Chemistry Reaction Schemes

Dehydration of an alcohol (E1): 2-methylpropan-2-ol to 2-methylpropene

Heating an alcohol with concentrated acid removes water to make an alkene. Tertiary alcohols react by E1: the OH is protonated, water leaves to give a carbocation, and loss of a neighbouring proton forms the C=C bond.

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Elimination (E1), acid-catalysed dehydration

Protonating the OH group turns it into a good leaving group (water). Loss of water gives a tertiary carbocation, and removal of a neighbouring proton forms 2-methylpropene and regenerates the acid catalyst.

Overall reaction

2-methylpropan-2-ol
conc. H₂SO₄ (catalyst)
heat
2-methylpropene
+
water
byproduct

Mechanism — 3 steps

  1. 1Protonation of the OH groupproton transfer · reversible
    2-methylpropan-2-ol
    +
    H₃O⁺ (from the acid)
    H⁺
    protonated alcohol
    +
    water

    Electron movement

    1. A lone pair on the alcohol oxygen takes a proton from the acid.
    2. The O–H bond of H₃O⁺ breaks; its electrons stay on that oxygen.

    OH⁻ is a very poor leaving group, but once the oxygen is protonated it can leave as a neutral water molecule.

  2. 2Loss of waterheterolysis (rate-determining) · reversible
    protonated alcohol
    −H₂O
    tert-butyl carbocation
    +
    water

    Electron movement

    1. The C–O bond breaks and water leaves, taking both bonding electrons.

    This is the slow step. It is fastest for tertiary alcohols because the resulting carbocation is stabilised by three alkyl groups.

  3. 3Loss of a proton to form C=Cproton transfer (elimination) · reversible
    tert-butyl carbocation
    +
    water
    −H⁺
    2-methylpropene
    +
    H₃O⁺

    Electron movement

    1. A water molecule (or HSO₄⁻) removes a proton from a carbon next to the positive carbon.
    2. The C–H bonding electrons form the C=C double bond, cancelling the positive charge.

    Losing a proton from the neighbouring carbon gives the alkene and regenerates H₃O⁺, so the acid is a catalyst.

Stereochemistry

2-methylpropene has no E/Z isomers. With unsymmetrical alcohols, more than one alkene can form; the more substituted alkene is usually the major product (Zaitsev's rule).

Key concepts

  • Acid catalysis
  • Turning OH into a good leaving group
  • Carbocation intermediate
  • Zaitsev's rule for the major alkene

Common mistakes

  • Having OH⁻ leave directly without protonation
  • Removing the proton from the positive carbon itself
  • Forgetting that the acid is regenerated

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