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Sn2 mechanism: stereospecificity

How nucleophilic attack in Sn2 reaction results in inversion of configuration at carbon with leaving group.

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Video transcript

- [Narrator] In the last 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 OH, the nucleophile, substitutes for our leaving group. So for our final product we now have an OH 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 R. We look at our product, this is that chiral center. But now we have our OH 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 configuration. 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 it with all these loan pairs of electrons. And that will 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 do a video where I'm gonna use the model set to show you how R2 bromobutane turns into S2 butanol, and hopefully it'll be more obvious. Here is R2 bromonbutane and I've made bromine yellow. And I've left the hydrogens 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, so 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 get our nucleophile, our nucleophile is trying to attack this carbon. So the hydroxide ion has to approach our substrate from the side that is 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. Now if we look at that central carbon there and we look at the three groups, the methyl, the hydrogen and the ethyl group, those 3 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. If I keep this model of the transition state up, we can compare it with the final product. So here's the final product, the alcohol. We still have the methyl group coming out at us space. With the hydrogen going away from us, our OH 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, so we can see our carbon chain. So if I turn it, if I turn it this way, now our OH is going out at us in space. So that's another way to view our final product. I can also turn this model set a little bit so if I turn it over to view the carbon chain this way, now the OH 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 off onto the bromine. Let's draw out the transition state. So we're gonna put some brackets in here. And I'm gonna put in my OH, I'll have some lone pairs of electrons here. And we saw that a partial bond forms between this oxygen and this carbon here. In 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. Our 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 our transition state. And we designate the transition state with this little symbol up here. So let me draw that in. Now the hydroxide ion over here was a full negative charge but 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 on to the bromine, to form the bromide anion, we're gonna get a little more negative charge on that bromine. So I'm 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 draw our product here and just like I did in the video, right, I'm gonna point out how our methyl group stays coming out at us here. So 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. So let me draw our ethyl group going down and now we have a bond to ROH. And we're back to tetrahedral geometry. So this carbon, right, our chiral center has tetrahedral geometry to start with. Here for our three groups we have planar geometry. But for our final product we're back to tetrahedral geometry. And this is one way that we drew our product in the video I showed moving the model set around and one of the view points that we looked at was if your carbon chain looks like this, that OH 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. Your OH is going away from you in space. So these are three different ways to represent our product which is S2 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 gonna 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 aside its 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 dash going away from us in space, we have to show the bond to our nucleophile here coming out as a wedge. So I'm gonna put the SH here. So this will be our final product. For simple systems you'll see inversion of configuration. If you're 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 compound. 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 have a CH3. We have a -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 loss 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 product. If you 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 should be number two, followed by carbon here which is number three and then hydrogen is the lowest priority group number four which is going away from us. So we're going around in this direction which is counter clockwise and so that's an S configuration to our starting compound. For our product again this carbon here, let me change colors, this carbon is our chiral center and when we assign priority this time, now the fluorine has 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.