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Pubchem_71310606

Base Information
  • Chemical Name:Pubchem_71310606
  • CAS No.:910048-20-3
  • Molecular Formula:C24H36BF8P2Rh-
  • Molecular Weight:652.2
  • Hs Code.:
  • DSSTox Substance ID:DTXSID10746260
  • Mol file:910048-20-3.mol
Pubchem_71310606

Synonyms:910048-20-3;PUBCHEM_71310606;(1Z,5Z)-cycloocta-1,5-diene;(2R,5R)-1-[2-[(2R,5R)-2,5-dimethylphospholan-1-yl]-3,3,4,4-tetrafluorocyclobuten-1-yl]-2,5-dimethylphospholane;rhodium;tetrafluoroborate;DTXSID10746260;1,2-Bis[(2R,5R)-2,5-dimethylphospholano]-3,3,4,4-tetrafluoro-1-cyclobutene(1,5-cyclooctadiene)rhodium(I) tetrafluoroborate, 95%;1,2-Bis[(2R,5R)-2,5-dimethylphospholanyl]-3,3,4,4-tetrafluoro-1-cyclobutene 1,5-cyclooctadiene rhodium(I) tetrafluoroborate

Suppliers and Price of Pubchem_71310606
Supply Marketing:
Business phase:
The product has achieved commercial mass production*data from LookChem market partment
Manufacturers and distributors:
  • Manufacture/Brand
  • Chemicals and raw materials
  • Packaging
  • price
  • Sigma-Aldrich
  • 1,2-Bis[(2R,5R)-2,5-dimethylphospholano]-3,3,4,4-tetrafluoro-1-cyclobutene(1,5-cyclooctadiene)rhodium(I) tetrafluoroborate 95%
  • 500mg
  • $ 376.00
  • American Custom Chemicals Corporation
  • 1,2-BIS-[(2R,5R)-2,5-DIMETHYL-PHOSPHOLANYL]-3,3,4,4-TETRAFLUORO-1-CYCLOBUTENE(1,5-CYCLOOCTADIENE)RHODIUM-(I)-TETRAFLUOROBORATE 95.00%
  • 5MG
  • $ 498.49
Total 3 raw suppliers
Chemical Property of Pubchem_71310606
Chemical Property:
  • Hydrogen Bond Donor Count:0
  • Hydrogen Bond Acceptor Count:9
  • Rotatable Bond Count:2
  • Exact Mass:652.13125
  • Heavy Atom Count:36
  • Complexity:530
Purity/Quality:

97% *data from raw suppliers

1,2-Bis[(2R,5R)-2,5-dimethylphospholano]-3,3,4,4-tetrafluoro-1-cyclobutene(1,5-cyclooctadiene)rhodium(I) tetrafluoroborate 95% *data from reagent suppliers

Safty Information:
  • Pictogram(s):  
  • Hazard Codes: 
MSDS Files:

SDS file from LookChem

Useful:
  • Canonical SMILES:[B-](F)(F)(F)F.CC1CCC(P1C2=C(C(C2(F)F)(F)F)P3C(CCC3C)C)C.C1CC=CCCC=C1.[Rh]
  • Isomeric SMILES:[B-](F)(F)(F)F.C[C@H]1P([C@@H](CC1)C)C2=C(C(C2(F)F)(F)F)P3[C@@H](CC[C@H]3C)C.C1/C=C\CC/C=C\C1.[Rh]
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