# Which is most stable conformation of optically inactive butane-2,3-diol?

The most stable conformation of optically inactive butane-2,3-diol is:

(A)

(B)

(C)

(D)

I choose C as the correct answer but, in the answer key it given to be B. C is the staggered form and the most stable one. It can also have hydrogen bonding between -OH and -H groups. B can also have hydrogen bonding but there would be repulsion between the two Me groups. So, which one is correct?

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Nice question and wow I just looked at Wikipedia for 2,3-butanediol and it is crazy wrong about how many stereoisomers there are. It apparently doesn't know tha (R,S)-2,3-butanediol is the same as (S,R)-2,3-butanediol because of symmetry (i.e. it is a meso compound). At least, that's what I think but I'm not a stereochemical expert. Can someone update? –  Curt F. Apr 2 at 23:07

Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on hydrogen boning, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value.