Abstract
Chemiluminescence is observed when alkaline solutions of luminol containing oxygen are subjected to pulses of 10 MeV electrons. It is shown that the first reaction in a series which leads finally to luminescence is that between luminol and the radical OH having k= 8.7 × 109 l. mol–1 s–1. The product of this reaction reacts with oxygen (k= 1.0 × 109 l. mol–1 s–1) to give a species which undergoes a first order change (k ca. 105 s–1) to the intermediate directly responsible for the luminescence. The kinetics of the luminescence are consistent with the conclusion that the emitting species is produced in the bimolecular reaction of this intermediate with itself (k= 3.8 × 108 l. mol–1 s–1). A reaction scheme is proposed which suggests that the aminophthalate excited state which gives rise to the luminescence is produced by the mutual reaction of peroxy free radicals in this last step of the sequence.