in this video we're going to go over sn1 sn2 e1 and e2 reactions so the first thing you need to know is if you have a methyl substrate for example like methyl bromide the reaction will proceed using an sn2 mechanism it really doesn't matter what type of solvent we're using or if we're using a base it's going to be an sn2 reaction now if we have let's say a primary substrate like uh ethyl bromide in this case it's going to be an sn2 reaction unless you decide to use a bulky base in which case it's going to be an e2 reaction now let's say if you use a strong unhindered base like hydroxide it's a strong base but it's not bulky the predominant mechanism will be sn2 however if the substrate is sterically hindered then you could get the e2 reaction sm2 reactions work with substrates that are not sterically hindered but if it is then the e2 can win in competition against the sn2 reaction so if you have a sterically hindered base and a primary substrate it's going to favor the e2 reaction but if the substrate is not hindered sterically hindered then it's going to be sn2 let's say if you use a strong base like hydroxide or even methoxide now ethoxide would also fit this category so we see et that's ethyl ch3ch2 one good example of a bulky base would be ter butoxide and so if you reacted with a primary alkyl halide you're going to get an e2 reaction now if we have a secondary alkyl halide that's when things can get interesting so if you use an aprotic solvent such as acetone or dmf or dmso with a very good nucleophile like iodide cyanide or a sulfur with a negative charge this is going to favor the sn2 reaction now if you use a protic solvent like water or methanol or even ethanol where you have hydrogen bonds typically this will favor the sn1 and the e1 mechanism but this can vary too sometimes it could favor an sn2 reaction so you got to pay special attention to this category but the protic solvent favors the sn1 and the e1 mechanism over the sn2 mechanism now what mechanism will occur if we use a secondary alkyl halide with a strong base in this case you can get both the sn2 mechanism and the e2 mechanism however the e2 mechanism will usually be the major product so if you have a bulky or sterically hindered secondary alkyl halide it's going to favor the e2 reaction over the sn2 reaction and using a bulky base will also favor e2 over sn2 however if you have let's say hydroxide and a secondary alkyl halide that's not sterically hindered then you can get a good mixture of the sn2 reaction and the e2 reaction but typically the e2 will slightly be you'll get the e2 in a better yield over the sn2 but you'll still get a good mix of both them so i'm going to put the e2 on top here because this is going to be the major product in this category but know that the sn2 reaction still exists in this category it's simply the minor product now if we have a tertiary alkyl halide we're not going to get the sn2 reaction because it's too sterically hindered and so we're going to get the sn1 and the e1 reaction regardless of the solvent that we choose to use protic solvents favor the sm1 reaction aprotic solvents favor the sn2 reaction and heat favors e1 over sm1 by the way so if you wish to increase the e1 yield over the sn1 yield you need to raise the temperature using a strong base with a tertiary alkyl halide will favor the e2 reaction you're not going to get the sn2 reaction so this table is not perfect there are exceptions but this is simply a general guideline to help you predict which mechanism will be the major pathway in a substitution or elimination reaction so let's go ahead and work on some examples so let's start with this one two bromo butane and let's react it with sodium cyanide in acetone so which mechanism will this reaction proceed with is it going to be sn1 sn2 e1 or e2 and then once we know the mechanism let's predict the major product and show the mechanism to get to that product so we have a secondary alkyl halide the carbon that bears the bromine atom that carbon is attached to two other carbon atoms so we have a secondary alkyl halide now the cyanide ion it's the nucleophile in this reaction and so the nucleophile is going to attack the carbon from the back also known as the backside attack and it's going to expel the leaving group because it attacks from the back we're going to get the inverted product so the sn2 reaction proceeds with inversion of configuration and so here on the left side we have the r isomer this is r two bromobutane and now we have the s isomer now let's move on to our next example let's say we have tert-butylchloride and let's react it with water go ahead and predict the major product for this reaction so first we need to identify the mechanism this carbon is tertiary so we have a tertiary alkyl halide and we have a protic solvent h2o so the dominant mechanism will be the sn1 and the e1 mechanism so let's draw the products for the sm1 mechanism and for the e1 mechanism sm1 stands for the first order nucleophilic substitution reaction and so we're going to replace the chlorine atom with an o h group in an e1 reaction or elimination reaction we're going to remove a hydrogen and the leaving group forming a double bond or an alkene now