Química Orgânica 2 · Enóis, enolatos e reações em alfa
Haloform reaction — acetophenone bromination and cleavage
Condições: Acetophenone in aqueous Br₂/base; target benzoate and bromoform without acidic workup. Use three Br₂ and four hydroxide equivalents: add one Br₂/base pair for each bromination, then the fourth hydroxide for cleavage. All waters and bromides remain explicit. Counterions are not drawn; no stereochemistry is assigned without bond markers.
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O mecanismo, passo a passo
Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
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Move the electrons
Hydroxide and Br₂ are present. Alpha deprotonation begins halogenation 1 of three; all earlier water and bromide products remain.
Nota: Remove the uppermost hydrogen drawn on the alpha carbon. One equivalent each of hydroxide and Br₂ is present.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
Alpha halogenation 1 replaces one hydrogen with bromine. One Br atom remains on the methyl-derived carbon and the other becomes bromide; carbon retains its octet.
As setas curvas
- Oxygen donation restores the ketone carbonyl.
- The enolate pi pair forms the alpha carbon–bromine bond.
- The Br–Br bond pair becomes bromide.
Move the electrons
The next equivalents of hydroxide and Br₂ are added. Alpha deprotonation begins halogenation 2 of three; all earlier water and bromide products remain.
Nota: Remove the uppermost hydrogen drawn on the alpha carbon. Add one equivalent each of hydroxide and Br₂; earlier byproducts remain.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
Alpha halogenation 2 replaces one hydrogen with bromine. One Br atom remains on the methyl-derived carbon and the other becomes bromide; carbon retains its octet.
As setas curvas
- Oxygen donation restores the ketone carbonyl.
- The enolate pi pair forms the alpha carbon–bromine bond.
- The Br–Br bond pair becomes bromide.
Move the electrons
The next equivalents of hydroxide and Br₂ are added. Alpha deprotonation begins halogenation 3 of three; all earlier water and bromide products remain.
Nota: Remove the last hydrogen on the alpha carbon. Add one equivalent each of hydroxide and Br₂; earlier byproducts remain.
As setas curvas
- The base removes an alpha hydrogen next to the carbonyl.
- The alpha C–H electrons form a π bond to the carbonyl carbon.
- The carbonyl π electrons move onto oxygen.
Move the electrons
Alpha halogenation 3 replaces one hydrogen with bromine. One Br atom remains on the methyl-derived carbon and the other becomes bromide; carbon retains its octet.
As setas curvas
- Oxygen donation restores the ketone carbonyl.
- The enolate pi pair forms the alpha carbon–bromine bond.
- The Br–Br bond pair becomes bromide.
Move the electrons
The fourth hydroxide equivalent is added after all three alpha halogenations. Addition to the tribromomethyl ketone gives the tetrahedral acyl intermediate.
Nota: Add the fourth hydroxide equivalent. All three Br₂/base cycles are complete; earlier waters and bromides remain.
As setas curvas
- The nucleophile uses a lone pair to form the new single bond.
- The π electrons move off carbon, keeping eight electrons around it.
Move the electrons
Carbonyl reformation cleaves the acyl–CBr₃ bond. The three electron-withdrawing bromines stabilize the departing tribromomethyl anion; every bromine remains explicit.
As setas curvas
- A lone pair forms the new π bond.
- The leaving group takes the electrons from the single bond.
Move the electrons
The tribromomethyl anion accepts the carboxylic-acid proton, producing neutral bromoform and benzoate; neither the carbanion nor the carboxylic acid is left as the final basic endpoint.
As setas curvas
- The base uses a lone pair to accept a proton from the O–H bond.
- The bond to hydrogen breaks, and its electrons stay on the atom losing the proton.
Products
The three methyl hydrogens end up in water; three Br atoms remain in CHBr₃ and three become Br⁻. The fourth hydroxide supplies the second benzoate oxygen and the hydrogen for the bromoform C–H bond. The net reaction is ketone + 3 Br₂ + 4 OH⁻ → benzoate + CHBr₃ + 3 Br⁻ + 3 H₂O, conserving every atom and total charge −4.
O que você vai praticar
Execute all three alpha brominations, nucleophilic acyl cleavage and the final proton transfer.
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