in this video we're going to go over the sn2 reaction mechanism so in this reaction we have a nucleophile reacting with the substrate the nucleophile is hydroxide it has a negative charge and the substrate is the alkyl halide bromine is more electronegative than carbon so bromine bears a partial negative charge carbon bears a partial positive charge so the nucleophile is attracted to the partially positive carbon atom and so it approaches it from the back and as it attacks the methyl group the bromine group gets expelled and so we get the product which is methanol and we also get a side product also known as the leaving group which is the bromide ion and so that's the basic idea behind the sn2 reaction mechanism to draw the transition state it looks like this so here we have the three hydrogen atoms attached to the carbon and we have a bond being created between the oxygen and the carbon atom and a bond that's breaking between the carbon and the bromine atom and so the transition state looks like this carbon bears a partial positive charge bromine bears a partial negative charge and the same is true for hydroxide now let's go over some other reactions that is associated with the sn2 mechanism what do you think the major product for this reaction will be let's say if we have butyl bromide and we wish to react it with sodium methoxide now what you need to realize is that the sodium ion is a spectator ion so it's not going to do anything in the reaction the nucleophile is the methoxy ion and so what it does is it attacks the carbon from the back that is away from the bromine atom and then it expels the bromine atom now it doesn't want to attack from the front because bromine has a partial negative charge and the oxygen atom has a negative charge so if it tries to approach from the front it's going to be repelled by the electronegative bromine atom so instead it attacks from the back therefore we're going to get this product which is an ether now let's work on some other examples let's say we have in this case two bromo butane and let's react it with potassium iodide and acetone acetone is a polar aprotic solvent there are no o h or nh bonds and it turns out that polar aprotic solvents they enhance sn2 reactions what do you think the major product for this reaction will be so first we need to identify the nucleophile the nucleophile is the iodide ion and so what it's going to do is it's going to approach the carbon from the back expel in the lemon group now notice that we have a chiral center and notice that the bromine is on the wedge so now because the iodide ion approach from the back it's going to be basically on a dash so this means the bromine atom is coming out of the page and the iodide ion or the iodine atom is going into the page and so this is the product of the reaction for an sn2 reaction we get inversion of configuration so as you can see the stereochemistry changes this was r 2-bromobutane now we have s two iodo butane now here's another example so let's say we have the chair conformation of one bromo three 3-methyl cyclohexane let's say the methyl group is in the equatorial position and the bromine atom is in the axial position draw the major sn2 product for this reaction let's use sodium cyanide in dmso dimethyl sulfoxide which is another type of polar aperitic solvent that enhances an s2 reaction so go ahead and predict the major product for that reaction so we know the cyanide ion which looks like this it's basically a triple bond between the carbon and the nitrogen atom but the carbon atom bears the negative charge so it's the nucleophilic center it's going to attack the carbon from the back expelling the leaving group now because the sn2 reaction proceeds with inversion of configuration the cyanide group is going to be on the equatorial bond because the bromine atom was on the axial bond and so you need to switch it and so the product let me put over here where i have more space the product will look like this so the method group is unaffected and now cyanide is in the equatorial position and so this is the sn2 product for this reaction the sn2 reaction is a second order nucleophilic substitution reaction the rate of the reaction depends on the concentration of the substrate and the concentration of the nucleophile so if we double the concentration of the substrate the rate will double if we triple the concentration of the substrate the rate will triple if we double the concentration of the substrate and double the concentration of the nucleophile 2 times 2 is 4 so the rate will increase by a factor of 4. if we triple the concentration of the substrate quadruple the concentration of the nucleophile the rate will increase by a factor of 12. so it's first order in the substrate first order in the nucleophile overall the rate is second order if you add one and one which gives you two now the energy diagram for the sn2 reaction looks like this there's only one transition state so here it is this is the energy of the reactants this is the energy of the products and on the y-axis we have the energy on the x-axis we have the progress of the reaction the activation energy is the difference between the energy of the reactants and the transition state now it's important to understand that the sn2 reaction is a