SN1 and E1 Reactions

hello class in reviewed remember when we want to do an sn2 reaction what do we need we need a primary or secondary alkyl halide and a strong nucleophile or if we wanted to do an E2 reaction we would need a alkyl halide it could be one two or three but we need a strong base let's now take a look at a situation where you have an alkyl halide hmm okay and in this particular case a tertiary alkyl halide and you treat it with a solvent all right let's use the solvent ethanol and the feature of this solvent is that you can see that it's neutral right so this is ethanol and this solvent is a weak nucleophile and a weak base and also this solvent is polar so when you have an alkyl halide and more particularly a tertiary or a secondary alkyl halide and you treat it with a polar solvent that is a weak base and a weak nucleophile you are going to get a mixture of products you are going to get the product that looks like this now what type of reaction is that that is a substitution right the alkyl halide has been replaced with the ethoxide piece right here so you can see here's this oxygen right here it's no longer here and I will talk about that in a minute but overall you can see that the bromide has been replaced with this piece and this piece right here came from the solvent or or and you are going to get something that looks like that you're going to have the ill uh elimination product so we have a substitution product an elimination product but they go by a different mechanism than the sn2 and the E2 and we've discussed that briefly that would be an sn1 product and this would be a E1 product and I'm going to go through the mechanism to show you what that is another thing okay so how does this come to be how does this happen the mechanism is going to help us figure this out so if I take and let's do the let's do the substitution problem first here so if we have our cells our alkyl halide like that the first step in an sn1 mechanism is the heterolysis step where the alkyl halide leaves and then we are left with a a tertiary carbocation plus our bromide that fell off and then in the next step we're going to have our ethanol like this and the ethanol is a weak nucleophile and it's going to come in and attack the carbocation and then so we are now going to have something that looks like this our oxygen atom attacked the carbon and so that's what I'm showing here you see how the auction atom attacked the carbon of interest and if we want to start the carbon of Interest I'll do that you see it's right there but what is still attached to this auction atom the hydrogen's still there did I draw an arrow anywhere showing that hydrogen oxygen Bond broke no I didn't the hydrogen is still there and they're still a lone pair this lone pair right there was used to form this Bond oxygen carbon Bond so now what has happened to the formal charge of this molecule we have an oxygen atom with three bonds and a lone pair so that makes that positively charged and then we have one more step all right now remember in this reaction here ethanol is the solvent as well all right so there's going to be excess amounts of ethanol large amounts and compared to our alkyl halide so what we're going to have is an excess amount of ethanol and we said ethanol is also a weak base so that ethanol another ethanol molecule can act as the weak base to do this proton transfer so this step right here is called the proton transfer step because we are transferring a proton and so our product is going to be fat Plus what's going to happen to our ethanol it's going to add a proton so now we have a protonated ethanol molecule but you can see this right here matches this product right there so that's our our sn1 mechanism now let's kind of block this off here let's see if I have enough board space okay to do the E1 mechanism down here let's give it a shot see what happens here so in a E1 mechanism the first step is the same as the sn1 the alkyl halide needs to leave and we're going to generate a carbocation plus our bromide the next step is where we're going to take ethanol which we've already seen right here acts as a weak base can take a look at this carbocation and the carbocation is on that Central carbon right so we'll call that Alpha and the beta carbons there's three beta carbons here has what a hydrogen atom and so ethanol can come in now and act as a weak base and grab that proton break the carbon hydrogen bond bringing those electrons there and what's that going to do that is going to generate are alkene Plus protonated ethanol which that would have to have what a positive charge now what's what's one driving force for this process to happen well look we have a a carbocation that's not very stable it wants to react to get to a more stable form and the more stable form here is the alkene for one reason it's neutral there's no formal charge of plus one and a neutral molecules are typically lower in energy and more stable so what we have here is a our E1 product as you can see here they match now what's interesting another interesting thing about this these two reactions is when you look at the rate okay so the rate for a E1 E1 mechanism and an sn1 foreign are dependent on the substrate all right the substrate concentration I'm trying to say a word and spell the other word and I just did not work for me substrate now what do I mean by the substrate the substrate is the alkyl halide so if you don't like using the word substrate to refer to the alkyl halide you could just say alkyl halide if you'd like but the text refers to the alkyl halide as substrate so when they were doing the kinetic studies for sn1 and E1 reactions they would look at this reaction scheme here and if we just did the same analysis as we did with the sn2 and E2 what you'll notice is that this species right here if they doubled this concentration nothing happened to the rate the rate did not adjust or change but if they increase the concentration of the substrate or the alkyl halide if you double let's say we doubled the concentration of the alkyl halide then the rate of the reaction increased by double so since E1 and S01 the rate is only dependent upon this the alkyl halide well that is going to be a first order first order reaction okay so E1 and sn1 reactions are first order the rates are only dependent on the substrate sn2 and E2 is a second Orbiter reaction where it's dependent upon the the nucleophile or the base and the substrate concentration okay the next thing that I want to do is look at the sn1 mechanism in a diagram now so I've written up the mechanism for the sn1 mechanism again up top yes I the arrows have been cut off a little bit so you see the arrows here cut off and it's just a little funky but you've seen the mechanism just a few moments ago and now I've Rewritten it so what we want to do is take a look at this energetically so if we draw ourselves a diagram here on the y-axis we have potential energy right there and then the reaction coordinate so we're going to track the energy of this reaction okay so I dropped my pen I need to go get it so what we're going to have is our starting material or say that starting point right there is our alkyl halide so I'm going to just number it right there so that's going to be molecule or alkyl halide and what what's the first step it is a heterolysis step and that takes some energy in order to do that and so it's going if this is the alkyl halide in order to break that bromine carbon Bond it's going to take some energy to do that all right and then what happens after it breaks it turns into let let's make this go down a little bit deeper oh we can see some contrast we'll go down a little bit okay so what do we have here this right here at the very top right there that's the transition state and down right here at the the the well that is the intermediate the carbocation is what's called an intermediate because after the Alka after the heterolus step there's a moment in time where you create this and so if we number that this molecule is going to be right here the carbocation is an intermediate this right here is the transition state for this process right here all right