so we've seen that alkal halides undergo substitution reactions and remember substitution of course is when you have that leaving group that's been substituted for another on um a alane and so when we are looking at these alkal halides they can undergo substitution but they can also undergo a process called elimination and in an elimination reaction you have the leaving group that has been removed from your reactant along with a hydrogen on an adjacent carbon and so therefore the product of an elimination reaction is going to be an alken and so we have the leaving group that is removed from one carbon and a hydrogen that's removed from an adjacent carbon to form then a double bond between those two carbon atoms and so therefore this base is going to get the hydrogen and of course your leaving group is still going going to be left okay so this is an elimination reaction and we're going to look at the two mechanisms by which an elimination reaction can occur so like we said there's two ways that elimination reaction can occur just like there are two nucleophilic substitution reactions so we know that there's sn1 reactions and sn2 reactions depending on how many molecules were involved in the raate determining step right sn1 reactions formed that carbocation and sn2 reactions had a direct substitution where the nucleophile came in and kicked off the uh leaving group so similarly there's E1 and E2 and so we're first going to look at E2 reactions e of course stands for elimination and two stands for bimolecular so when we think about an E2 reaction the rate of an E2 reaction is going to depend linearly on the concentrations of both the alkal halide and the base so the rate is affected by the alkal halide that's present and the base that's present so both of those two things are involved in the transition state of your rate determining step so you can see here that in this E2 reaction we have have our alkal halide and we have our base and when we have that E2 reaction both molecules are involved in forming then our products and so this is a one-step me mechanism a one-step reaction and so if we look at the mechanism of this reaction on the next slide we can see that it happens all in one step so you can see in the mechanism for an E2 reaction that what's going to happen is that this alkal halide has the leaving group that's eliminated at the same time that the base is going to remove a hydrogen from an adjacent carbon now we're calling it a base rather than a nucleophile because it is removing a hydrogen here right bases accept hydrogens and so that's why we're kind of calling it a base rather than a nucleophile but remember we've already kind of talked about the nuances of what a base is versus what a nucleophile is and so this base that's in solution that has a lone pair that can accept a hydrogen comes in and removes a hydrogen from an adjacent carbon to the one that has the Hy on it and so this proton comes off and now forms a bond with the lone pair of electrons from this base and then the electrons that were between the hydrogen and the adjacent carbon now go to form a double bond between these two carbon atoms and because it forms a double bond between those two carbon atoms this carbon now has then one two three four bonds and if it kept to the bromide it would have five and so because the bromide is the best leaving group out of what's attached to this carbon it's going to be the one that's kicked off it's the easiest to be removed and so as a result we have that alken that's formed along with the base that has gained a proton so water in this case and our leaving group our bromide ion so let's now go over a little bit of terminology that we need to be aware of with respect to these elimination reactions the first additional piece of terminology to be aware of is that the carbon that's directly attached to the hallogen has a special name this carbon is called the alpha carbon okay the alpha carbon because it's where the reaction is kind of happening we've got the bromine attached that's going to be coming off the next thing you need to be aware of is that the carbon that the base is going to take off a hydrogen from is going to be called the beta carbon okay so the beta carbon is a carbon that's directly adjacent to the alpha carbon and it's where the hydrogen is removed from so for example this is the alpha carbon and it's always going to be the alpha carbon because this is always going to be where the brone is attached now there can be two beta carbons at least you could have this one here or if there was another Carbon on the other side and that had the hydrogen removed from it then that could be the beta carbon as well okay so always one alpha carbon but the beta carbon is the one directly adjacent to the alpha carbon but a requirement for it to be a beta carbon is that the hydrogen needs to be removed from it by that particular base coming in Okay so so we can call this reaction not only an elimination reaction but you also might hear it called something like a dehydrohalogenation reaction let's break down that name for a minute dehydro right removing the hydrogen and the hogen dehydrohalogenation so ultimately this name of the reaction along with the idea that it's that it's an elimination reaction tells you exactly what we're removing that we're removing a hydrogen and a hallogen