Ring-opening reactions of epoxides: Strong nucleophiles

If oxides have a lot of ring or angle strain, this makes them very reactive towards ring opening. So in this video, we'll look at the ring opening reactions of epoxides using strong nucleophiles. So in the first step, we add a strong nucleophile to our epoxide. And in the second step, we add a proton source. And the nucleophile is going to end up opening the ring and adding, in an anti fashion, to the OH that is created here. And so in terms of regiochemistry, that nucleophile adds to the substituted carbon. So it's going to add to this carbon right here. If we go back and look at our epoxide on the left, that's this carbon. And this carbon is attached to two other carbons, so it would be classified as a secondary carbon, whereas the carbon on the right is attached to three other carbons-- and so that's a tertiary carbon. So the mechanism for this reaction is an SN2 type mechanism, which means that our strong nucleophile is going to attack the least sterically hindered carbon, which is the one on the left, since a hydrogen takes up less space than some bulky R group over here on the right. And so that's the regiochemistry here. The stereochemistry is an anti opening of the ring for an anti addition of the nucleophile. And this will affect possible chirality centers, as we'll see in later examples. Let's look at some examples of strong nucleophiles. So for your strong nucleophile, you could use an alkoxide anion-- so something like this over here. You could use a Grignard reagent, which we know are sources of carbanions. So we could have a negatively charged carbon like this. And you didn't have to write your Grignard reagent like this. You could have written it like RMgBr. Hydride itself is not a strong nucleophile, but if hydride is coming from something like lithium aluminum hydride, it functions like a strong nucleophile. So we could go ahead and show that hydride is really coming from lithium aluminum hydride. So if we go ahead and draw this in here, aluminum is going to have a negative 1 formal charge. And then lithium would have a plus 1 formal charge, like that. So lithium aluminum hydride is a source of a strong nucleophile. It's just that the hydride anion on its own, we've already talked about, is not the best nucleophile. And then, of course, you could have something like hydrogen sulfide anion or the hydrosulfide anion right here, which also functions as a strong nucleophile. Let's take a look at the mechanism to form our alcohols here. So we're going to, once again, start with our epoxide. And I'll use the same epoxide that we had before, meaning there's going to be an R prime here and R double prime going away from us, hydrogen coming out at us, and an R group going away from us, like that. So when I think about electronegativity differences, I know that the oxygen is more electronegative than this carbon, for example. So the oxygen is going to pull the electrons, and the bonds between the oxygen and that carbon, closer to the oxygen, giving it a partial negative charge. So pulling some electron density away from this carbon will give this carbon down here a partially positive charge. And so that carbon is going to function as our electrophile. So when the nucleophile comes along, it will have a lone pair of electrons and a negative 1 formal charge. It has two options. It could attack the carbon on the left or it could attack the carbon on the right. And since it is a strong nucleophile, which precedes via an SN2 type mechanism, it's going to attack the carbon on the left. So once again, the carbon on the left is the least sterically hindered. That hydrogen is not very big compared to some alkyl group over here on the right. So the nucleophile is going to attack the electrophile, like that. And when that happens, it's going to kick these two electrons in here off onto your oxygen. So we can go ahead and draw the product of that nucleophilic attack. So now, we broke our epoxide ring there, and we have an extra lone pair of electrons on that oxygen, giving that oxygen a negative 1 formal charge. These R groups on the carbon on the right haven't changed. So they are still there, like that. We've now added our nucleophile, like that. And that's going to push the hydrogen up relative to the plane. And then we have an R group over here, like that. So the nucleophile has to attack from below, because the oxygen is on top and the nucleophile is going to attack the least sterically hindered carbon. This ends up putting your nucleophile and your oxygen anti to each other. And in the second step, we can add a proton source-- so something like water. And we're going to get an acid-base reaction, where alkoxide is going to take a proton from water, which kicks these electrons in here off onto the oxygen. And that will form our final product, with an alcohol. So I'm going to have an alcohol right here. And once again, my R groups are still there. So I have an R prime and R double prime. Our nucleophile is added, and our hydrogen is pushed up, and our R group is still back here. So that is the mechanism for the SN2 type reaction. Let's do a problem which involves some stereochemistry, so we can see how the mechanism will determine the stereochemistry of the products here. So I'm going to start out and put my epoxide coming out at me in space, like that. And then at this top carbon here, I'm going to have a methyl group going away from me in space. So that's my reactant. And I'm going to use sodium ethoxide. So I'll go ahead and draw sodium ethoxide here. And we're going to use ethanol as the solvent. So we have ethanol here as our solvent, like that. So if we look at this epoxide, it's a little bit hard to see the stereochemistry. So let's go ahead and redraw this molecule and we'll look at it from a different vantage point here. So if I look at that epoxide from a different vantage point, I'm looking at it from above and down a little bit. And that would put my epoxide going up, like that. And then it was this carbon right here that's going to have the methyl group going down relative to the plane of the ring. And then this carbon over here on the left has a hydrogen. So my first step would be to identify my nucleophile, which, of course, is the ethoxide