Ring opening reactions of epoxides: Acid-catalyzed
How acid-catalyzed reactions can open the highly strained ring of an epoxide. Created by Jay.
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- Wouldn't the steric hinderance of the methyl group stop the weak base from attacking the anti-bond of the tertiary carbon. The acid seems to be very weak and bulky making it more likely to attack the secondary carbon for an SN2 reaction?(12 votes)
- The reaction is like SN1. It is NOT SN1 but it is carbocationic character driven. Think of it like this, The strong nucleophile will attack vigorously and will not be selective ie. it will attack the carbon which is easily available to it, the one less hindered. Whereas the weak nucleophile which won't be vigorous would be selective and would attack on carbon which is carbocationic driven so it gets more stability and the reaction proceeds even with the weak nucleophile.
In a nutshell, strong doesn't need carbocationic character's stability as it is strong enough and reacts quickly whereas the weak one needs it to compensate its weakness.(14 votes)
- How do you distinguish that a reaction is going to be a strong nucleophile ring-opening reaction or an acid-catalyzed ring-opening reaction?(2 votes)
- If it was an acid-catalyzed ring opening reaction, H+ would be one of the first reagents rather than added later to protonate the alkoxide first so that a strong nucleophile isn't required.(4 votes)
- primary alkyl halides react with alkoxide via E2 mechanism to form an alkene (here alkoxide is a strong base) and in Williamson reaction they again react to give an ether by SN2 mechanism. when what happens?(2 votes)
- In both cases I believe there will be significant side reactions by the other mechanism leading to unwanted products.
One way to bias the results is by choice of solvent. For example, using a polar aprotic solvent (e.g. an ether) to favors SN2, while a protic solvent favors E2 (since protic solvents interfere with nucleophilicity, but not basicity).(2 votes)
- At3:28, why won't oxygen break the bond from the carbon giving it a full carbocation character. Why do we only see a partial carbocation character.(2 votes)
- these videos and intro organic chemistry textbooks are confusing. I feel as if the mechanisms we are given are simplified to the point that when we try to follow the rules that we learn from them universally, we get confused because we get taught more mechanisms that violate those rules. anyone else feel the same way?(2 votes)
- same here
i feel (organic) chemistry is highly empirical than theory-based
which means when one found an exception breaking a grounded rule, it tends to add another rule to it as an exceptional rule rather than to explain with the given rule by digging it deeper or replace it with a new one
this way of thinking or practicing seems to give a birth of rows of rules and their exceptional rules, i feel(1 vote)
- @5:42, why did the ethanol get a plus charge after it attacked the epoxide? Is it because it gave an electron away to the epoxide?(2 votes)
- The Oxygen on the ethanol molecule has a plus-1 charge on it because it is bound to 3 other atoms: the Hydrogen, the Carbon belonging to the Benzene ring, and the remainder of the ethanol molecule. Hope that clears things up!(1 vote)
- how to do mechanism of base catalyzed reaction of epoxide(1 vote)
- The hydroxide ion attacks one of the carbon atoms in the epoxide in an SN2 displacement reaction. The C-O bond breaks to form an oxide ion on the adjacent carbon atom. The epoxide ion then picks up a proton from the water. The result is a 1,2-diol.(1 vote)
- If the methyl group was cis to the epoxide on a different carbon of the ring, what happens to the stereochemistry of that methyl group after the ring-opening hydrolysis occurs?(1 vote)
- Can this type of ring opening occur with H2O and H2SO4? Would H2O be a strong enough nucleophile since it would relive ring strain??(1 vote)
- In short, yes, but it depends on the substitution of the ring carbons. With greater subsitution, the ring can open (because the carbocation will be stabilised), and then water will attack(1 vote)
- I didn't understand why did the epoxide oxygen atom undergoes protonation first unlike other epoxide reactions.....
The fact is clear that since there is a weak nucleophile, it will attack the most substituted carbon atom.... hence favouring SN1 mechanism.........
But what does it explain on oxygen atom undergoing addition of hydrogen atom first.....?
