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Why is the most favorable conformation of ketones and aldehydes the one where the alkyl group is anti to the other alkyl group (or hydrogen atom in an aldehyde) as attached? Why isn't more stable the conformation where the bond between the alpha and beta carbon atoms can participate in hyperconjugation with the carbonyl pi antibonding MO?

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By contrast, the slightly more favorable conformation for butene is the right one: enter image description here

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  • $\begingroup$ You have to bear in mind that the acidity of the \alpha-hydrogens/enolisability of the ketone. There will be an equilibrium between the keto form and the enol form which will also play into the preferred conformation. Admittedly the equilibrium lies very much to the keto form... $\endgroup$ – Leeser Sep 2 '15 at 9:28
  • $\begingroup$ I would have expected the conformation shown to be the most stable for ketones - to minimize steric interactions between R groups. But for aldehydes, I would have expected the opposite, since H is smaller than O. What is the evidence? $\endgroup$ – iad22agp Sep 2 '15 at 11:50
  • $\begingroup$ pubs.acs.org/doi/abs/10.1021/ja01091a016 $\endgroup$ – Marko Sep 2 '15 at 12:17
  • $\begingroup$ It could be because of some weak hydrogen bonding between a hydrogen atom attached to one of the carbons of the R group and the carbonyl oxygen? $\endgroup$ – Marko Sep 2 '15 at 17:06
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I found in Ian Fleming's Molecular orbitals and Organic Chemical Reactions, page 93-94 the answer (the aldehyde is propanal):

The polarisation of the H-C bonds in the methyl group leaves a weak positive charge on the outside and this is attracted to the partial negative charge on the oxygen atom, lowering the energy of this conformation

(the difference in energy of the two eclipsed conformations of propanal is 8 kJ/mol).

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  • $\begingroup$ That is of course assuming R = Me. If you add a page number with your reference, I'll upvote you. $\endgroup$ – Jori Sep 3 '15 at 7:07
  • $\begingroup$ Of course, for a C(Me)3 group, the other conformer of the aldehyde would be more stable. $\endgroup$ – Marko Sep 3 '15 at 9:11

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