Why is cis-1,2-cyclohexadiol less polar than trans-1,2-cyclohexadiol? I know it has something to do with stability and chair conformations, but I'm not sure how it relates to polarity. I know cis-1,2-cyclohexadiol has a more stable chair conformation than trans.

  • $\begingroup$ Does the cis compound have an internal hydrogen bond which decreases the polarity of the molecule? $\endgroup$ Apr 13, 2017 at 7:31
  • $\begingroup$ Could you indicate what the difference in dipole is, and perhaps add pics of the structures? $\endgroup$
    – porphyrin
    Apr 13, 2017 at 12:38
  • $\begingroup$ Did someone say VDW? In some cis, trans (geometric) isomerism you may experience lack of polarity. But that doesn't mean there's none as much of polarity rise from intermolecular bonding composed of dipole-dipole and van der Waals dispersion forces. $\endgroup$
    – bonCodigo
    Oct 19, 2021 at 5:59

1 Answer 1


If I assume you refer to 1,2-cylohexanediols, the two isomers to consider were (cis), and (trans)

enter image description here

Assuming the cyclohexane ring as a plane, an (axial, equatorial) orientation of the two O-substitutents is puts them both on the same side of this reference plane; the relative orientation of the two substituents is (cis):

enter image description here

This contrasts to the case of the (trans)-configuration, where the two substituents may be either both in axial, or (likely preferred by thermodynamics) both in equatorial orientation:

enter image description here

From the later picture, taking into account i) the relative orientation of the two hydroxyl groups towards each other as well in respect to the cyclohexane moiety and ii) the (assumed) conformational preference for this form over the conformer with two axial oriented hydroxyl groups, I would assume the two individual vectorial contributions along $\ce{C -> O}$ are better lined up than in the instance of the (cis)-configuration to yield a larger (global) dipolar moment.

  • $\begingroup$ The dihedral angle of the cis- structure is -28.2° while that of the trans- is 77.6° when drawn in Avogadro and geometry optimised using the molecular mechanics software installed. Wouldn't vector algebra suggest that the cis- structure has a greater dipole moment? Hence my question about internal hydrogen bonds mitigating the polarity of C-O bond. $\endgroup$ Apr 14, 2017 at 7:35
  • $\begingroup$ I found some values on the web which are 2.33 D and 2.39 D for cis and trans, so probably identical within error (not quoted with data). I calculated (v. simply using bond dipoles from textbook and cyclohexane geometry) and got 1.99D cis and 2.1D trans (so also similar and surprisingly accurate considering model used) but only 0.16D for trans a-a isomer in your bottom figure. So it would seem trans o-e is most stable as expt values are similar. $\endgroup$
    – porphyrin
    Apr 15, 2017 at 10:01
  • $\begingroup$ @porphyrin I refrained from bringing all three structures into Avogadro. Beside it is currently not accessible for me, I am not aware if applying MMFF94 (as maybe more suitable than UFF) and its implementation in Avogadro are tools allowing to discern the situtation between the three instances (especially trans-configuration as in axial, axial vs. equatorial, equatorial orientation) and hence addresses the question "correctly" within its allowances by theory. The small differences (experimentally, and by calculation) you fond suggest me, here "no" is more likely than anticipated by mine. $\endgroup$
    – Buttonwood
    Apr 15, 2017 at 15:50
  • $\begingroup$ You switched the pictures. Cis isomer should have both subsituents on the same side, NOT on opposite. For instance, the 2-D-molecule you have drawn for the trans isomer has two directional bonds in front of the molecule, but you have drawn the correct conformer for the name. Note that on the top of the left hydroxy group there is a hydrogen which truely is in front! $\endgroup$
    – user115162
    Oct 18, 2021 at 7:01
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    $\begingroup$ @冰淇淋 Thank you very much for identifying the problem. The figures are redrawn from scratch, the text edited slightly to correct this entry. $\endgroup$
    – Buttonwood
    Oct 18, 2021 at 15:43

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