in this video we're going to focus on nucleophilic substitution reactions there's two common types that you need to know the sn1 reaction and the sn2 reaction but let's begin our discussion with the sn2 reaction this is a second order nucleophilic substitution reaction the rate of the reaction depends on the concentration of the substrate and it depends on the concentration of the nucleophile its first order with respect to the substrate and first order with respect to the nucleophile but it's second order overall that's why it's an asymptote reaction one plus one is two if you double the concentration of the substring the rate will double if you double the concentration of the nucleophile the rate will double if you triple the concentration of the substrate the rate is going to triple if you quadruple the rate of the nucleophile the rate will quadruple what's going to happen if you double the concentration of the substrate and triple the concentration of the nucleophile in that case the rate is going to increase by a factor of 6. if you triple the concentration of the substrate quadruple the rate of nucleophile the rate will increase by a factor of 3 times 4 which is 12. so what exactly is the substrate well the substrate is basically an alkyl halide so here we have two bromo butane and we're going to add a nucleophile let's use iodide iodide is a good nucleophile and in this reaction iodide will attack the carbon from the back expelling the bromide leaving group the s1 reaction occurs in a single step it's a conservative reaction mechanism all bond breaking processes and bond forming processes occur at the same time now this reaction occurs with inversion of stereochemistry the bromine was in the front but now the iodide is in the back since it attacked a carbon from the back so let's say if well let's find out what the configuration was before if br is in the front age is in the back so let's assign br a value of one the ethyl group two methyl three h is four so if we count from one two to three we can see that we have this is going uh clockwise so we have the r isomer therefore the product is the s isomer the hydrogen is now in the front and if you count it it's a one two three it appears to be r but because hydrogen is in the front you have to reverse it so it's us so the configuration of the cloud center it reverses an sn2 reaction as you can see no carbocations were formed and therefore the sn2 reaction is not subjected to carbocation rearrangements wherever the leaving group was located the nucleophile is going to be on that same carbon now why is it called a nucleophilic substitution reaction as you can see the leaving group was replaced or substituted with a nucleophile and so it's called the nucleophilic substitution reaction now sn2 reactions do they prefer primary substrates or tertiary substrates it turns out that an sn2 reaction works best with a methyl substrate it's most reactive with these and then it works well for primary substrates as well but for methyl substrates is better and it doesn't work very well for a tertiary substrate now let's compare a methyl substrate with a tertiary substrate let's see why so here we have methyl bromide and we're going to draw also terbutyl bromide let's use hydroxide as a nucleophile now if you react hydroxide with methyl bromide it's going to be an acid 2 reaction and it works pretty well but if you try to react hydroxide with terbuto bromide it doesn't work well why do you think that's the case why does the first reaction work very well but the second one does not work very well for an s2 reaction in order for the hydroxide to react with the carbon atom with the methyl bromide it has to attack the carbon so it can expel the bromine atom in order for it to do that the carbon has to be accessible hydroxide has a difficult time getting into this carbon because of these bulky methyl groups the methyl groups block they basically protect this carbon atom from the hydroxide group and so because of steric factors it's very difficult for hydroxide to access the carbon and that's why tertiary substrates don't work very well the carbon atom that has the leaving group is virtually inaccessible it's very difficult to get to so the sn2 reaction for a tertiary substrate is very very very very very slow so basically it's almost non-existent so sn2 reactions work well whenever you have an unhindered substrate or an accessible carbon atom to attack now let's talk about the mechanism of a sn1 reaction so let's say we have tert-butyl bromide and we're going to react it with a negatively charged nucleophile let's use iodide again so the first thing that happens is the leaving group leaves you have the formation of a carbocation now at this point carbocation rearrangements can occur but there's not going to be any rearrangements in this particular structure because the plus charge is on a tertiary carbon tertiary carbocations are more stable than secondary primary ones so after the carbocation is formed the nucleophile attacks it and so in this case we get butyl iodide so if you have a negatively charged nucleophile the sn1 reaction occurs in two steps the rate of an s1 reaction is equal to the concentration of the substrate times k it doesn't depend on the concentration of the nucleophile because the first step is very limited the slow step is the formation of the carbocation it takes a long time to produce this carbocation the second step the combination of these two ions is pretty fast so that's why sm1 reactions are known as first order nucleophilic substitution reactions overall it's first order now what if we have an s and one reaction where the nucleophile is not negatively charged it's neutral in this case the nucleophile is also the solvent whenever the nucleophile is the solvent you have a sulfolysis reaction so in the first step the leaving group is going to leave and so we're going to get a secondary carbocation which will not rearrange and water the nucleophile can attack it from the back or it can tack it from the front so because it can approach the carbocation from both sides we're going to get a racemic mixture if it attacks it from the back like the sn2 reaction this is going to produce the inverted product we'll talk about the final product which is an oh but i'll show you how to get there soon so the inverted product will look like this if it attacks let's say from the front then we're going to get the retention product now keep in mind this is not an equal racemic mixture we don't get 50 or 50 s rather it's more like sixty forty slash seventy thirty we get more of the uh inverted product than the retention product and the reason why that's the case is because there's a bromide ion that even though it's disassociated from the carbocation it's not always very far from it so the oxygen of water has a