the first step for the sn1 reaction is that the leaving group leaves and so we're going to get a tertiary carbocation and then the water molecule is going to add to this tertiary carbocation and so we're going to get this oxonium species and then we need to use another water molecule to take off a hydrogen so the end result is an alcohol and so this is the sn1 product for this reaction now let's draw the e1 product for that reaction so the first step in the s1 reaction and the e1 reaction is the same and that is ionization the leaving group is going to leave and we're going to get the same tertiary carbocation now for the e1 mechanism this is where the path changes compared to the sn1 mechanism for the e1 reaction the water is not going to act as a nucleophile attacking the carbocation instead it's going to act as a base going for the hydrogen and then the carbon hydrogen bond will break those two electrons will be used to form the pi bond giving us an alkene so that's the e1 mechanism now let's try this example let's react to this alkyl halide with methanol go ahead and predict the products of this reaction so on the left we have a tertiary alkyl halide and we have a protic solvent when the solvent acts as a nucleophile it's called a solvolysis reaction and when you see that it's typically an sm1 e1 reaction so let's focus on the sn1 product the s1 reaction is the substitution reaction so we're going to substitute the leaving group with this group and we're going to get rid of the hydrogen so we're going to get an ether however notice that this carbon is chiral and so we're going to get a mixture of stereoisomers so the och3 group it can be in the front it could be on the wedge or it can be on a dash so these are the s and one products for this reaction now let's talk about why we get a mixture of stereoisomers why we get this racemic mixture well we know the first step is ionization so this will produce a tertiary carbocation now the methanol molecule is going to act as a nucleophile and notice that it can approach the carbocation from either side it can go from the back leading to the inverted product or it can go to the front which is retention so let's say the bromine atom is in the front or on the wedge so if it attacks from the back we're going to get this the inverted product if it attacks from the front we're going to get retention now of course there's one more step we need to take off the hydrogen but will we get an equal amount of the inverted product and the retention product or will one stereoisomer have a greater yield over the other now the bromide ion is still in the vicinity of the carbocation and so when the methanol molecule approaches from the front it's going to be repelled by the negative charge of the bromide ion keep in mind that the oxygen atom has a partial negative charge and so it's repelled by the bromide ion therefore we're going to get a lower yield of the retention product than the inverted product so the backside attack is still more favorable so it could be 60 of the inverted product and 40 percent of the retention product or it could be a 70 30 ratio either case you get slightly more of the inverted isomer as opposed to the retention stereoisomer now for the last step of the mechanism all we need to do is use another methanol molecule to remove the hydrogen and so this will give us our ether product so one is going to be on the wedge and the other will be on a dash now let's focus on the e1 reaction feel free to pause the video and write a mechanism for the e1 reaction so the first step is ionization and this will give us our tertiary carbocation now in the next step for the e1 reaction methanol is going to act as a base instead of a nucleophile so it's going to abstract one hydrogen away from the carbocation so the hydrogen has to be one carbon away from the carbocation it can take the white hydrogen it can take the blue hydrogen or it can take the green hydrogen so if it takes the green hydrogen we can get this product now if it goes for the blue hydrogen we can get this product and if it goes for the white hydrogen we can get this product and so we get a mixture of e1 products for this reaction however not all of these alkenes are equally stable so this particular alkene is a disubstituted alkene there's two carbon atoms that are attached to the two double bonded carbon atoms this alkene is trisubstituted and the same is true for this one it has three r groups and so this is the least stable alkene so therefore the major product for the e1 reaction will be these two alkanes they will both form in good yield this will be the minor e1 product now let's say if we have two bromo 3-methylbutane now let's react it with sodium methoxide and methanol what is the major product in this reaction go ahead and try so we have a secondary alkyl halide and methoxide is a strong base it's not a sterically hindered base but it is a strong base and so whenever you have a secondary alkyl halide with a strong base the major product is typically the e2 product and the sn2 reaction will give you the minor product but we're going to draw both so let's start with the s and 2 reaction so the methoxide will attack this carbon kicking out