Elimination reactions often can yield multiple products. However, not all of these products will be of equal stability. Zaitsev’s Rule (also spelled Saytzeff’s Rule) helps us predict the major product.
Concept: Defining Zaitsev’s Rule2m
Now we have to talk about one of the 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 guys 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 elimination reactions what we're going to find is that there's multiple alkenes that are present at the end. So how do we determine are they 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. 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. 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.
Now the one with the less R groups around it or the least stable product, is going to be called the Hofmann product. All right, so that's just some basic vocabulary that you need to understand before we can even start using this rule.
Zaitsev’s Rule predicts that in most cases, the most substituted product will be favored. This is also known as the Zaitsev product. Draw the Zaitsev product of the following reaction:
Example: An example of a Zaitsev product.8m
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.
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 that's going to be a negatively charged nucleophile. That's going to go down the left-hand side of the flowchart.
So let's go to step two. Step two is is NaOEt one of my bulky bases. No. We have a list of bulky bases, NaOEt is not one of them. So I'm just going to say no.
Let's go to my third question. What type of alkyl halide do we have or what type of leaving group? Well, this carbon right there is attached to two other carbons, one, two, 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. Cool.
So now we're 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.
So, what have we got? We've got two different beta-carbons. Let's say this is beta one and let's say this is beta two. Those are my two 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 three. How many of them could actually react in an E2 reaction using the anti-coplanar rule? Remember you always have to think of that rule.
The answer is two 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.
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 would be three arrows, just like it always is. Let's say I'm taking off the red H. then what I'm going to do is I'm going to dump my electrons into the bond between the alpha and the beta and then I'm going to kick out my Cl.
So what that means is I'm going to have two different products possible. I'm going to have a product possible where I eliminate where I just did, so it would be a double bond right there with now no Cl. I don't have to draw that and a methyl group facing down like that.
Now, why is it important that I drew my methyl group on a stick? Because remember now this is trigonal planar If I draw it on a wedge that would look like I have no clue what I'm doing. I would kind of look like an organic noob and you don't want to look like a noob when you're trying to get points from your professor. So that's the way we would draw that product. Obviously, I would have my Cl negative leaving group.
But there's also another product I could have gotten. The other product would have been if I eliminated up to the green hydrogen and that would have given me a product that looks like this. A double bond there and – let me just straighten that out a little bit. And that methyl group would still be on the wedge. Why am I drawing it on the wedge here? Because that carbon is still tetrahedral because it still has an H on it. So that is a carbon with four different groups around it so that one should be drawn as a tetrahedral.
So now we've drawn our two different products. Are they both going to be formed in equal amounts? No. It turns out that Zaitsev's rule explains – just look down here – that we will always favor the more substituted thermodynamically stable product. Now, what the hell does thermodynamically stable mean? I know that whenever you bring thermodynamics or kinetics, things get confusing. It just means overall which one is going to have the least energy. Which one is going to be the overall most stable at the end?But guess what? We know how to figure this out because we can just use the alkene stability rules to determine that.
So out of these two double bonds which of them is going to be more stable and which one of them is less stable. What do you guys think? This is like its own question now. Well, this one here is trisubstituted. This one here is disubstituted. So which one going to be the overall more stable product? Red. Red is going to be more stable than green. Which one is the Zaitsev product? Red. Because Zaitsev is the more stable one. This one's the less stable, it's Hofmann.
So instead of saying less stable/more stable, from now on I'm just going to be using the words Zaitsev and Hofmann because guess what? Those are just synonyms of the less stable and more stable are just Zaitsev and Hofmann. I can just use those words.
So now we have to figure out which one is major and which one is minor. Which one am I going to form in higher amount than the other? What Zaitsev's rule explains is that I'm going to favor the Zaitsev product. That means that this is going to be my major product and this is going to be my minor product. Does that mean that I only get one of these? No, I still get both, but I'm going to get a lot more of the red and a lot less of the green.
Now sometimes your professor may ask a question that just says give the major product for the reaction. If they're asking for major product, then you would just draw red. But if your professor asks give all the products, then you would draw both of them. Does that make sense?
But many times especially in multiple choice exams, if you have any multiple choice component, many times your professor is just going to say draw the major product or select the major product. If you're doing online homework and it says select the major product, it would be this one. If you're just checking different resources out online, this would be the more stable, more favored product. Awesome guys.
So Zaitsev's rule's pretty easy, right? We just use the number of R groups to figure out which one is more stable, which one's less stable. Select the more stable one.
The exception to this rule comes with the use of bulky bases. These promote the formation of the les substituted product.
The less substitued product is called the Hofmann product. Let's draw the Hofmann product of the following reaction:
Example: An example of a Hofmann product.3m
Now one more thing, it turns out there's an exception to this rule because guess what, organic chemistry always has one exception.
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've very good at pulling off protons, but they're not very good at donating electrons. 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.
Let me show you guys how this works. Let's say that here I'm reacting with a bulky base tert-butoxide. So that's my tert-butoxide 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 here and then I have my red hydrogen here. So we have the same option to pull the green one and get the less stable product or the red one and get the most stable product.
