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Topic: selectivity in enolate formation  (Read 2188 times)

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Offline kriggy

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selectivity in enolate formation
« on: January 23, 2016, 10:53:55 AM »
Im reading some papers on stereoselective synthesis and one thing makes me wonder, how can i control which enol I make?
For example, if I have methy-ethyl ketone and I add base, I think two products will be formed: the one where the methyl group is deprotonated and the one where methylene group is deprotonated. I suppose the deprotonation at methyl is faster because the protons are more acidic, while the deprotonation at methylene gives more stable product because there are more hydrogens in hyperconjugation with the double bond. Is that right? Does it mean that if I let the deprotonation run longer, the enol where the methylene group is deprotonated will dominate in the mixture?
Or are there other ways how to force one or the other enol to dominate?

Online Babcock_Hall

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Re: selectivity in enolate formation
« Reply #1 on: January 23, 2016, 06:56:03 PM »
Have you done any reading on kinetic versus thermodynamic control of formation of enolates?  There is a little bit in Carey and Sundberg's 4th edition on this (they even use a similar example on p. 216), but I would imagine that there are more extensive treatments.  I am a little outside of my comfort zone, but this ("I suppose the deprotonation at methyl is faster because the protons are more acidic...") does not sound quite right. How would you define the work "acidic?"

Offline kriggy

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Re: selectivity in enolate formation
« Reply #2 on: January 24, 2016, 06:11:52 AM »
No I have not, but thanks for pointing it out, Ill check if I can find something in the books I have home

Online Babcock_Hall

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Re: selectivity in enolate formation
« Reply #3 on: January 27, 2016, 01:21:48 PM »
Check starting around p. 225 in Modern Organic Synthesis by Zweifel and Nantz.

Offline kriggy

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Re: selectivity in enolate formation
« Reply #4 on: January 27, 2016, 01:43:55 PM »
I dont have acces to that book but i checked other sources and I think I get it now (at least the basics): the kinetic enolate is formed at low temperature using stericaly hindered bases like LDA -the sterics of the base make it difficult to deprotonate the other proton and at the low temperature the the formation of the faster forming enolate is prefered, while the thermodynamic si formed at room temperature using non hindered bases like NaH

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