Zaitsev's Rule for E2 and E1 reactions. Created by Sal Khan.
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- Markovnikov - the rich get richer
Zaitsev - the poor get poorer(506 votes)
- I always remember this sentence, which is actually very true if you take a look at the hydrogens. :) very good, thank you!(8 votes)
- I noticed that in this E2 reaction you took the hydrogen off the beta carbon. This is not what i was questioning. My real question is concerning the sterochemistry of the reaction. In the E2 reaction above shouldn't the hydrogen that is preferentially selected be the one that is anti coplanar (180 degrees) to the the leaving group?? I thought this allows for easier departing of the leaving group and easier orbital overlap between the base and proton that is removed.(13 votes)
- I think this is true as well. Probably wasn't the point of this video, it was pretty general.(2 votes)
- What happens when there are two -CH2 groups? Which one gets preference to lose a hydrogen according to the saytzeff's rule?(8 votes)
- what is hyperconjugation?(6 votes)
- Conjugation is formed by the overlap of adjacent p orbitals.
Hyperconjugation is the overlap of a p orbital with an adjacent sp³ orbital, usually a C-H bond.
The overlap is poor, because the sp³ orbital is pointing away from the p orbital, but it is still there.(9 votes)
- When the alkene is formed, is there a way of telling whether it will be an E or Z alkene? I've been told that E1 reacitons generally favour an E alkene, but what about E2 reactions?(6 votes)
- Whether an E alkene or a Z alkene product is favored depends on how large the base is in the E2 reaction. Large bases (LDA, tert-butoxide) favor the least substituted product (Z) because there is too much steric hindrance to form the more thermodynamically favored E product. Smaller bases (-OH, -OCH3) will favor the E product.(1 vote)
- It's not really mentioned in this video, but in my textbook it says that the product of an E1 reaction cannot really be controlled in terms of regioselectivity. You can't use a sterically hindered base to produce the Hoffman product, but with E2 you can. Can anyone explain this difference?(5 votes)
- The product of an E1 reaction cannot be controlled in terms of regioselectivity because the first step of the mechanism that occurs is the leaving group leaving by itself and leaving a carbocation. There is no way to control this since it is an independent event and the rate determining step.
For an E2 reaction, a sterically hindered base will produce the Hoffman product (least substituted) because the base will abstract the most accessible proton (H). Ideally, the most thermodynamically favored product would form, but because the base is sterically hindered, it is unable to abstract the proton that would form this product. A smaller base will be able to abstract the proton that forms the most thermodynamically favored product.(3 votes)
- Does a E1 and E2 reaction always result in a double bond being formed?(4 votes)
- Yes, an elimination reaction involves the loss of a leaving group and an H on an adjacent atom.(4 votes)
- At10:45Sal mentions R'? Derivatives? o_O(1 vote)
- As you know,we designate alkyl groups by the symbol R. If there are different alkyl groups, we distinguish (differentiate?) among them by using primes as superscripts. Thus, R, R' and R'' refer to three different alkyl groups. The primes have an infinitesimally small relationship to calculus. Warning: Don't drink and derive.(8 votes)
- What happens if the beta carbons are all equally substituted?(2 votes)
- If both β carbons are equally substituted, you will get an approximately50:50mixture of both possible products.(4 votes)
- Second, due to statistical factor, it has more chances to remove the hydrogen attached to the carbon with more hydrogens.Third,after the first elimination in order to make the former product, the product at first stage would be CH3CH2CH=CBrCH3 / CH3CH2CBr=CHCH3 but it's difficult to remove the bromine atom as it has a partial double bond character. Are these three factors still weaker than the Zaitsev's rule (Hyper conjugation) ?(3 votes)
In the video on E2 reactions, I showed you how a strong base-- and all of this happened simultaneously-- a strong base can nab a hydrogen off of this carbon right here. And it's just nabbing the proton itself. It's not grabbing the hydrogen and the electron. And then that electron goes to this carbon right over there, and then that allows that carbon to give away the electron that was forming a bond with the chlorine to go with the chlorine. This all happened at once. The chloride got eliminated. Now, one thing that might pop out in your brain is why did I pick this hydrogen? Why did I pick the hydrogen right over here? Why couldn't I have picked that hydrogen over there? And I'm going to introduce you to a little bit of terminology and then I'll introduce you to a rule. And then I'll tell you a little bit about why people think this rule works. So in general, the carbon that has the functional group on it, that's the alpha carbon. So let me label it. This carbon right here is the alpha carbon. And in order to have an E2 reaction in this case, when we did the video on E2 reactions, but actually, the rule will hold as well for E1 reactions. But in order to have the E2 reaction, the