Química Orgánica 1 · Transformaciones de alquinos
Hydroboration–oxidation of hex-3-yne
Condiciones: BH₃·THF; then H₂O₂, NaOH, H₂O
Dibuja este problema tú mismo
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 8 pasos escritos
Concerted syn hydroboration
In one four-centre step the alkyne π pair bonds carbon to boron's empty p orbital while the B–H pair delivers hydride to the other carbon. B and H add to the same face (syn); boron goes to the less hindered carbon and no carbocation forms.
Las flechas curvas
- The π pair bonds C to the empty orbital on boron.
- The B–H pair forms the new C–H bond on the same face.
Hydroperoxide adds to boron
Under the basic oxidation conditions, HOO⁻ (from H₂O₂ and NaOH) fills boron's empty orbital, giving a negatively charged borate.
Las flechas curvas
- A lone pair of the negatively charged peroxide oxygen bonds to boron.
Alkenyl group shifts from boron to oxygen
The C–B pair migrates to the adjacent oxygen as the weak O–O bond breaks, expelling hydroxide. The C=C geometry is kept.
Las flechas curvas
- The C–B bond pair forms the new C–O bond.
- The O–O bond breaks; hydroxide leaves with the pair.
Hydroxide adds to boron
Boron is electron-deficient again, so hydroxide bonds to it, setting up cleavage of the B–O bond to the enol oxygen.
Las flechas curvas
- A hydroxide lone pair bonds to boron.
Release the enolate
The B–O bond breaks with both electrons going to oxygen, releasing a resonance-stabilized enolate and a neutral boron by-product.
Las flechas curvas
- The B–O pair moves onto oxygen, forming the enolate.
Ketone, not aldehyde
Internal vs terminal alkynes.
Protonate the enolate at carbon
Water protonates the enolate at the α-carbon while the oxygen lone pair forms the C=O bond: base-mediated tautomerization to the carbonyl compound.
Las flechas curvas
- The oxygen lone pair forms the C=O π bond.
- The C=C π pair takes a proton from water.
- The O–H pair stays on oxygen as hydroxide.
Carbonyl product
The keto form is the stable tautomer, so the carbonyl compound is isolated; boron ends up as a borate/boric acid derivative.
Qué vas a practicar
Draw syn hydroboration of the symmetric internal alkyne hex-3-yne, oxidation of the alkenylborane to the enolate, and protonation at carbon to hexan-3-one.
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