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The epoxide on the top of the picture, in aqueous acid hydrolyzes to the compound below. What is the mechanism of this reaction? Standard epoxide opening in acidic conditions would happen at the more substituted end, therefore the secondary -OH group would end up below the plane of the paper, and the tertiary center would racemize to a significant extent. What is happening here?
enter image description here

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    $\begingroup$ The tertiary carbon of the epoxide is more substituted, so I would expect the C-O bond length be slightly longer after protonation and the antibonding orbital having a larger coefficient than the carbon in the other bond. Therefore I would expect an SN2 like attack at the tertiary carbon, giving the depicted product. $\endgroup$ – Martin - マーチン Jun 28 '15 at 14:21
  • $\begingroup$ Isn't the quaternary C atom more substituted? $\endgroup$ – RBW Jun 28 '15 at 15:04
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Standard epoxide opening in acidic conditions would happen at the more substituted end

This premise is false. Usually $\ce{S_{N}2}$ is faster than the cation-stabilized pathway (most textbooks don't even bother to explain the subtleties involved), yielding mainly the less substituted (less sterically hindered) product. It is not clear cut which of the two mechanisms will be predominant in an acid catalyzed epoxide cleavage given the reactants and reaction conditions.

See also Regioselectivity of acid-catalyzed ring-opening of epoxides (especially the second answer and then the excerpt from Clayden et al.).

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