When the free-radical bromination of propane is carried out in the presence of chlorine, the bromination selectivity (ratio of 1-bromopropane to 2-bromopropane) is much lower than it is when chlorine is omitted from the reaction (In other words, chlorine changes the observed ratio of 1-bromopropane to 2-bromopropane). In fact, the bromination selectivity is similar to the chlorination selectivity. Explain. Attempt at solution: I know that 2-bromopropane is the major product in bromination (without the addition of chlorine), because it has the more subsituted carbocation intermediate. I also know that chlorination won't occur because the enthalpy for that reaction is around -10 kcal/mol. However, I'm not entirely sure as to how the chlorine lowers the ratio. I thought it was something along these lines: without the chlorine, 1-bromopropane is the least favorable product; in the presence of chlorine, a chlorinated product becomes the least favorable product, therefore moving the 1-bromopropane up to be the second least favorable product and thereby lowering the ratio. But I don't think this accounts for a difference in the ratios. Can anyone explain, or give me a hint as to why this occurs?