Zaitsev Rule: Study with Video Lessons, Practice Problems & Examples

In elimination reactions, Zaitsev's rule helps predict the major and minor products when multiple alkenes are formed. The Zaitsev product, more stable due to a higher number of alkyl substituents, contrasts with the less stable Hoffman product. Understanding the difference between thermodynamic control, which favors stability (Zaitsev), and kinetic control, which favors the pathway with lower activation energy (Hoffman), is crucial for predicting reaction outcomes. Energy diagrams illustrate these concepts, highlighting the relationship between activation energy and free energy changes in reactions.

Defining Zaitsev’s Rule

Now we have to talk about one of the most important aspects of elimination reactions. I know that you guys thought we were done learning about them, but we're not. There's something else that you need to know and that's how to predict if you have multiple products possible, how to get the major product and the minor product. And this relates to Zaitsev's rule. So many times in an elimination reaction what we're going to find is that there are multiple alkenes that are present at the end. So how do we determine if they are all made equally or are they made in different ratios? How do we know that? And we use Zaitsev's rule to figure that part out. Okay? So whenever you have more than one unique alkene as a product, that's when you use Zaitsev's rule. What does Zaitsev's rule say? Well, basically, at this point, you should already know how to tell when a double bond is more stable or less stable based on the number of R groups that are around it. And based on that rule, the most stable product is going to be called my Zaitsev product. And that is based on how many R groups it has around it. So the more R groups it has more than the other one, that's the Zaitsev product. Okay? Now, the one with fewer R groups around it or the least stable product is going to be called the Hoffman product. Alright? So that's just some basic vocabulary that you need to understand before we can even start using this rule.

Zaitsev product

Now I've given you an example here of basically an alkyl halide with a nucleophile. Let's use the big daddy flowchart to figure out what mechanism this would be. So let's go ahead and ask our first question. Notice that my nucleophile is NaOET. Is that negatively charged or is that neutral? For those of you that said neutral, you're forgetting that sodium can dissociate. So what it's actually going to look like is OET^-, so it's going to be a negatively charged nucleophile. That's going to go down the left-hand side of the flowchart. Okay? So let's go to step 2. Step 2 is NaOAT one of my bulky bases? No. Okay? We have a list of bulky bases. NaOAT is not one of them. So I'm just going to say no. Let's go to my 3rd question. What type of alkyl halide do we have or what type of leaving group? Well, this carbon right there is attached to 2 other carbons, 1, 2, so this would be a secondary alkyl halide. So do we know the mechanism now? No, we have to ask one more question. The last question is, I'm just going to put it down here, is my base a better nucleophile or a better base? So for this, you have to remember what were the strong bases. Is NaOET one of those strong bases? Yes, it is. Remember that one of the strong bases was oxides. Oxides have the general formula of OR^- and that's exactly what we have. We have OET, which is an ethyl group, negative. So this is an oxide, so this is going to favor E2. Alright, cool. So now we've got E2. Now we have to figure out how do we actually draw the mechanism for this and how do we predict the products. Do you guys remember what the first step of E2 is? Figure out how many different beta protons I have. Okay? So what do we got? We've got 2 different beta carbons. Let's say this is beta 1 and let's say this is beta 2. Okay? Those are my 2 different options. Do both of them have hydrogens on them? Yes, they both do. On the green one, I've got a hydrogen towards the front and a hydrogen towards the back. On the red one, I just have a hydrogen towards the back. So are we ready to eliminate yet? We have to ask ourselves one more question. Now that we know all of our beta protons, which is 3, how many of them could actually react in an E2 reaction using the anticoplanar rule? Remember you always have to think of that rule. The answer is 2 of these could react. I could eliminate in the green direction with this one right there and I could also eliminate in the red direction using that one right there. The reason is because my chlorine is facing towards the front, so I can only eliminate with a hydrogen that's facing towards the back. Okay? So now that I know that, let's go ahead and draw one of the mechanisms. We don't have to draw both, but let's just draw one of them and then predict the products. So the mechanism

Hofmann product

Now one more thing, it turns out that there's an exception to this rule because guess what? Organic chemistry always has one exception. Right? So the exception is going to be unless we're using a bulky base. Bulky bases, remember that they're not very nucleophilic. They're very basic. They're very good at pulling off protons, but they're not very good at donating electrons. So what that means is that a bulky base is going to promote the formation of the less substituted kinetic product. What does kinetic mean? It means it's the one with the lowest activation energy. The one that's the easiest to grab, even though it's not stable at the end, is going to be the one that is favored at the end. So let me show you guys how this works. Let's say that here I'm reacting with a bulky base terbutoxide. So that's my terbutoxide molecule. Notice that it's kind of bulky. It's got an O and it's got that tert-butyl group on the side.

Let's do the same reaction. I have my green hydrogen her

Using a Free Energy Diagram to explain thermodynamic vs. kinetic products.

