e2 reactions that'll be the topic of this lesson now in the last lesson we kind of had an introduction to elimination reactions and in this we're going to learn the first of two of the mechanisms that elimination occurs we call it e2 elimination and we'll go through that mechanism we'll go through the rate law and then we'll look at some characteristics of these two reactions we'll find out that they typically form what is known as the zaitsev product so but we'll also learn how to predict when the exception occurs when it forms what's known as either the anti-zaitsev or hofmann product instead now if this is your first time joining me my name is chad and welcome to chad's prep where my goal is to make science both understandable and even enjoyable now this is my brand new organic chemistry playlist i'm releasing these lessons weekly throughout the 2020-21 school year so if you don't want to miss one subscribe to the channel click the bell notifications you'll be notified every time i post a new lesson so we'll start with talking about e2 reactions and in this case we call this elimination bimolecular so we call it bimolecular because there's going to be two reactant molecules involved in the rate trimming step just like there was in sn2 so and in this case contrast with sn2 instead of having a nucleophile and a substrate we're going to have a base and the substrate and again find your leaving group and you've found your substrate so in this case we've got our substrate it's our alkyl halide so and sodium hydroxide you should definitely recall from gen chem as being a strong base so and just like sn2 and typically involved a strong nucleophile e2 is going to involve a strong base so definitely sodium hydroxide qualifies here in this case we definitely have an ionic compound here so what you'll find is that there are several strong bases out there but most of the strong bases we'll see in this chapter are going to have a negative charge on oxygen again there are other strong bases that exist but most the ones you'll talk about in the context of elimination reactions that are strong will have a negative on an oxygen all right way this is going to work is we're going to have sodium hydroxide come over here and he's going to deprotonate a beta hydrogen so that's going to free up these electrons hydrogen can only have one bond so if we're making a new one here the old one breaks and that's where the electrons come from to make the pi bond and so that we don't violate the octet rule this bond to bromine has to simultaneously break so we are doing a proton transfer here deprotonating we are forming a pi bond and we have the leaving group leaving and these all happen simultaneous simultaneously so when happening simultaneously just like in sn2 we call this a concerted mechanism cool so all happens in a single step just like sn2 typically happens in a single step and we can take a look at our rate law here so we see that the base is involved in this step so and we see that our substrate is involved in this step as well and so in this case first order with respect to the base first order with respect to the substrate second order overall adding those two individual reactant orders together so if you double the concentration of the base you'll double the rate of the reaction if you double the concentration of the substrate you will double the rate of the reaction as well so the rate is proportional to each of them independently here so uh we take a look and if we kind of take a look at our specific reaction so the base is sodium hydroxide you could just put hydroxide here so or and then our substrate is subromobutane here and so there's our specific rate law for this e2 elimination okay so it turns out the e2 elimination here has some stereochemical requirements we need to address as well so and i've kind of drawn it just such a way there's actually two hydrogens on this beta carbon right here one that's a wedge and one that's a dash and i made sure to draw the dashed one on purpose and so if you kind of take a look at a relationship here so this hydrant is pointing down and it is a dash so in fact let's put it right here down and a dash whereas the bromine is pointing up and is a wedge and if you kind of look at where they are relative to each other they're pointing 180 degrees apart and it turns out they have to point 180 degrees apart so and when they do we refer to that as being technically it means anti-coplanar but the word i'm going to write here is called anti-periplanar and it turns out it doesn't have to be exactly 180 it has to just be really really close to 180 so 180 degrees plus or minus a couple of degrees kind of sort of and that's what anti-periplanar mean it doesn't mean exactly 180 but it means really close to 180 gives you a little leeway so anti-coplanar is kind of the term i learned it with way back in the day but antiperiplane is kind of the more commonly accepted term now because it doesn't have to be exactly anti-coplanar which again anti-coplanar would be exactly 180 in the same plane but pointing opposite directions but anti-pillar planar means either exactly 180 or at least a little bit close all right so this is a stereochemical requirement of an e2 reaction so now a lot of people don't like to draw it like this they prefer you know let's rotate that around you know and let's get that bromine in the plane and that way we can get another hydrogen in the plane and they would actually rotate all the bonds so that we could put them in both in the plane and one i agree i mean that's easier to see we can see that the bromine points this way and the beta hydrogen points this way and obviously i haven't drawn the rest of the molecule and it's easy to see that yeah they point 180 degrees opposite in their antiperiplanar so but you can see it here as well so bromine here being a wedge and up 180 degrees from off the opposite of that would be a dash and down so cool we can see that based on the reaction happening just as we've drawn it here once we form that double bond we're going to lock it into a certain conformation here and i can see that these two methyl groups are trans to each other and when the reaction happens they get locked into that trans conformation now on the other hand we could actually rotate this around a little bit so in this case what i'm going to do is take and draw one more hydrogen in so we got i'll draw this one in on red and i'm going to rotate this