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I was building molecular models with my son who is in high school to teach him about molecular geometries and he asked me a question I had never considered before. We built Acetone, using plastic models and I mentioned that in reality there were not two equal bonds between C and O, but a Sigma bond running straight between them and a pi bond that actually existed both "above and below" the Sigma. He asked me if the Pi bond had to be in a plane perpendicular to the atoms in the molecule as it appears to be with the limitations of plastic ball and stick models. Does the Pi bond actually lie oriented in the plane perpendicular to the atoms, or is it in the plane of the atoms, or are BOTH "conformations" possible, and if so does that constitute two different isomers of Acetone with slightly different properties?

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    $\begingroup$ I think the differentiation between sigma and pi-molecular orbitals are done in terms of symmetry. In the real molecule, there is no separate orientation, there is just electron density between the two atoms (which we call bond). So, there can't be any "conformer" based on orientation of pi-bonds. $\endgroup$
    – S R Maiti
    Aug 2 at 11:34
  • $\begingroup$ Isomers are not possible in this example. At the high school/introductory college level the $\pi$-bond is taught to be orthogonal to the plane of the carbons and oxygen. The nonbonding pair of electrons on oxygen lie in the plane of the molecule. $\endgroup$
    – user55119
    Aug 2 at 17:23
  • $\begingroup$ @user55119 That is what I had remembered or assumed, having taken organic chemistry over 30 years ago, though your answer is not consistent with S R Maiti's response above. $\endgroup$ Aug 2 at 21:37

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