the bromine atom now we don't have to worry about the stereochemistry because it wasn't specified at this carbon even though we could get a mixture of steroid isomers because this compound could be r or s we don't know but we're not going to worry about stereochemistry the s and two reaction will give you the inverted products so let's say if this was in the front then the och3 group would have to be in the back or if this was in the back then the och3 group will have to be in the front so since the stereochemistry wasn't specified we're not going to specify it here either so this is going to be the sn2 product and this is going to be the minor product in this reaction it can happen but it's not going to be in good yield now for the e2 reaction no rearrangements can occur the base can take off a hydrogen one carbon away from the carbon that bears the bromine atom so it can take the white hydrogen or the blue hydrogen methoxide will preferentially take the blue hydrogen over the white hydrogen and so we're going to get this particular e2 reaction so this is going to be the major e2 product now it could take the white hydrogen put in a double bond here and this will be the minor e2 product now you need to know that internal alkenes are generally more stable than terminal alkenes so this alkene here is more stable than the one on the right and the reason being is it's more substituted here we have a tri substituted alkene on the right there's only one carbon attached to these two double bonded carbon atoms and so this is a mono-substituted alkane tri-substituted alkanes are more stable than mono substitute alkenes and so that's why this is the major product it's simply the more stable alkene the most stable alkene is known as the zaitsev product whereas the less stable alkene is known as the hofmann product and so this is the answer for this problem so the major mechanism for this reaction is the e2 mechanism and this is the major product overall keep in mind the sn2 reaction can occur for this particular problem but it will lead to a minor product now let's say if we have the same alkyl halide the same substrate but this time we're going to use a sterically hindered base such as terpetoxide in turbul what's going to happen so we have a protic solvent but we also have a strong base and so if you're using the trite that i mentioned in the beginning you need to use the fourth column because the strong base takes priority over the protic solvent because this has a negative charge it's far more reactive than terbianol so we have a sterically hindered base and also a relatively sterically hindered alkyl halide due to this methyl group on the top so whenever you have a bulky base an hysterically hindered substrate if both the substrate and the base are sterically hindered this will give you the hofmann product and so you're going to get the less stable alkene so because terpetoxide is so bulky it wants to go for the most accessible hydrogen and the hydrogen atoms on a primary carbon are more accessible than the hydrogen atom on a tertiary carbon terpetoxide doesn't want to have to struggle against these two methyl groups to get to that hydrogen and so it's not going to form the zaitsev product in good yield it can still work but it's going to be the minor product instead it's going to go for the most accessible hydrogen atom so we're going to get the hoffman product as the major product for this reaction now we can still get the zaitsev product but it's going to be the minor product in this case so the e2 reaction is the major mechanism in this example and so this is the major product now i want to compare these two reactions together two bromo pentane and 2-floral pentane so let's start with 2-bromopentane i'm going to use methoxide as the strong base dissolved in methanol what is the major product for this reaction so in this case methoxide is going to go for the hydrogen that's one carbon away from the carbon bromine bond and so we're going to get the zeta product as the major product you can still get the hoffman but that's going to be the minor product now if we use an alkyl fluoride something different is going to happen the methoxide is going to go for the primary hydrogen as opposed to the secondary hydrogen and so we're going to get the hofmann product as the major product the question is why why do we get the hofmann product as the major product if fluorine is a leaving group whereas if bromine or chlorine or iodine is the leaving group we're going to get the zetaphrodic for one thing chlorine bromine and iodine they're good leaving groups and so when you have a good leaving group it's going to favor the zeta product fluorine on the other hand it's a bad leaving group and so you're going to get the hoffman product now the reason for this is when you have a good leaving group the transition state resembles an alkene and so because of that you need to find the most stable alkene so the major product will have or will be associated with the most stable alkene because that's going to give you the most stable transition state in the case of an alkyl fluoride because it's a bad leaving group when the methoxide ion approaches to attack the hydrogen the fluorine