Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: xshadow on November 29, 2019, 05:00:22 AM
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My textbook says that alcohol(s)usually give conjugated addition with a "michael acceptor" because they are a weak base and good leaving group
Now later speaks about hard/soft and says that hard species tend to give direct addition to the Carbonyl C=O rather than C=C
BUT between hard species I can see RO- and ROH,H2O too
So alcohols what do they do usually?
Direct addition to C=O or the conjugated one (C=C addition)?
And for alcholate RO-??
They are strong bases + hard so they should do direct addition to carbonyl
Is it correct?
Thanks
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Please, see the reference below and take a closer look at the base catalysts.
Oxy-Michael Reaction
https://www.epfl.ch/labs/lspn/wp-content/uploads/2018/10/Oxy-Michael-Reaction.pdf
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Please, see the reference below and take a closer look at the base catalysts.
Oxy-Michael Reaction
https://www.epfl.ch/labs/lspn/wp-content/uploads/2018/10/Oxy-Michael-Reaction.pdf
Hi
Read it and I see that RO- gives conjugated addition to C=C with Michael acceptor
Didn't find a truly explanation so I've tried to write an explanation for this:
(https://i.imgur.com/mVAW4Rz.jpg)
Does this seem reasonable to you all??
Thanks!!! :)
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I think your explanation is great, except for one thing. Probably conjugate addition is also reversible. The whole thing could be an equilibrium, where the conjugate addition product is more stable than the 1,2 addition product, and is therefore favored.
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A closer look reveals that oxy-Michael reactions are also catalyzed by soft bases like phosphines.
Concerning hard base catalysts like t-BuOK, I agree with Spirochete because apart the hard/soft character of an acid/base, the mechanism of a chemical reaction also depends on other factors too, like equilibrium phenomena and thermodynamic stability of the products. Besides, it is clearly mentioned therein that oxy-Michael reactions are equilibrium ones.
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PGK, do you know the point of using the phosphine nucleophile in that situation? I looked on the slides in your link and I understand how it generates the alkoxide. But it seems like such a complicated way to create an alkoxide, when you could also just simply add some catalytic alkoxide right away and not add any phosphine at all.
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Indeed, the proposed mechanism in the link, seems complex, weird and exotic.
In my opinion, it is acidic catalysis because phosphine behaves as Lewis acid catalyst by complexation with the alcohol, in a similar way as with amines; which can also explain the H/D exchange, when deuterated alcohol is used.
Nevertheless, it is more convenient to add a catalytic amount of phosphine than preparing and working with a catalytic amount of t-BuOK, even if obtaining a little lower reaction yields.
1). New synthetic opportunities using Lewis acidic phosphines, Journal of the Chemical Society, Dalton Transactions, (23), 4307-4315, (2002)
https://pubs.rsc.org/lv/2002/dt/b208715d
2). Nitrogen Ligands on Phosphorus(III) Lewis Acceptors: A Versatile New Synthetic Approach to Unusual N−P Structural Arrangements, Inorganic Chemistry, 42(4), 1087-1091, (2003)
https://pubs.acs.org/doi/full/10.1021/ic026090l
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I think your explanation is great, except for one thing. Probably conjugate addition is also reversible. The whole thing could be an equilibrium, where the conjugate addition product is more stable than the 1,2 addition product, and is therefore favored.
So they are both reversible
So the thermodinamic says that the conjugated addition is the most stable....(C=O bond also there in conjugated addition, more stable than C=C)
Thanks