Sn1 and Sn2
Sn1 vs Sn2: Summary
- [Instructor] In this video, we're going to look at how to determine if a reaction proceeds via an S N 1 or an S N 2 mechanism and also how to draw the product or products for those reactions. To help us, we're gonna look at this S N 1 versus S N 2 summary and the first thing that we're going to look at is the structure of our substrate. For example, for this reaction down here, we have a primary alkyl halides, a primary substrate, so we need to think about an S N 2 reaction which requires decreased steric hindrance and that's what we have with a primary alkyl halide. From an earlier video, we know that an S N 2 mechanism has our nucleophile attack at the same time that we get loss of a leaving group. And for this reaction, our nucleophile must be our solvent here, which is ethanol. So, our nucleophile is going to attack and the oxygen is gonna form a bond with this carbon which I'll go ahead and highlight in red. So our nucleophile attacks, at the same time, we get loss of leaving group. These electrons come off to form the iodide and ion which is an excellent leaving group. Let's draw what we would form. Let's sketch in our carbon chain here and we know that a bond forms between the oxygen and the carbon in red, so the carbon in red is this carbon and let's make these electrons magenta. So those electrons form a bond between the oxygen and the carbon in red. The oxygen is still attached to this ethyl group here, so let's draw in those two carbons. And the oxygen is still bonded to this hydrogen. So let's put in this hydrogen. We still have a lone pair of electrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus one formal charge on the oxygen. Next, we need to make a neutral molecule for our product, so we need to have another molecule of ethanol come along, so let's draw that in here. So, ethanol is our solvent. And this time, the ethanol molecule is going to function as a base. We need to take this proton here and these electrons are left behind on the oxygen. Let's draw our final product. We sketch in our carbon chain. We have our oxygen. We have these two carbons. And now we have two lone pairs of electrons on this oxygen. Let's make these electrons blue. So, our second step is an acid-base reaction where we take a proton and these are, these electrons in blue here, to form our final product which is an ether. Notice we don't have to worry about any stereochemistry for our final product. We don't have any chiral centers to worry about. Let's look at another reaction. For this reaction, we're starting with a secondary alkyl halides. If I look at my summary over here with a secondary substrate, we could have either an S N 2 mechanism or an S N 1 mechanism, so we need to look at a few more things. First, let's look at the nucleophile. This is N a plus and SH minus, so let me draw in the SH minus here which that is going to be our nucleophile and that's a strong nucleophile. A negative charge on a sulfur would make a strong nucleophile. And for our solvent, we saw in an earlier video, DMSO is a polar A product solvent which favors an S N 2 reaction. So with a strong nucleophile and a polar A product solvent, we need to think about in S N 2 mechanism. So we know our nucleophile attacks at the same time that we get loss of our leaving group, so our nucleophile is going to attack this carbon. So again, I'll make this carbon red. At the same time that we get loss of leaving groups, so these electrons are gonna come off onto the bromine to form the bromide and ion. For this reaction, we need to think about the stereochemistry of our S N 2 reaction. Our nucleophile has to attack from the opposite side of our leaving group, so we get inversion of configuration. So if we have a chiral center, we have to worry about our stereochemistry for this reaction. So we have the bromine on a wedge, so drawing the final product here, we need to have the SH going away from us in space, so we put that on a dash. Again, we saw details about this in an earlier video. So we get inversion of configuration for this S N 2 reaction. First, let's look at our alkyl halides. The carbon that's bonded to our bromine is bonded to two other carbon. So, this is a secondary alkyl halide. And so we know we could have either S N 1 or S N 2. We need to look at the nucleophile and the solvent next to decide which mechanism it is. Our nucleophile will be formic acid which is a weak nucleophile and water is a polar product solvent. So we know that these two things favor an S N 1 type mechanism. The polar product solvent water can stabilize the carbocation that would result. So the first step should be loss of our leaving group to form our carbocations. These electrons come off onto our bromine to form the bromide ion and we're taking a bond away from this carbon in red. So the carbon in red is gonna have a plus one formal charge. Let's draw our carbocations. Let me put in this ring here with our pi electrons. And then let me highlight our carbon in red. The carbon in red is this one. So that should get a plus one formal charge. And let me highlight the other carbon over here. So this carbon in magenta, I moved it up to here to make it easier to see the mechanism. So this is our secondary carbocation, but this is actually a benzylic carbocation which makes it even more stable than we would normally expect. The pi electrons in the ring can actually provide us with other resonant structures to stabilize this positive charge. So I don't have the time or the space to show that here, but we have our secondary benzylic carbocation which will be our electrophile and our next step is to have our nucleophile attack our electrophile and our nucleophile is formic acid here. We have two oxygens in formic acid and one of them is more nucleophilic than the other, so it turns out that this carbonyl oxygen is more nucleophilic than this oxygen down here and we'll go into more detail in a couple of minutes, but for right now, let's show our nucleophile attacking our electrophile. So, a lone pair of electrons on our oxygen are gonna form a bond with this carbon in