Substitution and Elimination Reactions

hello class today&#39;s discussion is going to focus on these two words here we have What&#39;s called the substitution reaction and an elimination reaction two different uh what we call mechanisms and mechanisms are how the reaction occurs and you will see very shortly that mechanisms are going to be key for your success in organic chemistry if you can learn the mechanisms then you can understand the reactions so as we go through the course I&#39;m going to teach you 10 basic Elementary steps and these 10 basic Elementary steps are the 10 different types of mechanisms that if you understand those 10 Elementary steps which are basically just mechanisms then you will be able to understand 90 of all the reactions that you&#39;ll ever encounter because how many reactions are there there are hundreds of thousands maybe even millions of reactions I don&#39;t know for sure all I know is there are a lot of them and it is not possible to memorize all the reactions but what is more possible is learning and knowing those 10 Elementary steps and if you know those 10 Elementary steps you don&#39;t have to memorize as much so now I&#39;m going to get off my soapbox but that&#39;s key once you need to learn mechanisms it will really help you so what we&#39;re going to learn here is substitution reactions and elimination reactions with alkyl halides so for example we could have carbon here an alkyl halide let&#39;s put a chlorine and then we have some R groups here now these R groups could be uh hydrogens they could be alkyl groups doesn&#39;t really matter we just have an alkyl halide now this is this halogen right here which could be represented by X because we want to incorporate all alkyl halides is attached to what type of carbon okay that is a sp3 hybridized carbon in contrast if you look at something like this you see that this carbon that the halogen is directly attached to is SP2 hybridized so we&#39;re not going to talk about any examples where the halogen is attached to an SP2 hybridized carbon we will talk about that in chapters way down the road we want to focus on this type of alkyl halide and so this you have to understand can you see the difference how this is sp3 and that is SP2 if you don&#39;t then that&#39;s a good I it&#39;s a good idea to figure out the hybridization States now in substitution and elimination reactions we have different components here okay now all have different names so let&#39;s say if we look at an alkyl halide and when we take a look at this alkyl halide we could treat that alkyl halide with a nucleophile I&#39;m going to represent that nucleophile with a negative charge what&#39;s going to happen here when you look at the overall process what has happened we have taken this halogen right here and substituted it with the nucleophile so this is called a substitution reaction when we are substituting the alkyl halide for a nucleophile a nucleophile if you look at the word it&#39;s composed of two parts nucleo and then file what that&#39;s saying is a nucleophile represented by n u is nucleus loving it loves to attack the nucleus so what is going to happen here mechanistically let&#39;s clear off some space here is in a nucleophile we can have it the mechanism looks like this so we&#39;re going to take the nucleophile and we&#39;re going to take the lone pair and that lone pair is going to come and attack the carbon that is directly attached to the halogen and when that occurs What&#39;s Happening Here if I stop with that one orange arrow what have I done I formed that Bond that&#39;s what the orange arrow is showing but look at here how many hydrogens were here already there were two hydrogens so now what do we have we have a Texas carbon and that&#39;s not possible Right not possible so mechanistically what&#39;s going to happen is as the nucleophile comes in and attacks the carbon that&#39;s directly attached to the halogen the carbon halogen Bond breaks and so we already have the three lone pairs on the halogen but then it&#39;s going to break and dump both the electrons onto the halogen so now what we have the nucleophile attacked to that carbon and if we want to number them one two three one two three the nucleophile attacked carbon three you see it&#39;s a test carbon three and then we have the halogen came off so now that halogen is negatively charged and this is what a mechanism is all about it&#39;s showing so this is just an overall showing us the transformation and this is called a substitution reaction yeah the halogen is replaced with the nucleophile great how does that transformation occur that&#39;s where we come up here to the mechanism and if you can learn these Mech this mechanism and all the others that I&#39;m going to show you organic chemistry will be easier I promise you you cannot memorize your way through this course gotta understand the principles here&#39;s the mechanism here and this mechanism has a very special name it is called a s subscript capital n 2 and sn2 we will talk about what that means a little bit later I&#39;m just introducing this at this point so this is a sn2 mechanism when you have a alkyl halide and the nucleophile attacks the alkyl halide in this manner foreign ations well let&#39;s let&#39;s leave this up here so we can compare and contrast there so all the erased at the top now let&#39;s take another alkyl halide looks like that and let&#39;s treat it with a generic base represented by the letter B for base and when we have a base treated with a alkyl halide like this what we&#39;re going to generate is a compound that looks like this if we generate what what&#39;s functional group is that that&#39;s an Alka alkene and then we also generated a BH right there Plus the halogen right there all right so if we look at the lone pairs like so like that okay so what has happened well do you see how the bass did not substitute the X see how the the bass isn&#39;t even attached to the molecule of Interest this is the molecule of Interest the the base isn&#39;t attached to it but what has happened we&#39;ve eliminated the X off of the molecule and we&#39;ve eliminated a hydrogen do you see at this carbon right here if we want to number them so one two three one two three carbon 3 on this side on the alkyl halide we have what three hydrogens on this carbon come over here how many hydrogens do we have on carbon three only two so what have we done we have eliminated the halogen and one of the hydrogens and so when you take it alkyl halide and treat it with the base you get a elimination reaction which gives you a in this particular case an alkene you&#39;re going to form a double bond here now mechanistically how is this happening and we&#39;ll draw it we&#39;ll just use this right here let&#39;s draw the mechanism here how does this happen well the base is going to come and Abstract the proton that carbon hydrogen bond is going to break you see what that carbon hydrogen bond breaking it dumps its electrons right here so that Pi Bond right there is generated from the electrons that are found between the carbon and the hydrogen but if I only have two orange arrows then I have formed a Texas carbon I also have to get the halogen to leave like that so the carbon halogen Bond breaks dumping the electrons in that Bond onto the halogen as you can see here now we have four lone pairs and that is yet going to generate R alkene oftentimes now this is the crazy thing watch this oftentimes the base and the nucleophile can be the same molecule so for