Nomenclature and preparation of epoxides
Naming of epoxides (cyclic ethers with three atoms in the ring). Preparation of epoxides from alkenes. Created by Jay.
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- at1:47, shouldn't it be 2,3 - epoxy 2 - methyl butane? (alphabetical order)(20 votes)
- I thought so too at first since most sites (including this one, see the bottom of the page: http://www.chem.ucalgary.ca/courses/350/orgnom/ethers/ethers-02.html) seem to have "epoxy" listed before "methyl" in the name. However, I searched "epoxybutane" on ChemicalBook and some of the names that came up had "fluoro" or "hydroxyl" in front of "epoxy" (http://www.chemicalbook.com/Search_EN.aspx?keyword=epoxybutane). So it's all a bit confusing, but I would go with the general IUPAC rule of alphabetical order for substituents because that seems to be the most recognised as far as I can tell. You can also check with your professor to see what he/she prefers on exams.(4 votes)
- in the second minute he says that he talked about the general mechanism of synthesizing epoxides from alkenes..what video is that in? the previous one (Cyclic ethers & epox naming) did not include that and i need to learn it!(7 votes)
- C-electronegitivity 2.55
why does bromine-carbon create partial charges but carbon-hydrogen doesn't?(4 votes)
- I think it is not just about electronegativity as shown in this video....it is also about "being a good leaving group" and bromide, Br-, is an excellent one.....(2 votes)
- What about heterocyclic nomenclature? Don't these rules take precedence, making this ethylene oxide (common name) or oxirane (systematic name)?(2 votes)
- Epoxides are named in many ways.
1. Common name — as alkene oxides: ethylene oxide.
2. As oxygen-substituted cycloalkanes — oxacyclopropane.
3. As epoxy substituents — epoxyethane.
4. Hantzsch-Widman — oxirane.
You won't get chemists to give up their old familiar names. Ethylene oxide will always be with us.
The IUPAC recognizes both methods 3 and 4. Thus, propylene oxide should be called either 1,2-epoxypropane or 2-methyloxirane.(4 votes)
- Why isn't hydroxide acting as a nucleophile in the halohydrin mechanism ?(2 votes)
- It probably does, but it has to come from the solution and hit the intermediate from the correct direction and with enough energy.
The O⁻ is in just the right position for a backside attack on the carbon atom.(3 votes)
- In the mechanism involving the halohydrin, Jay uses the base sodium hydroxide to depronate the halohydrin leading to an alkoxide. The alkoxide, however, is a stronger base than the hydroxide. How is it possible that hydroxide, the weaker base, can depronate the alcohol, yielding an alkoxide, a much stronger base, at a reasonable pace?(2 votes)
- Actually, the bromo group increases the acidity of the alcohol by induction.
ethanol pKa = 15.9
2-bromoethanol pKa = 13.82
For a video on induction see:
- What makes a concerted mechanism "concerted"?(2 votes)
- in a concerted mechanism, all the steps happen at the same time.(2 votes)
- respected sir from reference of cengage learning book I have found the name of 1,2-epoxy ethane as oxirane and 1,3-epoxy propane as oxetane......so is there any type of another nomenclature?(2 votes)
- When/why does the ether itself (the bonds to oxygen) have wedges/dashes on them rather than the substituents?(2 votes)
- why is in the synthesis of epoxide does a sn1 reaction not occur. to be more clear why doesn't the halogen/cl/br leave since its a good leaving group and the strong base form a bond with the carbocation(1 vote)
- Remember there is a difference between nucleophilicity and basicity in molecules. A strong base does not denote that it will be a strong nucleophile.
