KINETIC AND EQUILIBRIUM STUDIES OF THE REACTIONS OF CYSTEINE AND PENICILLAMINE WITH AQUEOUS IRON(III)

The reactions of cysteine (cys) and penicillamine (H(3)L(+), pen) with aqueous iron(III) have been studied in dilute acid solutions (0.01-0.2 M H+) with iron(III) concentrations (0.01-0.1 M) in excess over the amino acid ((similar to 1.5-8.5) x 10(-4) M) at 25 degrees C in 1.0 M LiClO4/HClO4. Stopped-flow and standard spectrophotometry were used to measure the kinetics of the complexation and oxidation-reduction reactions, respectively. Complexation equilibrium constants (K-f = [MLH][H+](2)/[M][LH(3)]) have been determined from the absorbance immediately after the complexation reaction. Both monomeric (Fe(OH2)(6)(3+)) and dimeric (Fe-2(OH2)(8)(mu-OH)(2)(4+)) iron(III) species form complexes with K-f values of 0.063 and 0.38 M with penicillamine and of 0.025 and 0.07 M with cysteine, respectively. The dependence of the complexation rate on iron(III) and H+ concentrations indicates that the dominant reaction pathways and their rate constants (M(-1) s(-1)) are as follows: (H2O)(5)Fe(OH)(2+) + H(3)L(+), k = 1.4 x 10(3) (cys), 1.2 x 10(3) (pen); (H2O)(5)Fe(OH)(2+) + H(2)L, k = 7.4 x 10(3) (cys), 7.4 x 10(3) (pen); Fe-2(OH2)(8)(mu-OH)(2)(4+) + H(2)L, k = 4.3 x 10(3) (cys), 8.1 x 10(3) (pen). The oxidation of cysteine is second order in cysteine with terms in the rate law first, second, and third order in Fe(OH2)(6)(3+). From the [H+] dependence, these are assigned to reactions of (H2O)(4)FeLH(2+) + H(2)L, (H2O)(4)FeLH(2+) + (H2O)(4)FeLH(2+) or Fe-2(OH2)(6)(mu-OH)(2)(LH)(3+) + H(2)L, and Fe-2(OH2)(6)(mu-OH)(2)(LH)(3+) + (H2O)(4)FeLH(2+). Penicillamine reacts similarly but more slowly, and the first pathway is not observed, but one fourth order in iron(III) is found. Limited studies with excess penicillamine show that the reaction is second order in penicillamine and is inhibited by H+ and iron(II), and the reactive species is suggested to be Fe(LH)(2)(+). Spectrophotometric results under these conditions yield K-f12=[M(LH)(2)][H+](2)/[MLH]-[H-3] = 1.6 x 10(-2) M.