example let&#39;s say I show you a base let&#39;s say the base is going to be hydroxide hydroxide can behave as a base whereas it can also behave as a nucleophile so what is the difference between let&#39;s let&#39;s do it in subscript here that we&#39;re going to call this one a base and this one&#39;s going to be our nucleophile okay they&#39;re the same molecule but the way they behave mechanistically is different because in the base you see the mechanism arrows right the base comes and abstracts a proton and then kicks off the halogen by forming a double bond that&#39;s what a base does but a nucleophile is one where it comes in and does a substitution reaction let&#39;s draw on some lone pairs here so we can see the charge so because hydroxide can behave as a base and behave as a nucleophile when we do these reactions we are going to get a mixture of products we&#39;re going to get some of the substitution product and we&#39;re going to get some of the elimination product so this will be an example of an elimination product and this would be an example of a substitution product and so that makes organic synthesizing molecules a bit of a challenge because what if you only want the substitution product or you only want the elimination product it makes it a little difficult if there&#39;s going to be a mixture but there are techniques that we are going to learn that helps us to persuade the reaction to go in one route to form the product that you desire or the one that you want okay so make sure you understand the difference between what a base does and a nucleophile basis abstract protons nucleophiles attack the carbon nucleus so if we have a secondary alkyl halide like this why in order for us to do a substitution reaction why do we need the alkyl halide why do we need that and so we&#39;re going to take a look at that well what the halogen is what more electronegative than the carbon right so if it&#39;s more electronegative then it&#39;s going to pull electron density towards the halogen right and so what&#39;s what is that what&#39;s the effect of that well this carbon right here is going to be partially positive and then when we take a look at the nucleophile what is its charge State it&#39;s negatively charged so this negative charge is going to be attracted to this partial positive carbon and that partial positive is generated because of the electron withdrawing ability of the halogen and so when you have these two species in solution with one another the nucleophile is attracted to this carbon because of its partial positive charge and so we can see mechanistically it&#39;s going to be attracted to that carbon and then the halogen is going to leave and then we would get that product right there so the reason why substitution reactions are occurring is because of the electron withdrawing ability of the halogen to make the carbon that the halogen is directly attached to electrophilic it&#39;s electron pore so in a lot of organic reactions what we have going for us is we have a electron Rich species and then we&#39;re going to have a electron pore and so we have an electron for carbon electron rich nucleophile and it&#39;s just basic laws of attraction electron Rich it&#39;s going to be attracted to electron pore and when we have an electron pore site we call that the electrophile and when we have an electron Rich species we can call it a nucleophile remember nucleus loving what is an electrophile electron loving electrophiles loves to accept electrons now the reason why I said electron Rich species may be nucleophiles it&#39;s because we just recently saw that hydroxide could act as a base as well so elect but I&#39;m calling this a base but it is still electron rich so electron-rich species can be a nucleophile or a base but the point is that electron-rich species are attracted to electron poor species or the electrophile okay now the next reason and another thing that we need to understand as to why we need a halogen is because we have a phrase the halogen right here is a good leaving group and I&#39;m just going to abbreviate leaving group with LG halogens are good leaving groups so you see in a substitution reaction the halogen has to leave and if you go back to your notes and look at the the elimination reaction the halogen also has to leave halogens are good leaving groups if I contrast that why does this reaction right here let&#39;s put in a methyl group and treat it with a hydroxide why does this reaction not work to give us this product why does that not work the reason why this does not work is because methyl groups are not good leaving groups this mechanism Arrow right here or both these mechanism arrows don&#39;t occur it can&#39;t because the methyl group will never leave it&#39;s a horrible leaving group but in contrast halogens are really good leaving groups okay so how do you know if a group on a molecule is a good leaving group so here is the definition or the statement that will tell us if a leaving group is good or not and it says good leaving groups are the conjugate bases of strong acids okay that&#39;s the answer but now you have to translate what that sentence is trying to say so to help you understand that let&#39;s look at these two examples here so we want to do a reaction with these are is this iodine right here a good leaving group compared to this hydroxide group is that a good leaving group so what we want to ultimately do is can you do this reaction can you take a nucleophile and go in like this are both of these reactions possible or only one of them and if so which one well let&#39;s go back to this statement so this is how you do it you look at your leaving group or your proposed leaving group and you turn it into an acid okay so if we take the iodine we can turn that into hydroiodic acid just by adding that h we will then add it to some water okay and we will see what happens does and now we are going back to acid-base chemistry here so this is going to come in the iodine is going to leave you see how it&#39;s leaving off of the hydrogen right there and what are we going to make we&#39;re going to make our hydronium there&#39;s the conjugate acid and then there just like that we go to the pka table and we figure out what&#39;s the pka of the hydroiodic acid and it has a pka of a negative 11. all right and hydronium is what negative 1.7 I think let&#39;s see here and then we contrast that all right with the other one what was the other leaving group it was a hydroxide so we will take the hydroxide and add a hydrogen to it okay and then put that into water and see what happens right all right what&#39;s the pka of water about 16 15 or 16. hi so at equilibrium what&#39;s going to be favored let&#39;s go back to this guy well it&#39;s going to be favored to what the side where there&#39;s the weakest acid so equilibrium is going to look more like this let&#39;s go to favor clear over to the right but this guy on in contrast here that&#39;s the weaker acid so it&#39;s going to favor much more to the left right this is going to favor this way so if we come back to our statement good leaving groups are the conjugate basis of strong acids so you see that this compared to that the hydroiodic acid is a much stronger acid than water so when we look at the conjugate bases of these two here&#39;s the conjugate base and here&#39;s the conjugate base right there conjugate bases of strong acids are good leaving groups this is the conjugate base of this guy right here it&#39;s not a strong acid so it&#39;s not a good