 What about hydrohalogenation? This is the addition of a hydrogen halide across a double bond. So if we continue our sequence, this time, instead of adding hydrogen or a halogen, we're going to add one of each. Let's call it hydrogen bromide. In this case, the hydrogen bromide is going to add, as usual, across the double bond, and so we're going to have a single bond with our original hydrogens. But this time, one hydrogen and one bromine. So the product this time will become bromo ethane. Because there's only one bromine, it has to be on the first carbon, and therefore we don't need the number, we just call it bromo ethane. You can see this time because it's a hydrogen halide, so we could also have added hydrogen chloride, hydrogen iodide, or hydrogen fluoride, and the same thing would have happened. We would have had an addition across the double bond with the hydrogen going on to one atom and the halogen going on to the other. Now this has been quite easy in a sense when we've had equal numbers of hydrogens on either side, but there's a further complication that we can find when we have an unequal number of hydrogens on the two carbons where the double bond is sitting between them. So here's an example underneath. This would be propene. Propene. Okay, don't need a number because it's got to be on the number one carbon. Now what would happen if, for example, I added hydrogen iodide to this? Now if I added hydrogen iodide to my propene, then I can have the hydrogen adding, if I just switch the colorator, so you can see. I can either have the double bond disappearing and a hydrogen coming here and an iodine coming here, or it could be the other way around, iodine here and a hydrogen here. Now in our previous examples, the molecules were the same, whichever way I had drawn the two hydrogens, the two halogens or even the hydrogen halide wouldn't have mattered because there were the same number of hydrogens on each of the two original carbons, but in this case there isn't. There are two hydrogens already here on the terminal carbon, but only one on that middle carbon. So then we need to add an extra little rule here, and that extra little rule is called Markov-Nikov's rule, and Markov-Nikov's rule is basically when you're adding atoms in, they will favor the side where there's already a higher number of them. So if I have hydrogen iodide being added in here, my most likely product is going to be carbon, carbon, carbon. Now of course I have my single bonds. My hydrogens are already, are not going to change from where they were, so these are my original ones, but on the end carbon I have two hydrogens, and on the central carbon I have only one, so therefore I'm going to put my iodine off the central carbon and my hydrogen off the end carbon. So when I name the product here, it would be two iodo propane. Of course now the double bonds gone, so it's now gone from being an alkane to an alkane, and the iodine has gone onto the central carbon. Now if it went onto an end carbon we would simply call that one iodo propane, but we just need to remember that Markov-Nikov's rule is going to tell us a little something about the preferred atom to which an addition reaction might occur. This is also an important rule to look at for some later substitution reactions, and in fact it's more common when we look at substitutions, but it's worth just flagging now just so you know that where you've got a couple of isam as possible, often one of them is more likely than another.