Rifalazil

CAS: 129791-92-0

Ref. TR-R504000

Product Information

• Name: Rifalazil
• Synonyms: 1',4-Didehydro-1-deoxy-1,4-dihydro-3'-hydroxy-5'-[4-(2-methylpropyl)-1-piperazinyl]-1-oxorifamycin VIII, 2,7-(Epoxypentadeca[1,11,13]trienimino)-6H-benzofuro[4,5-a]phenoxazine, rifamycin VIII deriv., ABI 1648, Isobutylpiperazinyl rifa, KRM 1648, (2S,16Z,18E,20S,21S,22R,23R,24S,25S,26S,27S,28E)-5,12,21,23-tetrahydroxy-27-methoxy-2,4,16,20,22,24,26-heptamethyl-10-[4-(2-methylpropyl)piperazin-1-yl]-1,6,15-trioxo-1,2-dihydro-6H,13H-2,7-(epoxypentadeca[1,11,13]trienoazeno)[1]benzofuro[4,5-a]phenoxazin-25-yl acetate, (2S,16Z,18E,20S,21S,22R,23S,24S,25S,26S,27S,28E)-5,12,21,23-tetrahydroxy-27-methoxy-2,4,16,20,22,24,26-heptamethyl-10-[4-(2-methylpropyl)piperazin-1-yl]-1,6,15-trioxo-1,2-dihydro-6H,13H-2,7-(epoxypentadeca[1,11,13]trienoazeno)[1]benzofuro[4,5-a]phenoxazin-25-yl acetate, 1′,4-Didehydro-1-deoxy-1,4-dihydro-3′-hydroxy-5′-[4-(2-methylpropyl)-1-piperazinyl]-1-oxorifamycin VIII, 3'-Hydroxy-5'-(4-isobutylpiperazinyl)benzoxazinorifamycin, 5,12-Dihydroxy-2,4-dimethyl-10-[4-(2-methyl-propyl)-l-piperazinyl]-2,7-[oxy(5-acetoxy-7,9-dihydroxy-3-methoxy-4,6,8,10,14-pentamethyl-15-oxo-l,ll,13-pentadecatriene-l,15-diyl)imino]-6H-benzofuro[4,5-a]phenoxazine-l(2H),6-dione, Abi 1648, Ami 1648, Krm 1648, Rifamycin VIII, 1′,4-didehydro-1-deoxy-1,4-dihydro-3′-hydroxy-5′-[4-(2-methylpropyl)-1-piperazinyl]-1-oxo-
• CAS: 129791-92-0

Description

Applications Rifalazil has been used as a part of multi drug regimen in the treatment of tuberculosis.
References Rothstein D. M., at al., Expert Opin. Investig. Drugs: 12, 255-271 (2003)

• Brand: TRC

Chemical properties

• Molecular weight: 941.073
• Formula: C₅₁H₆₄N₄O₁₃
• InChI: InChI=1S/C51H64N4O13/c1-24(2)23-54-16-18-55(19-17-54)32-21-33(57)39-35(22-32)67-48-40(52-39)36-37-44(60)30(8)47-38(36)49(62)51(10,68-47)65-20-15-34(64-11)27(5)46(66-31(9)56)29(7)43(59)28(6)42(58)25(3)13-12-14-26(4)50(63)53-41(48)45(37)61/h12-15,20-22,24-2
• InChI key: InChIKey=SGHWBDUXKUSFOP-NQRKKCEWSA-N
• SMILES: CO[C@H]1/C=C/O[C@@]2(C)Oc3c(C)c(O)c4c(c3C2=O)C2=C(Oc3cc(N5CCN(CC(C)C)CC5)cc(O)c3N2)C(=NC(=O)/C(C)=C\C=C\[C@H](C)[C@H](O)[C@@H](C)[C@@H](O)[C@@H](C)[C@H](OC(C)=O)[C@@H]1C)C4=O