from that alkan um another name you might hear is called a beta elimination reaction or a one two elimination reaction because the atoms that are being removed are on adjacent carbon so that one and two refers to the relative positions not the actual positions or numbering of the molecule but the relative positions on which those atoms are being removed mov D so one carbon and the second carbon relative to each other okay so four total names elimination beta elimination dehydrohalogenation and then one two elimination are all names that refer to these particular types of reactions now the next thing we need to be aware of is the relative reactivities of alkal halides in these elimination reactions so if we think about how easy these elimination reactions are to occur what we can say is that tertiary alal halides are actually going to be more reactive in an E2 reaction than primary alkal halides the reason why is that tertiary alkal halides are going to be the more substituted um alkanes sorry alkenes that form so for example here you have a tertiary alkal halide and from this tertiary alkal halide you have a hydrogen that's eliminated from this beta carbon and what forms then is an alken with three alkal substituents on it notice these three are groups attached and so these tertiary alkal halides that form tertiary alkenes right alkenes with three carbon groups attached are going to be more stable alkenes than for example a secondary alkal halide that has two alkal groups attached and will form an alken with two alkal substituents on it and a primary alkal halide that would form a alken with only one alkal substituent on it and remember the more alkal groups that are attached to those SP2 hybridized carbons the more stable it is so the more stable the alkine the easier it's going to be to form the lower the the activation energy that's going to be required to get that reaction going and so that's why you have tertiary alkal halides being the most reactive in an E2 reaction and primary alkal halides being the least reactive in an E2 reaction because when you have tertiary alkal halides react you're going to form compounds that have more alkal groups attached to it you're going to form alken that are going to be more stable than alkenes that are going to be less stable from say a primary alkal halide so when we think about an E2 reaction an E2 reaction is Regio selective there are going to be more of one constitutional isomer formed than other constitutional isomers and so when you have an alkal halide such as two bromobutane notice that there's going to be two beta carbons there's this beta carbon that could have a hydrogen removed from it and in this beta carbon that could have a hydrogen removed from it and so there's two places where we could potentially remove a hydrogen in this particular case the two beta carbons are not identical right it's going to be easier to remove a hydrogen from one carbon than it is from the other carbon and so when we look at removal of the hydrogen from this particular beta carbon carbon one here we see one buttine that would be formed right because then a double bond would be formed between these two carbon atoms if we see a hydrogen being removed from this guy this beta carbon we'd have a double bond being formed here between carbons 2 and carbons 3 so because we have these two structurally different beta carbons we can have two potential products now the one that's formed in highest yield is this Tu buttine where we remove a hydrogen from this beta carbon this particular E2 reaction is Regio selective because you have more of one constitutional isomer formed than the other so it's Regio selective because you have not 50/50 but more of one than the other and why we have more of two buttine rather than one buttin is because the major product is going to be the most stable alken and so the most stable alken in general what we can say is that it's obtained by removing a hydrogen from the beta carbon that's bound to the fewest hydrogens right because here we're forming an alen that has one two alkal groups attached to it and in this situation we're forming an alken that only has one group attached to it so these reactions are Regio selective which results in the formation of the more stable alen the reason why an E2 reaction forms the more stable alken is because the transition state is somewhat alken likee because this all happens in a concerted mechanism you have bonds breaking and bonds forming at the same time and so the bond that's going to be formed is going to be um that of an alken and thus the more stable Aline which is this guy over here is going to have a lower energy for the transition state and have an easier Hill to climb than the transition state that leads to the less stable alken right so the same factors that stabilize an alken stabilize the transition state as well and as a result the more stable alken is going to be formed faster we've learned that the more stable alen is generally going to be formed from these E2 reactions and this is something called zit sev's rule so Zev rule says that the more stable alken is the one that's going to be formed in the highest concentrations now ziv's rule does have some limitations to it let's say we have an elimination reaction where we have our trary alkal halide right because it's going to form the