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Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate

Base Information Edit
  • Chemical Name:Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate
  • CAS No.:384835-46-5
  • Molecular Formula:C42H78NaO8P
  • Molecular Weight:765.0
  • Hs Code.:
  • DSSTox Substance ID:DTXSID30677027
  • Mol file:384835-46-5.mol
Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate

Synonyms:384835-46-5;Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate;sodium;[(2R)-2,3-bis[[(Z)-octadec-9-enoyl]oxy]propyl] propyl phosphate;DTXSID30677027;18:1 Phosphatidylpropanol, 1,2-dioleoyl-sn-glycero-3-phosphopropanol (sodium salt), powder;9-Octadecenoic acid (9Z)-, (1R)-1-[[(hydroxypropoxyphosphinyl)oxy]methyl]-1,2-ethanediyl ester, sodium salt;9-Octadecenoicacid(9Z)-,(1R)-1-[[(hydroxypropoxyphosphinyl)oxy]methyl]-1,2-ethanediylester,sodiumsalt

Suppliers and Price of Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate
Supply Marketing:Edit
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
  • American Custom Chemicals Corporation
  • 1,2-DIOLEOYL-SN-GLYCERO-3-PHOSPHOPROPANOL SODIUM SALT 95.00%
  • 5MG
  • $ 504.91
Total 2 raw suppliers
Chemical Property of Sodium (2R)-2,3-bis{[(9Z)-octadec-9-enoyl]oxy}propyl propyl phosphate Edit
Chemical Property:
  • Hydrogen Bond Donor Count:0
  • Hydrogen Bond Acceptor Count:8
  • Rotatable Bond Count:41
  • Exact Mass:764.53320072
  • Heavy Atom Count:52
  • Complexity:897
Purity/Quality:

99% *data from raw suppliers

1,2-DIOLEOYL-SN-GLYCERO-3-PHOSPHOPROPANOL SODIUM SALT 95.00% *data from reagent suppliers

Safty Information:
  • Pictogram(s):  
  • Hazard Codes: 
MSDS Files:
Useful:
  • Canonical SMILES:CCCCCCCCC=CCCCCCCCC(=O)OCC(COP(=O)([O-])OCCC)OC(=O)CCCCCCCC=CCCCCCCCC.[Na+]
  • Isomeric SMILES:CCCCCCCC/C=C\CCCCCCCC(=O)OC[C@H](COP(=O)([O-])OCCC)OC(=O)CCCCCCC/C=C\CCCCCCCC.[Na+]
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