and then what happens we have another step so from one to two we have the heterolysis step then now we have another step here indicated by this Arrow where we have a coordination step where the ethanol comes in and attacks the carbocation so it's going to be there so what do we have now okay that's not the smoothest looking line what what do we have here there's the transition state for the reaction of the carbocation with the ethanol to form this species so that's molecule three shown right there so from molecule 2 to molecule 3 that's the transition state so from two to three that's the transition state and then from one to two there's another transition state number three here is another intermediate you see how we have another Well right there and this reaction coordinates not big enough so we need to keep going where and then what happens next we have a proton transfer step and proton transfer steps are very low in energy that doesn't take very much so that is another step so that's the transition state from three to four and then right here is going to be our product and that's our product so what we have going on here is we have you can track the energy of the mechanism or the reaction you see we do our so this part right here would be our heterolysis step from here to here and then from here to here is our coordination step and then from here to here is our proton transfer now do you recall that I said that the rate of an sn1 mechanism is related to the concentration of our substrate right so when we look at this so it's only dependent on the concentration of this species and the reason why that's the case is because it is the rate determining step to remember that from General chemistry rate determining step when you look at an energy diagram in order to find the rate determining step you just need to find the hill that's the tallest and right here we see that that is clearly higher in energy than that Peak or saying it in another word another way it is the highest peak so that is our rate determining step okay so for this example whenever you find something with the tallest peak that's the rate determining step now there are some minor exceptions or there's an exception to that I'm going to show to you in a minute but I just need to make sure before I erase this there's nothing else that I need to say about this so when we take a look at these two graphs here it's the same idea we have potential energy on our Y and then we have our reaction coordinate on the x-axis so when you look at this graph here you need to be able to identify a few things where are the starting materials that would be the starting materials where's the products there's our products where's the transition States well there's one transition state there's a second transition state where's the intermediates it's going to be the well so that's the intermediates which is the uh rate determining step right determining step remember how I said that it's the tallest peak so the rate determining step would be this one and what step is that well look how many transition states there are this would be step one and this is the second transition state so that would be step two so it is step two that is the rate determining step because it has the uh tallest peak but then you are could encounter an energy diagram like this but here's our starting materials there's our products there's two transition States that would be step one you can't see that stuff can you this is step two right over there so this is step one right here what do we have we have an intermediate right there so where is the rate determining step where is that well you can see that both of these transition States they're at the same energy level and you may argue that they're not my intent was to draw these transition States at the same energy level okay so which one would you select as the rate determining step well you can't use using this logic over here right since they're the same what you have to do is look for the energy of activation well energy of activation and that is when you compare the steps so if this first step starts here and it takes that much of energy to get it going gonna do a lowercase a but then from the seconds for the Second Step the Second Step starts right here at the intermediate going to the product so what is the energy of activation for the Second Step well it's going to be from this point to there so when you have a situation like this what is the uh rate determining step it's the one with the largest energy of activation so the rate determining step would be step one so that's another thing that you could draw over here in this diagram here is that you have energy of activation as well right there and that distance right there so that's the energy of activation for the first step energy activation for the second step and the same thing over here okay so when you have uh Hills and Valleys the tallest one is going to be the rate determining step unless you come to a situation where all the hills are at the same energy level and so you have to look at the energy of activation to make that call right so next let's look at the E1 mechanism and look at it on the energy diagram here okay so I drew the E1 mechanism all over again showing the the steps here so the first step in an E1 is just the same as this in a sn1 in which our product right here goes to a high energy transition state to get to a intermediate so that's our one that's one here's our transition state for step one so that's step one right there there's a transition state this is our intermediate the carbocation so that could be molecule two so the intermediate is just the carbocation we then do a electrophile elimination step which isn't as hard and then we go to our product so let's just so that's two which is our intermediate so we have to make sure we understand that's our intermediate and then this is step three step three no not step three that's step two so this right here the reacting with the electrophile here that's the transition state and that is molecule three so that's three there so this is step two so where's the rate determining step well it's the this peak is taller than that Peak so this Step One is the rate determining step and like I said already the rate is dependent only upon the concentration of the substrate for the alkyl halide okay and that's because it has the tallest peak but in this particular case look at it it also has the largest energy of activation so as two things going for us energy of activation is the tallest or the largest and it's the tallest transition state all right but remember you only really need to compare the energy of activation to determine the rate determining step if the transition state Peaks are at the same level but since we have the tallest peak here that wins that's our rate determining step okay so there's our energy diagram there what else do we have to say about this one right here so when you look at an energy diagram here I want you to be able to say hey this right here is the heterolysis step this right here is the electrophile elimination step and identify every component of the reaction diagram so I forgot to put here that this is the same as everything that we've been seeing potential energy and then this would be the reaction coordinate okay so let's take a look at this secondary alkyl halide here and we're going to treat it with water so when we look at water water has no charge on it so we can look at it and say hey that's going to be a weak nucleophile or a weak base it could behave as much so when the reacting species so this can react as a base or a nucleophile and we've identified it as a weak one but when that attacking species is also the solvent that has a special name called cevolysis so cevolysis is simply the name for when the attacking species is the solvent itself so here we have a solvolysis reaction but what what is the product going to be I'm going to show you this product it's going to be this so what type of reaction have we done but we've done a substitution reaction because we've replaced this bromine with the o h now where did that oh come from well it came