is still sticking around it doesn't want to leave and so you have this buildup of negative charge on this uh molecule and so the transition state resembles more like a carbanion instead of an alkene and what you need to know is that primary carbonites are more stable than secondary carbanions and as a result it's better to put the negative charge on a primary carbon as opposed to a secondary carbon and this is why the hoffen product is the major product now i want you to compare these two reaction mechanisms let's say we have two bromo 3-methylbutane and we're going to react it with water now let's compare this to 2-bromobutane and we're going to react to this molecule with the acetate ion which is going to be dissolved in acetic acid so go ahead and determine the substitution mechanisms that these two reactions will proceed by ignore the elimination mechanism now let's focus on the reaction on the bottom so we have a secondary alkyl halide and a protic solvent so if we use the chart this would indicate that it should proceed by the sn1 mechanism however it turns out that this reaction actually goes by the sn2 mechanism and for one reason acetate is a decent nucleophile it has a negative charge so it's more nucleophilic than h2o here we have a secondary alkyl halide and a protic solvent and it turns out that this mechanism favors the sm1 or this reaction favors the sn1 mechanism now let's talk about why so as we said before we have a better nucleophile and so that favors the sn2 reaction the nucleophile has a negative charge whereas this nucleophile is neutral even though both solvents are protic now the second reason why this one favors an acid to reaction is because that substrate is not sterically hindered here we have a sterically hindered secondary alkyl halide due to this methyl group an hysterically hindered alkyl halide actually favors the sn1 reaction because once we get the secondary carbocation it can rearrange and form a more stable tertiary carbocation so whenever you're dealing with secondary alkyl halides you need to be careful because sometimes it can be sn1 sometimes it could be sn2 now we know it's not going to be an each reaction because we don't have a strong base so we can ignore that option but when you're between these two mechanisms ask yourself what type of secondary alkyl halide do i have is hysterically hindered if it is it's going to favor the sn1 mechanism if it's not it favors the sn2 mechanism is there a protic solvent or is there an aprotic solvent if there's an aprotic solvent it favors the sn2 reaction if there's a protic solvent typically it favors the sn1 mechanism but in this case this exception we have a good nucleophile in a protein solvent so that good nucleophile favors the sn2 mechanism and so you have to look at every detail and weigh all your options carefully now let's go over the mechanism of those two reactions so let's start with this one where we're going to react it with h2o and we said that this is going to be the sm1 mechanism so the first step the leaving group leaves and so we're going to get a secondary carbocation now whenever you have a secondary carbocation next to a tertiary carbon you're going to get a hydrate shift and so this is going to give us a more stable carbocation intermediate and so water is going to attack the carbocation giving us this intermediate and then in the next step we're going to use another water molecule to take off a hydrogen atom and so our final answer is a tertiary alcohol now this alcohol is not chiral because it has two methyl groups and so we don't have to worry about stereoisomers this is going to be the major sn1 product now let's go over the other example where we had two bromo butane reacting with the acetate ion so we said that this reaction will proceed by the sn2 mechanism and so the oxygen of the acetate ion is going to attack from the back expel in the leaving group and so we're going to get inversion of configuration so this is going to be our product now here's the interesting question for you let's say we have this substrate and let's react it with methanol what's going to be the major product in this reaction and what type of mechanism do we have is it the sn2 mechanism the e2 mechanism sm1 e1 what's going on here so the carbon that has the bromine atom is a primary carbon so we have a primary alkyl halide which usually favors sn2 reactions and we have a protic solvent so are we going to get an sn2 reaction notice that this primary alkyl halide is very hindered next to the primary carbon we have a quaternary carbon and so due to the fact that this primary carbon is sterically hindered due to the adjacent carbon it's safe to say that this reaction will not proceed by the sn2 mechanism it's too sterically hindered so what's going to happen can it proceed by the sm1 mechanism it can however we can't just kick out the br because if we do we're going to get a primary carbocation and primary carbocations are not stable they're very difficult to form and so that's not going to happen instead we're going to get a concerted reaction we're going to get a methyl shift before the br leaves to produce the unstable carbocation so this methyl group is going to