Química Orgánica 2 · Adición y condensación aldólica
Cyclopent-1-ene-1-carbaldehyde from hexanedial
Condiciones: NaOH, H2O, Δ
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Arrastra desde los electrones que se mueven hasta donde van, o toca uno y luego el otro. Cada flecha se revisa; Pista te ayuda cuando te atascas. No necesitas cuenta.
El mecanismo, paso a paso
Siguiente reproduce las flechas curvas de cada paso y las explica. Los pasos escritos están abajo.
Leer los 7 pasos escritos
Hydroxide forms the enolate
Hydroxide removes an α-hydrogen next to one aldehyde group. The two ends of the molecule are identical, so either end can serve as the donor.
Las flechas curvas
- Hydroxide takes an α-hydrogen.
- The C–H pair forms the enolate C=C bond.
- The C=O π pair moves onto oxygen.
Five-membered ring closure
The enolate carbon attacks the other aldehyde carbon of the same chain. The new bond closes a five-membered ring (the α-carbon, three CH2 groups and the attacked carbon).
Las flechas curvas
- The oxygen lone pair re-forms the aldehyde C=O.
- The enolate π pair forms the ring C–C bond.
- The aldehyde π pair moves onto oxygen.
Protonation of the alkoxide
Water protonates the alkoxide, giving 2-hydroxycyclopentane-1-carbaldehyde and regenerating hydroxide.
Las flechas curvas
- The alkoxide takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Enolization at the ring carbon
Hydroxide removes the ring hydrogen α to the aldehyde, which is also next to the carbinol carbon.
Las flechas curvas
- Hydroxide takes the α-hydrogen.
- The C–H pair forms the enolate C=C bond.
- The C=O π pair moves onto oxygen.
Where does the C=C end up?
E1cB places the new double bond between the α- and β-carbons.
E1cB loss of hydroxide
The enolate re-forms C=O and expels hydroxide from the β-carbon, giving the ring C=C conjugated with the aldehyde.
Las flechas curvas
- The oxygen lone pair re-forms the C=O.
- The enolate π pair forms the ring C=C bond.
- The C–O pair leaves as hydroxide.
Cyclopent-1-ene-1-carbaldehyde
Cyclopent-1-ene-1-carbaldehyde forms; five-membered rings like this one form readily in intramolecular aldols.
Qué vas a practicar
Draw the intramolecular aldol condensation of hexanedial: C2 enolate formation, five-membered ring closure onto the other aldehyde, protonation and E1cB dehydration to cyclopent-1-ene-1-carbaldehyde.
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