Química Orgánica 2 · Enoles, enolatos y reacciones en alfa
Iodoform reaction of acetone to acetate and iodoform
Condiciones: I2 (3 equiv), NaOH (4 equiv), H2O
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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 11 pasos escritos
Hydroxide removes an α-hydrogen
Hydroxide takes a proton from a methyl group of acetone, giving an enolate. In base, halogenation does not stop at one iodine.
Las flechas curvas
- The hydroxide lone pair takes the α-proton.
- The C–H pair forms the enolate C=C.
- The C=O π pair moves onto oxygen.
The enolate attacks iodine
The O⁻ lone pair pushes, the enolate C=C bonds to iodine, and iodide leaves, putting another iodine on the same carbon.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The C=C π pair forms the new C–I bond.
- The I–I pair leaves as iodide.
Remove the second α-hydrogen
The iodine already on this carbon withdraws electrons, so its remaining hydrogens are more acidic than before and hydroxide removes one faster.
Las flechas curvas
- The hydroxide lone pair takes the α-proton.
- The C–H pair forms the enolate C=C.
- The C=O π pair moves onto oxygen.
Second iodination
The O⁻ lone pair pushes, the enolate C=C bonds to iodine, and iodide leaves, putting another iodine on the same carbon.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The C=C π pair forms the new C–I bond.
- The I–I pair leaves as iodide.
Remove the last α-hydrogen
With two iodines attached, the last α-hydrogen is the most acidic of all; hydroxide removes it at once.
Las flechas curvas
- The hydroxide lone pair takes the α-proton.
- The C–H pair forms the enolate C=C.
- The C=O π pair moves onto oxygen.
Third iodination
The O⁻ lone pair pushes, the enolate C=C bonds to iodine, and iodide leaves, putting another iodine on the same carbon.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The C=C π pair forms the new C–I bond.
- The I–I pair leaves as iodide.
Why does it stop at CI3?
Each iodine makes the next deprotonation faster, so the methyl becomes CI3 and then cleaves.
Hydroxide adds to the carbonyl
A fourth hydroxide adds to the carbonyl carbon of 1,1,1-triiodoacetone, giving a tetrahedral alkoxide.
Las flechas curvas
- The hydroxide lone pair forms a bond to the carbonyl carbon.
- The C=O π pair moves onto oxygen.
Expel the triiodomethyl anion
The alkoxide re-forms C=O and pushes out ⁻CI3. Three iodines stabilize this carbanion enough for it to leave, which is why only methyl ketones give the haloform reaction.
Las flechas curvas
- An oxygen lone pair re-forms C=O.
- The C–C pair leaves with the CI3 carbon.
Proton transfer gives iodoform
The carboxylic acid is far more acidic than iodoform, so the triiodomethyl anion takes its proton, giving acetate and yellow iodoform, the positive test.
Las flechas curvas
- The carbanion lone pair takes the acidic O–H proton.
- The O–H pair stays on oxygen, giving carboxylate.
Acetate and iodoform
Acetone is cleaved to acetate and iodoform (CHI3), the yellow precipitate of the iodoform test. Water and iodide are by-products.
Qué vas a practicar
Draw three cycles of enolate formation and iodination at one methyl of acetone, hydroxide addition to the carbonyl, loss of the triiodomethyl anion, and the final proton transfer giving acetate and iodoform.
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