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Current time:0:00Total duration:6:07

Video transcript

this video we're going to look at the stereochemistry of the sn1 reaction on the left is our alkyl halides 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 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 +1 formal charge so let's draw the resulting carbo cation here so let me sketch that in the carbon in red is this carbon so that carbon should have a plus 1 formal charge and the next step of our mechanism our nucleophile will attack right so the nucleophile attacks the electrophile and a bond will form between the sulfur and the carbon in red they'll remember the geometry directly around that carbon in red the carbons that are bonded to it so this carbon in magenta this carbon 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'll 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 halides 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 the electrons coming off on to our bromine and leaving to form a carbo cation that's not what the carbo cation should look like we need planar geometry around that central carbon so here's another model which is more accurate now the nucleophile to the from the left or from the right and first let's look at what happens when the nucleophile attacks from the left 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 carbo cation here's our carbon cation 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 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 carbo cation so we can compare the two now let's compare this product with the product when the nucleophile attacked from the left sides on my left hand I'm holding the products with a nucleophile attacks on 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 there are 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 and nanti members now let's look at our products from a different viewpoint I'm going to take the product on the left and I'm going to turn it so that the SH is coming out at me in space so here we can see the SH 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 products and this time if we're going to keep the same carbon chain the methyl groups coming out at us in space the SH 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 but putting the SH on a wedge let me just draw that in here so here's our SH on a wedge which means the methyl group must be going away from us in space and if I was going to draw the other enantiomer I would have to show the SH going away from us in space which means the methyl group is coming out at us and notice that these two products match match the model sets that we drew here since there is 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 I'm going to say you're approximately 50% is this enantiomer and approximately 50% of our products is this enantiomer finally let's go through the hybridization states of this carbon in red one more time so for our starting alkyl halides the carbon in red is tetrahedral right sp3 hybridized who has tetrahedral geometry when we formed our carbo cation 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