understood what elimination is especially from compounds that have a double bond within them but we haven't really talked too much about elimination from cyclic compounds from things like six membered rings and so we're going to touch on what elimination looks like from these ring structures and so ultimately what you're going to see here is that when we do elimination from these six membered rings especially if E2 elimination here what we see is the both of the groups that are going to be eliminated need to be in axial positions the reason why it needs to be in axial positions is because in an E2 elimination reaction we generally have anti- elimination that's going to happen meaning that the halide and the hydrogen are going to be removed from opposite sides of that double bond and so for these two groups that are going to be removed to be parallel and able to be removed they have to be on opposite sides of the double bond and in axial positions if they're in Equatorial positions they're not parallel right they're not able to be removed actually what we see here is if we look at one chlorocyclohexane what we see is that the more stable conformer actually doesn't undergo an elimination reaction this is because the chloride is going to be in the equatorial position and in order for it to be eliminated it needs to be in the axial position so therefore the type of conformer that undergoes these E2 reactions is going to be the conformer of cyclohexane of one Cy of one chlorocyclohexane that's going to be the less stable of these ring structures the one where chlorine instead of being in the equatorial position is in the axial position because when it's in the axial position you have both the chlorine and hydrogen that are going to be on opposite sides of that uh cyclohexane and can be removed but and form a double bond from that so what does that mean for the rate of the reaction how fast is this going is this reaction going to occur and to answer that question we do have to consider the value of the equilibrium constant right K equilibrium so that value of the equilibrium constant depends on whether the reaction is going to take place through the more stable conformer or the less stable conformer K equilibrium is the extent of a reaction the more products there are versus reactant the more the higher the K equilibrium value is going to be right so if you have a large amount of the more stable conformer in the products and less of the less stable conformer in the reactants what you're going to get is a large value for K equilibrium and if you have the less stable conformer being in the products you're going to have the products here being a small number and the reactants being a large number and so the K equilibrium is going to be small so ultimately what we're saying here is we can consider the value of K equilibrium to consider whether the reaction is going to take place very quickly or it's going to take place very slowly so what we can say is that for a reaction that takes place to a more stable conformer the equilibrium is going to be large because we have more products that are able to participate in that reaction relative to the amount of reactants that are not the more stable conformer versus the less stable conformer versus a reaction that's going to have a small equilibrium constant and takes place through the um less stable conformer because we'll have a smaller amount of those relative to the equilibrium constant so let's look at an example of what we talked about on the previous slide so here we have uh a compound and this compound is called neome methyl chloride now notice that the more stable confirmation of this compound is going to have the methyl and the ethyl in these equatorial positions and the chlorine as a result then is going to be in the axial positions right because we're putting the bulkier substituents away from the bulk of the plane so chlorine then is therefore going to be in an axial position as a result the hydrogen here is going to be in also an axial position on the opposite side of that ring so the more stable conformer here is the one that has the hydrogen and the chlorine in axial positions on opposite sides of the ring so this reaction then is going to happen fairly quickly because elimination is going to occur through the more stable conformer we're going to have more of this product anyway and so therefore since it's already in this confirmation it's going to happen fairly quickly the left stable conformer the one where the methyl group is in the axial and ethyl is in the a or not ethyl um isopropyl is in the axial positions it's not as stable and so we don't have the chlorine being in the equatorial position and remember when the chlorine is in the equatorial position here it's not going to occur very quickly for an elimination reaction so the point of this slide is showing you here that we exist in a confirmation the more stable confirmation is the one where the chlorine and the hydrogen are in axial positions and E2 elimination can recur very quickly versus a situation where the reaction is slow so here we have menthal chloride again you can see these different positions but now chlorine is going to be in an equatorial position along with the methyl and the isopropyl group and so what we see here for this particular compound is this is the more stable confirmation but it's not the confirmation that can undergo these E2 reactions and so the less stable confirmation is the one that has those all of those groups now in the axial position positions so we need the chlorine in an axial position to undergo elimination but as a result of the chlorine being in the axial position you also have the methyl and isopropyl in axial positions and as a result this is the less stable conformer for that particular compound so this reaction is going to occur very slowly because we have much fewer molecules in the less stable conformer that are available to react with our nucleophile and so this particular E2 reaction is going to be very slow compared to the one we just looked at on the previous slide now what about E1 elimination right we've considered E2 elimination from six membered rings but what about E1 because