العربية
中国
Français
Deutsch
Ελληνική
हिन्दी
Italiano
日本語
Português
Русский
EspañolBETA-HYDRIDE ELIMINATION
'Beta-hydride elimination' is an organometallic reaction in which an alkyl group bonded to a metal centre is converted into a metal bonded hydride and an alkene.
The alkyl must have hydrogens on the beta carbon. For instance butyl groups can undergo this reaction but methyl groups cannot. The metal complex must have an empty ( or vacant) site cis to the alkyl group for this reaction to occur.
The beta-hydride elimination can either be a vital step in a reaction or an unproductive side reaction. For instance in the synthesis of RuHCl(CO)(PPh3)3 from ruthenium chloride, triphenylphosphine and methoxyethanol the alkoxide ligand needs to undergo a beta-hydride elimination to form the hydride ligand and the pi-bonded aldehyde which then is later converted into the carbonyl (carbon monoxide) ligand.
In the case of nickel and palladium catalyzed couplings of aryl halides with alkyl Grignards, the beta-hydride elimination can cause the yield to be lowered. As the reaction starts to form alkenes rather than the required product.
There are several strategies for avoiding β-hydride elimination. The first is for the alkyl ligand to lack a beta-hydrogen in the first place. Alternatively, the beta position may be blocked by non-hydrogen atoms. Fluorine is not suitable because metal-fluorine bonds are often strong, thus the abstraction of fluoride is thermodynamically favorable.
If due the ligands are bulky, such as t-butyl or trimethyl silyl, the hydrogen atom may not be able to approach a coplanar configuration with respect to the metal, and the α and β atoms.
If the metal center does not have empty coordination sites, for example, by the complex already being in the 18 electrons, β-hydride elimination is not possible as well.
For the above example, the unwanted beta-hydride elimination is prevented by using a diphosphine where the two phosphorus atoms are fixed apart in space. One way of doing this is to use a ferrocene unit, the nickel and palladium complexes of 1,1'-diphosphinoferrocenes are arranged such that the metal has two phosphorus atoms in the trans sites. As these metals form square planar complexes, no vacant site cis to the alkyl group can be formed. Hence the beta-hydride elimination is prevented.
The alkyl must have hydrogens on the beta carbon. For instance butyl groups can undergo this reaction but methyl groups cannot. The metal complex must have an empty ( or vacant) site cis to the alkyl group for this reaction to occur.
The beta-hydride elimination can either be a vital step in a reaction or an unproductive side reaction. For instance in the synthesis of RuHCl(CO)(PPh3)3 from ruthenium chloride, triphenylphosphine and methoxyethanol the alkoxide ligand needs to undergo a beta-hydride elimination to form the hydride ligand and the pi-bonded aldehyde which then is later converted into the carbonyl (carbon monoxide) ligand.
In the case of nickel and palladium catalyzed couplings of aryl halides with alkyl Grignards, the beta-hydride elimination can cause the yield to be lowered. As the reaction starts to form alkenes rather than the required product.
| Contents |
| Avoiding β-hydride elimination |
Avoiding β-hydride elimination
There are several strategies for avoiding β-hydride elimination. The first is for the alkyl ligand to lack a beta-hydrogen in the first place. Alternatively, the beta position may be blocked by non-hydrogen atoms. Fluorine is not suitable because metal-fluorine bonds are often strong, thus the abstraction of fluoride is thermodynamically favorable.
If due the ligands are bulky, such as t-butyl or trimethyl silyl, the hydrogen atom may not be able to approach a coplanar configuration with respect to the metal, and the α and β atoms.
If the metal center does not have empty coordination sites, for example, by the complex already being in the 18 electrons, β-hydride elimination is not possible as well.
For the above example, the unwanted beta-hydride elimination is prevented by using a diphosphine where the two phosphorus atoms are fixed apart in space. One way of doing this is to use a ferrocene unit, the nickel and palladium complexes of 1,1'-diphosphinoferrocenes are arranged such that the metal has two phosphorus atoms in the trans sites. As these metals form square planar complexes, no vacant site cis to the alkyl group can be formed. Hence the beta-hydride elimination is prevented.
This article provided by Wikipedia. To edit the contents of this article, click here for original source.
psst.. try this: add to faves
