Adding two hydroxyls to opposite faces of an alkene double bond via an epoxide intermediate. Created by Jay.
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- Is CH3CO3H what is commonly referred to as mCPBA?(4 votes)
- CH3CO3H is not mCPBA.
However, mCPBA is related to CH3O3H as they both fit under the RCO3H reactant that's used in the mechanism. mCPBA has a chlorobenzene as its R group, where as CH3CO3H has a methyl as its R group.(7 votes)
- You add hydronium to an epoxide and the epoxyl group breaks so you end up with those two hydroxyl groups. I get that but my question is, will the hydroxyl groups be trans or cis?(3 votes)
- It depends.
If the ring opening is SN2, the OH groups will be trans, because the nucleoplile can attack only from the back side of the oxonium ion.
If the ring opening is SN1, half of the product will be cis and the other half will be trans, because the intermediate carbocation is planar, so the nucleophile can attack either from the same side as the OH that is already present (to give the cis diol) or from the opposite side (to give the trans diol).(12 votes)
- at1:16jay says "concerted 8 electron mechanism " what does it mean and in which video does he introduce the word "concerted"(2 votes)
- A concerted mechanism is one in which all the bond breaking and bond formation occurs at the same time. If you watch the next 30 s of the video, you will see that he uses four electron arrows. Each arrow represents the motion of two electrons, so this is an "8 electron mechanism". Since all the movements happen at the same time, this is a "concerted 8 electron mechanism".
This may be the first video in which he uses the word "concerted", but SN2 displacements and E2 eliminations are also concerted reactions.(8 votes)
- When H3O+ is used to open the epoxide ring, it is not a reagent, but only a catalyst, am I right or wrong in this?(4 votes)
- ya catalysts are never consumed in reaction, they may change forms (such as with friedal-crafts alkylations) but they come out the same(3 votes)
- around7:25- why would the H2O nucleophilic attack the partially positive carbons when the oxygen in the epoxide has a formal positive charge?(4 votes)
- Formal charge is not necessarily the same thing as the actual charge. In this case much of the actual positive charge is on the carbons.
Also, the result of that interaction would be a peroxide with a highly unstable (high energy) bond between two oxygens.(1 vote)
- Why do the electrons in the bond between O and H want to move to form a bond with the carbon at01:30. Is this just due to the character of RCO3H? And a similar question applies to the electron movement in the bond between O and Os (of osmium tetroxide) to form a bond with C in the next video (syn dihydroxylation).(2 votes)
- It may make more sense if you consider the steps in a different order. The O in the OH group has a partial positive charge, so it is electrophilic. The pi electrons from the right hand carbon move to the electrophilic O atom, then the weak O-O bond breaks, etc. The reaction is concerted. That is, all the steps are happening at once in a daisy chain fashion.
It is the same with OsO4. The pi electrons of the double bond move to one of the O atoms in OsO4, and the electrons again move in a concerted mechanism to give a cyclic osmate ester.(3 votes)
I would think that the double carbon-oxygen bond would be pulled at by the oxygen, giving it a slight negative charge, and the hydrogen-oxygen bond would be pulled at by the oxygen, giving the hydrogen a slight positive charge. These two partial charges would create the effect shown.
