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  4. Wieland–Miescher ketone by intramolecular aldol

Organic Chemistry 2 · Aldol addition and condensation

Wieland–Miescher ketone by intramolecular aldol

Draw the mechanism · Challenge

Conditions: NaOH, MeOH, Δ

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← All Aldol addition and condensation problemsNext problem: Crossed aldol addition — acetone enolate and benzaldehyde →

The mechanism, step by step

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  1. Hydroxide forms the side-chain methyl enolate

    Hydroxide removes a hydrogen from the terminal methyl of the 3-oxobutyl side chain. The ring diketone has no α-hydrogen between its carbonyls (that carbon is quaternary), and this methyl enolate is the one that can reach a ring carbonyl through a six-membered ring.

    The curved arrows

    • Hydroxide takes a hydrogen of the side-chain methyl group.
    • The C–H pair forms the terminal enolate C=C bond.
    • The side-chain ketone π pair moves onto oxygen.
  2. Six-membered ring closure

    The enolate carbon attacks one of the two ring ketone carbonyls. The new ring holds the enolate carbon, the side-chain ketone, two CH2 groups, the quaternary carbon and the attacked carbonyl carbon, which becomes a tertiary alkoxide at the ring fusion.

    The curved arrows

    • The enolate oxygen lone pair re-forms the side-chain C=O.
    • The enolate π pair forms the ring C–C bond to a ring ketone carbon.
    • That ring ketone π pair moves onto its oxygen.
  3. Water protonates the alkoxide

    Water protonates the ring-fusion alkoxide, giving the bicyclic aldol (a ketol) and regenerating hydroxide.

    The curved arrows

    • The tertiary alkoxide takes a proton from water.
    • The O–H pair stays on oxygen as hydroxide.
  4. Enolization next to the ring fusion

    On heating, hydroxide removes the hydrogen on the new ring carbon between the ketone and the carbinol carbon at the ring fusion.

    The curved arrows

    • Hydroxide takes the α-hydrogen next to the ring fusion.
    • The C–H pair forms the enolate C=C bond.
    • The ketone π pair moves onto oxygen.
  5. E1cB dehydration

    The enolate re-forms the C=O and expels hydroxide from the ring-fusion carbon. The new C=C sits at the ring fusion, conjugated with the ketone, giving the bicyclic enone.

    The curved arrows

    • The oxygen lone pair rebuilds the C=O.
    • The enolate π pair forms the ring-fusion C=C.
    • The C–O pair leaves as hydroxide.
  6. Wieland–Miescher ketone

    The Wieland–Miescher ketone forms as the racemate, with water as the by-product; its quaternary methyl-bearing carbon is a stereocentre set by which ring carbonyl was attacked.

What you'll practice

Draw the aldol ring closure of 2-methyl-2-(3-oxobutyl)cyclohexane-1,3-dione: methyl enolate formation, six-membered ring closure onto a ring ketone, protonation, second enolization and E1cB dehydration to the Wieland–Miescher ketone.

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Tags

  • #Aldol
  • #RingStrain
  • #E1cB
  • #BaldwinRules
  • #MechanismDrawing

More Aldol addition and condensation problems

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  • The bond an aldol forms
  • How much base an aldol needs
  • Equilibrium aldol product from K
  • Which step makes the acetone aldol condensation go
  • Stopping at addition or going to the enone
  • Heating the aldol of acetaldehyde
  • Why aldol dehydration is easy

Keep practicing

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