I'm going to use the assignment in owl 11.8b to teach you about how alkanes react with chlorine. This reaction is called a chlorination reaction or more general it's called a halogenation reaction. First question reads, when two, three di methyl butane reacts with chlorine, gas, CL2, according to the equation, two possible constitutional isomers may form. Draw structural formulas for the two isomers. Before we start drawing structures, let's just look at the chemical equation and understand what happened at a fundamental level. What happened is that a halogen on the alkanes was substituted with one of the chlorines from the CL2 just through the accounting. On the left we have 14 hydrogens on the right. There are 13 hydrogens on the alkanes after the reaction. The hydrogen was removed from the alkanes and it actually hooked up with the other chlorine from the CL2. The two chlorines from the CL2 that split up between the alkanes and this new compound, HCl. Let me draw the structure of two, three di methyl butane. Here we have the structure of both reactants, two, three di methyl butane and the chlorine gas. Any hydrogen on a carbon will be replaced by chlorine. If we examine the structure carefully, these methyl groups are chemically identical. Any one of these hydrogens can be replaced and that will be one of the structural isomers. These two carbons with a single hydrogen almond are also chemically identical. Either one of these hydrogens can be replaced and from the other structural isomer. Here we have the two possible structures that would form as products. Now I like to point out something that the question doesn't bring up. That is the amount of either one of these constitutional isomers that are formed. An equal amount of each isomer is not formed. There is a preferential isomer that is formed and it's this one here. This bottom isomer is formed preferentially because the chlorine is bonded to a carbon that has many carbons bonded to it as opposed to these methyl groups out here. These methyl groups only have one carbon bonded to each of them. Here's another problem. We predict the number of constitutional isomers that can form when two di methyl butane can react with chlorine according to the given equation. So we understand the fundamental idea of the reaction that a hydrogen would be substituted with a chlorine. But let's see what different hydrogens we have here on this compound. These three methyl groups are bonded to this carbon. So chemically speaking, these three methyl groups are identical. So replacing any one of those hydrogens would form one of the isomers. Well this carbon here doesn't have any hydrogens so we'll exclude that. The CH2 is chemically different than this CH3 and this CH3 is chemically different than these CH3s out here because this CH3 on the end is bonded to a CH2 where these CH3s are all bonded to this one carbon that doesn't have any hydrogens on it. So one of these hydrogens on the CH3, one of these hydrogens on the CH2 and any one of these hydrogens on any one of these carbons will form three isomers. And here the three isomers the chlorine is on them. Now I'll ask you which one do you think would be the major product? It's going to be this one here in the middle because this carbon is bonded to two other carbons whereas this carbon is only bonded to one carbon and this carbon down here is also only bonded to one carbon. All three products would form but the majority the product would be of this structuralisomer here.