hydrogen has to get swiped off of a beta carbon. And a beta carbon is just a carbon that's one away from the alpha carbon. So this is a beta carbon and this is also a beta carbon. And so it's completely reasonable one would think that, well, I could swipe it from there or I could swipe it from there. And let's think about this reaction. Let's just draw it out, so you can visualize it a little bit better. So here we swiped it from this beta carbon. Let me redraw the reaction where we're swiping it from the other beta carbons. I want both reactions on the screen at the same time. Let me draw our methoxide. So we have our oxygen bonded to a CH3 to a methyl group. The oxygen has seven valence electrons: one, two, three, four, five, six, seven. I'll do the seventh one. The one that will bond with the hydrogen, I'll do it in green. Instead of attacking that hydrogen or taking that proton, I should say, because it's not taking the electron with it, it does it to this one. So what it does is it takes this proton or it bonds with that hydrogen proton, and then the hydrogen protons or that hydrogen's electron can then be taken by this molecule, so then it goes to the alpha carbon to form a double bond between this beta carbon and the alpha carbon. And so now, this alpha carbon got that electron. It doesn't need the electron that's bonded with the chloro group anymore, and so that goes to the chlorine to form chloride. Chlorine was already way more electronegative. It was already hogging it. Now it gets to go there. And now, when all is said and done, our products look like this. So we still have the methanol just like we had in the original reaction because this grabbed this hydrogen. Let me draw it. So you have your OCH3. Let me draw all of this in. You have that pair right over there. You have this pair right over here. And then you have this purple electron and now it's bonded with this green electron, which is now on the hydrogen. So it is now bonded with this green electron that has been given to the hydrogen. Oh, and make sure we don't forget. This oxygen over here had seven valence electrons, so it had a negative charge. Neutral oxygen would have six. So this had a negative charge. But now that it gave its electron to this hydrogen, it now has a neutral charge. It is now methanol. Exactly what we saw when we first learned about E1 reactions. We also know the chloro group, it took that electron. It's now a chloride anion. So let me draw that. And that's exactly the same as in the first video on E1 reactions. So it's now a chloride anion. It has grabbed this orange electron. Let me do it in orange. It has grabbed this orange electron, so it now has a negative charge. You can imagine that the negative charge has been transferred from the methoxide to the chloride anion. And now what's different this time is that the double bond is now between the one and the two carbon and not between the two and the three carbon. So now it's going to look like this. So now, the result of this product, if the reaction went this way would look like this. We have this carbon right here. It is bonded to two hydrogens. Now, it has a double bond with this carbon. It has a double bond, and I'll do the second bond. I'll do the pi bond of the double bond in purple right over there. I'll assume that that's the pi bond. That's the new double bond form. And now this carbon, which was the alpha carbon, is right over here. This was the alpha carbon. It's bonded to one hydrogen. And then let me draw everything else. So then you have a carbon, a carbon. This guy is bonded to three hydrogens. I could have just written it as CH3 if I wanted. This guy's bonded to two hydrogens. And we're done. So instead of, as we saw on the first video on E2 reactions, instead of forming but-2-ene, we now have-- it's still one, two, three four carbons, so it's still but-, but the double bond is on the one carbon. We'd start one, two, three, four, So we could call this but-1-ene or 1-butene, either way. So let's call this but-1-ene. So the question is, which is more likely to happen? Do both happen? Does one happen disproportionately? And the answer is, is that, yes, one happens disproportionately. This one, this is the dominant product. If you were to perform this reaction and you were to analyze in your beaker or wherever you're performing the reaction what you see most of, the majority product, the great majority, is going to be the but-2-ene, not the but-1-ene. Maybe you see very, very little of this. And the question is why? Or how would you even be able to determine that? Or how could you have predicted that? And to predict it, there's something called Zaitsev's rule. And I'm sure I mispronouncing it, but let me write it down. So Zaitsev's rule. And it's kind of analogous to Markovnikov's rule, but for elimination reactions. If you think about it, the addition reactions that we did many videos ago are the opposite of the elimination reactions. In the addition reactions, we're adding the chloro group, and in the elimination, we're taking it off. And so Zaitsev's rule is kind of analogous to Markovnikov's rule. Now first, I'll just tell you the rule, then we can think a little bit about why it works. The jury's not out on this. They think they know why it works, but they're not 100% sure. So Zaitsev's