Alkaline Hydrolysis of Oxaliplatin—Isolation and Identification of the Oxalato Monodentate Intermediate

Abstract

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 (k1) and the second step (k2) follow the equation k1 or k2 = k0 + kOH-[OH−], where k0 is the rate constant for the degradation catalyzed by water and kOH- represents the second‐order rate constant for the degradation catalyzed by the hydroxide ion. At 37°C the rate constants for the first step are kOH- = 5.5 × 10−2 min−1 M−1 [95% confidence interval (CI), 2.7 × 10−2 to 8.4 × 10−2 min−1 M−1] and k0 = 4.3 × 10−2 min−1 (95% CI, 4.0 × 10−2 to 4.7 × 10−2 min−1). For the second step the rate constants are kOH- = 1.1 × 10−3 min−1 M−1 (95% CI, −1.1 × 10−3 to 3.3 × 10−3) min−1 M−1 and k0 = 7.5 × 10−3 min−1 (95% CI, 7.2 × 10−3 to 7.8 × 10−3 min−1). Thus, the ring‐opening step is nearly six times faster than the step involving the loss of the oxalato ligand. © 2002 Wiley‐Liss Inc. and the American Pharmaceutical Association J Pharm Sci 91:2116–2121, 2002