leaving group so A good rule of thumb is in order for a leaving group to be a good leaving group you want to look at their asset forms and you want their PKA to be what Less Than Zero see if I make writing that down right yeah you typically want negative values for the acid component and then if it has a negative PKA then the conjugate base of that acid is going to be a good leaving group so if we come back to our example here which reaction will occur between this molecule and this molecule you would say hey hydroxide is not a good leaving group so that reaction cannot occur but iodine right there that is a good leaving group so the substitution reaction can occur and I used hydroxide as an example because just put that in your mind right now hydroxides this group right here hydroxides are not good leaving groups got it write that down not good leaving groups let&#39;s consider this reaction right here we can treat it with any nucleophile okay mechanistically we take the lone pair the electrons on the nucleophile and we take it and attack the electron poor carbon and then break the carbon halogen Bond that&#39;s going to give us this that would have to be a x right so what words would you use to describe what&#39;s happening here you would say hey this is a substitution reaction that is correct what else can we say about it well what type of mechanism is it hopefully you all see that it is a sn2 mechanism compare this substitution reaction to this one that I&#39;m going to draw now [Music] something like that look at this so we have lone pairs on that guy so I&#39;m going to make that leaf and then I will have something that looks like this you&#39;ll have a carbocation plus the halogen that just left and then we&#39;ll do another step right here where we will now introduce the nucleophile like that so when you look at this reaction in the bottom what could you what words could you use to describe what has happened in this reaction here is it an elimination or a substitution well it is a substitution reaction because we are replacing or substituting this halogen with the nucleophile so both reactions are a substitution but you can clearly see the mechanism between the two are different so this one on the bottom right here has a name and that&#39;s called a s and one mechanism so both are substitution reactions but the way they occur is different and that is the beauty of mechanisms so this name right here sn2 what it&#39;s referring to is s stands for substitution so you can see in the sn1 reaction you still have an S because they are substitution reactions now the N subscript right here is telling us that it is nucleophilic we have a nucleophile doing the attack so substitution nucleophilic now this 2 right here stands for bimolecular two means by molecular bimolecular or the 2 is referring to that in this sn2 reaction we have two molecules we have the alkyl halide and the nucleophile reacting at the same time and we call that a concerted reaction and since we have one two bimolecular so we put a 2 there sn2 and that&#39;s telling us that this substitution reaction is happening at this in a concerted fashion two molecules are coming together and reacting now can you see contrast what this one is going to stand for and you can you take a guess so what does one equal so the one stands for unimolecular and in a unimolecular reaction do you see these green arrows here that is an elemental step remember me referring to at the beginning of the video the elemental steps you can see for each Elemental step there&#39;s only one molecule involved and so when we have a alkyl halide and then the leaving group leaves in this manner we call this a heterolysis step so there this Elemental step right here we&#39;re showing I&#39;m showing you the mechanism the leaving group just leaves and generates a carbocation now in this step right here we call that a coordination step you can see that the nucleophile is negatively charged positively charged there&#39;s going to be an attraction there and they&#39;re coordinating with one another so that&#39;s a coordination step and then that forms our product okay so this sn1 for this one part right here is referring to this first part of the mechanism it&#39;s only using one molecule in the first step these the Second Step you can see it&#39;s using two molecules we&#39;re not worried about that in this definition okay substitution or sn1 is referring to this right there now we&#39;re going to learn more about why that&#39;s the case but from this point on we are going to focus our attention only on this concerted mechanism and learn more about that so you may be asking well what do we call this sn1 right here we could call this a stepwise it is a stepwise mechanism because it goes it uses one elemental step then another Elementary step that&#39;s what I&#39;m trying to say Elementary step than that another Elementary step okay so now let&#39;s delve into this concerted reaction sn2 there&#39;s a lot to talk about a lot to learn those many smart smart people were studying reactions and trying to learn the mechanisms of before they knew what it was called sn2 mechanisms they had to figure it out and one thing that they learned about this is if they took the alkyl halide and double this concentration but kept the nucleophile concentration constant what they noticed was that the rate of this reaction or the speed at which this reaction occurs doubled and then they said hey well what if we kept this the alkyl halide concentration constant and double the nucleophile concentration and what they found was that the rate also doubled and so for our sn2 mechanism the rate was simply uh the alkyl halide halide concentration multiplied by the nucleophile&#39;s concentration and going back to rate laws you will see that when that&#39;s the case when this doubles the rate doubles or when you double that and the rate doubles we have a second order reaction and the second order reaction helped side just understanding the kinetics here helped the scientists figure out the mechanism okay So based off of this observation that it was second order LED them to believe that this reaction had to have happen in a concerted fashion in one step so this is one just one bit of information to help us understand the mechanism but it is also a good to understand this and know this that in an sn2 reaction it is second order so if you double one of these is going to double the rate another very important thing about sn2 reactions is the stereo specificity what do I mean by that well what if I took a molecule that looked like the alkyl halide that looked like this now let&#39;s put an F an ethyl group there and we will keep that as a methyl okay now when you take a nucleophile and let&#39;s use this one right here okay we take this nucleophile before it even reacts we can look at this carbon here and say hey there&#39;s one two three four different groups so we know that is a stereocenter now what is the configuration of that stereocenter you will see that it has a r configuration now when we do this sn2 reaction when we do this sn2 reaction we know that means it&#39;s going to be a substitution so we&#39;re going to replace this bromine right here look at how it does it it&#39;s very very important to see this look the wrong molecule look at how it does it h chch3 and our ethyl can you see what happened we did a substitution reaction yes we did that so what should be over here there should be the bromide that&#39;s floating around right the leaving