bond around so i'm going to move and let's show this in green take advantage of all our colors here what i'm going to do is move the methyl group to where that hydrogen is the red hydrogen is going to go where the blue hydrogen is and the blue one's going to go back to where the methyl is i'm going to rotate this single bond right here around in a circle to get the other hydrogen antiperiplanar to the bromine let's see what that looks like cool so in this case the methyl group is now in the wedged position so the red hydrogen is now in the dashed position and then the blue hydrogen is now in the plane so and i've left that bromine right where he was cool and now i can see that it's actually the red hydrogen it's anti-periplanar to the bromine and so if the reaction happened in this conformation i can now see that these two methyl groups wouldn't be 180 degrees apart anymore and we can explain why we'd get the cis product forming as well and so when you've got a beta carbon with two hydrogens attached so assuming it's free to rotate then you're going to get both the cis and the trans the e and the z if it does have those stereoisomers as it does in this case also we could factor in that the other beta carbon over here has got three hydrogens and at any given moment in time one of those three is going to be in the anti-periplanar conformation so and in this case this particular alkene does not have cis and trans does not have e and z possible and you can kind of predict when that happens when either side of an alkene has two identical things there is no cis and trans so typically you can turn cis into trans and transient assist if you have either side and have the two groups trade places and so if either side has two of exactly the same group whether it be two hydrogens or two methyls or two ethyls just two identical groups on one side of the alkene there's no such thing as cis and trans no such thing as e and z cool so we can see here our three products and mr zaitsev says pick the top two but we also know that trans is more stable than cis and this is our major product and these would be our minor products in this case following mr zaitsev's rule cool so that's the stereochemical requirement on e2 reactions so and this is kind of something pretty important in a couple cases we'll see a couple examples here where we we see that importance all right a common type of question you'll see uh really requiring you to distinguish you know when this anti-periplanar hydrogen is present and when you've got to rotate a molecule to get there and stuff like that is going to be demonstrated in this following couple examples so in this case we're going to predict the major product in these e2 reactions i've got a nice strong base here so we said our strong base is typically going to have a negative charge on oxygen so in the last example it was naoh in this case it's going to be naoch3 so and again with metal and non-metal that's definitely ionic and got a negative charge on oxygen that's a good strong base it's essentially comparable to hydroxide all right so in this case i'm going to look at our alpha carbon in both cases and then we'll identify our beta carbons now the beta carbon on the left here does not have any hydrogens whatsoever so we cannot form a double bond here in fact if we tried we'd violate the octet rule for that carbon on the left so no options there so but then we've got a beta carbon on the right and with the beta carbon on the right there's only one hydrogen present and so in this case that hydrogen is on the dash and in this case that hydrogen is on the wedge cool and that's going to be important uh in the top example that actually already is anti-periplanar to the bromine so the hydro and the bromine are 180 degrees apart again a wedge and up versus a dash and down that's 180 degrees difference so we're ready to do this elimination reaction right now our base would come deprotonate that hydrogen we'd form the alkene kick off the leaving group and this would get locked in the conformation in which the reaction actually happens and in this case that means this is going to get locked in a trans conformation or e i guess is more proper as we'll see so but these two benzenes are going to end up trans to each other just as they're trans to each other so in the structure of the reactant when the reaction actually happens cool so it turns out this would be the e isomer so the major functional group on the left hand side is the benzene on this side and the bezel on this side when they're the two higher priority uh i shouldn't say function groups but substituents are trans to each other that's e cool and we don't observe the z product in this reaction at all so only the e and the idea is that when your beta carbon only has one hydrogen if e and z exist for that compound you're only going to get one of the two because that hydrogen's only going to be anti-periplanar in one of the possible conformations leading to only one of those products so if we look at this next one here there's a couple different ways to approach this but this wedged hydrogen is not 180 degrees away from the bromine right now it's just not and so we're going to have to rotate this around before the reaction can actually happen and so just like we did on the last example here what we're going to do is we're going to move this hydrogen to where the methyl group is we'll move the methyl group to where the benzene is and we'll move the benzene to where the hydrogen is we're going to rotate this bond right here between alpha and beta in this case 120 degrees so until the hydrogen is anti-periplanar to that bromine right there so let's rotate this around all right so in this case the benzene is now in that wedged position so our ant our hydrogen now is in the antiperiplanar position to the bromine so and then finally this methyl group is now in the plane over here and now this reaction is ready to go and we can see that it's not the two benzenes that are going to be trans to each other it's the benzene and a methyl group that are going to be trans to each other and so our base again is going to come and deprotonate the antiperiplanar hydrogen which frees up those electrons to form the alkene causing also the leaving group to simultaneously leave and so if we draw our product out now so we see that we actually get the z product now the two higher priority groups are on the same side we say so they're cis to each other