The players: a symmetric epoxide and two with a "more-substituted" carbon that regiochemistry will care about. Structures drawn live.
1. Epoxides Are Strained Three-Membered Ethers, So Nucleophiles Pry Them Open
Squeezing the C–O–C angle to ~60° stores about 27 kcal/mol of ring strain that is released when a C–O bond breaks, so nucleophiles that ignore a normal ether open an epoxide readily.
2. Opening Breaks One C–O Bond Into a 1,2-Difunctionalized Product
Cleaving one C–O bond leaves an –OH on one carbon and the nucleophile on the other — always a 1,2-difunctionalized product; the only question is which carbon the nucleophile takes.
Aqueous acid hydrolyzes propylene oxide to propane-1,2-diol — OH on both former ring carbons.
3. Under Base, the Nucleophile Attacks the Less-Hindered Carbon (SN2)
A strong nucleophile (hydroxide, alkoxide, azide, cyanide, Grignard) opens the ring by SN2, so steric access wins and it hits the less-hindered carbon with inversion.
Base: methoxide attacks the less-hindered CH2, so OMe lands there and OH on the substituted carbon.
4. Under Acid, the Nucleophile Attacks the More-Substituted Carbon
Protonating the oxygen builds carbocation character as the ring opens, so even a weak nucleophile is pulled to the more-substituted carbon that best stabilizes the positive charge — the attack still backside, still anti.
Acid: methanol attacks the more-substituted carbon, so OMe lands there and OH on the CH2.
Same epoxide and nucleophile, opposite regiochemistry — set entirely by acid vs base:
5. The Addition Is Always Anti — a Trans Product
In both regimes the nucleophile approaches opposite the breaking C–O bond, so the two new groups end up anti — a trans-1,2 product — no matter which nucleophile is used.
6. Summary
Strain drives the opening · base → SN2 at the less-hindered carbon (inversion) · acid → more-substituted carbon (carbocation character) · geometry is always anti (trans).
Worked example
- Under basic conditions the nucleophile does an SN2 on the epoxide.
- It attacks the less-hindered carbon (the CH2), relieving ring strain.
- The C–O bond breaks to give an alkoxide, then an alcohol on workup.
Answer. 1-methoxypropan-2-ol — attack at the less-substituted carbon (acidic conditions would flip this to the more-substituted carbon).
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The terminal, less-hindered CH₂ carbon. Basic conditions mean a concerted SN2, where steric access controls the outcome, so methoxide attacks the least crowded carbon and the –OH is left on the more-substituted carbon. The product is CH₃OCH₂CH(OH)CH₃, formed with inversion at the carbon attacked.
On the more-substituted carbon — the opposite of the base case. Acid protonates the oxygen, the opening transition state develops carbocation character, and that positive charge is more stable on the more-substituted carbon, so methanol is drawn there. The product is CH₃OCH(CH₃)CH₂OH.
Ring strain. The three-membered ring forces the bond angles to about 60°, storing roughly 27 kcal/mol of angle and torsional strain. Breaking a C–O bond relieves that strain, which provides the driving force an ordinary, unstrained ether completely lacks.
The stereochemistry is always anti. In both regimes the nucleophile attacks the backside of the C–O bond that is breaking, so the incoming group and the resulting –OH end up on opposite faces — a trans-1,2 product. Only the regiochemistry (which carbon) flips with conditions; the anti geometry does not.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, curved arrows, wedge/dash bonds and a clickable periodic table built in. No account needed.