attack this carbon kicking out the bromine atom and so now the methyl group is here by doing this we form a more stable tertiary carbocation intermediate without forming a primary unstable carbocation so now at this point the methanol could attack the carbocation giving us this intermediate and then in the next step we can use another methanol molecule to remove the hydrogen atom and so our final product is an ether and we don't have to worry about stereochemistry this is not a chiral center and so this is the answer so this is one of those rare situations where a sterically hindered primary alkyl halide can give you the sm1 mechanism now let's consider this example so once again we have another sterically hindered secondary alkyl halide but this time we're going to react it with sodium ethoxide in ethanol so go ahead and try this problem so we have a strong base that's ethoxide and based on the chart when you have a strong base like ethoxide and a secondary alkyl halide it can predominantly give you the e2 reaction mechanism but it can also give you the sn2 reaction however we do have a secondary sterically hindered halide and so we could say that the sn2 reaction is negligible if it does occur it's going to be in a very very low yield so we're going to ignore the sn2 reaction we're going to focus on each reaction because that's going to be the dominant mechanism if you have a sterically hindered substrate so notice that we can't get the zaitsev product in this case there is no hydrogen on its carbon because it's coronary so we have to remove the hydrogen on the primary carbon because we have no choice and so ethoxide will go for this hydrogen and we can only get one product for the e2 reaction there are no rearrangements and so this is one of those rare cases where we get the less stable alkene since we only have one choice we have no other choices and so that's it for this example what's going to happen if we mix methyl bromide with terpetoxide go ahead and predict the product of this reaction now whenever you have a methyl substrate the only mechanism that can work is the sn2 mechanism there are no other choices now even though we have a bulky base such as terpetoxide which typically favors elimination reactions over substitution reactions we can't form a double bond with one carbon atom you need at least two carbon atoms to form a double bond and so there's no way we can get the e2 reaction so the sn2 reaction will occur with a 100 yield what i'm going to do is draw the terpetoxide first and then the substrate after that so terpetoxide will attack the methyl group from the back expel in the bromide ion and so our product will be an ether and so this is the only product that we can get in this reaction now let's move on to this example so here we have butyl bromide and we're going to react it with methoxide and methanol so based on the chart if we have a primary alkyl halide that's not sterically hindered and we have a strong base what is going to be the dominant mechanism in this reaction so since we don't have a sterically hindered base the mechanism will predominantly be sn2 and so in this case the methoxide ion will attack the carbon that bears the bromine from the back expel in the leaving group and so we're going to get an ether now we don't have to worry about stereochemistry because this carbon is not chiral it has two hydrogen atoms and so that's it for that example now what if we have a primary alkyl halide but with a sterically hindered base like terpetoxide what's going to happen in this case well based on the chart this is going to favor the e2 reaction so the base this bulky base it doesn't like to behave as a nucleophile because it's so bulky unless it has to in the case of methyl bromide but it prefers to behave as a base abstract in a proton and so this is going to give us one butene now i want to compare the last two reactions with a new reaction and so you could see something so here we have a primary alkyl halide with an unhindered strong base and so this gave us the sn2 reaction in the last example we had a primary alkyl halide that was not sterically hindered but we had a sterically hindered base and so this gave us the e2 reaction now what if we have let's say a sterically hindered primary alkyl halide and let's use methoxide again will we get an sn2 reaction or an e2 reaction so in the first example if we don't have a sterically hindered base or a sterically hindered primary alkyl halide we get the sn2 reaction if we have a sterically hindered base we're going to get the each reaction if we have a primary sterically hindered substrate we're going to get the e2 reaction as well because any time you have a sterically hindered substrate it's going to weaken the yield of the sn2 reaction and so therefore we're going to get this alkane so make sure you understand that so the only time you get the sn2 reaction with a strong base is if you have a primary alkyl halide that is not sterically hindered if you have a sterically hindered base or a sterically hindered primary alkyl halide it's going to favor the e2 reaction over the sn2 reaction you can still get the s into the action but the each reaction will give you the major product you