partial negative charge bromide has a negative charge so if water attacks from the front as you can see it's going to be repelled by the bromide ion and so it's less likely to attack from the front but if it comes from the back there's no such repulsion in fact it's attracted to the carbon in the back because the carbon has a partial positive charge in fact well it has more than a partial positive charge once the bromide leaves it has a positive charge so it's definitely we want to attack it from the back we don't have the influence of this bromide and repelling the water molecule so that's why the inverted product is still slightly more than the retention product the bromide doesn't block access from the back it blocks it from the front but now let's finish the mechanism so let's say water attacks it from the back initially we didn't get a product that looks like this whenever oxygen has three bonds it has a plus charge and then we need to use another water molecule to get rid of this hydrogen giving us the alcohol so whenever you use a neutral nucleophile the sn1 reaction will occur in three steps if you use the negatively charged nucleophile it will occur it will happen in two steps the third step involves the removal of this hydrogen now what about the substrate we said that for an sn2 reaction it works better if you have a methyl or primary substrate because the carbon atom is more accessible what about for an s1 reaction what's the case here for an s1 reaction tertiary alkyl halides or tertiary substrates work better than secondary ones methyl substrates don't work very well now why is that the case now if you recall the rate limited step the slow step is the formation of the carbocation intermediate to increase the rate of that first step you need a more stable carbocation it's very difficult to form an unstable carbocation but it's easier to form a stable carbocation and because tertiary carbocations are more stable than secondary carbocations due to hypoconjugation and the inductive effect tertiary substrates work better for an sn1 reaction than secondary substrates so methyl substrates are the least stable because the methyl carbocations are let me take that back methyl substrates do not work well for an s1 reactions because the methyl substrates are the least stable so that's what i wanted to say and so that's it that's why tertiary alkyl halides work better for an s1 reaction is because the tertiary carbocation is fairly stable compared to the other carbocations now let's work on some more sn1 reactions and let's focus on rearrangements let's not worry about the stereochemistry of the reaction so let's say if we have two bromo 3-methylbutane and we're going to react it with methanol if we use water what is going to produce an alcohol if you react an alkyl halide with an alcohol for the sn1 product you'll get an ether but let's propose a mechanism for this process so the first step in an s1 reaction is the formation of the carbocation intermediate so the leave group have to leave and we're going to get a secondary carbocation the reason why this carbocation is secondary is because the carbon that bears the positive charge is attached to two other carbon atoms now notice that the secondary carbocation is adjacent to a tertiary carbon and we know tertiary carbocations are more stable than secondary ones so there's going to be a rearrangement a hydride shift that plus charge wants to be on the tertiary carbon and not on the secondary one so this hydride is going to glide towards the plus charge and basically they're going to trade places and so now the plus charge is on the tertiary carbon and the hydrogen is now on the secondary one but we don't need to show the hydrogen at this point methanol which has a partial negative charge on the oxygen is attracted to the carbocation and so it's going to attack the carbon with the plus charge and so we have an intermediate that looks like this since we have a neutral nucleophile the reaction will at least take three steps to occur but because of the carbocation rearrangement there's four steps in this reaction now once we form this intermediate we're going to use another methanol molecule to get rid of the hydrogen and so now we have the final product which is an ether notice that this carbon atom is not chiral we have two methyl groups in order for a carbon atom to be chiral it has to have four different groups so this is the product you do not need to indicate stereochemistry for this product in fact you can't because this carbon atom is not chiral so the product is an ether now let's try another example by the way if you want to find more examples on sn1 and sn2 reactions there's a lot of other videos i've created check out my channel there's one where it's basically a practice test of 75 multiple choice questions i created a while ago but if you uh type it in i do a search on youtube you should find it and you can get a lot of practice on these types of problems so hopefully if you watch the video it might be very useful to you so let's say if we have an intermediate that looks like this and we're going to use water in this case so we have a secondary alkyl halide with a protic solvent and this is going to favor an sn1 reaction by the way protic solvents favor sn1 reactions polar aprotic solvents favor sn2 reactions but let's propose a mechanism for this reaction so we know that the bromine atom is going to leave this is going to produce a secondary carbocation now this secondary carbocation is adjacent to a quaternary carbon which contains no hydrogen atoms so there's going to be a carbocation rearrangement but it's not going to be a hydrate shift instead it's going to be a methyl shift the methyl is going to trade places with the positive charge and this time we need to show the methyl it's not a hydrogen atom that we can ignore so now the plus charge is now on a tertiary carbocation or a tertiary carbon i should say so now we can use water to uh react with the carbocation let's not worry about stereochemistry in this example so we're going to have this intermediate and then the last step is to take off the hydrogen so this is going to give us an alcohol that looks like this so this is the product now if we decide to take stereochemistry into account do we get one product or two products do we need to show the stereoisomers so look at your final product determine if the carbon that has the oh if it's chiral or acaira so we have a methyl group here we have a secondary carbon and a tertiary carbon so this carbon the chiral carbon has one two and the left side is different from the right side because these two carbon atoms are different so it has four different groups so the carbon atom is chiral which means that the product consists of two stereoisomers the oh can be in the front or it can be in the back