 Let's look at another important type of reactions that haloalkanes undergo which is the elimination reactions. Now in order to undergo these elimination reactions, our haloalkanes must have a beta-hydrogen that is a hydrogen atom attached to the beta-carbon atom. Now alpha-carbon atom is the carbon to which halogen is directly attached and the carbon adjacent to alpha-carbon is the beta-carbon. Only when our haloalkanes have a beta-hydrogen can they undergo an elimination reaction. Now this is what a typical elimination reaction looks like. So when we heat our haloalkane in the presence of a strong base like an alcoholic KOH, then the base abstracts a hydrogen atom from the beta-carbon and produces an alkane as shown here. Because the base abstracts a beta-hydrogen, the elimination reaction is also called a beta-elimination reaction. Now another name of this reaction is also dehydrohalogenation. From the name itself you can see that we are eliminating a hydrogen atom as well as a halogen atom and that is why this elimination reaction is also called a dehydrohalogenation reaction. So now that you understood the framework of this reaction, let's quickly look at an example. Now here we have the alpha-carbon here where the carbon atom is directly attached to a bromine atom and we have two different beta-carbon atoms here, correct? In one beta-carbon we have a methyl group and two hydrogen atoms attached and in the other beta-carbon we have all three hydrogen atoms attached to it. Now when we heat this haloalkane in alcoholic KOH solution, the OH- can abstract a hydrogen atom from this beta-carbon in which case you will get this particular alkane. Now the double bond would be formed between alpha-carbon and this beta-carbon. On the other hand if our OH- abstracts this particular beta-hydrogen, in that case the double bond would be formed here. So in this case we get this particular alkane. Now the question is which among them would be formed in larger amounts or which among them is the major product formed? Well the major product would be formed based on Zaitsev rule. Zaitsev rule states that in dehydrohalogenation reaction the major product formed would be that one which has more number of alkyl groups attached to the double bond. In other words the more substituted alkane will be the major product. Among these two which is the more substituted alkane, yes it is this one and therefore this alkane would be a major product and would be formed in larger amounts. So based on this a tetra substituted alkane would be the most stable and therefore would be the most predominant one followed by a trisubstituted alkane followed by a disubstituted alkane which is followed by a monosubstituted alkane. Now just like substitution reactions elimination reactions can also occur by two different mechanisms even and E2 mechanisms. Now even can be compared to our SN1 reaction in the sense that both are unimolecular reactions. So even here stands for elimination unimolecular which means the rate of the reaction depends only on the substrate or our haloalkane. E2 on the other hand stands for elimination bimolecular and similar to SN2 reaction the rate of the reaction depends on both the substrate as well as the base. Now just like SN1 reaction the first step in an even reaction is the formation of a carbocation. So here a halogen atom leaves in order to form a carbocation and because a reactive carbocation intermediate is formed this is the rate limiting step and the slowest step in the reaction. In the second step deprotonation takes place such that the base abstracts the beta hydrogen and produces the alkene and this step is a fast one. Now just like SN2 reaction the E2 reaction is a concerted one in which the bonds break and new bonds form at the same time in the transition state. So this entire thing happens in one single step. Now depending on the reaction conditions and reagents that is involved the elimination can be E1 or E2. For example in the presence of a strong base the haloalkane is likely to undergo an E2 elimination reaction whereas in the presence of a weak base the reaction would follow an even mechanism. Now another important thing that you need to remember here is that an even elimination reaction will always compete with an SN1 substitution reaction because you see in both these cases the first step is the formation of carbocation right. So a nucleophile can attack this carbocation to give a substituted product in which case you will get an SN1 product or if you have a weak base it can also deprotonate and result in the formation of an alkene in which case we get an elimination or even product. And therefore we need to tweak the reagents and the reaction conditions in order to favor one product over the other. So let's learn more about even and E2 mechanisms and what are the different conditions which favor one type of reaction over the other in subsequent videos.