Química Orgânica 2 · Enóis, enolatos e reações em alfa
Iodoform reaction of acetone to acetate and iodoform
Condições: I2 (3 equiv), NaOH (4 equiv), H2O
Desenhe este problema você mesmo
Arraste dos elétrons que se movem até onde eles vão, ou toque em um e depois no outro. Cada seta é verificada; Dica ajuda quando você travar. Não precisa de conta.
O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
Ler todos os 11 passos 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
As setas 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.
O que você vai praticar
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.
Você acabou de desenhar cada seta deste problema.
Cada um dos 5349 problemas de Orgânica 1 e 2 funciona assim: feedback em cada seta, dicas quando você travar e um modo prova sem dicas. Grátis com uma conta, sem cartão.
Travou na sua própria lição?
Tire uma foto: o primeiro scan é grátis, sem cadastro
Você escaneia toda semana? Scan Standard: por cerca de problemas