concerted reaction mechanism everything happens in one step all the bond making and bond breaking processes occur together and so it's a concerted reaction mechanism now here's a question for you consider the following alkyl halides which of these alkyl halides is most reactive in an sn2 reaction and which one is the least reactive feel free to pause the video and answer that question so here we have a secondary alkyl halide because the carbon that bears the bromine atom is attached to two other carbon atoms and in the middle we have a tertiary alkyl halide because that carbon which is here is attached to three other carbon atoms and on the right we have a primary alkyl halide because the carbon that bears the bromine atom is attached to only one other carbon atom it turns out that the primary alkyl halide is the best for an sn2 reaction whereas the tertiary alkyl halide is the worst and the reactivity is as follows a methyl alkyl halide is better than a primary alkyl halide which is better than the secondary alkyl halide which is better than a tertiary alkyl halide now let's talk about why a tertiary alkyl halide is bad for an sn2 reaction so i'm going to use tert-butyl bromide as an example now in order for the sn2 reaction to take place the nucleophile let's use iodide as an example the nucleophile has to attack the carbon that bears the bromine atom so it has to attack this carbon but it's not easy to attack it because these three bulky methyl groups they basically provide a barrier against the iodide ion and so it's very difficult for the iodide ion to attack that carbon with those three methyl groups being in the way and so that's why the sn2 reaction virtually doesn't occur with tertiary alkyl halides it's too sterically hindered and so the reaction rate is very very very slow once in a while may happen but for the most part it's negligible so tertiary alkyl halides for the most part do not react in the sn2 reaction mechanism now compare these two alkyl halides which one is more reactive in an sn2 reaction is it the one on the left or the one on the right now this carbon atom is a secondary carbon and the same is true for that one so the substitution is the same but which one is going to work better on the right we have something called an arrow halide you need to know that the aero halide does not work well for an sn2 or an s1 reaction and the reason being in order for the sn2 reaction to work the iodide ion has to attack this carbon however the pi electron cloud of the benzene ring prevents that from happening and so the iodide ion can't approach from the back to attack that carbon due to the pi electrons so aero halides do not participate in the sn2 reaction therefore this one is better and the same is true if you compare let's say a vino halide with an aloe or an aluminum calide the veno halide doesn't work so if your leaving group is directly attached to a double bond that is in the ring or on a double bond like this it's not going to work so this doesn't work too well for the s and two reaction because once again the nucleophile has to approach from the back and the pi electrons they interfere with that the pi electrons have a negative charge and so they repel the negatively charged iodide ion so this is not going to work so the allylic halide works a lot better than the vanilla calide now here's another example for you consider these three alkyl halides a and determine which of these three is the best for an sn2 reaction now this carbon is a secondary carbon and this one is also secondary and this one is secondary so how can we determine which one is better for an sn2 reaction in a situation like this look at the adjacent carbons on the right all of these carbons are primary so there's no difference here but on the left this carbon is secondary this carbon is tertiary and this carbon is coronary just by looking at this a secondary alkyl halide is better than the tertiary or quaternary one so even though we all have a secondary alkyl halides for this example if you look at the adjacent carbon a secondary carbon is less sterically hindered than the tertiary or quaternary one so this is going to be the best for an sm2 reaction now it's easier for the nucleophile to approach this carbon rather than to approach this carbon because that coronary carbon has a lot more methyl groups than the secondary one and so this alkyl halide is more sterically hindered than this one so it's easier for the nucleophile to attack this particular alkyl halide so if you have alkyl halides with the same basically substitution look at the adjacent carbons and see which one is less sterically hindered and the sn2 reaction will work better for that one so this alkyl halide is the least sterically hindered which means the carbon is more accessible to a nucleophilic attack and that's why this one works the best now here's another question for you so let's say if we have one bromo butane reaction with hydroxide compared to one bromobutane reacting with water which reaction will proceed faster in an sn2 reaction is it the one with hydroxide or the one with water so the substrate is the same in both cases there are primary alkyl halides so the difference lies in the nucleophile if you recall for an essential reaction the rate depends on the