You can review episodes of Crash Course Organic Chemistry with the Crash Course App, available now for Android and iOS devices.

Hi! I’m Deboki Chakravarti and welcome to Crash Course Organic Chemistry!

A relatively slow reaction between sugar and yeast makes the ethanol in beer, while other compounds contribute all those hoppy, bitter, or citrusy flavors. But that sugar and yeast brew falls short if you want to make pure ethanol, which is an important solvent for industrial manufacturing. For that, people react ethene with steam, in the presence of a catalyst.

We’ve seen these arrows before: they mean the reaction is reversible. So we can also take ethanol and use it to make ethene (and some water). This type of reaction is known as the dehydration of ethanol, because we're removing water from the ethanol.

But, more generically, it's an elimination reaction: a reaction in which a group is “kicked out” of a larger molecule. Elimination reactions are very important, because they’re the main way we can make organic compounds containing double or triple carbon-carbon bonds. So let's have a look at how they work... [Theme Music].

Elimination reactions are related to substitution reactions, which we learned about in episodes 20 and 21. So it's worth remembering the SN1 and SN2 pathways before we dive in deeper. There are a few key terms we use to talk about substitution reactions.

The substrate contains an sp3 hybridized carbon. A leaving group accepts electrons as it, well, leaves the substrate. Then, there's the nucleophile or electron-pair donor.

When the leaving group departs the substrate in the SN1 mechanism, it forms a carbocation intermediate, and the stability of that carbocation is really important. Because of this, all things being equal,. SN1 reactions are faster with tertiary carbon substrates than they are with less substituted secondary carbons.

In the SN2 mechanism, the nucleophile attacks the substrate and simultaneously kicks out the leaving group. So steric hindrance plays an important role, and the opposite trend happens:. SN2 reactions are generally faster with primary carbon substrates than they are with more substituted secondary carbons.

The key to substitution reactions is: in a chemical reaction between a substrate and a nucleophile, the nucleophile acts as an electron donor. The nucleophile causes substitution. The key to elimination reactions is slightly different: in a chemical reaction between a substrate and a nucleophile, the nucleophile acts as a proton acceptor.

The nucleophile causes elimination. This idea that a nucleophile can either be an electron donor or a proton acceptor is subtle and kind of tricky, so let's go back to the basics... of acids and bases. A nucleophile is something with a non-bonded pair of electrons or pi bond.

It has some spare electrons hanging around that it can generously give to something that’s a bit electron-deficient. A Lewis base is something that can donate a pair of electrons, so all nucleophiles are Lewis bases. And if we consider Brønsted-Lowry bases, we know that nucleophiles are also potentially proton acceptors.

That "potentially" is important though: nucleophilicity and basicity aren’t interchangeable terms. There’s a subtle difference! Nucleophilicity refers to how readily a nucleophile attacks any non-hydrogen atom.

Since this is an organic chemistry course, that non-hydrogen atom is usually going to be carbon. On the other hand, we'll use basicity to describe   how readily a nucleophile attacks hydrogen atoms, specifically. When we talk about a nucleophile acting as a base, we mean it's acting as a proton acceptor.

Remember, hydrogen atoms have one proton, one electron, and no neutrons. So when we talk about protons like this, we mean a hydrogen atom without its electron. Different groups have different amounts of nucleophilicity and basicity, which lets us know what sort of reaction they’re likely to cause when considering substitution or elimination.

For example, something bulky like the tert-butoxide anion isn’t a great nucleophile, but is a strong base. Likewise, something can be a good nucleophile, but a weak base, such as the halide ions. And some things, like hydroxide ions, have it all.

They're a good nucleophile and a strong base. Like I mentioned before all this, an elimination reaction is where a nucleophile acts as a proton acceptor. In other words, a strong base.

It rips a proton off the substrate. Then, the electrons that formed a carbon-hydrogen bond in the substrate form a pi bond instead: the carbon-carbon double bond in the elimination product. Here's the general form of an elimination reaction, with placeholders.

We can use a merry-go-round analogy for elimination reactions too. It's kind of like a small kid (the nucleophile) crashing into the merry-go-round at full speed, grabbing one of the other kids (the hydrogen) and pulling them off completely. Then, another kid jumps off the merry-go-round too, and decides they’ve had enough of this dangerous, not-fun playground, and heads home.