from water but how did it come from water so if we take a look at a mechanism let's let's invoke an sn2 mechanism okay to kind of understand what's going on here I could possibly say we have water here and I could say sn2 okay so that means it attacks backside attack kicks that off all in one step and then I'm going to get this the oxygen attacked that carbon and it still has the two hydrogens attached right I did not remove those and it still has a lone pair and it's going to have a positive charge so then I could do a a proton transfer reaction in which I can take water again because it's the solvent there's a lot of it that can come in grab that proton do that to give us our product right there now at first I'm feeling pretty good about myself I'm like hey this to this how did that happen I proposed a mechanism right and all the arrow pushing looks good but guess what I'm going to throw a wrench in that right here because guess what when you do the solvolysis reaction you also get this product let's see here how do I want to draw it like this so do you see whoa whoa whoa what have we done here if we number the carbons one two three four what do we see over here one two three four so that makes sense the bromine was attached to carbon three so that's where you do the substitution but here one two three four you see how carbon 3 no longer has the bromine or the oh but the oh is now over here so if this is what's happening these are the products that we're going to get the mechanism doesn't align with the products because if this reaction proceeded via an sn2 and you follow it here the only product that you can get is this one product we'll call it product a versus product B the mechanism says you would only get a but we are getting B so that tells us the mechanism is wrong so if we take a look at a different mechanism maybe we can figure out what's going on here so let's look at asn1 so what if we had a sn1 mechanism so what's the first step that would be heterolysis step so that will give us what that will give us a secondary carbocation and then what others what's the next step here we have water here so now that can do a coordination step attack there see everything still good can't forget all our carbons okay and then we would then do a proton transfer where we would take another molecule of water and we would do our proton transfer step to give us our product okay so that gives us our product a so that looks legit but then how can we have product B4 how is this a correct mechanism well it all comes down to this intermediate right here that intermediate is key because what have we learned in the previous chapter we've learned about hydride shifts right so look at what can happen here is I'm now going to erase the last part of this mechanism okay I'm just going to start over here so I already showed you how to get a but now that is way too long I don't like that okay let's see here you know I am just messing this up let's see it's like this boom boom there we go so what we could do also is do a hydride shift so I see that on the adjacent carbon from the carbocation I have a hydrogen so I could take that hydrogen and do a shift you see how I'm drawing it from the center of the bond there and if I do that hydride shift I'm going to generate this I'm going to add the hydrogen there now the carbocation disappears but where does it reappear it is not going to be on this carbon so what why would this do this because we went from a secondary carbocation 2A tertiary carbocation and we've learned that tertiary carbocations are more stable so whenever you generate a carbocation it is a good rule of thumb to look for rearrangements and so we have our hydride shift or hydride rearrangement and now we can have our water come in and attack and so that would be our coordination step so that would give us this [Music] that hydrogen is not visible so I'm gonna violate geometry rules so you can see the hydrogen and then you can see we will do a proton transfer mechanism with some excess water and so that excess water will come and grab that proton to give us our product and that matches product B right so what's cool about this reaction right here because we know for a fact that we get two different products it supports that there is such thing as an sn1 mechanism it supports the generation of this carbocation intermediate if it wasn't for this hydride shift step right here we would never form product B and that is one reason why an sn2 mechanism is totally out of consideration because if it was sn2 we would only get this but another thing that kind of alerts you to the fact that this can't be sn2 is the fact that this is a weak nucleophile and in sn2 and E2 mechanisms it has to be a strong nucleophile or base in order for it to work right so let's look I'll erase this and then we'll look at another type of shift now here is an example of another secondary alkyl halide and it's going to give us two different products but look at what's happened here we see that the bromine has been substituted with an o h okay that's a simple substitution reaction but then when you look at this other product on the right look it's what's what it's been replaced with it's been replaced with a methyl group how does that happen well the mechanism helps us to answer that question so in order to get two different products here we have to invoke a the first step as the heterolysis step in order to generate a carbocation and what type of carbocation did I generate that's a secondary carbocation so that's stable but can it get more stable and that's what you always have to look for and so what you do is you look at the adjacent carbons and you see if there's a hydrogen okay so if I look at if I number if I look at this carbon right here I see that there's a hydrogen adjacent to the carbocation so I'm like hey that could work so I'm like okay let's try it so let's just do it so hydride or hydride shift boom so what's that going to give me I'll put the new hydrogen there then that generates what that generates a primary carbocation so that was not that's not going to work it's not a bad idea to try it because you're always looking at the adjacent hydrogens and there it is when you hydrate shift it what does it do gives you a primary carbocation so that's not going to work it's not going to happen but it was a good try good analysis okay so that's not going to happen so what if I looked at the other adjacent carbon so here's our carbon adjacent to its right here and I'm like are there any hydrogens there's not but remember if they're methyl groups you can have a methyl shift so this methyl right here or this one or this one it doesn't matter just one of the methyls can shift over and when that shifts over it's going to shift like that we will generate our carbocation now was that a good investment of our time yeah because we went from a secondary to what a tertiary carbocation and so that will work and so we just do the next two steps where we take water and attack in and then do the proton transfer we will generate that product right there so more one key takeaway that you have to always remember whenever you generate a carbocation can you do rearrangements to get a more stable carbocation Ary versus tertiary tertiary definitely the stronger or the more stable and more favorable carbocation okay class now it's your turn to put all these ideas together that we've been discussing to see if you fully understand what's Happening Here so here's a reaction that I put on the board I'm going to tell you that it is a sn1 reaction which one of these products is the correct product there's only one so which is the correct and only product and once you've figured that out you always have to explain why so like if this is a multiple choice question this is a flip the coin if you did not know it would just flip a coin in that one right that's not going to do you much good on my exams because for every choice that you make you have to rationalize and justify why you did that so this question could be something more like a free response where it says which is the major product or which is the only product why you have to