Consider the reactions below
a) What is the name of the 3-membered ring functional group in the starting material?
b) Provide a mechanism for the reaction to the right
c) Provide a mechanism for the reaction to the left
d) Explain why the regioselectivity is different for this reaction in acidic and basic conditions.
e) We are taught that OH- is a bad leaving group for Sn2. Yet, in the reaction to the right, we have a negative oxygen as a leaving group and no one seems to care. Explain why it is okay to have a negative oxygen as the leaving group for Sn2 in this example, with epoxide opening in basic conditions.
Solution
a) Epoxide
b)
c)
d) In basic conditions, the reaction is governed by sterics. The methoxide ion attacks the least sterically hindered carbon.
In acidic conditions, the reaction is governed by electronics.
Methanol attacks the carbon (on the protonated epoxide) that is the most partially positive. How do we know the more substituted carbon is the one with the most positive charge here? Well, when the epoxide is protonated, you can draw 2 other resonance structures of this positive ion. (This is strange since you’re breaking bonds, I know). But it is analogous to some other reactions you may have learned: Bromination of alkenes, or halohydrin formation. In these reactions, there is a 3-membered positively charged intermediate (just like the protonated epoxide).
Since the middle resonance contributor is more significant than the rightmost contributor, the more substituted carbon has more partial positive charge in the resonance hybrid (i.e. the actual molecule) of the protonated epoxide. Hence, MeOH attacks the more substituted carbon.
e) The epoxide ring has lots of ring strain, so opening it up relieves this ring strain. This relief of ring strain compensates for the negative oxygen being the leaving group.