[miaou5]

This is what I've read regarding opening up a bromonium ion:

If the alkene is substituted, let's say with the methyl group, the bromonium ion is no longer symmetric. The more substituted carbon (2o) can accommodate positive charge better than the less substituted carbon (1o), or in other words, resonance structure C contributes more to the description of the molecule than resonance structure A (see below). This situation is just the reflection of the increased stability of the secondary carbocations over the primary carbocations that is due to hyperconjugation and inductive effects. Thus the nucleophile will be more likely to attack the more electropositive, more substituted carbon.

However, I'm recently reading about opening up an epoxide (which looks exactly like a bromonium ion, except swapping O for Br), and people are telling me that a nucleophile will actually attack the last substituted carbon due to steric hindrance effects. Which one is correct? Thank you so much in advance!

[Dan]

an epoxide (which looks exactly like a bromonium ion, except swapping O for Br)

Not exactly. A bomonium is positively charged, an epoxide is neutral. The positive charge on the bromonium leads to development of carbocation character in the transition state. This will also apply for epoxides only if the epoxide is protonated - i.e. in acidic conditions. The mechanism is somewhere between S_N1 and S_N2, sometimes referred to as "loose S_N2".

In basic conditions, the epoxide will open by S_N2 at the least hindered end.

This thread may be useful for you: http://www.chemicalforums.com/index.php?topic=47329.0

[miaou5]

that makes things a lot clearer. thank you so much!