At pH 4.0, the radical ˙CH(OH)SO3− formed by reaction of ˙OH with hydroxymethanesulfonate (HMS) reacts rapidly with oxygen to form the peroxyl radical, ˙OOCH(OH)SO3−, with k4 = (2.6 ± 0.3) × 109 d mol−1 s−1. This radical decays by a first-order process with k = 2.4 × 104 s−1 and Ea = (29 ± 1.2) kJ mol−1. The decay is by a split pathway, reactions (6) and (7), with k6 = 1.7 × 104 s−1 and k7 = 7.0×103
在pH4.0,自由基CH(OH)SO 3 -被OH的反应与羟
甲基磺酸盐形成的(
HMS)迅速与
氧气以形成
过氧化氢自由基,OOCH(OH)反应SO 3 - ,其中ķ 4 =(2.6 ±0.3)×10 9 d mol -1 s -1。该自由基通过k = 2.4×10 4 s -1和E a =(29±1.2)kJ mol -1的一阶过程衰减。衰减是通过分裂路径,反应(6)和(7)进行的,其中k 6 = 1.7×10 4 s -1和k 7= 7.0×10 3 s -1。反应(6)涉及直接消除HO 2。反应(7)最初形成
甲酰基过氧自由基,但是在反应(8)中将其
水解为
甲酸和HO 2,其中k 8 = 2.46×10 3 s -1。CH(OH)SO 3 - + O 2 →OOCH(OH)SO 3 -(4)OOCH(OH)SO 3 - →HO 2 ˙+ CHOSO 3 -(6)˙OOCH(OH)SO 3 − →˙OOCHO+