There are two specific ways this can happen though, called E1 and E2. And to talk about the difference between these elimination mechanisms, we need some new terms. The alpha carbon of the substrate is the carbon with the leaving group, attached to it.

Alpha hydrogens are bonded to the alpha carbon. The beta carbons of the substrate are any carbons attached to the alpha carbon. Any hydrogens attached to a beta carbon are called beta hydrogens.

In E1 reactions, the first step is that a carbocation forms on the alpha carbon, the same as in SN1. Both of these mechanisms are unimolecular, which is why they have a 1. Their reaction rates depend on one molecule (the substrate).

Next, the nucleophile acts as a base and takes a proton from a beta carbon. That's what makes it elimination! Then, the electrons that were in that beta carbon-hydrogen bond form a pi bond, and a double bond forms between the alpha and beta carbons.

As an example of an E1 reaction, we can look at 2-chloro-2-methylpropane reacting under neutral conditions which forms an alkene. There's actually a low yield here because the alkene is the minor product of this reaction. SN1 and E1 mechanisms often compete, giving us a mix of products!

We’ll talk more about this idea in the next episode, and what reaction conditions favor different mechanisms. On to E2 reactions! Like SN2 reactions, everything happens at once and there’s no carbocation to worry about.

The nucleophile acts as a base, takes a beta proton, and the leaving group does its thing and… leaves. This mechanism is bimolecular, which is why there's a 2. And its reaction rate depends on two molecules: the substrate and the base.

Because the base is involved in the rate, stronger bases, like negatively charged oxygen and nitrogen anions, favor E2 reactions. For most E2 reactions, the hydrogen and the leaving group must be antiperiplanar, which means they need to be in the same plane, but on opposite sides, of the bond between the alpha and beta carbon. We can check for this by looking at the substrate's Newman projection, using what we learned way back in episode 6!

For example, let's look at 2-bromobutane reacting under basic conditions. This molecule has several different staggered conformations, which are where the groups on the alpha carbon aren't eclipsing or overlapping the groups on the beta carbon. There's a 60 degree angle between them.

And to do an E2 reaction, we need to make sure the beta hydrogen and bromine are antiperiplanar– right across from each other. The most energetically stable conformation is when the methyl groups aren't right next to each other, because there's only one gauche interaction between one of the methyls and the bromine. We have our antiperiplanar arrangement with a beta hydrogen and bromine, so this leads to the major product, the E-alkene. (Where the methyls are E-cross from each other.) The conformation with the methyl groups right next to each other has two gauche interactions, so it's less stable.

We still have our antiperiplanar arrangement with a beta hydrogen and bromine, so this produces the minor product, which is the Z-alkene. (Where the methyls are on “ze Zame Zide” of each other.) An E2 reaction simply can’t happen with this third staggered conformation, because we don't have that antiperiplanar arrangement at all! Let's take a closer look at the E2 reaction on our more stable conformer. The nucleophile base attacks, the bromide leaving group leaves, and a double bond forms.

We can convert our structure back from the Newman projection, and see the C-H-three groups are on opposite sides! Specifically, this E2 mechanism is called anti elimination, because of the antiperiplanar arrangement that made it possible. To be totally honest, there's another E2 mechanism called syn elimination, which involves a much-less-stable eclipsed conformation, where the leaving group overlaps a hydrogen (or in this case, a hydrogen isotope) in our Newman projection.

It's rare, but can happen with rigid structures. You don't need to worry too much about it. So we took 2-bromobutane, did our E2 reaction, and have major and minor alkene products.

We're done… right? Well, if you look closely at 2-bromobutane, we have one alpha carbon (which is attached to the bromine), but two beta carbons. So there are two places the carbon-carbon double bond could form.

How do we decide? Never fear, Zaitsev’s rule is here! It describes this pattern: in an elimination reaction, the most substituted alkene is the most stable, and is the major product.

