Alkaline hydrolysis of oxaliplatin—Isolation and identification of the oxalato monodentate intermediate

✍ Scribed by Elin Jerremalm; Pernilla Videhult; Gunvor Alvelius; William J. Griffiths; Tomas Bergman; Staffan Eksborg; Hans Ehrsson

• Publisher: John Wiley and Sons
• Year: 2002
• Tongue: English
• Weight: 93 KB
• Volume: 91
• Category: Article
• ISSN: 0022-3549
• DOI: 10.1002/jps.10201

✦ Synopsis

The alkaline degradation of the chemotherapeutic agent oxaliplatin has been studied using liquid chromatography. The oxalato ligand is lost in two consecutive steps. First, the oxalato ring is opened, forming an oxalato monodentate intermediate, as identified by electrospray ionization mass spectrometry. Subsequently, the oxalato ligand is lost and the dihydrated oxaliplatin complex is formed. The observed rate constants for the first step (k(1)) and the second step (k(2)) follow the equation k(1) or k(2) = k(0) + k(OH(-) )[OH(-)], where k(0) is the rate constant for the degradation catalyzed by water and k(OH(-) ) represents the second-order rate constant for the degradation catalyzed by the hydroxide ion. At 37 degrees C the rate constants for the first step are k(OH(-) ) = 5.5 x 10(-2) min(-1) M(-1) [95% confidence interval (CI), 2.7 x 10(-2) to 8.4 x 10(-2) min(-1) M(-1)] and k(0) = 4.3 x 10(-2) min(-1) (95% CI, 4.0 x 10(-2) to 4.7 x 10(-2) min(-1)). For the second step the rate constants are k(OH(-) ) = 1.1 x 10(-3) min(-1) M(-1) (95% CI, -1.1 x 10(-3) to 3.3 x 10(-3)) min(-1) M(-1) and k(0) = 7.5 x 10(-3) min(-1) (95% CI, 7.2 x 10(-3) to 7.8 x 10(-3) min(-1)). Thus, the ring-opening step is nearly six times faster than the step involving the loss of the oxalato ligand.