red. So let's draw the result of our nucleophilic attack. We have our benzene ring, so I'll draw that in here and our carbon in red is this one. And let's highlight the electrons on our oxygen. So, this lone pair of electrons in blue forms a bonds with our carbon in red. So now, let's draw in our oxygen. Our oxygen still has a lone pair of electrons on it and our oxygen is double bonded to a carbon which is bonded to a hydrogen. And on the right, we have our oxygen and a hydrogen. Now we have a plus one formal charge on our oxygen. So this oxygen has a plus one formal charge. And this product is resonance stabilized, so we can move in a lone pair of electrons here and push these electrons off onto our oxygen. And let's go ahead and show the results of that. So we'll put in our resonance brackets here and our resonance arrow. So we have our benzene ring, so let's draw that in. And we have our carbons. We have a bond to this oxygen and I'll drawn on that lone pair of electrons on the top oxygen here. Should only be one bond now to this carbon. I'll put in my hydrogen, and now actually, we have a double bond over here. Let me draw everything in. And let me show the movement of electrons. So first, let's start with these electrons which I will make magenta. Those electrons moved into here to form a double bond. And then let's highlight these electrons here in red. So these electrons came off onto our oxygen right here. And now, we have a plus one formal charge on this oxygen. So the product of nucleophilic attack by the carbonyl oxygen is resonance stabilized. And then to go to our product, just think about a base coming along, something like water, and we could take this proton and then these electrons would be left behind. These electrons would be left behind on that oxygen. So let's draw in our product. We would have a benzene ring. So here's our benzene ring. We would have an oxygen here. And then we would have our carbonyl and then our hydrogen. So this oxygen has two lone pairs of electrons and so does this one. So we could highlight some of those electrons. Let me make them green. So these electrons in green here came off onto this oxygen and that gives us our product. Notice that in our products, we have a chiral center. So this carbon right here is a chiral center. We have four different groups attached to it. So we would expect to get a racemic mixture. We should get both in the enatiomers here because remember, going back to our carbocation, this carbocation is planer and our nucleophile can attack from either sides. If you're wondering why this oxygen is not as nucleophilic as the carbonyl oxygen, let's show the result of what would happen if this oxygen attacked our positively charged carbon. Let me draw in our benzene ring. Let's sketch that in. And we would form a bond between the oxygen and, let me highlight our carbon which I made red below, so this is our carbon in red. And I'm gonna show this lone pair of electrons forming a bond with that carbon. So, that's bonded to this oxygen and the oxygen is bonded to our carbonyl which is bonded to our hydrogen. And our oxygen still has a hydrogen on it and a lone pair of electrons. So there's still a hydrogen on it and there's still a lone pair of electrons which gives this oxygen a plus one formal charge. Notice this positively charged oxygen is right next to this carbonyl carbon and we know that this carbonyl carbon is partially positive because this carbonyl oxygen withdraws electron density from it. So you have a positively charged oxygen next to a partially positively charged carbon and we know that like charges repel. So having these two positive charges next to each other would destabilized this structure. And so that's the reason why this oxygen is not the nucleophile. The carbonyl oxygen is. Let's look at one final example. So for this alkyl halide, this is a tertiary alkyl halide and a tertiary substrate means, think about an S N 1 mechanism. So our first step here would be loss of a leaving group. These electrons come off to form the iodide and ion and we're taking a bond away from this carbon in red to form a carbocation. Let's get some more room. Let's go down here and let's draw our carbocation. So we have a six-membered ring, so let me draw in our six-membered ring here. And our carbon in red is this carbon, so that carbon gets a plus one formal charge. We have a tertiary carbocation which is relatively stable for a carbocation. And in our next step, our nucleophile attacks our electrophile and our nucleophile is our solvent, which is methanol. So our nucleophile is going to attack our electrophile. Our oxygen is gonna form a bond with this carbon in red. So let's draw the results of our nucleophilic attack. I'll put in this ring here, move the methyl group over to make room for the oxygen. So here's that oxygen. Let's highlight the carbon in red. And let's also highlight some electrons. So, these electrons in magenta form the bond between the oxygen and the carbon in red. The oxygen is still bonded to a hydrogen and a methyl group, so let's put those in. So here is our hydrogen and here is our methyl group. We still have a lone pair of electrons on the oxygen, so let's put in that lone pair of electrons and that gives the oxygen a plus one formal charge. To get to our product, we need a neutral product, so another another molecule of methanol comes along but this time, instead of acting like a nucleophile, it's gonna act as a base and take off that proton. So here is our molecule of methanol and we're gonna take this proton and leave these electrons behind on the oxygen. So let's draw in our final product up here. So, here is our ring and let's move up a little bit. So, let's move up. In our ring, we have this methyl group here and we have our oxygen and another methyl group, two lone pairs of electrons on the oxygen. Let's show those electrons. These electrons in here in blue come off onto the oxygen to form our final product. Notice that for our final product, we don't have any chiral centers to worry about, so we don't need to worry about specifying any stereochemistry.