explain your reasoning so if you just picked the correct answer because you flipped a coin if I had no explanation you are going to lose the majority of the points so I would definitely pause the video now and try to figure this one out on your own and then come back to the video to see what happens so if we come back here we have I told you it's an sn1 mechanism so we have to do it so if we do an sn1 mechanism here can we the first step is to get that guy to leave right so that's going to give us what foreign that's going to give us a primary carbocation but how is that even possible because we said we don't generate primary carbocations when we do this right it doesn't happen so if we know that does not happen then we have to invoke just a slightly different mechanism here and that's where you do the out the methyl shift and the heteroysis step all in the same step so that would be taking this methyl group right here and shifting it to that carbon at the same time the halogen leaves and so that's going to give us this species right here because we know we cannot generate primary carbocations and now we can then add our water here and then do a proton transfer step to get us to our product so if that attacks there you know never do that that's a bad idea never draw through the bonds that's a bad idea all right if you find yourself with a crazy Pathway to get there you can maybe go like this all right there now from there can you see what the product is going to be product B is going to be the only product formed when you do this reaction all right that is I'm putting a stipulation on this though I'm saying hey this is an sn1 mechanism all right now could you get to this product by a different mechanism you could if you invoked an sn2 mechanism if you invoke an sn2 mechanism then you would generate this product but remember with an sn2 you have to use a strong nucleophile if you did that reaction then you would start seeing a as your major product here but then again I'm not going to say we're going to see a lot of this product why because remember from sn2 a material that when you look at the beta carbon if the beta carbons vary sterically hindered then an sn2 mechanism is going to be very difficult remember that how to put it all together here all right so I take a look at these four different alkyl halides and if I'm going to take each one of these and do the same reaction treat them with water all right to generate our final product as this okay or this so I'm invoking a E1 or an sn1 mechanism which one of the the which one's going to react the fastest if I have a fluorine or an iodine all right and the answer is as you go down right here this is going to be the slow end this is going to be the fast so if you have an alkyl halide with an iodine it's going to react the fastest whereas when you have a fluorine it is so slow that it's totally impractical that we basically say that there's no reaction when we have fluorine it's really just these three right here that we use for alkyl halides to do E1 or to do our reactions with when we're talking about E1 and sn1 now why is this the case well it all comes down to this idea of leaving group ability and we've we've discussed that before when you just look at the leaving group of abilities so that's just like this right fluorine doesn't leave very well iodine leaves very well and we rationalize this due to the fact that when it leaves the fluorine then becomes a fluoride and then iodine and when we compare these two which one's the most stable anion and when we go down a group it is about size and polarizability this is larger more polarizable more stable so this iodine when it kicks off the alkyl halide it's chill it's okay being by itself the fluoride ion is I hate this I need to be attached to the alkyl halide so it never wants to leave huh another way we've looked at this is if we look at the iodide and then look at it in water foreign which ones then it's going to generate our hydronium plus our iodine right and if you compare that to if you did hydrofluoric acid and water you would get h3o Plus our fluoride right and we just ask ourselves which one's the stronger acid and we clearly know this is the stronger acid and so it's the strongest acid which makes this a very weak base and in this context a very very good uh leaving group so stronger the acid means it's conjugate base is better leaving group that's another way of looking at it and remembering it when looking at sn1 and E1 reactions let's take a look at what type of substrates can work for these reactions well we definitely want a tertiary alkyl halide that is the best best one that will work secondary alkyl halides can also work but we don't see they're a little bit more rare but they can still happen primary alkyl halides no way why won't that work because they generate a primary carbocation and that's not possible okay but there are so not it's really not an exception but if we take a molecule that looks like this okay like that okay and I put a carbocation right there how many people would say no way we can't form that and how many of you would say that cat form because that's a primary carbocation and you would be right based off of this definition but we don't call that a primary carbocation when you have a carbon that's a cation directly attached to a Benzene ring we call that a benzylic carbocation and benzylic carbocations even though they look like they're primary they're not they're benzyl at carbocations and they are stable and why do you think they're going to be stable maybe if I draw something like this you see the rest of the structure it's rested and stabilized that's why that guy so that's why that carbocation is stable what's another one we have something called an allylic cation let's see if I spelled that right so we have an allylic carbocation which looks like this when you take a first look at it you're like that can't form no way that's a primary carbocation it's not it is a allylic carbocation now why is that stable why does that form because as you can guess we have a resonance structure and due to Resonance that's going to stabilize this allylic carbocation enough that it's us it can form and so you can have a reaction with uh these two type of carbocations so for example if you had a reaction like this let's see here x2o okay can that undergo an sn1 yes yes it can keep consistent here that that is very plausible right okay let's see what's another thing that we need to talk about I know we're a little bit over but I want to say one more thing teach one more principle then we will conclude this video all right that is can you do can you have this happen all right and we have a halogen there remember that carbon that the halogen is attached to is SP2 hybridized can this process happen can the the halogen act as a good leaving group and leave and generate this no it cannot it cannot happen all right and the reason why it cannot happen is because this carbocation that we're proposing is so so unstable and the reason why it's so so unstable is due to the fact to effective electronegativity effective electronegativity activity yeah okay so that does not happen and if you need to review effective electronegativity that's the reason why that doesn't form and so the same reason this does not happen okay look at this one put a halogen right there right on the Benzene ring all right can this happens no it cannot okay that cannot happen because why we have a SP2 hybridized carbon and the reason why same idea effective electronegativity okay so that's where we'll end on this video

