Sn1 mechanism: stereochemistry

- [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 first step of our SN1 mechanism should be loss of a leaving group. So if these electrons come off onto the bromine, we would form the bromide anion. And we're taking a bond away from the carbon in red. So the carbon in red should get a plus one formal charge. So let's draw the resulting carbocation here. So let me sketch that in. The carbon in red is this carbon. So that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack, alright? So the nucleophile attacks 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 plane as the carbon in red, and so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using 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 coming out at us in space is going to be red in the video. On the right side this ethyl group here will be yellow, and finally on the left side this propyl group will be gray. So here's our alkyl halide 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 step is loss of our leaving group so I'm going to show these electrons coming off onto our bromine and leaving to form a carbocation. But that's not what the carbocation should look like. We need planar geometry around that central carbon. So here's another model which 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 nucleophile attacks from the left. So we form a bond between the 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 say the 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 model of the product that forms when the nucleophile attacks 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 this product with the product when the nucleophile attacked from the left side. So in my left hand I'm holding the product when the nucleophile attacked from the left, and on the right I'm holding when the nucleophile attacks from the right. So what's the relationship between these two? Well they're mirror images of each other, but if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm going to take the 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-H coming 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 going to keep the same carbon chain, the methyl group's coming out 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 products that we got from the video, but since you won't always have a model set let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer 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 draw the other enantiomer, I would have to show the S-H going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm going to say here approximately 50% is this enantiomer, and approximately 50% of our product is this enantiomer. Finally let's go through the 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 it has 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 back to an sp3 hybridized carbon with tetrahedral geometry. So we have to think about the stereochemistry.