group but what happened to that stereocenter is it still r you will notice that it is not still r it is an s configuration so what has happened we have inverted stereocenter whenever you have a an alkyl halide attached to a carbon that is a stereocenter and you do a sn2 reaction you are going to invert the stereocenter if it&#39;s if it starts as an R it will go to an s and vice versa if it started with an S it would be inverted to an R every single time now the reason for this is the mechanism of the sn2 the way it approaches okay so I&#39;m going to show you two approaches and show you which one&#39;s the correct one so the first approach that I&#39;m going to show you no yes the first approach I&#39;m going to show you is called backside attack so we&#39;ll say back side attack and backside attack means that when you look at the molecule you find the alkyl halide and the nucleophile is going to attack 180 degrees from it so that means the alkyl halide has to approach from this back side so where&#39;s the back side with respect to this bromine it&#39;s over here so this hell or this nucleophile has to come in and go all the way around and attack backside and when it attacks backside it&#39;s going to invert the stereocenter what does not happen is front side attack the nucleophile does not come in and attack from the front you see how I&#39;m approaching it from the front side of this bromine and then it leaves it does not do that so I missed one arrow with the backside attack when the nucleophile comes in attacks backside yes the leaving group does have to leave okay so we have two ways to attack this alkyl halide backside attack and front side attack front side attack does not happen backside does now why would you say front side attack why would that not work so that&#39;s a good question good thought we need to clean this board so we can answer that so if we take and look at front side attack first let&#39;s look at it in the transition state so the transition state would look something like this um there&#39;s our methyl Group which I&#39;ll put right there me can stand for our methyl uh and then there&#39;s my ethyl group here so if this happened to be front side attack what&#39;s going to happen then the bromine bond is going to start to break right at the same time that the nucleophile bond is starting to form now this is kind of weird because this guy&#39;s trying to come in and this guy&#39;s trying to come out at the same time it&#39;s a concerted reaction and so it&#39;s just going to get really really crowded they&#39;re going to like be getting in the way of one another like hey I&#39;m coming no you&#39;re leaving and it&#39;s just like a little chaos here and another thing is that there are a lot of electrons here major electron clouds partially negative fully negative and they&#39;re just going to repel each other it is a electronic argument that there&#39;s just too many electrons around here it&#39;s not going to work but when you contrast the transition state to Backside attack the way it really happens you will see foreign something that looks like that now since these ball these groups are 180 degrees from one not one another you can see that the sulfur carbon bond is forming at the same time the carbon bromine bond is breaking they&#39;re far away from each other no electron clouds are going to interfere this guy&#39;s coming in at the same time that guy&#39;s leaving it is just so much better foreign so that&#39;s one explanation as to why backside attack is the only way it occurs a more sophisticated way is to look at molecular orbitals and so I&#39;m going to have to clean up this board and draw those molecular orbitals so you can see what&#39;s going on okay molecular orbital Theory here what we have is I change the molecule a little bit on the alkyl halide I replace the ethyl group and the methyl groups with hydrogens just to make the drawing easier okay but even if I replace these back with the ethyl and methyl group The principle would still be the same so why do we not have front side attack well in MO Theory we realized that the whole molecule right is a molecular orbital and you see in green here we have our methyl bromide here here&#39;s our carbon there&#39;s our bromine so this in green matches this now what you see in blue and pink right here is the lumo of the alkyl halide and chemistry is really simple when we say it chemistry is just the reaction of or the interaction between the homo and the lumo it&#39;s really that&#39;s all it is okay so if we take so this right here right there is the lumo so if we take the homo of the nuclear file and try to attack from the front side right there to attack that molecular orbital why is that a bad thing well because I&#39;m not I don&#39;t have my Mo image 100 correct yet because right here right down there is a node no electron density so if we&#39;re proposing that this nucleophile is going to come an attack that&#39;s impossible because there&#39;s a node right there no electrons can exist right there so you can&#39;t attack from the front because a node is a place where you&#39;ll never find electrons so it&#39;s impossible but if you look at backside attack all right you can see that hey look at this guy right here lots and lots of orbital overlap for the electrons in the homo to be placed into the lumo right so ml Theory helps us to explain why backside attack is a per is the only way to go and then the um what&#39;s it called the transition state drawing that I drew let&#39;s see well how much time do we have here so let&#39;s just wrap this uh idea by doing one example so if I have a reaction that looks something like this okay let&#39;s put chlorine okay and I want to treat this alkyl halide with hydroxide and I tell you from the beginning that this is going to be a sn2 mechanism what is the product going to be now you could memorize it which I I&#39;m telling you right now is a horrible idea because you&#39;re going to get answer choices like this if we call that a we call that b okay so you could see on an exam you could have an answer choice of a B or a and b and the problems can get even more complicated we could have answer choices of a b c and d and then it could be is it A and B A and C and it&#39;s just going to get very confusing if you&#39;re trying to memorize it but if we do the mechanism I told you it is an sn2 so just do the mechanism foreign you&#39;re like sn2 concerted it has to invert the stereocenter because we know it goes by backside attack so it can only be this product because the mechanism says that it can only attack backside attack it&#39;s going to invert it so the only answer is that this is not going to be a mixture between a and b the mechanism doesn&#39;t say that okay very very important to understand that the mechanism is going to help you predict the product now if I told you this was an sn1 mechanism then the answer is going to be different but we don&#39;t want to talk about sn1s yet because we are focusing our energy on sn2 only one product one answer it is inverted the stereocenter okay we&#39;ll end there