hello class in reviewed remember when we want to do an sn2 reaction what do we need we need a primary or secondary alkyl halide and a strong nucleophile or if we wanted to do an E2 reaction we would need a alkyl halide it could be one two or three but we need a strong base let&#39;s now take a look at a situation where you have an alkyl halide hmm okay and in this particular case a tertiary alkyl halide and you treat it with a solvent all right let&#39;s use the solvent ethanol and the feature of this solvent is that you can see that it&#39;s neutral right so this is ethanol and this solvent is a weak nucleophile and a weak base and also this solvent is polar so when you have an alkyl halide and more particularly a tertiary or a secondary alkyl halide and you treat it with a polar solvent that is a weak base and a weak nucleophile you are going to get a mixture of products you are going to get the product that looks like this now what type of reaction is that that is a substitution right the alkyl halide has been replaced with the ethoxide piece right here so you can see here&#39;s this oxygen right here it&#39;s no longer here and I will talk about that in a minute but overall you can see that the bromide has been replaced with this piece and this piece right here came from the solvent or or and you are going to get something that looks like that you&#39;re going to have the ill uh elimination product so we have a substitution product an elimination product but they go by a different mechanism than the sn2 and the E2 and we&#39;ve discussed that briefly that would be an sn1 product and this would be a E1 product and I&#39;m going to go through the mechanism to show you what that is another thing okay so how does this come to be how does this happen the mechanism is going to help us figure this out so if I take and let&#39;s do the let&#39;s do the substitution problem first here so if we have our cells our alkyl halide like that the first step in an sn1 mechanism is the heterolysis step where the alkyl halide leaves and then we are left with a a tertiary carbocation plus our bromide that fell off and then in the next step we&#39;re going to have our ethanol like this and the ethanol is a weak nucleophile and it&#39;s going to come in and attack the carbocation and then so we are now going to have something that looks like this our oxygen atom attacked the carbon and so that&#39;s what I&#39;m showing here you see how the auction atom attacked the carbon of interest and if we want to start the carbon of Interest I&#39;ll do that you see it&#39;s right there but what is still attached to this auction atom the hydrogen&#39;s still there did I draw an arrow anywhere showing that hydrogen oxygen Bond broke no I didn&#39;t the hydrogen is still there and they&#39;re still a lone pair this lone pair right there was used to form this Bond oxygen carbon Bond so now what has happened to the formal charge of this molecule we have an oxygen atom with three bonds and a lone pair so that makes that positively charged and then we have one more step all right now remember in this reaction here ethanol is the solvent as well all right so there&#39;s going to be excess amounts of ethanol large amounts and compared to our alkyl halide so what we&#39;re going to have is an excess amount of ethanol and we said ethanol is also a weak base so that ethanol another ethanol molecule can act as the weak base to do this proton transfer so this step right here is called the proton transfer step because we are transferring a proton and so our product is going to be fat Plus what&#39;s going to happen to our ethanol it&#39;s going to add a proton so now we have a protonated ethanol molecule but you can see this right here matches this product right there so that&#39;s our our sn1 mechanism now let&#39;s kind of block this off here let&#39;s see if I have enough board space okay to do the E1 mechanism down here let&#39;s give it a shot see what happens here so in a E1 mechanism the first step is the same as the sn1 the alkyl halide needs to leave and we&#39;re going to generate a carbocation plus our bromide the next step is where we&#39;re going to take ethanol which we&#39;ve already seen right here acts as a weak base can take a look at this carbocation and the carbocation is on that Central carbon right so we&#39;ll call that Alpha and the beta carbons there&#39;s three beta carbons here has what a hydrogen atom and so ethanol can come in now and act as a weak base and grab that proton break the carbon hydrogen bond bringing those electrons there and what&#39;s that going to do that is going to generate are alkene Plus protonated ethanol which that would have to have what a positive charge now what&#39;s what&#39;s one driving force for this process to happen well look we have a a carbocation that&#39;s not very stable it wants to react to get to a more stable form and the more stable form here is the alkene for one reason it&#39;s neutral there&#39;s no formal charge of plus one and a neutral molecules are typically lower in energy and more stable so what we have here is a our E1 product as you can see here they match now what&#39;s interesting another interesting thing about this these two reactions is when you look at the rate okay so the rate for a E1 E1 mechanism and an sn1 foreign are dependent on the substrate all right the substrate concentration I&#39;m trying to say a word and spell the other word and I just did not work for me substrate now what do I mean by the substrate the substrate is the alkyl halide so if you don&#39;t like using the word substrate to refer to the alkyl halide you could just say alkyl halide if you&#39;d like but the text refers to the alkyl halide as substrate so when they were doing the kinetic studies for sn1 and E1 reactions they would look at this reaction scheme here and if we just did the same analysis as we did with the sn2 and E2 what you&#39;ll notice is that this species right here if they doubled this concentration nothing happened to the rate the rate did not adjust or change but if they increase the concentration of the substrate or the alkyl halide if you double let&#39;s say we doubled the concentration of the alkyl halide then the rate of the reaction increased by double so since E1 and S01 the rate is only dependent upon this the alkyl halide well that is going to be a first order first order reaction okay so E1 and sn1 reactions are first order the rates are only dependent on the substrate sn2 and E2 is a second Orbiter reaction where it&#39;s dependent upon the the nucleophile or the base and the substrate concentration okay the next thing that I want to do is look at the sn1 mechanism in a diagram now so I&#39;ve written up the mechanism for the sn1 mechanism again up top yes I the arrows have been cut off a little bit so you see the arrows here cut off and it&#39;s just a little funky but you&#39;ve seen the mechanism just a few moments ago and now I&#39;ve Rewritten it so what we want to do is take a look at this energetically so if we draw ourselves a diagram here on the y-axis we have potential energy right there and then the reaction coordinate so we&#39;re going to track the energy of this reaction okay so I dropped my pen I need to go get it so what we&#39;re going to have is our starting material or say that starting point right there is our alkyl halide so I&#39;m going to just number it right there so that&#39;s going to be molecule or alkyl halide and what what&#39;s the first step it is a heterolysis step and that takes some energy in order to do that and so it&#39;s going if this is the alkyl halide in order to break that bromine carbon Bond it&#39;s going to take some energy to do that all right and then what happens after it breaks it turns into let let&#39;s make this go down a little bit deeper oh we can see some contrast we&#39;ll go down a little bit okay so what do we have here this right here at the very top right there that&#39;s the transition state and down right here at the the the well that is the intermediate the carbocation is what&#39;s called an intermediate because after the Alka after the heterolus step there&#39;s a moment in time where you create this and so if we number that this molecule is going to be right here the carbocation is an intermediate this right here is the transition state for this process right here all right and then what happens we have another step so from one to two we have the heterolysis step then now we have another step here indicated by this Arrow where we have a coordination step where the ethanol comes in and attacks the carbocation so it&#39;s going to be there so what do we have now okay that&#39;s not the smoothest looking line what what