The last thing I want to do is show you guys an energy diagram that I'm going to sketch up really quick, explaining what these words mean between thermodynamics and kinetics because this is going to come up more in Orgo 1 and in Orgo 2. So I just want to show you guys, remember that you have an energy diagram, and the way that it works is that you have some kind of spontaneity here, and you have a reaction coordinate here where, basically, at the end, I have a double bond, and at the beginning, I have just my alkyl halide plus the nucleophile. Okay? So what I want to show you guys is that this is a concerted reaction, so it all happens at the same time. I'm only going to have one hump, I'm going to have one transition state. Remember that E2 just has a transition state. The thing is that what it looks like, what the kinetic versus the thermodynamic energy diagrams look like, are going to be different. So, for the thermodynamic one, I'm going to start up here at this energy level. I'm going to pass through a pretty big hump in energy and then I'm going to gain a lot of energy at the end because my double bond is overall going to save me some energy. So that's what the elimination product would look like for the first one. Okay? Are you guys following so far? Cool.

Now for the second one, what I would find is that my energy level is at the same place at the beginning. Okay, so it's right here; there's my kinetic pathway. But it turns out that for the kinetic pathway, I'm going to have a much lower activation energy, but then I'm also going to have a much lower gain in stability. Okay? So what you can tell is that check out the enthalpy for a second or the spontaneity. Okay? Overall, I'm going to gain, I'm sorry, it's supposed to be change, I'm going to gain, I'm sorry, it's supposed to be change in ΔG. Okay? Overall, my product for the red, for the Zaitsev product, is going to be overall more stable at the end. I'm going to gain more ΔG or more free energy by going in that direction than by going in this direction. Is that making sense so far? So basically, the red one is overall more stable than the green. Okay? But what we're also going to notice is that the activation energy of the first one is much higher and the activation energy of the second one is much lower. Okay? So the green is what we would call kinetic control. Kinetic control means that all I'm looking at is the one with the lowest activation energy. Okay? I'm saying whichever one is the easiest one to form, that's the one that's going to be favored. Okay?

Whereas thermodynamic control, I'm just going to put here thermal, is the one that looks at the overall lowest ΔG. The one that changes, that gets the most free energy at the end, that's the most stable at the end, that's going to be the one that I favor. Okay? And that's the difference between Zaitsev and Hoffman. Basically, Zaitsev is thermodynamic control, where all I care about is the stability of the end product. Okay? Whereas Hoffman is going to be kinetic control because all I'm going to care about is the one that's the easiest to form or the one with the lowest activation energy. Is that difference kind of making sense? Now the reason I'm telling you guys this is because this is going to come up later when we talk about other reactions in Orgo 2. There's going to be kinetic control and thermodynamic control, and it's going to be the same kind of principle where I'm looking at either the stability beginning. Alright. So I hope I didn't confuse you guys here. I just really want you guys to understand the difference between and how one is thermodynamic and one is kinetic between Zaitsev and Hoffman. Alright. So let's go ahead and do some practice problems based on Zaitsev's rule.

What is the major product of the reaction?

• A.
• B.
• C.
• D.

What is the major product of the reaction?

• A.
• B.
• C. Both A and B
• D. None of these

What is the major product of the reaction?

• A.
• B.
• C.
• D.

Here’s what students ask on this topic:

What is Zaitsev's rule in organic chemistry?

Zaitsev's rule states that in an elimination reaction, the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) will be the major product. This is because more substituted alkenes are generally more stable due to hyperconjugation and inductive effects. Therefore, when multiple alkenes can be formed, the one with the greater number of substituents around the double bond will predominate.

How do you determine the major product in an elimination reaction using Zaitsev's rule?

To determine the major product in an elimination reaction using Zaitsev's rule, identify all possible alkenes that can form. Then, count the number of alkyl groups attached to the carbons of the double bond in each alkene. The alkene with the most alkyl groups (i.e., the most substituted) is the Zaitsev product and will be the major product. This product is more stable due to hyperconjugation and inductive effects.

What is the difference between Zaitsev and Hoffman products?

The Zaitsev product is the more substituted and more stable alkene formed in an elimination reaction, while the Hoffman product is the less substituted and less stable alkene. Zaitsev products are favored under thermodynamic control, where the stability of the final product is the main factor. Hoffman products are favored under kinetic control, where the pathway with the lowest activation energy is preferred.

How does thermodynamic control relate to Zaitsev's rule?

Thermodynamic control in elimination reactions relates to Zaitsev's rule because it favors the formation of the most stable product, which is the Zaitsev product. Under thermodynamic control, the reaction conditions allow the system to reach the most stable state, resulting in the major product being the more substituted and stable alkene. This is due to the lower free energy of the Zaitsev product compared to other possible products.

Can you explain the energy diagram for Zaitsev and Hoffman products?

In an energy diagram for elimination reactions, the Zaitsev product has a higher activation energy but results in a more stable final product with lower free energy. The diagram shows a higher peak for the transition state but a lower final energy level. In contrast, the Hoffman product has a lower activation energy, making it easier to form initially, but it results in a less stable final product with higher free energy. The diagram for the Hoffman product shows a lower peak for the transition state but a higher final energy level.