hello class today&amp;#39;s discussion is going to focus on these two words here we have What&amp;#39;s called the substitution reaction and an elimination reaction two different uh what we call mechanisms and mechanisms are how the reaction occurs and you will see very shortly that mechanisms are going to be key for your success in organic chemistry if you can learn the mechanisms then you can understand the reactions so as we go through the course I&amp;#39;m going to teach you 10 basic Elementary steps and these 10 basic Elementary steps are the 10 different types of mechanisms that if you understand those 10 Elementary steps which are basically just mechanisms then you will be able to understand 90 of all the reactions that you&amp;#39;ll ever encounter because how many reactions are there there are hundreds of thousands maybe even millions of reactions I don&amp;#39;t know for sure all I know is there are a lot of them and it is not possible to memorize all the reactions but what is more possible is learning and knowing those 10 Elementary steps and if you know those 10 Elementary steps you don&amp;#39;t have to memorize as much so now I&amp;#39;m going to get off my soapbox but that&amp;#39;s key once you need to learn mechanisms it will really help you so what we&amp;#39;re going to learn here is substitution reactions and elimination reactions with alkyl halides so for example we could have carbon here an alkyl halide let&amp;#39;s put a chlorine and then we have some R groups here now these R groups could be uh hydrogens they could be alkyl groups doesn&amp;#39;t really matter we just have an alkyl halide now this is this halogen right here which could be represented by X because we want to incorporate all alkyl halides is attached to what type of carbon okay that is a sp3 hybridized carbon in contrast if you look at something like this you see that this carbon that the halogen is directly attached to is SP2 hybridized so we&amp;#39;re not going to talk about any examples where the halogen is attached to an SP2 hybridized carbon we will talk about that in chapters way down the road we want to focus on this type of alkyl halide and so this you have to understand can you see the difference how this is sp3 and that is SP2 if you don&amp;#39;t then that&amp;#39;s a good I it&amp;#39;s a good idea to figure out the hybridization States now in substitution and elimination reactions we have different components here okay now all have different names so let&amp;#39;s say if we look at an alkyl halide and when we take a look at this alkyl halide we could treat that alkyl halide with a nucleophile I&amp;#39;m going to represent that nucleophile with a negative charge what&amp;#39;s going to happen here when you look at the overall process what has happened we have taken this halogen right here and substituted it with the nucleophile so this is called a substitution reaction when we are substituting the alkyl halide for a nucleophile a nucleophile if you look at the word it&amp;#39;s composed of two parts nucleo and then file what that&amp;#39;s saying is a nucleophile represented by n u is nucleus loving it loves to attack the nucleus so what is going to happen here mechanistically let&amp;#39;s clear off some space here is in a nucleophile we can have it the mechanism looks like this so we&amp;#39;re going to take the nucleophile and we&amp;#39;re going to take the lone pair and that lone pair is going to come and attack the carbon that is directly attached to the halogen and when that occurs What&amp;#39;s Happening Here if I stop with that one orange arrow what have I done I formed that Bond that&amp;#39;s what the orange arrow is showing but look at here how many hydrogens were here already there were two hydrogens so now what do we have we have a Texas carbon and that&amp;#39;s not possible Right not possible so mechanistically what&amp;#39;s going to happen is as the nucleophile comes in and attacks the carbon that&amp;#39;s directly attached to the halogen the carbon halogen Bond breaks and so we already have the three lone pairs on the halogen but then it&amp;#39;s going to break and dump both the electrons onto the halogen so now what we have the nucleophile attacked to that carbon and if we want to number them one two three one two three the nucleophile attacked carbon three you see it&amp;#39;s a test carbon three and then we have the halogen came off so now that halogen is negatively charged and this is what a mechanism is all about it&amp;#39;s showing so this is just an overall showing us the transformation and this is called a substitution reaction yeah the halogen is replaced with the nucleophile great how does that transformation occur that&amp;#39;s where we come up here to the mechanism and if you can learn these Mech this mechanism and all the others that I&amp;#39;m going to show you organic chemistry will be easier I promise you you cannot memorize your way through this course gotta understand the principles here&amp;#39;s the mechanism here and this mechanism has a very special name it is called a s subscript capital n 2 and sn2 we will talk about what that means a little bit later I&amp;#39;m just introducing this at this point so this is a sn2 mechanism when you have a alkyl halide and the nucleophile attacks the alkyl halide in this manner foreign ations well let&amp;#39;s let&amp;#39;s leave this up here so we can compare and contrast there so all the erased at the top now let&amp;#39;s take another alkyl halide looks like that and let&amp;#39;s treat it with a generic base represented by the letter B for base and when we have a base treated with a alkyl halide like this what we&amp;#39;re going to generate is a compound that looks like this if we generate what what&amp;#39;s functional group is that that&amp;#39;s an Alka alkene and then we also generated a BH right there Plus the halogen right there all right so if we look at the lone pairs like so like that okay so what has happened well do you see how the bass did not substitute the X see how the the bass isn&amp;#39;t even attached to the molecule of Interest this is the molecule of Interest the the base isn&amp;#39;t attached to it but what has happened we&amp;#39;ve eliminated the X off of the molecule and we&amp;#39;ve eliminated a hydrogen do you see at this carbon right here if we want to number them so one two three one two three carbon 3 on this side on the alkyl halide we have what three hydrogens on this carbon come over here how many hydrogens do we have on carbon three only two so what have we done we have eliminated the halogen and one of the hydrogens and so when you take it alkyl halide and treat it with the base you get a elimination reaction which gives you a in this particular case an alkene you&amp;#39;re going to form a double bond here now mechanistically how is this happening and we&amp;#39;ll draw it we&amp;#39;ll just use this right here let&amp;#39;s draw the mechanism here how does this happen well the base is going to come and Abstract the proton that carbon hydrogen bond is going to break you see what that carbon hydrogen bond breaking it dumps its electrons right here so that Pi Bond right there is generated from the electrons that are found between the carbon and the hydrogen but if I only have two orange arrows then I have formed a Texas carbon I also have to get the halogen to leave like that so the carbon halogen Bond breaks dumping the electrons in that Bond onto the halogen as you can see here now we have four lone pairs and that is yet going to generate R alkene oftentimes now this is the crazy thing watch this oftentimes the base and the nucleophile can be the same molecule so for example let&amp;#39;s say I show you a base let&amp;#39;s say the base is going to be hydroxide hydroxide can behave as a base whereas it can also behave as a nucleophile so what is the difference between let&amp;#39;s let&amp;#39;s do it in subscript here that