Samarium(II) bromide
Chemical compound
From Wikipedia, the free encyclopedia
Samarium(II) bromide is an inorganic compound with the chemical formula SmBr
2.[6] It is a brown solid that is insoluble in most solvents but degrades readily in air.[2]
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| Names | |
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| IUPAC name
samarium(II) bromide | |
| Other names
samarium dibromide dibromosamarium | |
| Identifiers | |
3D model (JSmol) |
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| ChemSpider | |
PubChem CID |
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CompTox Dashboard (EPA) |
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| Properties | |
| SmBr2 | |
| Molar mass | 310.17 g/mol[1] |
| Appearance | Brown crystals |
| Melting point | 669 °C (1,236 °F; 942 K)[2] |
| Boiling point | 1,880 °C (3,420 °F; 2,150 K)[citation needed] |
| +5337.0·10−6 cm3/mol [3][4] | |
| Structure | |
| SrBr2[5] | |
| Hazards | |
| GHS labelling: | |
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| Warning[1] | |
| H315, H319, H335[1] | |
| P261, P305+P351+P338[1] | |
| Related compounds | |
Other anions |
Samarium(II) chloride Samarium(II) iodide |
Other cations |
Samarium(III) bromide |
Except where otherwise noted, data are given for materials in their standard state (at 25 °C [77 °F], 100 kPa).
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Structure
In the gas phase, SmBr
2 is a bent molecule with Sm–Br distance 274.5 pm and bond angle 131±6°.[7]
History
Samarium(II) bromide was first synthesized in 1934 by P. W. Selwood, when he reduced samarium tribromide (SmBr3) with hydrogen (H2). Kagan also synthesized it by converting samarium(III) oxide (Sm2O3) to SmBr3 and then reducing with a lithium dispersion in THF. Robert A. Flowers synthesized it by adding two equivalent of lithium bromide (LiBr) to samarium diiodide (SmI2) in tetrahydrofuran. Namy managed to synthesize it by mixing tetrabromoethane (C2H2Br4) with samarium metal, and Hilmerson found that heating the tetrabromoethane or samarium greatly improved the production of samarium(II) bromide.[8]
Reactions
Samarium(II) bromide has reducing properties reminiscent of the more commonly used samarium diiodide.[9] It is an effective for pinacol homocouplings of aldehydes, ketones, and cross-coupling carbonyl compounds. Reports have shown that samarium(II) bromide is capable of selectively reducing ketones if it is in the presence of an alkyl halide.[8]
Samarium(II) bromide forms soluble adducts with hexamethylphosphoramide. This species reduces imines to amines and alkyl chlorides to hydrocarbons.[10] For example, SmBr2(hmpa)x converts cyclohexyl chloride to cyclohexane.[11]
Samarium(II) bromide will reduce ketones in tetrahydrofuran if an activator is absent.[12]