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Acid-catalysed hydration of an alkene: the mechanism, step by step

Acid gives the alkene a proton, leaving a positively charged carbon; water attacks it and then loses a proton, and the acid is returned.

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. The alkene takes a proton from the acid

    The electrons of the double bond reach out and take a hydrogen from the acid. That leaves the other carbon short of electrons and positively charged.

    Protonation of the π bond. Which carbon ends up charged decides the product, and it is the one that holds the charge better — the more substituted one.

  2. Water attacks the charged carbon

    Water has electrons to spare and the carbon is short of them, so water attacks and a bond forms. The oxygen is left carrying the positive charge.

    Capture of the cation by water. Both faces are open, so a carbon that becomes a stereocentre here is left as a mixture.

  3. The acid is given back

    The oxygen has one hydrogen too many. Something takes it away, the alcohol is left, and the acid that started the whole thing is back.

    Loss of the proton to the solvent, regenerating the catalyst. This is why the acid is used in a small amount rather than an equivalent.

What this drawing assumes

Read these before you quote the mechanism
  • The cation is drawn as a plain flat carbon. Where a neighbouring group can shift across to it, the alcohol ends up somewhere else, and this does not show that.
  • The proton that starts it and the proton that ends it are the same catalyst returning, which is why so little acid is needed.

Other pathways this class runs by

  • Hydroboration adds water the other way about, putting the oxygen on the less substituted carbon, because nothing there goes through a cation at all.
  • A carbon that cannot hold a positive charge does not react this way, and the alkene simply sits.

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.