(In other reactions it was protonated in the second step, ). plz help(1 vote)
- This is an acid-catalysed reaction, which means that there will be protons in solution that can participate in an acid-base reaction with the oxygen (from epoxide), and therefore protonate the epoxide.(1 vote)
Here's the general reaction for a ring opening of epoxides when everything is acid-catalyzed. So if I first start by looking at my epoxide over here on the left, I can classify this carbon, and I can see this carbon is attached to two other carbons, so this carbon would be secondary. Or I could think about a hydrogen replacing the R group, and then it would be a primary carbon, so either primary or secondary. If I look at the carbon on the right side, so this carbon right here, it's attached to three other carbons, so this carbon on the right would be tertiary. So if I react this epoxide with a weak nucleophile so H-Nu is going to refer to my weak nucleophile, and I make it an acid-catalyzed reaction, I can see the product over here on the right. The nucleophile has added to the most substituted carbon, so here i can see my nucelophile. I add it to my tertiary carbon over here on the right. So that's the regiochemistry, a nucleophilic attack of the most substituted carbon. And I can see, when the ring opens, the OH group is going to be anti- to the nucleophile. Let's take a look at a reaction. And we can run through the entire mechanism for this ring opening. So if I start with my epoxide, so I'll go ahead and make it like this. I'm going to have my epoxide coming out at me in space. And then, at this top carbon here, I'm going to have a methyl group going away from me in space. So here's my methyl group. I'm going to react this epoxide with ethanol. So I'm going to go ahead and put ethanol in here. Ethanol is going to function as my week nucleophile. And it's acid-catalyzed, so I'll make sure and put my protons in there like that. OK, let's go ahead and redraw the epoxide so we can see it a little bit better. So we're going to be looking down on our epoxide, like this, so, therefore, the oxygen is going to be up relative to the plane of the ring. A lone pair of electrons on the oxygen, like that, and then, at this carbon, there's going to be a methyl group going down in space. The first step of the mechanism is an acid-base reaction. This is an acid-catalyzed reaction, so they're H plus protons floating around. Lone pair of electrons on oxygen are going to pick up that proton. So I'm going to protonate my epoxide. So when I draw the product of that acid-base reaction. And I now have, on my oxygen, I still have one lone pair of electrons. The other lone pair formed a bond with that proton, so now it is a protonated epoxide, which gives the oxygen a plus one formal charge. I still need to have my methyl group down relative to the ring, like that. Now when I think about the next step, I need to think about the classification of the carbon atoms. So if I look at this top carbon right here, I can see that this one is attached to three other carbon, so this is my tertiary carbon. The one down here is only secondary. And I know the nucleophile is going to attack the more highly substituted carbon, which would be this one down here, like that, so this is the carbon that's going to get the attack. If I think about that oxygen being positively charged, oxygen doesn't like to be positively charged, oxygen is very electronegative. So it's going to do its best to pull some of these electrons and the bond between that carbon and that oxygen a little bit closer to the oxygen there. So I'm going to withdraw some electron density from this carbon. So this carbon is going to end up having a partial, positive charge, like that. And if I think about what kind of a carbo cation would that be, it has a little bit of partial cargo cation character, and I can think about the fact that that carbon is attached to those three other carbons-- it's a little bit easier to see here --there's one, two, and three other carbons. And, in earlier videos, we've seen that tertiary carbo cations are the most stable, much more stable than secondary carbo cations. And if I have a weak nucleophile attacking, it's going to attack the most substituted carbon, so it's going to attack the one that would be the most stable carbo cation. So again, we have more of a partial carbo cation character here. But when the weak nucleophile comes along, that lone pair of electrons is going to attack this carbon right here. And that is going to kick these electrons in here, the magenta ones, off onto your oxygen. So let's go ahead and draw the results of that nucleophilic attack. So I have my ring right here. And I've opened my epoxide. So now this oxygen is going to swing over to the left here, and it's bonded to a hydrogen. It had one lone pair of electrons. It just picked up one more lone pair of electrons, the magenta ones. And, in the course of that nucleophilic attack, the methyl group that's down relative to the ring here gets pushed up. So the methyl group will end up relative to the ring. And I have a lone pair of electrons from the ethanol molecule that are now bonded to this carbon. So I can go ahead and show those electrons. And there's an oxygen, a hydrogen, and then two carbons, and then there's still a lone pair of electrons left in that oxygen, which gives that oxygen a plus one formal charge. So let's just go ahead and highlight those electrons really fast. So, let's see, I'll make them red here. So these electrons right here, those are the ones that attack the carbon, and those are the ones that formed these covalent bonds, like that. So we're almost done. The last step will be an acid-base reaction to get rid of that plus one formal charge on the oxygen. So another molecule of ethanol comes along. This one's going to act as a base. So I go ahead and draw another molecule of ethanol. And it has lone pairs of electrons on it. One of those lone pairs of electrons are going to pick up that proton which kicks these two electrons off onto this oxygen. And we're going to end up with our products. So let's go ahead and draw the product right up here. And, like usual, we're going to be looking down on our molecule this way, and so we're going to look down right here and draw what we see. So we're going to have our ring, like that. And, at the top carbon, which would correspond to this one right here, I can see that there's a methyl group coming out at me. So I can go ahead and put my methyl group coming out at me like that. And then, going away from me, I'm going to have my oxygen, so I'm of course talking about this oxygen down here like that. So I can go ahead and draw that oxygen going away from me. And connected to that oxygen of course would be two carbons. So I can go ahead and do that. And when I look at the carbon on the left over here-- now I'm talking about this carbon, I can see there's an OH group coming out at me in space, so I can go ahead and put that OH group coming out at me in space right there. So that's going to be our product. So, again, I went ahead and showed-- I'm going to go ahead and put my two lone pairs of electrons on the oxygen, because I showed that proton being removed in the last step of that mechanism there. So let's look at the stereochemistry. And let's first start with this top carbon here. So I go back to my reactants and. I look at this top carbon. And I think about the fact that I have a wedge to an oxygen, so oxygen's coming out at me, and a carbon's going away from me in space. When I look at my products-- this would be the carbon that corresponds to that carbon, it's the opposite. Now I have a carbon coming out at me in space, and an oxygen going away from me. So that's inversion of configuration, so at that top carbon there, so at this carbon right here, I'm going to get inversion of absolute configuration here. So let me go ahead and write that-- inversion of configuration. So there's some stereochemistry involved. And when I look at the other carbon-- so when I look at this carbon down here, I can see that there's an oxygen coming out at me in space. And when I look at that carbon for my products, I see there's still an oxygen coming out at me in space. So it's the same absolute configuration at this carbon. So we only see inversion of configuration at the carbon which is undergoing the nucleophilic attack, which is the most substituted carbon.