Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding even though there is no 6-membered ring system? If we don't consider one of the $\ce{H}$ (which is what is usually done), then there is only a 5-membered ring formed. How then does hydrogen bonding take place?


(Also, I feel that there must be some reason apart from 6-member ring for hydrogen bonding which I'm not aware of)

  • $\begingroup$ What's the deeper reason for you to go looking for a six-membered and only a six-membered ring system? $\endgroup$
    – Jan
    Commented May 14, 2015 at 15:35
  • $\begingroup$ Ok thats what i've learnt from my teacher till now.If you know of any other reason please let me know! $\endgroup$
    – user14857
    Commented May 14, 2015 at 15:46
  • $\begingroup$ How is it that your reputation never increases? $\endgroup$ Commented May 16, 2017 at 15:12

1 Answer 1


Cyclopentane has a relatively low strain energy, something on the order of 6.5 kcal/mol more than cyclohexane (reference). In the case of ethane-1,2-diol, the strain energy in the 5-membered ring formed by hydrogen bonding will be even less because some of the $\ce{H-C-C-H}$ and $\ce{C-C-C-C}$ interactions that produce the strain in cyclopentane are no longer present; the oxygens don't have the hydrogens attached like a methylene group does and the hydrogen used in the hydrogen bond is also much smaller than a methylene group.

A typical $\ce{O-H..H}$ hydrogen bond can stabilize a system by ~7 kcal/mol (reference). So to form a 5-membered ring in ethane-1,2-diol and create a hydrogen bond it will cost something less than 6.5 kcal/mol to overcome the strain energy in the ring, but we get 7 kcal/mol back due to hydrogen bond formation - a net stabilization.


  • $\begingroup$ Updated your first reference. $\endgroup$
    – user55119
    Commented Aug 18, 2020 at 16:55

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