The rate limiting step for an Sn1 reaction is the leaving group leaving forming a carbocation, this is because it takes some time for the leaving group to actually leave on its own accord. A strong base is not going to wait around for the leaving group to get out of there when it can take a perfectly decent acidic hydrogen (the hydrogen in the hydroxyl group). This transforms the hydroxyl into an alkoxide (O-) which is a decent nucleophile and in the perfect position to perform an Sn2 (it is attached to the same molecule meaning it will be an intramolecular Sn2), hence the formation of an epoxide and not the base reacting with the leaving group.(2 votes)
It's possible to have ethers in a ring system. And there are many different types of ring systems that you can have with ethers in them. The one that's studied most of the time would be the epoxides, due to their reactivity. Here we have the simplest epoxide. And one name for this would be ethylene oxide because this molecule is made from ethylene and that's where you get your two carbons from, like that. So you could call this ethylene oxide, or you could give this an IUPAC name. And since there are two carbons in it, for the IUPAC name we start with ethane as our parent name. And we know that the epoxide forms between carbons one and two. So we can go ahead and write 1,2-epoxyethane for the IUPAC name. Let's do another one here. So let's put some more carbons on here, so we'll put a few extra carbons, and we'll name this using IUPAC nomenclature. So once again, find your longest carbon chain, right? So, we can go ahead and find how many carbons are along this carbon chain. That would be one, two, three, and four, like that. So we can go ahead and write butane for our parent name. So we go ahead and put butane in here. I want to next number my carbon chain to get the lowest number possible to the substituents. So in this case, it makes more sense to number from the left. So I get one, two, three, and four, to give my substituents the lowest number possible. I can see now that my epoxide side forms between carbons two and three. So I'm going to write 2,3-epoxybutane like that. And I know that I also have a methyl group coming off of carbon two. So to complete the name all I have to do is write 2-methyl on the front here. So now I have 2-methyl-2,3-epoxybutane for my IUPAC name. So how do we make epoxides? We've already seen one way to do it. And in an earlier video, we started with our alkene, and to the alkene, we added a peroxy acid. And a peroxy acid looks a lot like a carboxylic acid, except it has an extra oxygen in there like that. And in the mechanism for the epoxidation of alkenes we saw was a concerted mechanism, where one of those oxygens was added in here to form our epoxide like that. So check out the earlier video to see the mechanism for epoxidation of alkenes. So in this video, we're going to cover another way to make epoxides, and that is using halohydrins. So to make a halohydrin, you also start out with an alkene. And we also saw this mechanism in an earlier video, when you add a halogen and water. And we're going to add bromine and water. And we end up adding the OH and one bromine across our double bonds. So we ended up getting an anti-addition of an OH and a bromine. OK, so they're going to add on opposite sides from each other like that. OK, so this molecule is called a halohydrin. And again, check out an earlier video for the mechanism to form a halohydrin. Once you form a halohydrin, you can use that halohydrin to form an epoxide. So let's go ahead and take that halohydrin, and let's see the mechanism of how we can form an epoxide from that. So I'm going to redraw that halohydrin. So I'm going to go ahead and put in the OH here like that, put my lone pairs of electrons. And then I have my bromine over here, and I'll go ahead and put in my lone pairs of electrons on bromine as well. And for right now, we can say anything could be attached to this, and we'll go into stereochemistry in the next video. So what we need to do is add a base. So something like sodium hydroxide will work. So we're going to add in sodium hydroxide, Na plus OH minus. OK, so the hydroxide anion is going to function as a base. All right, so a lone pair of electrons on the oxygen are going to take this proton on our alcohol. And these electrons in here are going to kick off onto our oxygen. So let's go ahead and draw the result of that acid base reaction. All right, so what do we make from that? Well, now we have our oxygen with three lone pairs of electrons around it like that, which gives the oxygen a negative 1 formal charge. And we still have our bromine here, like that. And then we still have these other groups attached to our carbon. So in the next step, we need to think about, again, the polarization in the bond between carbon and our halogen, right? Our halogen is more electronegative, so the halogen is going to take a little bit of this electron density in the bond between carbon and bromine and therefore give the bromine a partial negative charge. Right, this carbon's going to lose a little bit of electron density. So this carbon is actually partially positive. And so the alkoxide that we formed when the alcohol was deprotonated has a negative charge. It's going to function as a nucleophile. The partially positive carbon wants electrons. It's going to function as an electrophile. And we're going to get a nucleophilic attack by our alkoxide anion on our partially positive carbon. So this is actually an intramolecular Williamson ether synthesis. So if you think about it, right, if these electrons in here are going to attack this carbon, that would kick these electrons off onto your bromine like that. And it's an intramolecular Williamson either synthesis, where your alkoxide is the nucleophile in an SN2-type mechanism. So if we go ahead and draw the product, right now, this oxygen was bonded to the carbon on the right. Now, it's also a bonded to the carbon on the left, and the bromine left. That was our leaving group. And so we can see that we're going to end up forming an epoxide with this mechanism. So let's go ahead and do a quick problem here. We're reforming an epoxide from an alkene. And we'll start with cyclohexane. So here is our cyclohexane molecule. And we'll make an epoxide two ways, right? So in the first way, we'll add a peroxy acid, and there are several that you can use. One of the most common ones would be peroxyacetic acid. So peroxyacetic acid looks very similar to acetic acid, except you have an extra oxygen in there, like that. So it's epoxidation of an alkene. And when we draw our product, right, so let's go ahead and draw our product to form an epoxide. And I'm going to go ahead and draw the product with a wedge here. All right, so there's an oxygen coming out relative to that plane. And if we go ahead and name our product, right, so the parent name would be cyclohexane, and our epoxide would form between carbons one and two, so we could go ahead and name this as 1,2-epoxycyclohexane, like that. Let's go ahead and to cyclohexane, let's do another reaction. Let's start with cyclohexane, and this time in the first step, we'll add some bromine and some water, and that will form our halohydrin. And in the second step, we'll add sodium hydroxide to act as our base. And we get an intramolecular Williamson ether synthesis, and so we're going to end up with the same product, We're going to end up with 1,2-epoxycyclohexane. Now for this reaction, we don't have to worry about stereochemistry. OK, so if you think about the oxygen adding from the other side of the ring, we don't have to worry about stereochemistry for our products because if the oxygen added from the other side of the ring, they would actually be the exact same molecule, 1,2-epoxycyclohexane. So we'll save stereochemistry for the next video, where we can focus in on what happens when you're adding an oxygen to different sides of a plane.