[Gulo]

http://imgur.com/zbIN5 (Image tag didn't work...)
The product is the answer to a practice final exam problem. I get that the reaction with peracetic acid will form an epoxide, but then how does the O get protonated and why does the ^-OCH₃ anion attack the carbon with fewer substituents?

[orgopete]

The protonation occurs because a proton source was present but not specified. Two possible solutions, the reaction was run in methanol so methanol was the proton source or the reaction was worked up with acid to neutralize the reaction mixture. Let's make a guess, how much NaOCH3 is required for this reaction?

The opening of epoxides are generally electrophile driven (acid) or nucleophile driven (base). In electrophile driven reactions, the (weak) nucleophile generally attackes the carbon best able to support a carbocation. It may be advantageous to draw a carbocation intermediate even if the chemistry is consistent with a protonated epoxide. You may do the same with bromination reactions.

If the reaction takes place because a strong base initiates the reaction, it will be more SN2 like and therefore sensitive to steric hindrance. In this case, just as SN2 reactions fail on tertiary carbons, opening of an epoxide also fails to take place on the tertiary carbon.