we&amp;#39;re going to call this one a base and this one&amp;#39;s going to be our nucleophile okay they&amp;#39;re the same molecule but the way they behave mechanistically is different because in the base you see the mechanism arrows right the base comes and abstracts a proton and then kicks off the halogen by forming a double bond that&amp;#39;s what a base does but a nucleophile is one where it comes in and does a substitution reaction let&amp;#39;s draw on some lone pairs here so we can see the charge so because hydroxide can behave as a base and behave as a nucleophile when we do these reactions we are going to get a mixture of products we&amp;#39;re going to get some of the substitution product and we&amp;#39;re going to get some of the elimination product so this will be an example of an elimination product and this would be an example of a substitution product and so that makes organic synthesizing molecules a bit of a challenge because what if you only want the substitution product or you only want the elimination product it makes it a little difficult if there&amp;#39;s going to be a mixture but there are techniques that we are going to learn that helps us to persuade the reaction to go in one route to form the product that you desire or the one that you want okay so make sure you understand the difference between what a base does and a nucleophile basis abstract protons nucleophiles attack the carbon nucleus so if we have a secondary alkyl halide like this why in order for us to do a substitution reaction why do we need the alkyl halide why do we need that and so we&amp;#39;re going to take a look at that well what the halogen is what more electronegative than the carbon right so if it&amp;#39;s more electronegative then it&amp;#39;s going to pull electron density towards the halogen right and so what&amp;#39;s what is that what&amp;#39;s the effect of that well this carbon right here is going to be partially positive and then when we take a look at the nucleophile what is its charge State it&amp;#39;s negatively charged so this negative charge is going to be attracted to this partial positive carbon and that partial positive is generated because of the electron withdrawing ability of the halogen and so when you have these two species in solution with one another the nucleophile is attracted to this carbon because of its partial positive charge and so we can see mechanistically it&amp;#39;s going to be attracted to that carbon and then the halogen is going to leave and then we would get that product right there so the reason why substitution reactions are occurring is because of the electron withdrawing ability of the halogen to make the carbon that the halogen is directly attached to electrophilic it&amp;#39;s electron pore so in a lot of organic reactions what we have going for us is we have a electron Rich species and then we&amp;#39;re going to have a electron pore and so we have an electron for carbon electron rich nucleophile and it&amp;#39;s just basic laws of attraction electron Rich it&amp;#39;s going to be attracted to electron pore and when we have an electron pore site we call that the electrophile and when we have an electron Rich species we can call it a nucleophile remember nucleus loving what is an electrophile electron loving electrophiles loves to accept electrons now the reason why I said electron Rich species may be nucleophiles it&amp;#39;s because we just recently saw that hydroxide could act as a base as well so elect but I&amp;#39;m calling this a base but it is still electron rich so electron-rich species can be a nucleophile or a base but the point is that electron-rich species are attracted to electron poor species or the electrophile okay now the next reason and another thing that we need to understand as to why we need a halogen is because we have a phrase the halogen right here is a good leaving group and I&amp;#39;m just going to abbreviate leaving group with LG halogens are good leaving groups so you see in a substitution reaction the halogen has to leave and if you go back to your notes and look at the the elimination reaction the halogen also has to leave halogens are good leaving groups if I contrast that why does this reaction right here let&amp;#39;s put in a methyl group and treat it with a hydroxide why does this reaction not work to give us this product why does that not work the reason why this does not work is because methyl groups are not good leaving groups this mechanism Arrow right here or both these mechanism arrows don&amp;#39;t occur it can&amp;#39;t because the methyl group will never leave it&amp;#39;s a horrible leaving group but in contrast halogens are really good leaving groups okay so how do you know if a group on a molecule is a good leaving group so here is the definition or the statement that will tell us if a leaving group is good or not and it says good leaving groups are the conjugate bases of strong acids okay that&amp;#39;s the answer but now you have to translate what that sentence is trying to say so to help you understand that let&amp;#39;s look at these two examples here so we want to do a reaction with these are is this iodine right here a good leaving group compared to this hydroxide group is that a good leaving group so what we want to ultimately do is can you do this reaction can you take a nucleophile and go in like this are both of these reactions possible or only one of them and if so which one well let&amp;#39;s go back to this statement so this is how you do it you look at your leaving group or your proposed leaving group and you turn it into an acid okay so if we take the iodine we can turn that into hydroiodic acid just by adding that h we will then add it to some water okay and we will see what happens does and now we are going back to acid-base chemistry here so this is going to come in the iodine is going to leave you see how it&amp;#39;s leaving off of the hydrogen right there and what are we going to make we&amp;#39;re going to make our hydronium there&amp;#39;s the conjugate acid and then there just like that we go to the pka table and we figure out what&amp;#39;s the pka of the hydroiodic acid and it has a pka of a negative 11. all right and hydronium is what negative 1.7 I think let&amp;#39;s see here and then we contrast that all right with the other one what was the other leaving group it was a hydroxide so we will take the hydroxide and add a hydrogen to it okay and then put that into water and see what happens right all right what&amp;#39;s the pka of water about 16 15 or 16. hi so at equilibrium what&amp;#39;s going to be favored let&amp;#39;s go back to this guy well it&amp;#39;s going to be favored to what the side where there&amp;#39;s the weakest acid so equilibrium is going to look more like this let&amp;#39;s go to favor clear over to the right but this guy on in contrast here that&amp;#39;s the weaker acid so it&amp;#39;s going to favor much more to the left right this is going to favor this way so if we come back to our statement good leaving groups are the conjugate basis of strong acids so you see that this compared to that the hydroiodic acid is a much stronger acid than water so when we look at the conjugate bases of these two here&amp;#39;s the conjugate base and here&amp;#39;s the conjugate base right there conjugate bases of strong acids are good leaving groups this is the conjugate base of this guy right here it&amp;#39;s not a strong acid so it&amp;#39;s not a good leaving group so A good rule of thumb is in order for a