anion. So I can go ahead and draw my ethoxide anion in here. And next, I need to think about where that nucleophile will attack. So I know that this top oxygen here, being more electronegative than the carbons, is going to have a partial negative charge. And so I have two options. I could be the carbon the left. I could be the carbon on the right. And if I think about the fact that this is a strong nucleophile SN2 type mechanism, it's going to attack the less sterically hindered carbon, which, of course, is the carbon on the left, which is going to have a partial positive charge. And so we're going to get nucleophilic attack. My nucleophile is going to attack this carbon right here, which would kick these electrons in here off onto the oxygen. So I can go ahead and draw the product of that. So I have my ring, like that. And I opened up the epoxide ring, so now this bond is to an oxygen, and it has three lone pairs of electrons, like that, which give it a negative 1 formal charge. There's still a methyl group going down at this carbon. And I've added my nucleophile to this carbon. So I can go ahead and draw my oxygen and then an ethyl, like that. And then I pushed up the hydrogen. So when this reaction occurred, the hydrogen that was down relative to the plane of the ring, got pushed up. So now there's a hydrogen up at that carbon. And in my final step, I'm going to protonate the alkoxide. So a lone pair of electrons will pick up a proton from ethanol, which kicks these electrons off onto here. And I form my alcohol as my product. So when I'm drawing my product over here, now I have an OH and I still have a methyl group going down. And over here, I have an oxygen and an ethyl, like that. So if I wanted to draw my product as a two dimensional dot structure, I could look this way. I could look down on it, like that, and just go ahead and draw what I see when I'm looking down. So I can draw my ring. So here is my ring. And if I look at this carbon right here, there's an OH coming out at me. So I can represent the OH coming out at me with a wedge right here. And there's a methyl group going away from me at that carbon. So this methyl group is going away from me. So I can go ahead and put in a dash for that methyl group. And when I look at the other carbon here, I have an O and an ethyl going away from me. So this guy would be a dash going away from me, like that. So that, of course, is this portion of the molecule. So that's my product. When I look at what happened to the stereochemistry-- let's first start with this top carbon right here-- this top carbon had a methyl group going away and an oxygen coming out at you at that carbon. And if we look over here, that's the same thing that we have. We have a methyl group going away from us, and then an oxygen coming out at us, like that. So it's the same absolute configuration at that carbon. If we look at the other carbon-- if we look at this carbon down here-- on the left, we have an oxygen coming out at us in space. And on the right, at this carbon now, the oxygen is going away from us in space. So we actually see inversion of configuration. So right here, we have inversion of configuration in terms of the stereochemistry. We went from a wedge to a dash. And that's, of course, due to the mechanism. We had a nucleophilic attack at a chirality center. And so we observe inversion of configuration at that chirality center. Let's do one more example-- of a strong nucleophile attacking an epoxide. And so this time, let's use a Grignard reagent. So I'm going to start with phenylmagnesium bromide. So we've already made phenylmagnesium bromide as our Grignard reagent. And in the first step, we're going to add ethylene oxide, the simplest epoxide. In our second step, we're going to add hydronium-- so H3O plus. And that'll be our source of protons. But that has to be in a second step. You can't add that at the same time. Because there would be an acid-base reaction from your Grignard reagent with your hydronium ions. So if I'm looking at phenylmagnesium bromide, I could draw it like that, or I could redraw it to identify my carbanion. So if I think about what's going to happen, the electrons in the bond between this carbon and magnesium in here, carbon is more electronegative than magnesium. So you could think about those two electrons as being on that carbon, like this, which would give that carbon a negative 1 formal charge. And so now we have a carbanion and then we have MgBr plus, like that. And that's going to be the nucleophile. When we have our ethylene oxide that comes along-- so same idea-- oxygen is more electronegative, so partial negative on the oxygen, and a partial positive on the carbons that are symmetrical. So when I think about nucleophilic attack, these electrons in blue are going to be my nucleophile. They're going to attack my electrophile, kicking these electrons off. So when I draw the product of the nucleophilic attack, I have my ring here. And now, I'm going to end up with an alkoxide that looks like this. So let me go ahead and highlight the electrons in blue. The electrons in blue formed a new bond, a carbon-carbon bond right here. So the electrons in blue formed a new carbon-carbon bond, which is why organometallics are so useful. You can form these carbon-carbon bonds here. And then I could see that the two carbons over here on the right, this carbon and this carbon, those are the ones that came from my ethylene oxide, like that. So in my last step, it's an acid-base reaction. And this alkoxide is going to pick up a proton from H3O plus in the second step. So a lone pair of electrons picks up a proton, kicks these electrons off, and I can go ahead and draw the product. So once again, here's my ring. And I now protonated my alkoxide to form an alcohol. And the name of this alcohol-- so if I think about this as being carbon 1 and this as being carbon 2, I have a phenyl group coming off of ethanol-- so 2-phenylethanol would be the name of the product that's formed. And 2-phenyl ethanol is famous for being a component of rose oil. So this molecule smells a lot like roses, and it's used in the perfume industry. So this is one way to make it 2-phenylethanol. There are actually several ways to do it. But I think perfume chemistry is extremely interesting.