epoxide opening destroyer?

Started by glk2101
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glk2101

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i know in mcmurry it says acidic opening will attack the less hindered side for primary or secondary epoxides, but will attack the more highly substituted side if the epoxide is tertiary..whereas destroyer says acidic opening always attacks the more substuted side..
anyone know which is right for the exam?
 
i know in mcmurry it says acidic opening will attack the less hindered side for primary or secondary epoxides, but will attack the more highly substituted side if the epoxide is tertiary..whereas destroyer says acidic opening always attacks the more substuted side..
anyone know which is right for the exam?


TLDR: Acidic opening always "attacks" the more substitued side.

Detailed explanation:
Epoxide opening in acidic conditions proceeds through an SN1-like mechanism, where there exists some small equilibrium between the open and closed forms of the molecule. Because there is an abundance of protons in solution, the epoxide oxygen is protonated, so it can equilibrate to the alcohol form and leave a positive charge on the alkane. Since alkyl carbocation stability increases the more substituted the carbocation is (methyl< 1 < 2 < 3), the ring will open towards the more substituted side. In acidic conditions, you usually don't have the greatest nucleophiles. That means that they are less inclined to proceed via the SN2 backside attack mechanism, which favors attacking the less substituted side.

Something to consider:
Even when the epoxide hasn't opened, there will be a delta positive charge on both the expoxide carbons connected to the oxygen. Since alkyl groups are electron donating, the more substituted carbon will donate more electron density to the oxygen, and thus, have a greater partial positive charge.