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Epoxide opening under acid: the mechanism, step by step

Acid protonates the ring oxygen, which stretches the bond to the carbon that can best hold a positive charge, and the nucleophile attacks that one.

rule-derived3 stepsNot a record of what was observed

The pathway runs in 3 steps. What sits between them is a real structure: it is what the reaction passes through, and on a bad day it is what you isolate instead of the product.

Below is the sequence as it is taught. On the platform it is drawn on whatever structures you supply, with the curly arrows resolved to the actual atoms involved rather than to a general case with R groups on it.

The steps

  1. Acid protonates the ring oxygen

    The acid gives a hydrogen to the oxygen of the three-membered ring. The ring is now under even more strain and one of its bonds starts to break.

    Protonation of the epoxide oxygen. The positive charge draws electron density out of both C-O bonds, and the one that breaks more easily is the one at the carbon that holds a positive charge best.

  2. The nucleophile attacks the more substituted carbon

    The nucleophile attacks the carbon that is better at holding a positive charge, which is the more substituted one, and the ring opens.

    Attack at the carbon with the greater positive character. The transition state has substantial cation character there, which is why the crowding that decides the base-catalysed case does not decide this one.

  3. A base takes the proton

    The oxygen that attacked is carrying a positive charge and a spare hydrogen. Something takes it, and the neutral product is left.

    Proton transfer to the solvent or to the conjugate base of the acid, restoring the catalyst.

What this drawing assumes

Read these before you quote the mechanism
  • Drawn as an attack on the protonated ring rather than through a free cation. The truth is in between: the bond is well on its way to breaking when the nucleophile arrives, which is why the more substituted carbon is attacked.
  • The proton transfers are fast compared with the attack.

Other pathways this class runs by

  • Under base there is no protonation, nothing stretches either bond, and the nucleophile goes to the less crowded carbon instead. That is the other pathway and gives the other product.

A class is not one pathway. Which of these runs depends on the substrate, the solvent and what else is in the flask, and the drawing above does not decide that for you.

Run it on your own structures

Enter the reactants and the product you expect. If the transformation is one the engine can perform and this pathway reaches that product from those structures, it is drawn on them — with the arrows on the right atoms and the intermediates you would actually pass through. If it does not reach the product, you are told that instead of being shown a drawing that does not apply.