Química Orgánica 2 · Enoles, enolatos y reacciones en alfa
α-Methylation of cyclohexanone to 2-methylcyclohexanone
Condiciones: LDA, THF, −78 °C; then CH3I
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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 5 pasos escritos
LDA removes an α-hydrogen
The amide nitrogen of LDA takes a proton from the carbon next to the carbonyl. The C–H electrons become the C=C π bond of the enolate and the C=O π electrons move onto oxygen; diisopropylamine (pKa ≈ 36) is so weak an acid that the step is complete.
Las flechas curvas
- The nitrogen lone pair of LDA takes the α-proton.
- The C–H bond pair forms the C=C π bond to the carbonyl carbon.
- The C=O π electrons move onto oxygen, which now carries the negative charge.
The enolate carbanion contributor
The negative charge of the enolate is shared by oxygen and the α-carbon. The carbon contributor shows the lone pair that forms the new C–C bond; it is the same ion, not a new molecule.
Las flechas curvas
- An oxygen lone pair re-forms the C=O π bond.
- The C=C π pair moves onto the α-carbon as a lone pair.
Why LDA?
LDA is a strong, hindered, non-nucleophilic amide base. Its conjugate acid, diisopropylamine (pKa ≈ 36), is far weaker than a ketone α C–H (pKa ≈ 19–20), so the enolate forms completely before the alkyl halide is added.
SN2 on methyl iodide
The nucleophilic α-carbon attacks the methyl carbon from the back side and iodide leaves in the same step. Methyl iodide is an ideal SN2 partner: unhindered, with an excellent leaving group.
Las flechas curvas
- The α-carbon lone pair forms the new C–C bond to the methyl carbon.
- The C–I bond pair leaves with iodide.
2-Methylcyclohexanone
One methyl group is now on the α-carbon: 2-methylcyclohexanone (racemic, since the planar enolate is attacked on either face). Diisopropylamine and iodide are by-products.
Qué vas a practicar
Draw LDA removing an α-hydrogen of cyclohexanone, the carbanion contributor of the enolate, and SN2 attack of the α-carbon on methyl iodide to give 2-methylcyclohexanone.
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