leaving group to be a good leaving group you want to look at their asset forms and you want their PKA to be what Less Than Zero see if I make writing that down right yeah you typically want negative values for the acid component and then if it has a negative PKA then the conjugate base of that acid is going to be a good leaving group so if we come back to our example here which reaction will occur between this molecule and this molecule you would say hey hydroxide is not a good leaving group so that reaction cannot occur but iodine right there that is a good leaving group so the substitution reaction can occur and I used hydroxide as an example because just put that in your mind right now hydroxides this group right here hydroxides are not good leaving groups got it write that down not good leaving groups let&amp;#39;s consider this reaction right here we can treat it with any nucleophile okay mechanistically we take the lone pair the electrons on the nucleophile and we take it and attack the electron poor carbon and then break the carbon halogen Bond that&amp;#39;s going to give us this that would have to be a x right so what words would you use to describe what&amp;#39;s happening here you would say hey this is a substitution reaction that is correct what else can we say about it well what type of mechanism is it hopefully you all see that it is a sn2 mechanism compare this substitution reaction to this one that I&amp;#39;m going to draw now [Music] something like that look at this so we have lone pairs on that guy so I&amp;#39;m going to make that leaf and then I will have something that looks like this you&amp;#39;ll have a carbocation plus the halogen that just left and then we&amp;#39;ll do another step right here where we will now introduce the nucleophile like that so when you look at this reaction in the bottom what could you what words could you use to describe what has happened in this reaction here is it an elimination or a substitution well it is a substitution reaction because we are replacing or substituting this halogen with the nucleophile so both reactions are a substitution but you can clearly see the mechanism between the two are different so this one on the bottom right here has a name and that&amp;#39;s called a s and one mechanism so both are substitution reactions but the way they occur is different and that is the beauty of mechanisms so this name right here sn2 what it&amp;#39;s referring to is s stands for substitution so you can see in the sn1 reaction you still have an S because they are substitution reactions now the N subscript right here is telling us that it is nucleophilic we have a nucleophile doing the attack so substitution nucleophilic now this 2 right here stands for bimolecular two means by molecular bimolecular or the 2 is referring to that in this sn2 reaction we have two molecules we have the alkyl halide and the nucleophile reacting at the same time and we call that a concerted reaction and since we have one two bimolecular so we put a 2 there sn2 and that&amp;#39;s telling us that this substitution reaction is happening at this in a concerted fashion two molecules are coming together and reacting now can you see contrast what this one is going to stand for and you can you take a guess so what does one equal so the one stands for unimolecular and in a unimolecular reaction do you see these green arrows here that is an elemental step remember me referring to at the beginning of the video the elemental steps you can see for each Elemental step there&amp;#39;s only one molecule involved and so when we have a alkyl halide and then the leaving group leaves in this manner we call this a heterolysis step so there this Elemental step right here we&amp;#39;re showing I&amp;#39;m showing you the mechanism the leaving group just leaves and generates a carbocation now in this step right here we call that a coordination step you can see that the nucleophile is negatively charged positively charged there&amp;#39;s going to be an attraction there and they&amp;#39;re coordinating with one another so that&amp;#39;s a coordination step and then that forms our product okay so this sn1 for this one part right here is referring to this first part of the mechanism it&amp;#39;s only using one molecule in the first step these the Second Step you can see it&amp;#39;s using two molecules we&amp;#39;re not worried about that in this definition okay substitution or sn1 is referring to this right there now we&amp;#39;re going to learn more about why that&amp;#39;s the case but from this point on we are going to focus our attention only on this concerted mechanism and learn more about that so you may be asking well what do we call this sn1 right here we could call this a stepwise it is a stepwise mechanism because it goes it uses one elemental step then another Elementary step that&amp;#39;s what I&amp;#39;m trying to say Elementary step than that another Elementary step okay so now let&amp;#39;s delve into this concerted reaction sn2 there&amp;#39;s a lot to talk about a lot to learn those many smart smart people were studying reactions and trying to learn the mechanisms of before they knew what it was called sn2 mechanisms they had to figure it out and one thing that they learned about this is if they took the alkyl halide and double this concentration but kept the nucleophile concentration constant what they noticed was that the rate of this reaction or the speed at which this reaction occurs doubled and then they said hey well what if we kept this the alkyl halide concentration constant and double the nucleophile concentration and what they found was that the rate also doubled and so for our sn2 mechanism the rate was simply uh the alkyl halide halide concentration multiplied by the nucleophile&amp;#39;s concentration and going back to rate laws you will see that when that&amp;#39;s the case when this doubles the rate doubles or when you double that and the rate doubles we have a second order reaction and the second order reaction helped side just understanding the kinetics here helped the scientists figure out the mechanism okay So based off of this observation that it was second order LED them to believe that this reaction had to have happen in a concerted fashion in one step so this is one just one bit of information to help us understand the mechanism but it is also a good to understand this and know this that in an sn2 reaction it is second order so if you double one of these is going to double the rate another very important thing about sn2 reactions is the stereo specificity what do I mean by that well what if I took a molecule that looked like the alkyl halide that looked like this now let&amp;#39;s put an F an ethyl group there and we will keep that as a methyl okay now when you take a nucleophile and let&amp;#39;s use this one right here okay we take this nucleophile before it even reacts we can look at this carbon here and say hey there&amp;#39;s one two three four different groups so we know that is a stereocenter now what is the configuration of that stereocenter you will see that it has a r configuration now when we do this sn2 reaction when we do this sn2 reaction we know that means it&amp;#39;s going to be a substitution so we&amp;#39;re going to replace this bromine right here look at how it does it it&amp;#39;s very very important to see this look the wrong molecule look at how it does it h chch3 and our ethyl can you see what happened we did a substitution reaction yes we did that so what should be over here