the reaction doesn't happen at the same time it's not concerted the hydrogen and chlorine actually aren't limited to axial positions so for example here we see the chlorine in an equatorial position but because we're forming a carbocation in an E1 reaction it doesn't matter and so once we form this carbocation then the base can remove a hydrogen from an adjacent carbon forming the elimination product so this reaction is not a concerted reaction and therefore the reaction is not limited to the chlorine being in an axial position for it to be removed and undergo elimination reactions so we're now going to look at the competition between E2 and E1 reac reactions so thinking back to the previous lecture we've seen that both the tertiary alkal halides can undergo both E2 and E1 elimination reactions so what's going to happen is tertiary alkal halides going to undergo E2 or E1 reactions now primary and secondary alcoholes are only going to be able to undergo E2 reactions because they don't form these stable carbocations but for tertiary because they're able to undergo both E2 and E1 reactions we have to think about the rates of these reactions is it going to undergo a mechanism via E2 mechanisms or E1 mechanisms and so what we can say is that the rate of the reaction of tertiary alkal halides when they're undergoing elimination is going to depend on the rate that an E2 reaction occurs and the rate at which an E1 reaction occurs and so we can combine these rates to form the actual rate of a reaction considering both E2 and E1 reactions so what you notice here hopefully is that this part of the rate law is going to be because of the E2 reaction and it's because it's a bimolecular reaction where it involves two molecules it's going to involve both the alkal halide and the base so the nucleophile is going to be important here when we undergo an E2 reaction but because we're forming a carbocation for E1 reactions the actual base doesn't matter for E1 reactions and so it's only the concentration of the alkal halide that affects the rate in an E1 reaction now remember these elimination reactions are going to produce the same product you can start with the same thing these tertiary aloh halides and produce the same product the more stable alen the only difference is the mechanism by which these reactions are going to be completed so ultimately what you should be able to see from this rate law is that for tertiary alkal halides those where the nucleophile is going to be important where you're going to have a strong base that's going to be present you're going to favor a mechanism via an E2 reaction where the base is weak it doesn't really matter you're going to favor more of an E1 reaction so to summarize that for tertiary alkal halides that can undergo both E2 and E1 reactions to form the same product what we can say is that high concentrations of a strong base ones where the nucleophile is going to be important in the rate law is going to cause the rate to recur faster for an E2 reaction but Reactions where the base is weak or it's not very strong or doesn't have a very high tendency to grab a hydrogen ion it's going to occur via an E1 reaction more of an E1 reaction because that nucleophile isn't going to be as a significant part of that particular rate law so if you need to undergo an E2 reaction for a tertiary aloh alide you'd want to use a high concentration of a strong base and if you'd want to undergo an E1 reaction for a tertiary alkal halide that would be favored by a low concentration of a weak base so the next thing we need to consider is well all four reactions together you guys are smart enough to realize that when we have a tertiary alkalide well we talked about that it could undergo sn1 reactions but we also just talked about that it could undergo sn2 or not sorry not sn2 excuse me E2 or E1 reactions right we need to say well when we have an alkal halide what type of reaction is it going to undergo they can undergo sn1 sn2 E1 or E2 reactions um and so before we can predict the products of a reaction we have to decide whether or which type of reaction that it's going to undergo now conditions that favor sn2 reactions also favor E2 reactions and conditions that favor sn1 reactions are also going to favor E1 reactions and so our choices are between whether a reaction is going to undergo sn2 or E2 or sn1 or E1 reactions you don't have sn1 being combined with E2 or sn2 being combined with E1 those combinations aren't possible so a reaction is either going to undergo sn2 or E2 or SN 1 or E1 okay so to decide whether a reaction undergos sn2 E2 or sn1 E1 the first thing we need to consider is the structure of the alkal halide so primary and secondary alkalides of course are going to undergo sn2 and E2 reactions you don't have an alkal halide that's able to undergo sn1 or E1 reaction ctions because we'd have to form a a carbocation and primary and secondary carbocations aren't very stable and because they're not very stable they're not formed readily and so if you have a primary or secondary alkoh halide the only choices that it has are sn2 or E2 um reactions a tertiary alkalide on the other hand can't under undergo sn2 reactions but a tertiary alkal halide could undergo sn1 E1 or E2 reactions right so the first thing we need to look at when we're trying to decide what products a reaction forms is what type of alkal halide do we have Okay so let's think about that that's the first thing we need to consider and then once we've identified what type of alkal halide we have if it's tertiary we need to decide whether the conditions are going to favor E2 or if the conditions are going to favor sn1 um or E1 reactions so for tertiary alkal halides if you have a weak base it's going to favor those sn1 and E1 reactions if you have a strong base then it's going to favor those E2 reactions okay so let's look at the next slide now that we've got kind of a background of what types of reactions we're going to be looking at for what alkal halides let's look at the kind of situations that we're going to see and what potential product can be formed for these for these substances so the first thing we're going to look at are conditions that have strong bases conditions