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Current time:0:00Total duration:7:21

Video transcript

let's look at the mechanism for an sn2 reaction on the left we have an alkyl halide and we know that this bromine is a little bit more electronegative than this carbon so the bromine withdraws some electron density away from that carbon which makes this carbon a little bit positive so we say partially positive that's the electrophilic Center so this on the left is our electrophile on the right we know that this hydroxide ion which we could get from something like sodium hydroxide it's a negative 1 formal charge on the oxygen which makes it a good nucleophile so let me write down here this is our nucleophile on the right and on the left is our electrophile which I'm also going to refer to as a substrate in this video so this alkyl halide is our substrate we know from an earlier video that the nucleophile will attack the electrophile because opposite charges attract this negative charge is attracted to this partially positive charge the lone pair of electrons on the oxygen will attack this partially positive carbon at the same time the two electrons in this bond come off on to the bromine let me draw the bromine over here the bromine had three lone pairs of electrons on it it's going to pick up another lone pair of electrons let me show those electrons in magenta so this bond breaks and these two electrons come off onto the bromine which gives the bromine a negative one a formal charge so this is the bromide anion and we're also forming a bond between the oxygen and this carbon and this bond comes from this lone pair of electrons which I've just marked in blue here so those two electrons in blue form this bond and we get our product which is an alcohol so the sn2 mechanism is a concerted mechanism because the nucleophile attacks the electrophile at the same time we get loss of a leaving group there's only one step in this mechanism let's say we did a series of experiments to determine the rate law for this reaction so remember from general chemistry rate laws are determined experimentally so capital R is the rate of the reaction that's equal to you the rate constant K times the concentration of our alkyl halides and it's determined experimentally this is to the first power times the concentration of the hydroxide ion also to the first power so what does this mean this means if we increased the concentration of our alkyl halide so if we increase the concentration of our alkyl halide by a factor of two what happens to the rate of the reaction well the rate of the reaction is proportional to the concentration of the alkyl halides to the first power so 2 to the first is equal to 2 which means the overall rate of the reaction would increase by a factor of 2 so doubling the concentration of our alkyl halide while keeping this concentration the hydroxide ion concentration the same should double the rate of the reaction and also if we kept the concentration of alkyl halide the same and we doubled the concentration of hydroxide that would also increase the rate by a factor of 2 and this experimentally determined rate law makes sense with our mechanism so if we increase the concentration of the nucleophile or we increase the concentration of the electrophile we increase the frequency of collisions between the two which increases the overall rate of the reaction so the fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our ID of a one-step mechanism finally let's take a look at where this sn2 comes from so we keep on saying an sn2 mechanism an sn2 reaction the S stands for substitution so let me write in here substitution because our nucleophile is substituting for our leaving group we can see in our final product here the nucleophile has substituted for the leaving group the N stands for nucleophilic because of course it is our nucleophile that is doing the substituting and finally the 2 here refers to the fact that this is by molecular which means that the rate depends on the concentration of two things the substrate and the nucleophile so that's different from an sn1 mechanism where the rate is dependent only on the concentration of one thing the rate of the reaction also depends on the structure of the alkyl halide on the structure of the substrate on the Left we have a methyl halide followed by a primary alkyl halide the carbon bonded to our bromine is directly attached to one alkyl group followed by a secondary alkyl halide the carbon bonded to the bromine is bonded to two alkyl groups followed by a tertiary alkyl halide this carbon is bonded to three alkyl groups turns out that the methyl halides and the primary alkyl halide react the fastest in an sn2 mechanism secondary alkyl halides react very slowly and tertiary alkyl halides react so so slowly that we say they are unreactive toward an sn2 mechanism and this makes sense when we think about the mechanism because remember the nucleophile has to attack the electrophile the nucleophile needs to get close enough to the electrophilic carbon to actually form a bond and steric hindrance would prevent that from happening so something like a tertiary alkyl halide has these big bulky methyl groups which prevents a nucleophile for attacking so let's look at a video so we can see this a little bit more clearly here's our methyl halide with our carbon directly bonded to a halogen which I'm saying is yellow and here's our nucleophile which could be the hydroxide ion the nucleophile approaches the electrophile for the side opposite of the leaving group and you can see with a methyl halide there's no steric hindrance when we move to a primary alkyl halide the carbon bonded to the halogen has only one alkyl group bonded to it it's still easy for the nucleophile to approach when we move to a secondary alkyl halide so for a secondary you can see that the carbon bonds of the halogen has two methyl groups attached to it now it gets a little harder for the nucleophile to approach in the proper orientation so these bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic Carbon we go to a tertiary alkyl halide so three alkyl groups there is one there's two and there's three there's a lot more steric hindrance and it's even more difficult for our nucleophile to approach as we saw in the video for an sn2 reaction we need decreased steric hindrance so if we look at this alkyl halide the carbon that is directly bonded to our halogen is attached to only one alkyl group so this is a primary alkyl halide and that makes this a good sn2 reaction so the decreased steric hindrance allows the nucleophile to attack the electrophile