Nicoson Jeffrey S, Perrone Thomas F, Huff Hartz Kara E, Wang Lu, Margerum Dale W
Department of Chemistry, Purdue University, West Lafayette, IN 47907-2084, USA.
Inorg Chem. 2003 Sep 22;42(19):5818-24. doi: 10.1021/ic0301223.
The reaction between BrO2(-) and excess HOCl (p[H+] 6-7, 25.0 degrees C) proceeds through several pathways. The primary path is a multistep oxidation of HOCl by BrO(2)(-) to form ClO(3)(-) and HOBr (85% of the initial 0.15 mM BrO(2)(-)). Another pathway produces ClO(2) and HOBr (8%), and a third pathway produces BrO(3)(-) and Cl(-) (7%). With excess HOCl concentrations, Cl(2)O also is a reactive species. In the proposed mechanism, HOCl and Cl(2)O react with BrO(2)(-) to form steady-state species, HOClOBrO(-) and ClOClOBrO(-). Acid facilitates the conversion of HOClOBrO(-) and ClOClOBrO(-) to HOBrOClO(-). These reactions require a chainlike connectivity of the intermediates with alternating halogen-oxygen bonding (i.e. HOBrOClO(-)) as opposed to Y-shaped intermediates with a direct halogen-halogen bond (i.e. HOBrCl(O)O(-)). The HOBrOClO(-) species dissociates into HOBr and ClO(2)(-) or reacts with general acids to form BrOClO. The distribution of products suggests that BrOClO exists as a BrOClO.HOCl adduct in the presence of excess HOCl. The primary products, ClO(3)(-) and HOBr, are formed from the hydrolysis of BrOClO.HOCl. A minor hydrolysis path for BrOClO.HOCl gives BrO(3)(-) and Cl(-). An induction period in the formation of ClO(2) is observed due to the buildup of ClO(2)(-), which reacts with BrOClO.HOCl to give 2 ClO(2) and Br(-). Second-order rate constants for the reactions of HOCl and Cl(2)O with BrO(2)(-) are k(1)(HOCl) = 1.6 x 10(2) M(-1) s(-1) and k(1)(Cl)()2(O) = 1.8 x 10(5) M(-)(1) s(-)(1). When Cl(-) is added in large excess, a Cl(2) pathway exists in competition with the HOCl and Cl(2)O pathways for the loss of BrO(2)(-). The proposed Cl(2) pathway proceeds by Cl(+) transfer to form a steady-state ClOBrO species with a rate constant of k(1)(Cl2) = 8.7 x 10(5) M(-1) s(-1).
溴酸根离子（BrO₂⁻）与过量次氯酸（HOCl，pH值为6 - 7，25.0℃）之间的反应通过多种途径进行。主要途径是BrO₂⁻对HOCl进行多步氧化，生成氯酸根离子（ClO₃⁻）和次溴酸（HOBr，占初始0.15 mM BrO₂⁻的85%）。另一条途径生成二氧化氯（ClO₂）和HOBr（8%），第三条途径生成溴酸根离子（BrO₃⁻）和氯离子（Cl⁻）（7%）。在次氯酸浓度过量的情况下，二氯一氧化二氯（Cl₂O）也是一种活性物种。在所提出的反应机理中，HOCl和Cl₂O与BrO₂⁻反应形成稳态物种，即次氯酸次溴酸根离子（HOClOBrO⁻）和二氯一氧化二氯次溴酸根离子（ClOClOBrO⁻）。酸促进HOClOBrO⁻和ClOClOBrO⁻转化为次溴酸次氯酸根离子（HOBrOClO⁻）。这些反应需要中间体具有链状连接性，其中卤素 - 氧键交替出现（即HOBrOClO⁻），而不是具有直接卤素 - 卤素键的Y形中间体（即HOBrCl(O)O⁻）。HOBrOClO⁻物种分解为HOBr和氯酸根离子（ClO₂⁻），或者与一般酸反应形成溴氯酸根（BrOClO）。产物分布表明，在过量HOCl存在的情况下，BrOClO以BrOClO·HOCl加合物的形式存在。主要产物ClO₃⁻和HOBr由BrOClO·HOCl水解形成。BrOClO·HOCl的一条次要水解途径生成BrO₃⁻和Cl⁻。由于氯酸根离子（ClO₂⁻）的积累，观察到二氧化氯形成过程中有一个诱导期。ClO₂⁻与BrOClO·HOCl反应生成2个ClO₂和溴离子（Br⁻）。HOCl和Cl₂O与BrO₂⁻反应的二级速率常数分别为k₁(HOCl) = 1.6×10² M⁻¹ s⁻¹和k₁(Cl₂O) = 1.8×10⁵ M⁻¹ s⁻¹。当大量过量添加Cl⁻时，存在一条与HOCl和Cl₂O途径竞争消耗BrO₂⁻的Cl₂途径。所提出的Cl₂途径通过Cl⁺转移进行，形成稳态的氯溴酸根（ClOBrO）物种，速率常数为k₁(Cl₂) = 8.7×10⁵ M⁻¹ s⁻¹。