that favor sn2 and E2 reactions and so let's think about these sn2 and E2 reactions okay so first off like we said we have a strong base we have hydroxide for example and that strong base favors these two conditions sn2 E2 and if we think about these two different types of reactions we know that in an sn2 reaction primary alkal halides are going to react the fastest and tertiary alkal halides are going to react the slowest whereas in an E2 reaction tertiary alkalides are going to react the fastest and primary alkal halides are going to react the slowest so we've seen in an sn2 and E2 reaction right we've seen that a hydroxide ion can attack your nucleophile kick off the leing group and form the substitution product or we've also seen the elimination product we've seen the base the strong base removing a hydrogen from the beta carbon to form the elimination product the alken and so what we're saying is that these two react reactions can take place at the same time those two reactions compete with each other because both of them take place for the same reason right you have the hallogen that accepts those electrons and leaves and a base from solution comes in and replaces that if it replaces the leaving group it's a substitution product if it pulls something else away it's that elimination product okay so so we can't take out both of these reactions they're going to happen under the same condition but what you're going to see is that in certain situations one of these reactions is going to be favored okay and it's probably going to be related to what we see down here of course so let's think about primary alkal halides under these strong base sn2 E2 conditions if you have a primary alkal halide you're going to have primarily substitution occurring so because a primary alkal halide is the most reactive in an sn2 E2 condition oh sorry I said that wrong because a primary alkal halide is the most reactive in an sn2 reaction and the least reactive in an E2 reaction the primary alkalide are going to form primarily substitution products so in other words we can say that the substitution product is going to win the competition that makes sense because you have a primary alkalide and substitution is favored for primary alkalides because the rate of the reaction happens faster for primary alkalides versus the elimination product where it's actually tertiary alkalides that are going to react the fastest there and primary alkal halides that are going to react the slowest so if you have a primary alkal halide you'll have primarily substitution let's think about secondary alkalides in substitution our order was primary followed by secondary and then tertiary in terms of how fast and how easy those reactions would occur under sn2 E2 conditions and for elimination it was the opposite tertiary alkal halides were the fastest followed by secondary alkal halides which were in the middle and primary alkal halides that were the slowest in an elimination reaction so secondary alkal halides fall in the middle for both of these situations right they are equally favored for substitution and elimination reactions so what's happening are we going to get like 50/50 50% of the substitution product and 50% of the elimination product not exactly here what it depends on is the base okay your nucleophile so if you have a weak base and a secondary alkal halide you're going to have substitution being favored if you have a strong base elimination is going to be favored so for example the Seth oxide ion is a fairly strong base and so we see the secondary alkalide of two chlorop propane forming mostly the elimination product okay because we've got a strong base present so we've got about 75% of the elimination product and about 25% of the substitution product because we have a stronger base we can also see this on the next slide as well so under these conditions we have a weak base here like the acetate ion and a strong base here like the ethoxide ion the ethoxide ion is a strong base and like we saw in the previous slide forms much more of the elimination product rather than a substitution product whereas a weak base forms actually none of the elimination product and 100% of the substitution product so even though secondary alkal halides under these sn2 E2 conditions can undergo both substitution and elimination there's various things that we can do to kind of push the reaction more towards the direction that we want so like we said it depends on the strength of the base and there's one other thing that it depends on it also depends on how bulky the nucleophile SL base is the stronger and bulkier the nucleophile the greater the amount of the elimination product that makes sense because a really bulky base is really hard to get in here for the substitution product to be formed and so if you have something like T butoxide it's really strong and there's a lot of methyl groups around that nucleophile and so that's going to have a hard time getting in to form a substitution product so therefore we can form much more of the mination product with these bulkier bases so if you're um trying to get a substitution product use a weaker base that's less stly hindered if you're trying to get the elimination product use a stronger base that's a lot more sterically hindered okay so um we see like I said two different factors that affect whether for a secondary alcohal substitution or elimination is going to be formed the strength of the base and the bulkiness of the base matter when we're trying to push a secondary alkoh halide towards either substitution or elimination Okay so we've looked at primary alkal halides and secondary alkal halides under sn2 E2 condition now let's look at what ter happens to tertiary alkalides under sn2 E2 conditions so tertiary alkal halides are only going to undergo elimination well that's just because we said already when we learned about substitution reactions that tertiary alkal halides can't undergo sn2 reactions they only undergo elimination reactions um when the when the um when the alkal halide excuse me the tertiary alkalide reacts with a strong base and again that's because substitution can't happen here those sn2 reactions can't happen where the