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Chemistry 15 Online
OpenStudy (anonymous):

Can someone help me with an organic chemistry problem? Please see attached question

OpenStudy (anonymous):

|dw:1449250332357:dw| How would the reaction go if in the presence of -Oet?

OpenStudy (anonymous):

@ParthKohli

Parth (parthkohli):

I'm sorry, I'll tag @Kainui for this one

Parth (parthkohli):

See I don't want to mislead you or anything

OpenStudy (anonymous):

It's okay. Thank you!

Parth (parthkohli):

Do you have any ideas for this one?

OpenStudy (anonymous):

Not sure... It could form an enolate with the ketone... But then it would want to attack the aldehyde which is unfavorable. since it would form a 3 member ring... Usually it would be a reaction in which the -Oet deprotonates a H from the B-carbon position, but there isn't any (this would cause the double bond of the ketone to break then the aldehyde oxygen can be eventually kicked out.... The methyl is what is confusing me

OpenStudy (anonymous):

Or the -Oet could directly attack the carbonyl, but I don't think it would help an overall reaction I am trying to do.

OpenStudy (anonymous):

@Zale101

OpenStudy (kainui):

Hmmmm I looked at this for a while, specifically cause I wasn't sure what would happen. The strange thing here is there are two ketone groups here and the only difference between their alpha carbons is one is a primary and the other is a secondary carbon. Generally speaking, aldol reactions tend to not go forward with ketones, only aldehydes, and the fact that there are two here is weird. I was thinking about how there's no beta hydrogen there, and trying to imagine some kinda methyl shift instead, or I think like you said maybe this thing cyclizes intramolecularly? All these seem weird and unlikely. Another thing is maybe it tries to polymerize and create a chain. My best guess is that the aldol reaction happens intermolecularly between the molecules at the least hindered ketones. In all, there are 4 possible different intermolecular products we could get here. I'm still looking into it, any brainstorming or questions are good.

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