Why active forms of butane-2,3-diol is more stable than the meso form?
Answers
Answer:
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 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.
Explanation:
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.
The correct answer cannot be A or D since they are representing a chiral diastereomer.