and so we get the z product cool and we only get the z we're not going to get the e in this case at all so cool we had to do some rotation stuff like this here but there is a little bit of a cheat here what we could have done is said if the reaction happens right now in this conformation well then we get the e just like we did above right so is it ready to happen well it's not because the hydrogen's not antiperiplanar so then you're not going to get the e confirmation you'll get the z instead so i took the time to rotate the bond and show us you know show the hydrogen and the bromine the antiperiplanar conformation but the truth is if the reaction is ready to happen great like the one above so with the h and the br antiperiplanar great then you know what the conformation's gonna look like so in this case though if the reaction is not gonna happen then you don't get the e you get the opposite in this case the z if it happened in this confirmation we'd get e since it's not ready to happen we'll get z instead so just a little trick little shortcut but again big key here is if you only have one beta hydrogen uh then if e and z are possible you only get one of the two depending on the arrangement you've got all right in the next example here we definitely want to cover cyclohexane because there's something special with cyclohexanes when it comes to pair antiperiplanar hydrogens and stuff so first of all using a strong base here so we'll know we're doing an e2 reaction we'll find out that e1s use weak bases just like sn1 use weak nucleophiles so in this case got a negative charge in oxygen in this case it's sodium methoxide sometimes written as naoet but a nice good strong base that we commonly see when dealing with elimination reactions so and if we look here our bromine here is on the alpha carbon and in this case we have got two beta carbons and if we draw in their relevant hydrogens here we got one here and over on this one we've got two hydrogens one a wedge and one a dash and that's going to be important now most single bonds are free to rotate except when they're part of a ring so a single bond that's part of a ring here any one of these single bonds that's part of the ring might have a little bit of rotation as possible like when we're switching chair conformation stuff but it just can't rotate freely like a single bond that's not part of a ring can and that's important because if a hydrogen is not anti-periplanar so or then it's never going to be so and that's important and ultimately what that's going to mean here and i'll just kind of give the give it away is the hiding that's going to be anti-periplanar has to be trans to the leaving group so if i look at bromine being on a wedge to be trans my beta hydrogen has to be a dash that one's not going to work that one's not going to work but that one will so out of our three beta hydrants only one of them is even possible it's only only for only one of them it's possible to react in an e2 elimination reaction so cool mr zaitsev would have said hey look down here this beta carbon only has one hydrogen this one has two use the beta carbon with fewer hydrogens i.e use the more substituted beta carbon so but we're not going to follow mr hoffman's rule here so when it comes to e2 reactions this idea of having antiperiplanar hydrogen takes precedence and this is one example where we'll find out that we'll do hoffman elimination instead of zaitsev elimination all right one thing we should do to kind of map this out is draw the chair conformation cool and what i'm going to do so bromine is a wedge so i'm going to point that bromine straight up so and as we come over here the methyl group is going to be straight down and that hydrogen there is going to be equatorial over here so and then on the adjacent carbon we've got a hydrogen pointing straight down and then an equatorial one slanted off over here and you can see that it's this hydrogen right here that'll kind of circle in red that one points 180 degrees away from the bromine here and so in this case that's our anti-periplanar hydrogen that's the one involved in the reaction we can see that this hydrogen over here on the more substituted beta-carbon it's not anti-periplanar bromine point straight up i needed it to point straight down where the methyl group was but it doesn't point there so that's why this hydrogen over here it's not possible for it to react in this particular reaction one other thing you should note is that when you've got a chair conformation you typically have two chairs in equilibrium so and it turns out only when the leaving group is axial and we'll we'll find out that it and the hydrogen that's antiplane will be transaxial but only when the leaving group's axial is this reaction even possible so in this case we arrange this this bromine is now going to be over here and be equatorial so and on the adjacent carbon let's get that methyl in first and draw in the relevant hydrogens so and here's that trans hydrogen here but when they're equatorial they don't point 180 degrees apart and the reaction can't happen and so we've got this equilibrium going we need this conformation with the leaving group axial for the reaction to actually happen anything that hinders the formation of this particular chair conformation would slow this reaction down so if i put a t-butyl group somewhere that forces it all to be in this conformation or mostly all in this conformation you can actually shut down an e2 reaction completely so but the key is only one of the two chairs is actually going to be usable in an e2 elimination reaction here all right so if we go back to our product then in this e2 elimination so we'll form the double bond right there so and once again we'll call this the hoffman product or some people say anti-zaitsev take your pick but hoffman's a little more oh and i can't spell apparently today either so because that is not how you spell hoffman's name it is a little bit of funky spending spelling though one f two ends at the end so and that's your hoffman product uh in this case the zaitsev product wasn't even possible all right so we've already seen one example where we might not actually preferentially follow mr zaitsev's rule and do hofmann elimination so and here i want to show the most common example of where you're actually going to do hoffman elimination to get the major product and it involves what we call a bulky base so and you can kind of see why they call