do we have here there&#39;s the transition state for the reaction of the carbocation with the ethanol to form this species so that&#39;s molecule three shown right there so from molecule 2 to molecule 3 that&#39;s the transition state so from two to three that&#39;s the transition state and then from one to two there&#39;s another transition state number three here is another intermediate you see how we have another Well right there and this reaction coordinates not big enough so we need to keep going where and then what happens next we have a proton transfer step and proton transfer steps are very low in energy that doesn&#39;t take very much so that is another step so that&#39;s the transition state from three to four and then right here is going to be our product and that&#39;s our product so what we have going on here is we have you can track the energy of the mechanism or the reaction you see we do our so this part right here would be our heterolysis step from here to here and then from here to here is our coordination step and then from here to here is our proton transfer now do you recall that I said that the rate of an sn1 mechanism is related to the concentration of our substrate right so when we look at this so it&#39;s only dependent on the concentration of this species and the reason why that&#39;s the case is because it is the rate determining step to remember that from General chemistry rate determining step when you look at an energy diagram in order to find the rate determining step you just need to find the hill that&#39;s the tallest and right here we see that that is clearly higher in energy than that Peak or saying it in another word another way it is the highest peak so that is our rate determining step okay so for this example whenever you find something with the tallest peak that&#39;s the rate determining step now there are some minor exceptions or there&#39;s an exception to that I&#39;m going to show to you in a minute but I just need to make sure before I erase this there&#39;s nothing else that I need to say about this so when we take a look at these two graphs here it&#39;s the same idea we have potential energy on our Y and then we have our reaction coordinate on the x-axis so when you look at this graph here you need to be able to identify a few things where are the starting materials that would be the starting materials where&#39;s the products there&#39;s our products where&#39;s the transition States well there&#39;s one transition state there&#39;s a second transition state where&#39;s the intermediates it&#39;s going to be the well so that&#39;s the intermediates which is the uh rate determining step right determining step remember how I said that it&#39;s the tallest peak so the rate determining step would be this one and what step is that well look how many transition states there are this would be step one and this is the second transition state so that would be step two so it is step two that is the rate determining step because it has the uh tallest peak but then you are could encounter an energy diagram like this but here&#39;s our starting materials there&#39;s our products there&#39;s two transition States that would be step one you can&#39;t see that stuff can you this is step two right over there so this is step one right here what do we have we have an intermediate right there so where is the rate determining step where is that well you can see that both of these transition States they&#39;re at the same energy level and you may argue that they&#39;re not my intent was to draw these transition States at the same energy level okay so which one would you select as the rate determining step well you can&#39;t use using this logic over here right since they&#39;re the same what you have to do is look for the energy of activation well energy of activation and that is when you compare the steps so if this first step starts here and it takes that much of energy to get it going gonna do a lowercase a but then from the seconds for the Second Step the Second Step starts right here at the intermediate going to the product so what is the energy of activation for the Second Step well it&#39;s going to be from this point to there so when you have a situation like this what is the uh rate determining step it&#39;s the one with the largest energy of activation so the rate determining step would be step one so that&#39;s another thing that you could draw over here in this diagram here is that you have energy of activation as well right there and that distance right there so that&#39;s the energy of activation for the first step energy activation for the second step and the same thing over here okay so when you have uh Hills and Valleys the tallest one is going to be the rate determining step unless you come to a situation where all the hills are at the same energy level and so you have to look at the energy of activation to make that call right so next let&#39;s look at the E1 mechanism and look at it on the energy diagram here okay so I drew the E1 mechanism all over again showing the the steps here so the first step in an E1 is just the same as this in a sn1 in which our product right here goes to a high energy transition state to get to a intermediate so that&#39;s our one that&#39;s one here&#39;s our transition state for step one so that&#39;s step one right there there&#39;s a transition state this is our intermediate the carbocation so that could be molecule two so the intermediate is just the carbocation we then do a electrophile elimination step which isn&#39;t as hard and then we go to our product so let&#39;s just so that&#39;s two which is our intermediate so we have to make sure we understand that&#39;s our intermediate and then this is step three step three no not step three that&#39;s step two so this right here the reacting with the electrophile here that&#39;s the transition state and that is molecule three so that&#39;s three there so this is step two so where&#39;s the rate determining step well it&#39;s the this peak is taller than that Peak so this Step One is the rate determining step and like I said already the rate is dependent only upon the concentration of the substrate for the alkyl halide okay and that&#39;s because it has the tallest peak but in this particular case look at it it also has the largest energy of activation so as two things going for us energy of activation is the tallest or the largest and it&#39;s the tallest transition state all right but remember you only really need to compare the energy of activation to determine the rate determining step if the transition state Peaks are at the same level but since we have the tallest peak here that wins that&#39;s our rate determining step okay so there&#39;s our energy diagram there what else do we have to say about this one right here so when you look at an energy diagram here I want you to be able to say hey this right here is the heterolysis step this right here is the electrophile elimination step and identify every component of the reaction diagram so I forgot to put here that this is the same as everything that we&#39;ve been seeing potential energy and then this would be the reaction coordinate okay so let&#39;s take a look at this secondary alkyl halide here and we&#39;re going to treat it with water so when we look at water water has no charge on it so we can look at it and say hey that&#39;s going to be a weak nucleophile or a weak base it could behave as much so when the reacting species so this can react as a base or a nucleophile and we&#39;ve identified it as a weak one but when that attacking species is also the solvent that has a special name called cevolysis so cevolysis is simply the name for when the attacking species is the solvent itself so here we have a solvolysis reaction but what what is the product going to be I&#39;m going to show you this product it&#39;s going to be this so what type of reaction have we done but we&#39;ve done a substitution reaction because we&#39;ve replaced this bromine with the o h now where did that oh come from well it came from water but how did it come from water so if we take a look at a mechanism