Why does zaitsev rule work

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 butene is a more likely product than butene. And if you look at butene, 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 butene. I just put an R here, but this is how it could be represented. Now, the butene, 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 butene-- let me draw it down here where I have more real estate. The butene 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 butene, 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.

Of the 4 atoms directly attached to the alkene in the major product, 3 of them are carbon and 1 is hydrogen. In the minor product, 2 carbon atoms and 2 hydrogen atoms are directly attached to the alkene. As an alkene becomes more substituted i. The major product of an elimination reaction tends to be the more substituted alkene.

This is because the transition state leading to the more substituted alkene is lower in energy and therefore will proceed at a higher rate. DOI : For leaving groups Cl, Br, and I the dominant elimination product was pentene i. For fluorine, the non-Zaitsev product is formed preferentially!

Polar Aprotic? Are Acids! What Holds The Nucleus Together? Thank you so much! This is a lot clearer than my textbook. Is it only for bases that are strong and bulky i. Conjugation indeed plays a role in alkene stability, but this post focuses on the relationship between carbon substitution and alkene stability. It can be hard to predict situations where multiple variables are in play and conflict, such as deciding whether a tetrasubstituted non-conjugated alkene will form as opposed to a disubstituted conjugated alkene.

Hence this post tries to keep things simple. Your email address will not be published. So carbon and stability is the exact opposite of carbocation stability. And again that transition state being carbanion- like, I'd rather have it on the primary beta carbon than on the tertiary beta carbon. And that's what's kind of behind us getting the Hoffmann product. And forming the Hofmann product, we'd have that carbanion-like structure on a primary carbon. In the Zaitsev product, we would have had it on a tertiary carbon instead.

So that's kind of the deal here again. This is a kinetic product. It's not the thermodynamic product. It's not the more stable product. But it does have the lower activation energy, so it's the kinetic product. You may or may not encounter this one. It's the least common of the four exceptions. So if it doesn't look familiar, a good chance it wasn't presented in your course.

All right. So here's our fourth exception to Zaitsev's rule and it's not going to make much sense yet. But if you don't have an anti-periplanar hydrogen on your beta carbon, then you can't form an alkene there.

So this is going to be a little bit strange here. We'll talk about what anti-periplanar means here in a little bit. But the most common place that it shows up is on cyclohexane. So, in this case, here's our alpha carbon. So, and we've got a beta carbon right here, and a beta carbon right here. And if we draw the relevant hydrogens in.

So, here we've got one on a wedge. So and, on the beta one here, we've got one on a wedge and one on a dash. So in this case, on a cyclohexane ring, what ultimately peri- planer is going to need to mean, is that the hydrogen I take from the beta carbon has to be trans to the leaving group. So here's our leaving group.

It's on a wedge. So the hydrogen we take has to be on a dash. Trans, on your cyclohexane ring. And these two wedge hydrogens will not work, it turns out. So there's some stereospecificity of sorts with the E2 that we haven't discussed yet. But suffice to say, I just want to point out in this case, this is your other exception to Zaitsev's rule. In this case, we would not be able to use that hydrogen at all and we would not be able to form the alkene in that location.

So it turns out we only get it with the other beta hydrogen. This anti-periplanar arrangement here is absolutely mandatory. And we're to take a much deeper look at what it means to be anti-periplanar here. But that's your fourth exception. So, bulky base, bulky substrate, and then this anti-periplanar hydrogen. These are the two most common examples of when you might violate Zaitsev's rule. Zaitsev's Rule also spelled Saytzeff's Rule is used to distinguish the major elimination product s when more than one are possible.

The elimination reactions we will specifically consider here are dehydrohalogenations resulting in an alkene. Dehydrohalogenation involves losing a hydrogen and a halogen from adjacent carbon atoms usually in a molecule while forming a pi bond. The carbon atom bonded to the halogen is referred to as the alpha carbon.

All adjacent carbon atoms are referred to as beta carbons and the attached hydrogen atoms as beta hydrogens. Zaitsev's Rule states that when there is more than one possible beta carbon that can be deprotonated while performing an elimination reaction, the more substituted one the one with fewer hydrogen atoms attached is preferred. Deprotonating the more substituted beta carbon will lead to a more substituted alkene.

And a more substituted alkene is a more stable alkene Take for instance the alkyl halide below:. There are three beta carbons: two are secondary carbons and are equivalent and the third is a primary carbon. Zaitsev's Rule predicts that the alkene formed when deprotonating either of the secondary carbons will be the major product referred to as the Zaitsev product whereas the alkene formed when deprotonating the primary carbon will be a minor product referred to as the anti-Zaitsev or Hofmann product as is the case in the E2 Elimination reaction below:.

While both E1 and E2 Elimination reactions generally follow Zaitsev's rule, there are four notable exceptions for E2 reactions:. Using a Bulky Base. No antiperiplanar beta hydrogen on cyclohexane. Poor Leaving Group. The majority of undergraduate organic chemistry courses are presented the idea that the use of a bulky base always leads to the anti-Zaitsev Hofmann product as the major product.

While this is the case with a bulky halide there are examples of some secondary halides still leading to the Zaitsev product as the major product. It is for these exceptions that I qualified the title of this section to include " with a Bulky Halide. While there are several bulky bases that are used in E2 reactions in practice, the most common by far is t-butoxide. For many classes, this is the only bulky base they are presented with and I show three ways it is often represented below:.

Below we can see the difference in the major product in the E2 elimination reaction when a tertiary halide is treated with NaOCH 3 typical strong base vs t-butoxide bulky base. The Zaitsev product is the more stable product and is therefore the Thermodynamic Product of this reaction a term you may learn later.

Due to the size of t-butoxide, it is considerable 'easier' for it to deprotonate the less substituted beta carbon and therefore the anti-Zaitsev product will form with a lower activation energy.

This makes it the faster-forming product and therefore the Kinetic Product another term you may learn later. So we see that the less substituted alkene is the major product with a bulky base due to steric effects. I'll conclude this section by showing some of the less common bulky bases.

For a cycloalkane rotation around the C-C bonds is restricted limiting when a beta hydrogen may be antiperiplanar to the leaving group. Consider the following reaction:. For this E2 elimination there are two beta carbons having hydrogens: one 3 o and one 2 o. Zaitsev's rule would form the alkene using the 3 o carbon. However, this Zaitsev product is not observed and the anti-Zaitsev product is the major product. This can be best understood by examining the cyclohexane chair conformations.

It turns out that an antiperiplanar relationship between the leaving group and a beta hydrogen is only possible when the leaving group is in an axial position. We can see that in the chair conformation on the left that the Br is in an axial position and the E2 reaction only takes place when the reactant molecule is in this conformation.

As the Br points directly up, an antiperiplanar beta hydrogen will point directly down. We can see that there is such an antiperiplanar hydrogen on the 2 o beta carbon highlighted in red but not on the 3 o beta carbon. This explains why the E2 elimination is only possible with the 2 o beta carbon and why the Zaitsev product is not observed.