this a bulky base this is here uh potassium tbutoxide and sodium or potassium tbtoxide is the most common bulky base in fact for many students it's the only one they're ever presented with so if we look at it here the oxygen here again this is ionic so potassium's got a positive charge being a metal oxygen non-metals got a negative charge and in this case the negative charge oxygen is bonded to a tertiary carbon so and that gives it some bulkiness associated with it and as a result it turns out it looks over at our beta carbons over here we've got in this case our alpha carbons where the halogen again is bonded and it's a tertiary carbon which means there are three beta carbons two that are secondary and one that is primary so let's give myself a little room here and if we look here the secondary ones each have two hydrogens but the primary one out here have three hydrogens and the idea is that our bulky base is like looking over here and saying well i gotta deprotonate one of these beta carbons but man i'm big and i'm just so big you know i just think i want to not get too close here because you know that would be a little uncomfortable let me just deprotonate the primary carbon cool instead of the secondaries and so it turns out he's just due to his bulkiness his sheer size that's what he prefers to do typically and as a result we don't get the product we expect as the major product we're going to d disobey zaitsev's rule here and form the double bond using the primary beta carbon so rather than following zaitsev's rule and using either one of these secondaries it turns out due to symmetry they're equivalent so whether you form the alkene here or here it's exactly the same thing so we call the one on the left the hoffman product or the anti-zaitsev product and in this case that's going to be the major product now we'll call the one on the right still the zaitsev product but this time around it's going to be the minor product now sometimes you know somewhere down the line we'll present you with something we'll talk about the most stable product being the thermodynamic product so but in certain cases you can actually have what's called the kinetic product be the major product like we're doing here so this is one example of that we'll definitely teach you some further down the road when that's the case but this is the kinetic product here this hoffman product that's going to be the major when we use this bulky base now it turns out that not every time you use a bulky base you get the hoffman product as the major product but it's a little confusing to teach you when exactly you do and when you don't so most of the time most textbooks and most professors just kind of oversimplify this and they just say anytime you use a bulky base you're going to get the hoffman product as the major product so well in this case the truth is this so for those of you that might have been presented with the truth if you use a tertiary halide which i'll call a bulky halide then yes you're going to get the hoffman product as the major product but if you use a secondary halide well some secondaries that are bulkier still form the hoffmann as the major but some secondary halides that are less bulky form the zaitsev is the major and it gets a little confusing which is why for most of you you're just going to be simply told to memorize that when you use a bulky base the hoffman product is the major product the least substitute alkene is the major product all right i want to present to you one more example of where you might actually get the hoffman product instead of the zaitsev as the major product here so in that case it turns out when you have a bad leaving group in fact uh hoffman actually got his name on the hoffman elimination which is properly a reaction of amines for specifically a reaction involving a bad leaving group now this wasn't the bad leaving group he dealt with so but it's the one we're going to deal with here in this chapter the most common one presented now for many of you you may never actually see fluorine as a leaving group so but if you do fluorine is not a good leaving group it is not a weak base like chloride bromide and iodide or not a super super weak base anyways and being a little bit stronger base he's not a good leaving group and when you don't have a leaving a good leaving group you often get the hoffman product as the major product here as well uh so in this case i still got a good strong base and he's not a bulky base this time you don't we don't need the bulky base here we just need a bad leaving group that's another way to get the hofmann as the preferred product and in this case once again our major product is the less substituted alkene and once again this is a kinetic product so it turns out if we can't take a look at what's going on here that's not an o let's make it an h so we're going to come and deprotonate that beta hydrogen so and that frees up these electrons to form the pi bond and so we're like okay let's form that pi bond so that fluorine can leave and fluorine is like whoa whoa whoa whoa slow down guys i suck at leaving let's just take this a little slower and let's back things up so let's undo this and uh yeah hold up on those electrons and this carbon's like dude hydron's gone already uh you need to leave and so we kind of get a little buildup of charge on that carbon now this is not the proper mechanism the proper mechanism this still all happens in a concerted fashion so but it turns out in the transition state along the way there is a buildup of charge on that carbon and it is on the beta carbon specifically and it's uh very anion-like so we have a little bit of a carbanion-like transition state and i'd rather have a carbanion on a less substitute carbon so for carbocations they're more stable with more substituted carbons but carbanions are more stable with less substitute carbons and that's why we chose to have this carbanion-like transition state on a primary carbon instead of on these secondary carbons so we're still not forming the most stable product this is still a kinetic product so but due to having a bad leaving group a little buildup of negative charge on that beta carbon again if you're going to properly show the mechanism though it's still all a concerted mechanism it still all happens in one step but we just know that in the transition state there is that buildup of negative charge on the 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