there should be the bromide that&amp;#39;s floating around right the leaving group but what happened to that stereocenter is it still r you will notice that it is not still r it is an s configuration so what has happened we have inverted stereocenter whenever you have a an alkyl halide attached to a carbon that is a stereocenter and you do a sn2 reaction you are going to invert the stereocenter if it&amp;#39;s if it starts as an R it will go to an s and vice versa if it started with an S it would be inverted to an R every single time now the reason for this is the mechanism of the sn2 the way it approaches okay so I&amp;#39;m going to show you two approaches and show you which one&amp;#39;s the correct one so the first approach that I&amp;#39;m going to show you no yes the first approach I&amp;#39;m going to show you is called backside attack so we&amp;#39;ll say back side attack and backside attack means that when you look at the molecule you find the alkyl halide and the nucleophile is going to attack 180 degrees from it so that means the alkyl halide has to approach from this back side so where&amp;#39;s the back side with respect to this bromine it&amp;#39;s over here so this hell or this nucleophile has to come in and go all the way around and attack backside and when it attacks backside it&amp;#39;s going to invert the stereocenter what does not happen is front side attack the nucleophile does not come in and attack from the front you see how I&amp;#39;m approaching it from the front side of this bromine and then it leaves it does not do that so I missed one arrow with the backside attack when the nucleophile comes in attacks backside yes the leaving group does have to leave okay so we have two ways to attack this alkyl halide backside attack and front side attack front side attack does not happen backside does now why would you say front side attack why would that not work so that&amp;#39;s a good question good thought we need to clean this board so we can answer that so if we take and look at front side attack first let&amp;#39;s look at it in the transition state so the transition state would look something like this um there&amp;#39;s our methyl Group which I&amp;#39;ll put right there me can stand for our methyl uh and then there&amp;#39;s my ethyl group here so if this happened to be front side attack what&amp;#39;s going to happen then the bromine bond is going to start to break right at the same time that the nucleophile bond is starting to form now this is kind of weird because this guy&amp;#39;s trying to come in and this guy&amp;#39;s trying to come out at the same time it&amp;#39;s a concerted reaction and so it&amp;#39;s just going to get really really crowded they&amp;#39;re going to like be getting in the way of one another like hey I&amp;#39;m coming no you&amp;#39;re leaving and it&amp;#39;s just like a little chaos here and another thing is that there are a lot of electrons here major electron clouds partially negative fully negative and they&amp;#39;re just going to repel each other it is a electronic argument that there&amp;#39;s just too many electrons around here it&amp;#39;s not going to work but when you contrast the transition state to Backside attack the way it really happens you will see foreign something that looks like that now since these ball these groups are 180 degrees from one not one another you can see that the sulfur carbon bond is forming at the same time the carbon bromine bond is breaking they&amp;#39;re far away from each other no electron clouds are going to interfere this guy&amp;#39;s coming in at the same time that guy&amp;#39;s leaving it is just so much better foreign so that&amp;#39;s one explanation as to why backside attack is the only way it occurs a more sophisticated way is to look at molecular orbitals and so I&amp;#39;m going to have to clean up this board and draw those molecular orbitals so you can see what&amp;#39;s going on okay molecular orbital Theory here what we have is I change the molecule a little bit on the alkyl halide I replace the ethyl group and the methyl groups with hydrogens just to make the drawing easier okay but even if I replace these back with the ethyl and methyl group The principle would still be the same so why do we not have front side attack well in MO Theory we realized that the whole molecule right is a molecular orbital and you see in green here we have our methyl bromide here here&amp;#39;s our carbon there&amp;#39;s our bromine so this in green matches this now what you see in blue and pink right here is the lumo of the alkyl halide and chemistry is really simple when we say it chemistry is just the reaction of or the interaction between the homo and the lumo it&amp;#39;s really that&amp;#39;s all it is okay so if we take so this right here right there is the lumo so if we take the homo of the nuclear file and try to attack from the front side right there to attack that molecular orbital why is that a bad thing well because I&amp;#39;m not I don&amp;#39;t have my Mo image 100 correct yet because right here right down there is a node no electron density so if we&amp;#39;re proposing that this nucleophile is going to come an attack that&amp;#39;s impossible because there&amp;#39;s a node right there no electrons can exist right there so you can&amp;#39;t attack from the front because a node is a place where you&amp;#39;ll never find electrons so it&amp;#39;s impossible but if you look at backside attack all right you can see that hey look at this guy right here lots and lots of orbital overlap for the electrons in the homo to be placed into the lumo right so ml Theory helps us to explain why backside attack is a per is the only way to go and then the um what&amp;#39;s it called the transition state drawing that I drew let&amp;#39;s see well how much time do we have here so let&amp;#39;s just wrap this uh idea by doing one example so if I have a reaction that looks something like this okay let&amp;#39;s put chlorine okay and I want to treat this alkyl halide with hydroxide and I tell you from the beginning that this is going to be a sn2 mechanism what is the product going to be now you could memorize it which I I&amp;#39;m telling you right now is a horrible idea because you&amp;#39;re going to get answer choices like this if we call that a we call that b okay so you could see on an exam you could have an answer choice of a B or a and b and the problems can get even more complicated we could have answer choices of a b c and d and then it could be is it A and B A and C and it&amp;#39;s just going to get very confusing if you&amp;#39;re trying to memorize it but if we do the mechanism I told you it is an sn2 so just do the mechanism foreign you&amp;#39;re like sn2 concerted it has to invert the stereocenter because we know it goes by backside attack so it can only be this product because the mechanism says that it can only attack backside attack it&amp;#39;s going to invert it so the only answer is that this is not going to be a mixture between a and b the mechanism doesn&amp;#39;t say that okay very very important to understand that the mechanism is going to help you predict the product now if I told you this was an sn1 mechanism then the answer is going to be different but we don&amp;#39;t want to talk about sn1s yet because we are focusing our energy on sn2 only one product one answer it is inverted the stereocenter okay we&amp;#39;ll end there

Transcript for:Substitution and Elimination Reactions

Transcript for:
Substitution and Elimination Reactions