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The most stable conformation of optically inactive Butane–2,3–diol is:

(A)enter image description here

(B)enter image description here

(C)enter image description here

(D)enter image description here

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 their would be repulsion between the two Me groups. So, which one is correct?

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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.

However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct.

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