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?
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).