rules says the carbon that is going to lose the hydrogen is the one that has fewer hydrogens. So let me write it down over here. Carbon more likely to lose hydrogen is-- I should say the hydrogen proton because it keeps the electron still-- is the one with fewer hydrogens. So if you were to look at this reaction right here, we have our alpha carbon. Either this beta carbon or this beta carbon could lose its hydrogen. This one has three hydrogens on it. This one only has two. So Zaitsev's rule tells us that this is the hydrogen, or actually the proton, that is more likely to be reacted with the base. You could almost view it as it is the more acidic proton. It is a lower-hanging fruit for this strong base to capture. Now, a more interesting question-- and that's a pretty easy rule to follow. And if they both have the same number, then you'd see equal products depending on which side it gets. Now the question is why is this happening? And here, something called hyperconjugation comes into effect. I'm not going to go into details in it and to the quantum mechanics of it. And hyperconjugation is the notion that the fact-- so we said that the one with the fewer hydrogens is the one that's less likely to lose. Or the one with fewer hydrogens is the one more likely to lose the hydrogen proton. But the one with fewer hydrogens is also bonded to more carbons. This guy's bonded to one carbon outside of the alpha carbon. He's actually bonded to two: the alpha and this carbon. This guy right here is only bonded to the alpha carbon. And hyperconjugation is the notion that not the beta carbon, but the carbons one over from that help stabilize the double bond that eventually forms. I almost think of it you have more electrons over here because carbons have more electrons to offer than hydrogens. At the end of the day, this guy is more likely able to donate electrons to form from the right-hand side to make a double bond than from the left-hand side. I won't go into the detail of hyperconjugation, but it's all based on the notion that the more stable double bond will be formed if we have other carbons near the double bond. Now, another way to think about is to look at the products. So we saw or Zaitsev's rule tells us that but-2-ene is a more likely product than but-1-ene. And if you look at but-1-ene, we could rewrite it like this. We could draw the double bond like this. This carbon is what was the alpha carbon. We could draw a carbon right here. And then it is bonded to a hydrogen. It's bonded to this hydrogen. And then it's bonded to just a chain of carbons. We'll just write R on that. And then this guy is just bonded to two hydrogens. Well, this isn't necessarily but-1-ene. I just put an R here, but this is how it could be represented. Now, the but-2-ene, if we wanted to draw it like this would look like this. We could call this right here R prime. It's just a chain of carbons. And then we could call this R prime prime. It's not even a chain. It's just one carbon. But if we call it that, then the but-2-ene-- let me draw it down here where I have more real estate. The but-2-ene would look like this. You have your carbon-carbon double bond. Now, the left-hand carbon is bonded to a hydrogen, that hydrogen right there, and to R prime. And the right-hand carbon is bonded to hydrogen and R prime prime. So it's bonded to a hydrogen and R prime prime. All I did here-- let me see if I can fit it all on the same screen-- is I just redrew this and I just abstracted away the chain as it goes away from the double bond. And I did that so that we can look at it this way. We can just have the double bond kind of as our focus of attention and think about what's going on around it. Over here, for the but-1-ene, and we already said this is the lesser product, so this is the greater product. This is the greater, or the dominant product. In the lesser product, if we go off of the double bond, we only have one alkyl group right there. That R right here. Over here, we have two. And we say that this is more substituted. And when w say substituted, you're imagining that you're substituting hydrogens with carbon chains, with alkyl groups. So this one right here is more substituted. And hyperconjugation, so the idea would have it, is that these carbon chains that are near the double bond help stabilize it. Some of their sigma electrons and sigma orbitals are there to somehow help stabilize the pi orbitals. And now this is getting into quantum mechanics and all of that, so it's a little bit-- you know, the world isn't 100% clear whether that's definitely the mechanism, although people have run the experiment, and they know that the more substituted product is what you're going to see more of as opposed to the less substituted, and that all comes from Zaitsev's rule. So, hopefully, you at least get Zaitsev's rule. The hyperconjugation, that's kind of a deeper concept. You know, the jury's not even out on it. That's a belief of why Zaitsev's rule works. But the rule itself is pretty straightforward. If you're trying to pick between two beta carbons, the one that's going to lose the hydrogen is the one that already has pure hydrogens or the one that's bonded to more carbons. And this is true. I drew it. Everything we focused on right now was in an E2 reaction. But it's just as true in an E1 reaction.