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  1. Practice problems
  2. Orgo 2
  3. Aldol addition and condensation
  4. (E)-Chalcone from acetophenone and benzaldehyde

Organic Chemistry 2 · Aldol addition and condensation

(E)-Chalcone from acetophenone and benzaldehyde

Draw the mechanism · Application

Conditions: NaOH, EtOH, rt

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The mechanism, step by step

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  1. Hydroxide forms the enolate of acetophenone

    Hydroxide removes an α-hydrogen of acetophenone. Only acetophenone has α-hydrogens; benzaldehyde cannot form an enolate, so it can only be the electrophile.

    The curved arrows

    • Hydroxide takes an α-hydrogen.
    • The C–H pair forms the enolate C=C bond.
    • The C=O π pair moves onto oxygen.
  2. Why does dehydration happen so easily?

    Base-mediated dehydration of an aldol goes through the enolate, not through a carbocation.

  3. The enolate attacks benzaldehyde

    The enolate α-carbon adds to the aldehyde carbon of benzaldehyde, which is more electrophilic than the ketone carbonyl, forming the new C–C bond and an alkoxide.

    The curved arrows

    • The oxygen lone pair re-forms the C=O.
    • The enolate π pair forms the new C–C bond.
    • The aldehyde π pair moves onto oxygen.
  4. Water protonates the alkoxide

    The alkoxide takes a proton from water, giving the β-hydroxy ketone (aldol) and regenerating hydroxide.

    The curved arrows

    • The alkoxide takes a proton from water.
    • The O–H pair stays on oxygen as hydroxide.
  5. Hydroxide forms the enolate of the aldol

    The remaining α-hydrogen, now flanked by the carbonyl and the benzylic carbinol carbon, is removed by hydroxide to give a new enolate.

    The curved arrows

    • Hydroxide takes the second α-hydrogen.
    • The C–H pair forms the enolate C=C bond.
    • The C=O π pair moves onto oxygen.
  6. E1cB loss of hydroxide

    The enolate re-forms C=O and expels hydroxide from the β-carbon (E1cB). The new C=C is conjugated with both the carbonyl and the benzene ring, which makes this dehydration fast and irreversible under the conditions; the E isomer forms because it keeps the aryl ring and the carbonyl apart.

    The curved arrows

    • The oxygen lone pair re-forms the C=O.
    • The enolate π pair forms the new C=C bond.
    • The C–O pair leaves as hydroxide.
  7. (E)-chalcone

    The (E)-chalcone forms as the E isomer, with hydroxide regenerated and one water released overall.

What you'll practice

Draw the base-catalysed condensation of acetophenone with benzaldehyde to (E)-chalcone, including the aldol addition, the second enolization and the E1cB loss of hydroxide.

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Tags

  • #Aldol
  • #Carbonyls
  • #Arenes
  • #E1cB

More Aldol addition and condensation problems

  • (E)-Cinnamaldehyde from acetaldehyde and benzaldehyde
  • (E)-Benzalacetone from benzaldehyde and acetone
  • (E)-α-Methylcinnamaldehyde from propanal and benzaldehyde
  • 2-(Hydroxymethyl)cyclohexan-1-one from cyclohexanone
  • 2-Cyclohexylidenecyclohexan-1-one from cyclohexanone
  • Ethyl trans-3-phenyloxirane-2-carboxylate by the Darzens reaction
  • Ethyl 1-oxaspiro[2.5]octane-2-carboxylate by Darzens
  • (E,E)-Dibenzalacetone: the second condensation

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