Video transcript
- [Narrator] In this video, we're going to look at the stereochemistry of the SN1 reaction. On the left is our alkyl halide, on the right is our nucleophile with a negative charge on the sulfur. We know that the firststep of our SN1 mechanism should be loss of a leaving group. So if these electronscome off onto the bromine, we would form the bromide anion. And we're taking a bondaway from the carbon in red. So the carbon in red shouldget a plus one formal charge. So let's draw theresulting carbocation here. So let me sketch that in. The carbon in red is this carbon. So that carbon should havea plus one formal charge. In the next step of our mechanism, our nucleophile will attack, alright? So the nucleophileattacks the electrophile and a bond will form between the sulfur and the carbon in red. But remember the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta are in the same planeas the carbon in red, and so the nucleophile couldattack from either side of that plane. At this point, I think it's really helpful to look at this reactionusing the model set. So here's a screenshot from the video I'm going to show you in a second, and in that video I make bromine green, so here you can see this green bromine, this methyl group comingout at us in space is going to be red in the video. On the right side this ethylgroup here will be yellow, and finally on the left sidethis propyl group will be gray. So here's our alkylhalide with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint, and we know that the first stepis loss of our leaving group so I'm going to showthese electrons coming off onto our bromine and leavingto form a carbocation. But that's not what thecarbocation should look like. We need planar geometryaround that central carbon. So here's another modelwhich is more accurate. Now the nucleophile could attack from the left or from the right, and first let's look at what happens when the nucleophileattacks from the left. So we form a bond betweenthe sulfur and the carbon, and let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again, and this time let's saythe nucleophile approaches from the right side. So we're going to form a bond between this sulfur and this carbon. And let's make a modelof the product that forms when the nucleophileattacks from the right. So here is that product. And then we hold up the carbocation so we can compare the two. Now let's compare thisproduct with the product when the nucleophileattacked from the left side. So in my left hand I'm holding the product when the nucleophileattacked from the left, and on the right I'mholding when the nucleophile attacks from the right. So what's the relationshipbetween these two? Well they're mirror images of each other, but if I try to superimposeone on top of the other, you can see I can't do it. So these are non-superimposablemirror images of each other. These are enantiomers. Now let's look at our productsfrom a different viewpoint. So I'm going to takethe product on the left, and I'm going to turn it so that the S-H is coming out at me in space. So here we can see the S-Hcoming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product, and this time if we're goingto keep the same carbon chain, the methyl group's comingout at us in space, the S-H is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two productsthat we got from the video, but since you won'talways have a model set let's go back to the drawings over here and pretend like wedon't have a model set. We know that our nucleophileattacks our electrophile and a bond forms betweenthe sulfur and that carbon. So if I draw in my carbon chain here I know a bond formed betweenthe sulfur and the carbon. Let me highlight the electrons. So let's say a lone pairof electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbonis a chiral center. There are four differentgroups attached to that carbon. So if you think about thestereochemistry of this mechanism, with the nucleophileapproaching the electrophile from either side of that plane, you should get a mixture ofenantiomers as your product. So if I draw in my carbon chain here, I could represent oneenantiomer by putting the S-H on a wedge, so let me just draw that in here. So here's our S-H on a wedge, which means the methyl group must be going away from us in space. And if I'm going to drawthe other enantiomer, I would have to show the S-Hgoing away from us in space, which means the methylgroup is coming out at us. And notice that these twoproducts match the model sets that we drew here. Since there's an equallikelihood that the nucleophile could attack from one side or the other, we would expect to see anequal mixture of our products. So I'm going to say here approximately 50% is this enantiomer, and approximately 50% of ourproduct is this enantiomer. Finally let's go throughthe hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral, right? It's sp3 hybridized so ithas tetrahedral geometry. When we formed our carbocation, the carbon in red is now sp2 hybridized, so it has planar geometry. But for our products we're backto an sp3 hybridized carbon with tetrahedral geometry. So we have to think aboutthe stereochemistry.
