I am aware they both occur under different conditions and that water attaches in a secondary position in elimination but primary position in hydration.

(Stage 2)  [

(Stage 3) enter image description here

But both appear to be the same intermediate, why exactly do different reactions follow this stage?

  • $\begingroup$ dehydration requires conc. acid, whereas hydration requires dilute acid, it is mostly the conc. of the reagents used , that determine the path of a reaction, in case of multiple paths possible. The conditions provided, like activation energy through heating or some other such means also play a major role, in such scenarios $\endgroup$
    – SubZero
    Mar 22, 2017 at 16:24
  • 1
    $\begingroup$ This is one reversible reaction, you only change equilibrium with suitable conditions. $\endgroup$
    – Mithoron
    Mar 22, 2017 at 17:03

1 Answer 1


As you have spotted, the reaction is reversible and can proceed both ways. Which way it takes depends on the conditions and by implication on Le Chatelier’s principle.

In concentrated $\ce{H2SO4}$, any water will be immediately ‘captured’ by the strongly hygroscopic sulphuric acid. Thus, water is actively drawn out of the equilibrium and the alkene is produced by elimination.

If the reaction is run in aquaeous solution (or at least with water as a cosolvent) and only catalytic quantities of acid, then the nucleophilic attack of water is more likely and the alcohol hydration product is produced.

(In modern synthetic lab-scale chemistry, neither of the reactions are used due to their rather poor selectivity (especially hydration) and harsh conditions (especially elimination). Instead, hydration is typically performed with boranes (anti-Markovnikov) or Markovnikov-selective reagents while elimination is typically achieved by transformation of the alcohol functionality into better leaving groups such as mesylate.)


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