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Current time:0:00Total duration:10:22

Video transcript

nós video we looked at the mechanism for the sn2 reaction the hydroxide ion will function as a nucleophile in this case and attack our electrophile so right here at this carbon and since the sn2 mechanism is concerted the nucleophile attacks the electrophile at the same time that our leaving group leaves so bromine leaves as the bromide anion and the o h the nucleophile substitutes for our leaving group so for our final products we now have an OHA attached to our carbon chain in this video we're going to look at the stereo specificity of the sn2 reaction and that just means that the stereochemistry of the reactant determines the stereochemistry of the product for example if we look at our substrate we know that this carbon is a chiral Center and our bromine is on a wedge it's coming out at us in space so the configuration at this chiral Center is our when we look at our products this is that chiral Center but now we have our Oh H going away from us in space and the configuration at this chiral center is s so the stereochemistry of our product is determined by the stereochemistry of our reactant and that's because of our sn2 mechanism we observe inversion of configurations so let me write that down here so inversion of configuration we're going from an R configuration here to an S configuration and that means that the nucleophile can only attack from the side that's opposite of the leaving group and that's consistent with our sn2 mechanism one way of thinking about this is your bromine is relatively large and it has a large amount of electron density around with all these lone pairs of electrons and that would repel your negatively charged nucleophile so your nucleophile has to approach from the side opposite of your leaving group and that requirement means that you get inversion of configuration so next we're going to go to a video we're going to use the model set to show you how our 2-bromobutane turns into s2 butanol hopefully it'll be more obvious here is our 2-bromobutane and I've made a bromine yellow and I've left the hydrogen's off the alkyl groups just to make it easier to see so if I turn the model a little bit and we look at our chiral Center we have tetrahedral geometry around this carbon there's a methyl group coming out at us in space a hydrogen going away from us in space an ethyl group going down and our bromine is off to the right so we'll get our nucleophile our nucleophile is trying to attack this carbon so the hydroxide ion has to approach our substrate from the side that's opposite of our leaving group so next let's look at the transition state notice that we still have our methyl group coming out at us in space our hydrogen going away from us in space our ethyl group going down and our halogen off to the right but now there's a partial bond that forms between this oxygen and this carbon there's also a partial bond between this halogen and this carbon and if we look at that central carbon there and we look at the three groups the methyl the the hydrogen and the ethyl group those three groups are all in the same plane so if I rotate this a little bit right you can see they're all planar there so the transition state has a bond forming at the same time a bond is breaking and I keep this model of the transition state up we can compare it with the final product so here is the final product the alcohol we still have the methyl group coming out at us in space with the hydrogen going away from us roh on the left and our ethyl group going down so this is one way to represent our product but if i rotate this model set a little bit we can see our carbon chain so if I turn it if I turn it this way now our H is going out at us in space so that's another way to view our final product I could also turn this model set a little bit so if I turn it over to view the carbon chain this way now the O H is going away from us and that's how I drew it in the original reaction so let's draw out what we saw in the video so our hydroxide ion our nucleophile attacks this carbon at the same time these electrons come on to the bromine let's draw out the transition state they're going to put some brackets in here and I'm going to put in my OHL have some lone pairs of electrons here and we saw that a partial bond forms between this oxygen and this carbon here and the video I pointed out how the methyl group is still coming out at us in space so our methyl group is coming out at us or hydrogen is going away from us and our ethyl group is going down and those three things the methyl the hydrogen the ethyl group are in the same plane and at the same time right we have a partial bond between our carbon and our leaving group which is our bromine here so let me put in these electrons here on the bromine so this is this is our transition state and we we designate the transition state with this little symbol appears let me draw that in now the hydroxide ion over here was a full negative charge but if it's forming if it's forming a partial bond with this carbon we would give it a partial negative so it's losing some of its electron density as that bond is forming and as the electrons are coming off onto the bromine to form the bromide anion we're going to get a little more negative charge on that bromine from putting a partial negative here so here's our transition state where the nucleophile is in the process of forming a bond at the same time the bond to the leaving group is breaking so finally let's let's draw our product here and just like I did in the video all right I'm going to point out how our methyl group stays coming out at us here's let me draw that on a wedge so here's our methyl group our hydrogen is still going away from us and our ethyl group is going down let me draw our ethyl group going down and now we have a bond to our o H and we're back to tetrahedral geometry so this carbon right our chiral Center has tetrahedral geometry to start with here here for our three groups we have planar geometry but for our final products we're back to tetrahedral geometry and this is one way that we drew our products are in the video I showed moving the model set around and one of the viewpoints that we looked at was if your carbon chain looks like this that OAH group is coming out at you in space and that's the same thing as if you turn the model set around where you would have your carbon chain like this euro H is going away from you in space so these are three different ways to represent our product which is s 2 butanol let's look at another sn2 reaction with stereochemistry and let's figure out the product so first I would say okay this ion here is negatively charged so that must be my nucleophile and for my substrate for my alkyl halides I know that chlorine is withdrawing electron density from this carbon since chlorine is more electronegative which means that carbon is partially positive so my nucleophile is going to attack my electrophile at the same time these electrons are going to come off onto the chlorine to form the chloride anion so I know that's what happens in my sn2 mechanism and now that we've looked at stereochemistry I know that the nucleophile has to attack from a situs opposite of our leaving group and we get inversion of configuration so when I draw my product here if I'm drawing the carbon chain the same way that we started with since this chlorine is on a - going away from us in space we have to show with the show the bond to our nucleophile here coming out as a wedge and so I'm going to put the SH here so this would be our final product for simple systems you'll see inversion of configuration if your starting compound as R you will get s for your product and if you start with s you will get R however it doesn't always have to be that way if we look at this reaction on the left is our starting compounds and this is an sn2 reaction and our nucleophile would be the methoxide anion so let me go ahead and draw that in so we have an oxygen we with ch3 and we have a negative 1 formal charge on the oxygen so our nucleophile is going to attack our partially positive carbon which is right here so our nucleophile attacks our electrophile at the same time that we get lost of a leaving group and the bromide ion would be the best leaving group here so these electrons come off onto the bromine to form the bromide ion and we form a bond between the oxygen and the carbon so this lone pair of electrons on the oxygen forms this bond and we get our products if we look at the stereochemistry let's first start on the left here so we know that this carbon is a chiral Center and if I'm assigning priority to the four groups attached to that chiral Center bromine has the highest atomic number so that would be priority number one followed by fluorine which would be number 2 followed by carbon here which is number 3 and then hydrogen is the lowest priority group number 4 which is going away from us so we're going around in this direction which is counterclockwise and so that's an S configuration to our starting compound for our products again this carbon here let me change colors this carbon is our chiral Center and when we assign priority this time now the fluorine at the highest atomic number followed by this oxygen here followed by this carbon and then of course with the hydrogen so when we go around in this direction again the configuration is still s so here's an example of an sn2 mechanism where our nucleophile has to attack from the opposite side of our leaving group so we still get inversion in terms of the mechanism but we don't get inversion of configuration because of how we assign priority to our groups in this case when our bromine leaves fluorine becomes the highest priority group and that's the reason why the configuration stays s