Why does the gauche conformation of ethane-1,2-diol show hydrogen bonding?

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)

• What's the deeper reason for you to go looking for a six-membered and only a six-membered ring system? – Jan May 14 '15 at 15:35
• Ok thats what i've learnt from my teacher till now.If you know of any other reason please let me know! – user14857 May 14 '15 at 15:46
• How is it that your reputation never increases? – Aaron John Sabu May 16 '17 at 15:12

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.
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