substrate and the concentration of the nucleophile so the strength of the nucleophile is important so which nucleophile is stronger hydroxide or water because hydroxide has a negative charge it's a stronger nucleophile in water therefore the first reaction will work faster in an sn2 reaction so this reaction is better because hydroxide is a stronger nucleophile than water on the periodic table we have elements such as carbon nitrogen oxygen and fluorine and then below that chlorine bromine and iodine so going towards the right nucleophilic strength increases so for instance nh2 minus is a stronger nucleophile than oh minus which is stronger than fluoride now in a protic solvent a good example of a protic solvent would be like water or methanol so you can write methanol as ch3oh sometimes you might see it as meoh or ethanol ch3ch2oh or etoh so typically in a protic solvent you'll see hydrogen bonds in a protic solvent nucleophilic strength increases going down the periodic table so what that means is that iodide is a better nucleophile than bromide which is better than chloride and which is better than fluoride to keep things short fluorine or rather the fluoride ion is a stronger base than iodide but in a protic solvent it forms hydrogen bonds with water and methanol and so it stabilizes fluoride which means that it's less accessible or it's less able to behave as a nucleophile because it's solvated by the solvent the solvent stabilize fluoride because it's a better base iodide being a weaker base is not stabilized by the solvent as much and so iodide is free to behave as a nucleophile in a protein environment and that's why iodide works better than fluoride even though fluoride is a stronger base than iodide it's because the solvent stabilizes the fluoride ion and so it's not free to behave as a nucleophile iodide being a weaker base is not held back by the solvent and so iodide behaves as a stronger nucleophile in a protic environment now in an a protic solvent the situation is different nucleophilic strength is the same as base strength so in the aprotic environment fluoride is a better nucleophile than chloride which is better than bromide which is better than iodide so let's say like acetone that's a polar apronic solvent as you can see there's no hydrogen bonds in acetone and so if you were to dissolve let's say sodium fluoride and acetone all of the acetone molecules will orient themselves towards let me draw this better all of these acetone molecules will orient themselves towards the sodium cation sodium has a positive charge and the oxygen atoms they all have a partial negative charge and so they're going to stabilize the sodium cation now the fluoride ion is not held back or solvated by the polar apic solvent and so and it's also not held back by the sodium ion so therefore the fluoride ion is free to react and so this enhances the potential energy level of the nucleophile and because it's free to react it's going to behave as a strong nucleophile and that's why polar aprotic solvents work very well for sn2 reactions polar protic solvents like water and methanol that work better for sn1 reactions and so keep that in mind so for sn2 reactions you want to use a polar aprotic solvent because they enhance the strength of the nucleophile they solvate the cation allowing the nucleophile to be free to react and remember fluoride is better than iodide in the aprotic environment in an sn1 reaction it's best to use a protic solvent and in a protic environment iodide is a better nucleophile than fluoride now let's work on some example problems let's say if we have one bromobutane what is the major product if we react it with water so feel free to draw the sn2 product of this reaction now water is going to behave as a nucleophile attacking the carbon atom expelling the leaving group and so that's the basic idea behind the sn2 reaction but this reaction occurs in two steps the first step is an sn2 reaction and the second step involves deprotonation of the hydrogen now when water attacks the carbon it behaves as a nucleophile however when water abstracts a proton it behaves as a base so water can behave as both a base or a nucleophile and so the end result is that we get an alcohol one butanol and so that's how you can convert an alkyl halide into an alcohol now let's look at another example let's say if we have an alcohol and we wish to react with hydrobromic acid what's going to happen so this two will also be an essential reaction but the first step will be protonation the oxygen is going to grab a hydrogen from the acid kicking out the bromine atom and so right now the oh group has been turned into a good leaving group in this form the oh group is a bad leaving group because let's say if you try to take a nucleophile and kick it out you're going to get a hydroxide ion which is a strong base and the strong base is not stable and so strong bases are bad leaving groups but if you protonate the alcohol put a positive charge on the oxygen and then it becomes a good leaving group so now bromide can attack from the back expelling water and water is a stable base and so water is a good lean group and so this gives us the original alkyl halide so that's how you can convert a primary alcohol into an alkyl halide using an acid you