nucleophile comes in for like a backside attack right because the backside attack is too sterically hindered for a tertiary alkalide so the only option with these strong bases then is to undergo E2 reactions and have the leaving group leave and a hydrogen removed by this strong base at the same time okay so primary alcohal primarily substitution secondary alkal halide both we can kind of get one versus the other to be favored depending on the strength of the base and how bulky it is and tertiary alkalide only only E2 because it doesn't undergo these sn2 reactions Okay so so these are for sn2 E2 conditions these strong base conditions now let's think about sn1 E1 conditions so under a weak base sn11 conditions we only have tertiary alkal halides that are able to undergo these types of reactions and remember tertiary alkal halides can only undergo sn1 and E1 reactions because they have a they have to form a carbocation and only tertiary alkalides can form a stable carbocation and so we see these sn1 E1 reactions happening when we have a weak base and so poor nucleophiles or weak bases favor these sn1 E1 conditions now let's think about what happens with an alkal halide we dissociate to form a carbocation and then once we form the carbocation we can either add our nucleophile to form the substitution product or we can take away an adjacent hydrogen to form the elimination product okay so we either react to form substitution or lose a proton to form elimination now what we're going to see here is that generally the substitution product is going to be the one that's favored here and the substitution product is going to be the one that's favored because think about how easy it is to do you've got something that's willing to accept that positive charge or you've got a pair of electrons that they're willing to share with the substance that has a positive charge but for elimination you've got to remove that hydrogen right you've got to remove the hydrogen form a bond and that's going to take more energy to do so what we see and we'll go on to the next slide here is that under sn1 and E1 conditions you have mostly substitution occurring you have very little elimination product forms because for the substitution product you don't have to break any more bonds you just got to take one put it with the other right it's a lot more easy to do a substitution reaction here than it is to do an elimination reaction now this is good this is a good thing because if they were to able to undergo E1 reactions here rather than sn1 we'd have no way to do substitution right we would have no way to do substitution because we know that tertiary alkalides don't undergo sn2 reactions so the only way they can form a substitution product is to undergo sn1 reactions but luckily tertiary alkal halides have another way to form the elimination product in fact if you want the elimination product it's better to use a strong base and undergo an e an E2 reaction and so when you're trying to decide with a tertiary alkalide whether it's going to undergo substitution or elimination if it's a weak base it's under the sn1 E1 conditions and we favor substitution because there's no need to break any more bonds to form the substitution product if it's under a strong base conditions it's going to form the elimination product but by an E2 reaction okay because when a strong basee is present you have E2 favored over sn2 because tertiary aloh alides don't undergo these sn2 reactions so that's a lot right let's summarize this on the next slide and put it into a into a chart it's a lot easier to understand when we put it in this chart format okay so let's summarize that chart we have primary aloh alides when we have strong base conditions strong nucleophile primary alkalides will undergo primarily substitution if we have a weak base you cannot have sn1 or E1 reactions happening for a primary alkal halide the reason being that they can't form a stable carbocation which is one of the the first step of an sn1 or E1 reaction okay so if you have a primary alkalide it's going to undergo substitution under strong base conditions if you have a secondary alkal halide under strong base conditions it's going to undergo both substitution and elimination weaker bases favor substitution and bulkier bases favor elimination so there's different things that we can do to kind of push the reaction one way or the other depending on what potential product we want okay so those are sn2 E2 conditions under weak based conditions sn1 versus E1 secondary alkal halides cannot undergo these types of reactions because we can't form stable carbocations right so that's the idea here if it's primary or secondary you have to have these sn2 E2 conditions if it's primarily primary you get primarily substitution if it's secondary you get both now tertiary is the only one where we can have these sn2 E2 conditions or sn1 E1 conditions and so tertiary aloh halides under the strong base conditions only undergo elimination because they can't undergo substitution under weak base conditions it can go undergo both substitution and elimination but substitution is favored because that base isn't very strong it can't easily remove a proton whereas once the carbocation has dissociated right once the alkalide has dissociated into a carbocation all that weak base has to do to form the substitution product is just come in and bind so we can have elimination products for tertiary alkalides and we can have substitution products for tertiary alkal halides so I hope this chart was helpful and it makes sense summarizing now what we know about the mechanisms of a reaction and just putting them all together right putting together what types of reactions Prim alkal halides can undergo and what conditions we need secondary and tertiary tertiary is going to be the trickiest one here so we'll work on a couple more homework problems in class and I'll try to have a worksheet for you I found a couple good ones online that deal with predicting which type of reaction is going to occur whether it's going to be sn2 E2 and then deciding between those or sn1 E1 conditions and then deciding um if the tertiary alal halide then will undergo substitution or elimination