So overall, we know that if we pull the red proton, we're going to get the more stable product, but this tert-butyl group is also pretty big. So what that means is that it can either come in here and fight through this methyl group and through the leaving group to try to get to that H or it can just easily pull off this green one where there's a lot of space.
It turns out that even though the red one is more stable, the tert-butoxide is going to be such a strong base and it's going to be so bulky that it will prefer to take the easy way out. It's going to prefer to take the less substituted H, just because it can get to that one better. Isn't that interesting?
So I'm going to get the same exact products here, where I'm going to get an elimination product that looks like this plus I'm going to get an elimination product that looks like this. In fact, the Zaitsev and Hofmann hasn't changed either. This is still going to be my Zaitsev product and this is still going to be my Hofmann product.
So nothing has changed. The only thing that's changed is that since I'm using a bulky base, I'm actually going to favor the Hofmann so that means my major product will actually be the Hofmann and my minor product is going to be the Zaitsev. Even though the minor product is the one that's more stable, but it's the one that's harder for the tert-butoxide to get to, so it's just going to pick the easiest way possible.
Concept: Using a Free Energy Diagram to explain thermodynamic vs. kinetic products.4m
Now the last thing I want to do is I just want to show you guys with an energy diagram that I'm going to sketch up really quick what these words mean between thermodynamics and kinetics because this is going to come up more in orgo one and in orgo two.
So I just want to show you guys, remember that you have an energy diagram. And the way that it works is 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.
What I want to show you guys is that this is a concerted reaction, so it all happens at the same time. So 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 a thermodynamic energy diagram look like are going to be different.
So for the thermal dynamic one, I'm going to start up here at this energy level. I'm going to pass through a pretty big – I'm sorry, 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.
That's what the elimination product would look like for the first one. Are you guys following so far? Cool.
Now for the second one what I would find is that my energy level's at the same place at the beginning, so it's right here. I'm still – 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.
So what you can tell is that – check out the enthalpy for a second, or the spontaneity. Overall, I'm going to gain – I'm sorry, that's supposed to be change in delta G. 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 delta G 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 one.
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. So the green is what we would call kinetic control. Kinetic control basically means that all I'm looking at is the one with the lowest activation energy. I'm saying whichever one is the easiest one to form, that's the one that's going to be favored.
Whereas thermodynamic control, I'm just going to put here thermo, is the one that looks at the overall lowest delta G. The one 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.
And that's the difference between Zaitsev and Hofmann. Basically, Zaitsev is thermodynamic control, where all I care about is the stability of the end product. Whereas, Hofmann 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 two, there's going to be kinetic control and thermodynamic control. It's going to be the same kind of principle where I'm looking at either the stability of the end product or I'm looking at whichever one's just the easiest one to form at the beginning.
I hope I didn't confuse you guys here. I just really wanted you guys to understand the difference between Zaitsev and Hofmann and how one is thermal dynamic and one is kinetic. So let's go ahead and do some practice problems based on Zaitsev's rule.
Problem: What is the major product of the reaction?3m
Problem: What is the major product of the reaction?2m
Problem: What is the major product of the reaction?6m
What is the major product of the elimination reaction?
Which of the following would be the best base for performing the following elimination? Select all that apply.
Be sure to answer all parts.
Draw both the SN1 and E1 products of the following reaction.
Provide the major organic product of the following reaction.
What is the major organic product of the following reaction?
a. 2, 3-dimethyl-1-hexene
b. 2, 3-dimethyl-2-hexene
d. (Z)-2, 3-dimethyl-3-hexene
e. (E)-2, 3-dimethyl-3-hexene
For the following dehydrohalogenation (E2) reaction, draw the major organic product(s), including stereochemistry.
For the following dehydrohalogenation (E2) reaction, draw the major organic product(s), including stereochemistry.
Draw the major organic product formed when the structure shown below undergoes a reaction with t-BuONa.
Draw an alkyl bromide that, upon treatment with base, will give the compound shown. Indicate the proper relative stereochemistry.
Draw the major organic product of the following reaction. Indicate the stereochemistry, if appropriate. wavy bond to indicate a mixture of configurations.
Provide structures for the major and minor products of the following reaction and indicate which type of reaction it follows: SN1, SN2, E1, or E2. In addition, also provide the IUPAC names of both the major and minor product. As a bonus, indicate if the products (major/minor) and reaction would be different if LDA was used instead.
Provide the major product for the following compound.
Write structural formulas for all the alkene products that could reasonably be formed from each of the following compounds under the indicated reaction conditions. Where more than one alkene is produced, specify the one that is the major product. (b) 1-Methylcyclopentyl chloride (sodium ethoxide, ethanol, 70°C)
When 2-bromo-2-methylbutane is treated with a base, a mixture of 2-methyl-2-butene and 2-methyl-1-butene is produced. When potassium hydroxide is the base, 2-methyl-1-butene accounts for 45% of the mixture. But when potassium tert-butoxide is the base, 2-methyl-1-butene accounts for 70% of the mixture. What would you predict for the percent of 2-methyl-1-butene in the mixture if potassium propoxide were the base?
a) less than 45%
c) between 45% and 70%
d) more than 70%
Which compound would most readily form the depicted alkene through an E2 mechanism?
Predict the major elimination product for the following reaction: