Bromate (BrO3–) in drinking water is traditionally seen as an ozonation byproduct from the oxidation of bromide (Br–), and its formation during chlorination is usually not significant. This study shows enhanced bromate formation during chlorination of bromide-containing waters in the presence of cupric oxide (CuO). CuO was effective to catalyze hypochlorous acid (HOCl) or hypobromous acid (HOBr) decay (e.g., at least 104 times enhancement for HOBr at pH 8.6 by 0.2 g L–1 CuO). Significant halate concentrations were formed from a CuO-catalyzed hypohalite disproportionation pathway. For example, the chlorate concentration was 2.7 ± 0.2 μM (225.5 ± 16.7 μg L–1) after 90 min for HOCl (Co = 37 μM, 2.6 mg L–1 Cl2) in the presence of 0.2 g L–1 CuO at pH 7.6, and the bromate concentration was 6.6 ± 0.5 μM (844.8 ± 64 μg L–1) after 180 min for HOBr (Co = 35 μM) in the presence of 0.2 g L–1 CuO at pH 8.6. The maximum halate formation was at pHs 7.6 and 8.6 for HOCl or HOBr, respectively, which are close to their corresponding pKa values. In a HOCl–Br––CuO system, BrO3– formation increases with increasing CuO doses and initial HOCl and Br– concentrations. A molar conversion (Br– to BrO3–) of up to (90 ± 1)% could be achieved in the HOCl–Br––CuO system because of recycling of Br– to HOBr by HOCl, whereas the maximum BrO3– yield in HOBr–CuO is only 26%. Bromate formation is initiated by the formation of a complex between CuO and HOBr/OBr–, which then reacts with HOBr to generate bromite. Bromite is further oxidized to BrO3– by a second CuO-catalyzed process. These novel findings may have implications for bromate formation during chlorination of bromide-containing drinking waters in copper pipes.