let&#39;s let&#39;s invoke an sn2 mechanism okay to kind of understand what&#39;s going on here I could possibly say we have water here and I could say sn2 okay so that means it attacks backside attack kicks that off all in one step and then I&#39;m going to get this the oxygen attacked that carbon and it still has the two hydrogens attached right I did not remove those and it still has a lone pair and it&#39;s going to have a positive charge so then I could do a a proton transfer reaction in which I can take water again because it&#39;s the solvent there&#39;s a lot of it that can come in grab that proton do that to give us our product right there now at first I&#39;m feeling pretty good about myself I&#39;m like hey this to this how did that happen I proposed a mechanism right and all the arrow pushing looks good but guess what I&#39;m going to throw a wrench in that right here because guess what when you do the solvolysis reaction you also get this product let&#39;s see here how do I want to draw it like this so do you see whoa whoa whoa what have we done here if we number the carbons one two three four what do we see over here one two three four so that makes sense the bromine was attached to carbon three so that&#39;s where you do the substitution but here one two three four you see how carbon 3 no longer has the bromine or the oh but the oh is now over here so if this is what&#39;s happening these are the products that we&#39;re going to get the mechanism doesn&#39;t align with the products because if this reaction proceeded via an sn2 and you follow it here the only product that you can get is this one product we&#39;ll call it product a versus product B the mechanism says you would only get a but we are getting B so that tells us the mechanism is wrong so if we take a look at a different mechanism maybe we can figure out what&#39;s going on here so let&#39;s look at asn1 so what if we had a sn1 mechanism so what&#39;s the first step that would be heterolysis step so that will give us what that will give us a secondary carbocation and then what others what&#39;s the next step here we have water here so now that can do a coordination step attack there see everything still good can&#39;t forget all our carbons okay and then we would then do a proton transfer where we would take another molecule of water and we would do our proton transfer step to give us our product okay so that gives us our product a so that looks legit but then how can we have product B4 how is this a correct mechanism well it all comes down to this intermediate right here that intermediate is key because what have we learned in the previous chapter we&#39;ve learned about hydride shifts right so look at what can happen here is I&#39;m now going to erase the last part of this mechanism okay I&#39;m just going to start over here so I already showed you how to get a but now that is way too long I don&#39;t like that okay let&#39;s see here you know I am just messing this up let&#39;s see it&#39;s like this boom boom there we go so what we could do also is do a hydride shift so I see that on the adjacent carbon from the carbocation I have a hydrogen so I could take that hydrogen and do a shift you see how I&#39;m drawing it from the center of the bond there and if I do that hydride shift I&#39;m going to generate this I&#39;m going to add the hydrogen there now the carbocation disappears but where does it reappear it is not going to be on this carbon so what why would this do this because we went from a secondary carbocation 2A tertiary carbocation and we&#39;ve learned that tertiary carbocations are more stable so whenever you generate a carbocation it is a good rule of thumb to look for rearrangements and so we have our hydride shift or hydride rearrangement and now we can have our water come in and attack and so that would be our coordination step so that would give us this [Music] that hydrogen is not visible so I&#39;m gonna violate geometry rules so you can see the hydrogen and then you can see we will do a proton transfer mechanism with some excess water and so that excess water will come and grab that proton to give us our product and that matches product B right so what&#39;s cool about this reaction right here because we know for a fact that we get two different products it supports that there is such thing as an sn1 mechanism it supports the generation of this carbocation intermediate if it wasn&#39;t for this hydride shift step right here we would never form product B and that is one reason why an sn2 mechanism is totally out of consideration because if it was sn2 we would only get this but another thing that kind of alerts you to the fact that this can&#39;t be sn2 is the fact that this is a weak nucleophile and in sn2 and E2 mechanisms it has to be a strong nucleophile or base in order for it to work right so let&#39;s look I&#39;ll erase this and then we&#39;ll look at another type of shift now here is an example of another secondary alkyl halide and it&#39;s going to give us two different products but look at what&#39;s happened here we see that the bromine has been substituted with an o h okay that&#39;s a simple substitution reaction but then when you look at this other product on the right look it&#39;s what&#39;s what it&#39;s been replaced with it&#39;s been replaced with a methyl group how does that happen well the mechanism helps us to answer that question so in order to get two different products here we have to invoke a the first step as the heterolysis step in order to generate a carbocation and what type of carbocation did I generate that&#39;s a secondary carbocation so that&#39;s stable but can it get more stable and that&#39;s what you always have to look for and so what you do is you look at the adjacent carbons and you see if there&#39;s a hydrogen okay so if I look at if I number if I look at this carbon right here I see that there&#39;s a hydrogen adjacent to the carbocation so I&#39;m like hey that could work so I&#39;m like okay let&#39;s try it so let&#39;s just do it so hydride or hydride shift boom so what&#39;s that going to give me I&#39;ll put the new hydrogen there then that generates what that generates a primary carbocation so that was not that&#39;s not going to work it&#39;s not a bad idea to try it because you&#39;re always looking at the adjacent hydrogens and there it is when you hydrate shift it what does it do gives you a primary carbocation so that&#39;s not going to work it&#39;s not going to happen but it was a good try good analysis okay so that&#39;s not going to happen so what if I looked at the other adjacent carbon so here&#39;s our carbon adjacent to its right here and I&#39;m like are there any hydrogens there&#39;s not but remember if they&#39;re methyl groups you can have a methyl shift so this methyl right here or this one or this one it doesn&#39;t matter just one of the methyls can shift over and when that shifts over it&#39;s going to shift like that we will generate our carbocation now was that a good investment of our time yeah because we went from a secondary to what a tertiary carbocation and so that will work and so we just do the next two steps where we take water and attack in and then do the proton transfer we will generate that product right there so more one key takeaway that you have to always remember whenever you generate a carbocation can you do rearrangements to get a more stable carbocation Ary versus tertiary tertiary definitely the stronger or the more stable and more favorable carbocation okay class now it&#39;s your turn to put all these ideas together that we&#39;ve been discussing to see if you fully understand what&#39;s Happening Here so here&#39;s a reaction that I put on the board I&#39;m going to tell you that it is a sn1 reaction which one of these products is the correct product there&#39;s only one so which is the correct and only product and once you&#39;ve figured that out you always have to explain why so like if this is a multiple choice question this is a flip the coin if you did not know it would just flip a coin in that one right that&#39;s not going to do you much good on my exams because for every