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Is a totally eclipsed Newman projection the most stable conformation for a 2 carbon compound, with one carbon bonded to more electronegative atoms (making them partially negative), and the other carbon with something less electronegative (making it partially positive)?

For example, let's imagine $\ce{CF3CH3}$. If they are eclipsed, the fluorine side is partially negative and the hydrogen side is partially positive, so wouldn't they attract?

Or do the electron clouds always repel each other, no matter what atom is used (making a staggered conformation the most stable, not eclipsed)?

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  • $\begingroup$ Electron clouds are always negatively charged and repel if they come too close, no matter if the atoms they belong to have an overall attraction to each other. $\endgroup$ – Karl Sep 29 '18 at 6:39
  • $\begingroup$ Hmmm, how do hydrogen bonds (or intermolecular forces) exist in that case? Wouldn't the atoms repel there too? I'm getting confused haha. $\endgroup$ – F16Falcon Sep 30 '18 at 17:08
  • $\begingroup$ All bonds have the lenght where long range attraction and short range repulsion cancel each other. $\endgroup$ – Karl Sep 30 '18 at 17:21
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For $\ce{H3C-CF3}$, the stable conformation is staggered. I would expect this to be case for any similar, freely rotating molecule (the exception being conformations imposed by rings etc.), with the reason being the repulsion between the $\ce{C-F}$ and $\ce{C-H}$ bonds.

To confirm, I have optimized the molecular geometry on different levels of quantum chemistry, starting from a conformer halfway between staggered and eclipsed. In all cases, I have observed convergence to the staggered conformer. The levels are

  • HF/def2-TZVP
  • PBE-D3/def2-TZVP
  • B3LYP-D3/def2-TZVP
  • PW6B95-D3/def2-TZVP
  • PW6B95-D3/def2-QZVP

In response to comment: I have performed PW6B95-D3/def2-TZVP optimizations for propene, acetaldehyde, and propanaldehyde (the latter starting from two different conformers). The result is that in all cases, the double bond is eclipsed by a single bond (and two non-equivalent conformers exist for propanaldehyde).

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  • $\begingroup$ So, the attraction between the fluorine and hydrogen is dwarfed by the repulsion by the C-F and C-H bonds? $\endgroup$ – F16Falcon Dec 28 '19 at 16:08
  • $\begingroup$ @F16Falcon That's my observation, yes. $\endgroup$ – TAR86 Dec 28 '19 at 16:15
  • $\begingroup$ @TAR86 AFAIK eclipsed confirmation of propene, acetaldehyde (ethanal) and propanaldehyde is more stable than staggered and gauche form. Can you check for them too, if I stand correct or corrected... $\endgroup$ – Zenix Dec 28 '19 at 16:31
  • $\begingroup$ @TAR86 I am sorry but I didn't get what you mean by .... and two non-equivalent conformers exist for propanaldehyde. And that means eclipsed confirmation of propene and acetaldehyde is the most stable. $\endgroup$ – Zenix Dec 28 '19 at 19:14
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    $\begingroup$ @Zenix Propanadehyde has three stable conformers. Two of them are mirror images of each other and thus equivalent. $\endgroup$ – TAR86 Dec 28 '19 at 19:18
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For 1,1,1-trifluoroethane the two extremes are the staggered and the eclipsed conformations. I have done some molecular mechanics calculations which indicate that the staggered conformation has an energy of 5.7 kCal per mole while the eclipsed it is 1.31 kCal per mole.

This is the result for the molecular mechanics, if we consider the charges calculated in extended Huckel theroy then the flourines and hydrogens do not have large partial charges.

The carbon bearing the three flourines has a charge of +0.865378, the flourines have a charge of -0.261741, the other carbon has a charge of -0.209224 and the hydrogens have a charge of + 0.0430

I have not done the calculations for the electrostatic attraction, but I think in this molecule that if we were to treat the atoms as point charges that the main effect would the flourine atoms and the carbon which they are attrached to. I think that a change of the conformation will make very little difference to the electrostatic attraction energy.

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  • $\begingroup$ AFAIK eclipsed confirmation of propene, acetaldehyde (ethanal) and propanaldehyde is more stable than staggered and gauche form. Can you check for them too, if I stand correct or corrected... $\endgroup$ – Zenix Dec 28 '19 at 16:31

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