Epoxide opening under basic conditions: practice and mechanism
Under basic conditions methoxide attacks the less substituted epoxide carbon (SN2, sterically more accessible). The OH group ends up on the more substituted carbon.
Example
Under basic conditions the strong nucleophile (methoxide) attacks via SN2: at the less hindered carbon and from the back. Cycloalkene oxides therefore give the trans product.
Mechanism step by step
- 1
The methoxide ion attacks the less substituted epoxide carbon from the back (SN2).
- 2
The ring C–O bond breaks; the ring strain (about 110 kJ/mol) makes this possible despite the poor RO⁻ leaving group.
- 3
The alkoxide takes a proton from methanol: a β-methoxy alcohol forms, with OCH₃ and OH trans.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- That would be the 1,2-diol from opening with water. Here methanol or methoxide is the nucleophile, so one carbon gets a methyl ether and the other an OH group.
- The epoxide is opened only once: one nucleophile attacks and the epoxide oxygen becomes the OH group. Two methoxy groups do not form.
- Wrong stereochemistry: the nucleophile attacks from the back (SN2-like), opposite the epoxide oxygen. OCH₃ and OH therefore end up trans, not cis.
Often confused with: Epoxide opening under acidic conditions
Frequently asked questions
Why does the nucleophile attack the less substituted carbon?
Under basic conditions there is no positive charge favouring one site. Only sterics count, as in any SN2 reaction.
Why do epoxides react but ordinary ethers do not?
The three-membered ring is highly strained. Opening releases this strain; ordinary ethers have none and need strong acids like HI.