to predict a reaction's enantioselectivity, so I cannot make those predictions. I'd simply use the example from your notes as a guide. However, you've made a significant change in the dienophile by moving the double bond. I'd doubt you'd get that product.
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Orgopete,
I agree. The dieneophile is no longer electron poor. Some thought must be taken.
I will post my answer to the question. It may be completely wrong.