So for the E2 reaction we just did with 2-bromobutane, we mostly get Zaitsev products, plus a little bit of this other alkene with the double bond in a different spot. To make sure we understand Zaitsev’s rule, here’s another example with cis-1-chloro-2-methylcyclohexane. If we show cyclohexene in its chair conformation, we can see that for the hydrogen and the chlorine leaving group to be antiperiplanar, both of these groups have to be axial.

Only anti elimination happens here, and syn elimination is impossible! Following Zaitsev’s rule, the major product is the one with the most substituted double bond:

1-methylcyclohex-1-ene. The minor product is 3-methylcyclohex-1-ene.

Phew! Those are the basics of elimination reactions, so let's return to that dehydration of ethanol that I mentioned right at the start. A great way to eliminate alcohols is by treating them with sulfuric acid or phosphoric acid.

And… hang on a moment. Haven’t we been talking about bases all this time? Well, remember, acids form conjugate bases.

For example, sulfuric acid forms the resonance-stabilized bisulfate anion, which is a poor nucleophile but is able to grab a proton, so it can help with elimination! If we mix ethanol with sulfuric acid and really heat it up, we can do this dehydration reaction of ethanol. But is it E1 or E2?

The first step in an E1 reaction is forming a carbocation. Ethanol is a primary alcohol, so the primary carbocation that forms from the alpha carbon would be fairly unstable, and there would be a very high activation energy for the reaction. So it’s much more likely that we're dealing with an E2 reaction here.

A proton from the sulfuric acid bonds to one of the lone pairs of the oxygen in the alcohol, making an oxonium salt and creating a great leaving group: neutral water. The newly-formed bisulfate anion acts as a base, grabs a beta proton, and kicks out water all at once in a single step. It’s an E2 mechanism!

E2 reactions with sulfuric acid are a bit unusual, they really need heat to make them go, which we’ll talk about more next episode. That gave us a little bit of practice with elimination reactions, but let's do some rapid fire problems to really make sure we understand what's going on. We’re going to put three elimination problems on screen and predict the likely mechanism and the products.

Then, we'll work through the answers, so pause right after the question if you want to solve them yourself. Ready? Here's problem number one.

We’re using N-H-2 minus as our nucleophile, which is a strong base. We remember from episode 21 that OTs is tosylate, one of those sulfonates – a good leaving group. And the alpha carbon is a secondary carbon, which is slightly less substituted and could do either E1 or E2.

Because of the strong base, though, we'd expect this to be an E2 reaction. And there’s only one place the double bond can go to form a product. Okay!

Here's problem number two. This time we have a tertiary alcohol and phosphoric acid. Like our dehydration of ethanol reaction, the phosphoric acid loses a proton to the alcohol group.

Then, the water leaves to form a carbocation. And in this case, unlike with ethanol, the tertiary carbocation that forms on the alpha carbon is fairly stable, so an E1 mechanism is favored. However, there are two possible products this time: a major one with a more substituted double bond per Zaitsev’s rule, and a minor one where the double bond is less substituted.

And here's our final problem, number three. The alpha carbon is secondary, which is slightly less substituted, but that doesn't clearly tell us if it’s E1 or E2. But sodium methoxide is a strong base, so I'm thinking E2.

Let’s draw our chair conformation… and oops, the chloride is equatorial. Remember, the chloride needs to be axial, so we can get an antiperiplanar hydrogen, so let’s do a chair flip! Now, we can see there's just one hydrogen that is also axial and antiperiplanar to the chloride.

And when we eliminate it, we get the less substituted double bond. That's the non-Zaitsev product! Super tricky.

So a rule can sometimes be broken, and it's really important to not take shortcuts when predicting products. With E2 elimination products, we have to check out our Newman projections and chair conformations to make sure a product can actually form. In this episode we learned… a lot!

But specifically that:. A nucleophile can also be a base. The rate-determining step in E1 reactions, like SN1, is the formation of a carbocation.

E2 reactions, like SN2, are bimolecular, with everything happening in one step. The antiperiplanar transition state is super important in E2 reactions, and The position of the double bond can usually be predicted with Zaitsev’s rule, but not always! In the next episode we’ll look more closely at elimination vs. substitution, and make sense of these competing mechanisms.

Until then, thanks for watching this episode of Crash Course Organic Chemistry. If you want to help keep all Crash Course free for everybody, forever, you can join our community on Patreon.