Química Orgânica 1 · Adição a alcenos e regioquímica
Open a bromonium ion at the tertiary carbon
Condições: 1-Methylcyclohexene with Br₂ in water. Water is the nucleophile and a second water is the base. The wedges show bromine bridging from the top face; attack of Br₂ on the bottom face gives the enantiomer by the same arrows.
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Próximo mostra as setas curvas de cada passo e as explica. Os passos escritos estão abaixo.
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Form the bromonium ion
The alkene attacks Br₂ to give a three-membered bromonium ion on the top face. The bridge is unsymmetrical: the bond to the methyl-bearing carbon is longer and weaker because that carbon, with three carbon neighbors, carries more of the positive charge.
As setas curvas
- The alkene π pair supplies one of the two carbon–bromine bonds.
- The Br–Br pair leaves with one bromine as bromide.
- A lone pair on the bridging bromine supplies the other bond; exchanging the two carbon assignments gives the same bridge.
Water opens the bridge
Water is far more plentiful than bromide, so water opens the ring. It bonds to the tertiary carbon, the carbon with more positive character, even though that carbon is more crowded. Backside attack puts oxygen on the face opposite bromine: anti addition.
Nota: Water is the nucleophile; determine which ring carbon it bonds to.
As setas curvas
- A lone pair on water bonds to the methyl-bearing carbon from the bottom face, opposite the bromine bridge.
- The breaking C–Br bond pair returns to bromine, which stays bonded to the neighboring carbon.
Deprotonate the oxonium ion
Loss of a proton to water gives the neutral bromohydrin plus hydronium; bromide remains in solution.
Nota: Remove the hydrogen drawn above the positive oxygen.
As setas curvas
- A lone pair on the second water accepts the hydrogen drawn above the positive oxygen.
- The O–H pair stays on oxygen, which becomes a neutral hydroxyl oxygen.
trans-2-Bromo-1-methylcyclohexan-1-ol
OH bonds to the more substituted carbon, and Br to its neighbor. The wedge (Br up) and hash (OH down) show the trans relationship from anti addition. Both enantiomers form equally, because the bromonium ion forms equally on either face of the alkene.
O que você vai praticar
Combine bromonium-opening regiochemistry and anti stereochemistry to draw the trans bromohydrin from a trisubstituted ring alkene.
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