Am I correct in my thinking?(2 votes)
- i don't understand how the final compound in the cyclohex-1-ene example (the last one) which gave cyclohean-1,2-diol are enantiomers (11:57) . I thought that they both were identical compounds.Please correct me if i am wrong
Thank you(1 vote)
- They are enantiomers. For the molecule on the right, the OH at the top is into the page, and the OH at the bottom is out of the page. If you flipped the molecule up (so that the OH from the bottom was now where the OH on the top had been), the "new" top OH will still be going into the page and the "new" bottom OH will still be coming out of the page. That is, no matter how you spin it, this right hand molecule will not look like the left hand molecule which has the opposite wedges and dashes. If you are having trouble seeing it, try building yourself a model to convince yourself that these two molecules are not the same and are in fact enantiomers.(3 votes)
If you start with an alkene and add to that alkene a percarboxylic acid, you will get epoxide. So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed, and we're going to talk about an acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you could assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right. Let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid except it has an extra oxygen. And the bond between these two oxygen atoms is weak, so this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular confirmation that it's in. So this hydrogen ends up being very close to this oxygen because there's a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this confirmation, the mechanism will begin. This is a concerted eight electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight electron mechanism. So, of course, at the bottom here we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen here. And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There is a new bond that formed between that oxygen and that hydrogen, and there is an R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here, those electrons are going to form the bond on the left side between the carbon and the oxygen like that. All right. Let's follow these electrons next. So now let's look at these electrons in here, the electrons in this pi bond. Those are the ones that are going to form this side of our epoxide ring like that. And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here, the electrons in this bond right here. These are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction for the formation of epoxide, and then we'll talk about how to form a diol from that. So if we start with cyclohexene-- let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? Well, it's based on acetic acid. But it has one extra oxygen in there, so it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid, we're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example, but we'll go ahead and put in our epoxide using wedges here. And that must mean going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this up epoxide using acid. So just to refresh everyone's memory, go back up here. Now we're going to look at this second part where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide. And I'm going to redraw our epoxide to give us a better view point here. So I'm going to put my oxygen right here, and then that's bonded to our two carbons like this. And then we see if we can draw the rest of the ring. And so in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. Now I have my H3O plus in here like this, so my hydronium ion is present with a lone pair of electrons, giving us a plus one formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. So let's go ahead and draw our oxygen here, and it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed the bond on that proton, so this is my structure now. And this would give this oxygen a plus one formal charge, so it's positively charged now. So this is the same structure that we saw in the earlier videos, like with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos-- check out the halohydrin video-- you're going to get a partial carbocation character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, the lone pair of electrons on water are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon, and you're going to get nucleophilic attacks. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the result of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack, so it has two hydrogens on it. It has one lone pair of electrons now, and it formed a plus one formal charge. Our epoxide opened. The electrons kicked off onto the top oxygen and that means that the top oxygen moves over here like that. So that would be your structure. Well, this lone pair electrons could have attacked this carbon, right? This carbon could have been the partially positive one in the resonance hybrid, which would kick these electrons off onto this oxygen. So let's go ahead and draw the result of that nucleophilic attack. So I'll go ahead and put in my cyclohexane ring like that. This time our oxygen is going to bond with a carbon on the left, two hydrogens attach to it, a lone pair of electrons, a positive one formal charge. And then this time the oxygen on top is going to kick off onto this oxygen over here in the left like that. So you're going to get an OH over there. So in the next step of the mechanism-- we're almost done, we've almost formed our diol. We're going to have water come along. And this time, instead of water acting as a nucleophile, water is going to act as a base. It's going to take a proton. So let's look at the product on the left, here. So this lone pair of electrons would take one of these protons, kick these electrons off onto your oxygen like that. So let's go ahead and draw the result of that acid-base reaction. So let's draw our cyclohexane ring like that, and now we have an OH down here. So this is now an OH, and this was an OH. So we've achieved our product. We've added 2 OHs anti to each other. Same thing can happen over here. You could grab these. You could grab this proton, kick these electrons off onto your oxygen like that. And so on the right, after we draw our cyclohexane ring, we're going to have an OH right here. And then we're going to have an OH over here like that. So we have two products. And if you look at them, they are mirror images of each other. If I were to put a mirror right here, you would see that they would be reflected in a mirror. And they are nonsuperimposable. So nonsuperimposable mirror images, enantiomers. So you're going to get two products for this reaction. So just to summarize this reaction, let's do it one more time. Let's start with the cyclohexene. And in the first step of our mechanism, the first of our reaction here, we added peroxyacetic acid. So we added CH3CO3H. And in the second step of our mechanism, in the second step of our reactions here, we added H3O plus. And so that opened up the epoxide that formed to form our diol, and we get two products. So this product over here on the left, let's go ahead and redraw this product over here on the left in a way that's a little bit more familiar. So once again, I put my eye right here. I stare down at this top carbon. If this is the top of my head, this OH is coming out at me. So I would draw that product. I would draw my cyclohexane ring, and at that top carbon I would show the OH coming out at me. And then, of course, at this carbon down here, the OH would be going away from me. So I go ahead and draw my OH as a dash here, down here. And then I do the same thing with this one right here. So if I stare down at that molecule, once again, if I stare down here and look this way, this time at this top carbon-- if my head's right here, top of my head, this OH would be down. So I go ahead and put a dash right here and put my OH. And then, over here, at this carbon it would be going up. So it might be easier to see that these are enantiomers when you look at them drawn like this. A different absolute configuration at both carbons. So this is how to form an epoxide and one way to make a diol. In the next video we will see another way to make a diol, although it will add in a different way to give you a slightly different product.