choice that you make you have to rationalize and justify why you did that so this question could be something more like a free response where it says which is the major product or which is the only product why you have to explain your reasoning so if you just picked the correct answer because you flipped a coin if I had no explanation you are going to lose the majority of the points so I would definitely pause the video now and try to figure this one out on your own and then come back to the video to see what happens so if we come back here we have I told you it&#39;s an sn1 mechanism so we have to do it so if we do an sn1 mechanism here can we the first step is to get that guy to leave right so that&#39;s going to give us what foreign that&#39;s going to give us a primary carbocation but how is that even possible because we said we don&#39;t generate primary carbocations when we do this right it doesn&#39;t happen so if we know that does not happen then we have to invoke just a slightly different mechanism here and that&#39;s where you do the out the methyl shift and the heteroysis step all in the same step so that would be taking this methyl group right here and shifting it to that carbon at the same time the halogen leaves and so that&#39;s going to give us this species right here because we know we cannot generate primary carbocations and now we can then add our water here and then do a proton transfer step to get us to our product so if that attacks there you know never do that that&#39;s a bad idea never draw through the bonds that&#39;s a bad idea all right if you find yourself with a crazy Pathway to get there you can maybe go like this all right there now from there can you see what the product is going to be product B is going to be the only product formed when you do this reaction all right that is I&#39;m putting a stipulation on this though I&#39;m saying hey this is an sn1 mechanism all right now could you get to this product by a different mechanism you could if you invoked an sn2 mechanism if you invoke an sn2 mechanism then you would generate this product but remember with an sn2 you have to use a strong nucleophile if you did that reaction then you would start seeing a as your major product here but then again I&#39;m not going to say we&#39;re going to see a lot of this product why because remember from sn2 a material that when you look at the beta carbon if the beta carbons vary sterically hindered then an sn2 mechanism is going to be very difficult remember that how to put it all together here all right so I take a look at these four different alkyl halides and if I&#39;m going to take each one of these and do the same reaction treat them with water all right to generate our final product as this okay or this so I&#39;m invoking a E1 or an sn1 mechanism which one of the the which one&#39;s going to react the fastest if I have a fluorine or an iodine all right and the answer is as you go down right here this is going to be the slow end this is going to be the fast so if you have an alkyl halide with an iodine it&#39;s going to react the fastest whereas when you have a fluorine it is so slow that it&#39;s totally impractical that we basically say that there&#39;s no reaction when we have fluorine it&#39;s really just these three right here that we use for alkyl halides to do E1 or to do our reactions with when we&#39;re talking about E1 and sn1 now why is this the case well it all comes down to this idea of leaving group ability and we&#39;ve we&#39;ve discussed that before when you just look at the leaving group of abilities so that&#39;s just like this right fluorine doesn&#39;t leave very well iodine leaves very well and we rationalize this due to the fact that when it leaves the fluorine then becomes a fluoride and then iodine and when we compare these two which one&#39;s the most stable anion and when we go down a group it is about size and polarizability this is larger more polarizable more stable so this iodine when it kicks off the alkyl halide it&#39;s chill it&#39;s okay being by itself the fluoride ion is I hate this I need to be attached to the alkyl halide so it never wants to leave huh another way we&#39;ve looked at this is if we look at the iodide and then look at it in water foreign which ones then it&#39;s going to generate our hydronium plus our iodine right and if you compare that to if you did hydrofluoric acid and water you would get h3o Plus our fluoride right and we just ask ourselves which one&#39;s the stronger acid and we clearly know this is the stronger acid and so it&#39;s the strongest acid which makes this a very weak base and in this context a very very good uh leaving group so stronger the acid means it&#39;s conjugate base is better leaving group that&#39;s another way of looking at it and remembering it when looking at sn1 and E1 reactions let&#39;s take a look at what type of substrates can work for these reactions well we definitely want a tertiary alkyl halide that is the best best one that will work secondary alkyl halides can also work but we don&#39;t see they&#39;re a little bit more rare but they can still happen primary alkyl halides no way why won&#39;t that work because they generate a primary carbocation and that&#39;s not possible okay but there are so not it&#39;s really not an exception but if we take a molecule that looks like this okay like that okay and I put a carbocation right there how many people would say no way we can&#39;t form that and how many of you would say that cat form because that&#39;s a primary carbocation and you would be right based off of this definition but we don&#39;t call that a primary carbocation when you have a carbon that&#39;s a cation directly attached to a Benzene ring we call that a benzylic carbocation and benzylic carbocations even though they look like they&#39;re primary they&#39;re not they&#39;re benzyl at carbocations and they are stable and why do you think they&#39;re going to be stable maybe if I draw something like this you see the rest of the structure it&#39;s rested and stabilized that&#39;s why that guy so that&#39;s why that carbocation is stable what&#39;s another one we have something called an allylic cation let&#39;s see if I spelled that right so we have an allylic carbocation which looks like this when you take a first look at it you&#39;re like that can&#39;t form no way that&#39;s a primary carbocation it&#39;s not it is a allylic carbocation now why is that stable why does that form because as you can guess we have a resonance structure and due to Resonance that&#39;s going to stabilize this allylic carbocation enough that it&#39;s us it can form and so you can have a reaction with uh these two type of carbocations so for example if you had a reaction like this let&#39;s see here x2o okay can that undergo an sn1 yes yes it can keep consistent here that that is very plausible right okay let&#39;s see what&#39;s another thing that we need to talk about I know we&#39;re a little bit over but I want to say one more thing teach one more principle then we will conclude this video all right that is can you do can you have this happen all right and we have a halogen there remember that carbon that the halogen is attached to is SP2 hybridized can this process happen can the the halogen act as a good leaving group and leave and generate this no it cannot it cannot happen all right and the reason why it cannot happen is because this carbocation that we&#39;re proposing is so so unstable and the reason why it&#39;s so so unstable is due to the fact to effective electronegativity effective electronegativity activity yeah okay so that does not happen and if you need to review effective electronegativity that&#39;s the reason why that doesn&#39;t form and so the same reason this does not happen okay look at this one put a halogen right there right on the Benzene ring all right can this happens no it cannot okay that cannot happen because why we have a SP2 hybridized carbon and the reason why same idea effective electronegativity okay so that&#39;s where we&#39;ll end on this video

Transcript for:SN1 and E1 Reactions

Transcript for:
SN1 and E1 Reactions