most stable alken and a base now what we would expect to happen is we'd expect this is the alpha carbon because it has the bromine on it and we'd expect this to be our beta carbon um because when that happens we're going to form a uh tertiary right a carbon an alken excuse me with three alkal groups attached to it right it'd be this guy over here you can see there's one two three alkal groups attached to these carbons of the double bond okay so we'd expect that to be our product because that's the one that's most stable and that's the one that should be formed in higher concentrations however if we look at our base this is T butoxy this is a t butoxy ion there's a lot of lot of atoms around here it's very bulky and so because it is so bulky it's actually really hard for this base to get to this hydrogen because it's kind of in the middle of the molecule so what actually happens here is that this bulky base is going to actually remove a hydrogen from a carbon that's more easily able to get to so here these hydrogens on these two carbons are actually more accessible to it it has to go through less trouble less steric interactions for it to get to these hydrogens less unfavorable interactions right because for it to get to this hydrogen it is going to be an unfavorable interaction and that's going to increase the energy so instead what it does is it goes to the more accessible hydrogens and removes a hydrogen from that so you can see here that if we removed a hydrogen from this carbon we would expect to form this product now this product only has two alkal groups attached to those SP2 carbons so this product is less stable however it is formed in higher percentages higher concentrations than the product that is more stable because of the bulkiness of the alkal halide and the bulkiness of the base okay it's actually too unfavorable not too unfavorable because it still happens in 28% of it but it's it's more highly unfavorable to remove the hydrogen from the carbon that we expect it to be removed from that forms the more stable alkine and instead it removes a hydrogen from the Le U more accessible carbon that is easier to get to because there's less steric interactions less negative charges that are trying to interact with each other less electron clouds that are pushing against each other to to get to that hydrogen so overall you can see you can actually Force the reaction to form more of the unfavorable product if you use a bulkier base so if you're trying to use an elimination reaction to form the product that's less stable instead of using a small base you might want to use a base that's bigger and bulkier in order to accomplish that so what we've said is that the steric properties of the of the base do affect what types of products we get now we have this reaction here and you can see this reaction is an elimination reaction and what we can say is that for this compound we could remove a hydrogen from either one of any one of these three beta carbons right this guy this guy or this guy and here we would have a alken as a product if we removed a hydrogen from here that would have four alkal groups attached right this is our SP2 hybridized carbon then this is our SP2 hybridized carbon and we'd have 1 2 3 4 alkal groups we'd form a more stable product because we have a greater number of alkal groups attached to that to that alken or we could remove a hydrogen from either of these two beta carbons and you can see here that if we remove a hydrogen from either of these two beta carbons we'd form an alken that only has one and two um alkal groups attached to it so we'd form the less stable alken here because it only has one two rather than four 1 2 3 four alkal groups attached and so what we're considering here is the base and so for example you see different bases and the relative percentages of the more stable Al and the less stable alen that you would form so here we have the ethoxide ion and the ethoxide ion isn't too bulky it's a little bulky but it's not too bad so what we would expect is it's fairly easy to get to this hydrogen for that ion and it was going to form more of the more stable alken and less of the less stable alken and so we have 79% and 21% relatively as we increase in the bulkiness of the base you can see these percentages decreasing right up until the point where we get to this guy here um we don't have a name for it but well we don't have a name that we know for a common name but you can see it's a lot more bulkier than what we had for the ethoxide ion and this guy is way too bulky to pull off this hydrogen and so what happens is that we only have 8% of our product being the more stable Al because it is so bulky it's so sterically hindered that it can't pull off the hydrogen to create the more stable alen and as a result we have 92% of the less stable alken that's going to be formed and if we want the less stable alken that's the way to go if we don't want the less stable alken then we'd want to use the ALCO oxide ion okay but for these elimination reactions there's ways to force the reaction or at least for these E2 reactions there's way ways to force the product to form more of the less stable alken just by using a bulkier base so we've got these three problems to work through in class dealing with these E2 elimination reactions and let me know if you have any questions