Tuesday, June 28, 2022

E1 and E2 Reactions: Crash Course Organic Chemistry #22

You can review episodes of Crash Course OrganicChemistry with the Crash Course App, available now for Android and iOS devices Hi! I’m Deboki Chakravarti and welcome toCrash Course Organic Chemistry! A relatively slow reaction between sugar andyeast makes the ethanol in beer, while other compounds contribute all thosehoppy, bitter, or citrusy flavors. But that sugar and yeast brew falls shortif 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 tomake ethene (and some water). This type of reaction is known as the dehydrationof ethanol, because we're removing water from the ethanol. But, more generically, it's an eliminationreaction: 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 doubleor triple carbon-carbon bonds. So let's have a look at how they work… [Theme Music].

Elimination reactions are related to substitutionreactions, which we learned about in episodes 20 and 21. So it's worth remembering the SN1 and SN2pathways before we dive in deeper. There are a few key terms we use to talk aboutsubstitution 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-pairdonor. When the leaving group departs the substratein the SN1 mechanism, it forms a carbocation intermediate, and the stability of that carbocationis really important. Because of this, all things being equal,.

SN1 reactions are faster with tertiary carbonsubstrates than they are withless substituted secondary carbons. In the SN2 mechanism, the nucleophile attacksthesubstrate and simultaneouslykicks out the leaving group. So steric hindrance plays an important role,and the opposite trend happens: SN2 reactions are generally fasterwith primary carbon substrates thanthey are with more substituted secondary carbons. The key to substitution reactions is: in a chemical reaction between a substrateand a nucleophile, the nucleophile acts as an electron donor. The nucleophile causes substitution. The key to elimination reactions is slightlydifferent: in a chemical reaction between a substrateand a nucleophile, the nucleophile acts as a proton acceptor.

The nucleophile causes elimination. This idea that a nucleophile can either bean electron donor or a proton acceptor is subtle and kind of tricky, so let's go backto the basics… of acidsand bases. A nucleophile is something with a non-bondedpair of electrons or pi bond. It has some spare electrons hanging around that itcan generously give to somethingthat’s a bit electron-deficient. A Lewis base is something that can donatea pairof 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 nucleophileattacks 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 describehow readily a nucleophile attackshydrogen atoms, specifically. When we talk about a nucleophile acting asa 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, wemean a hydrogen atom without its electron. Different groups have different amounts ofnucleophilicity and basicity, which lets us know what sort of reaction they’relikely to causewhen considering substitution or elimination.

For example, something bulkylike the tert-butoxideanionisn’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, haveit all. They're a good nucleophile and a strong base. Like I mentioned before all this, an eliminationreaction 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-hydrogenbond in the substrate form a pi bond instead:.

The carbon-carbon double bond in the elimination product. Here's the general form of an eliminationreaction, with placeholders. We can use a merry-go-round analogy for eliminationreactions 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 (thehydrogen)and pulling them off completely. Then, another kid jumps off the merry-go-roundtoo, and decides they’ve had enough of this dangerous,not-funplayground, and heads home. There are two specific ways this can happenthough, called E1 and E2. And to talk about the difference between theseelimination mechanisms, we need some new terms.

The alpha carbon of the substrate is the carbonwith the leaving group, attached to it. Alpha hydrogens are bonded to the alpha carbon. The beta carbons of the substrate are anycarbons attached to the alpha carbon. Any hydrogens attached to a beta carbon arecalled beta hydrogens. In E1 reactions, the firststep is that acarbocationforms 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 takesa proton from a beta carbon.

That's what makes it elimination! Then, the electrons that were in that betacarbon-hydrogen bond form a pi bond, and a double bond forms between the alpha and betacarbons. As an example of an E1 reaction, we canlookat 2-chloro-2-methylpropane reactingunder neutral conditions which forms an alkene. There's actually a low yield here becausethe alkene is the minor product of this reaction. SN1 and E1 mechanisms often compete, givingus a mix of products! We’ll talk more about this idea in the nextepisode, and what reaction conditions favor different mechanisms. On to E2 reactions!.

Like SN2 reactions, everything happens atonce and there’s no carbocation to worry about. The nucleophile acts as a base, takes a betaproton, and the leaving group does its thing and… leaves. This mechanism is bimolecular, which is whythere'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 theleaving group must be antiperiplanar, which means they need to be in the same plane, buton opposite sides, of the bond between the alpha and beta carbon. We can check for this by looking at the substrate'sNewman projection, using what we learned way back in episode 6!.

For example, let's look at 2-bromobutane reactingunder basic conditions. This molecule has several different staggeredconformations, which are where the groups on the alpha carbonaren'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 makesure the beta hydrogen andbromine are antiperiplanar– right across from each other. The most energetically stable conformationis whenthe 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 witha 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 methylgroups rightnext to each other hastwo gauche interactions, so it's less stable. We still have our antiperiplanar arrangementwith a beta hydrogen and bromine, so this produces the minor product, which is the Z-alkene. (Where the methyls are on “zeZame Zide”of each other.) An E2 reaction simply can’t happen withthis third staggered conformation, because we don't have that antiperiplanar arrangementat all!.

Let's take a closer look at the E2 reactionon our more stable conformer. The nucleophile base attacks, the bromideleaving group leaves, and a double bond forms. We can convert our structure back from theNewman projection, and see the C-H-three groups are on opposite sides! Specifically, this E2 mechanism is calledanti elimination, because of the antiperiplanar arrangement that made it possible. To be totally honest, there's another E2 mechanismcalled syn elimination, which involves a much-less-stable eclipsed conformation, where the leaving group overlaps a hydrogen(or in this case, ahydrogen 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-carbondouble bond could form. How do we decide? Never fear, Zaitsev’s rule is here! It describes this pattern:.

In an elimination reaction, the most substitutedalkene 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 bondin 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 chlorineleaving group to be antiperiplanar, both of these groups have to be axial. Only anti elimination happens here, and synelimination is impossible! Following Zaitsev’s rule, the major productis 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, solet's return to that dehydration of ethanolthat I mentioned right at the start. A great way to eliminate alcohols is by treatingthem with sulfuric acid or phosphoric acid. And… hang on a moment. Haven’t we been talking about bases allthis time? Well, remember, acids form conjugate bases.

For example, sulfuric acid forms the resonance-stabilizedbisulfate 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 reallyheat it up, we can do this dehydration reaction of ethanol. But is it E1 or E2? The first step in an E1 reaction is forminga carbocation. Ethanol is a primary alcohol, so the primarycarbocationthat forms from the alpha carbonwould be fairly unstable, and there would be avery high activationenergy for the reaction. So it’s much more likely that we're dealingwith an E2 reaction here. A proton from the sulfuric acid bonds to oneof the lone pairs of the oxygen in the alcohol,.

Making an oxonium salt and creating a greatleaving 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 bitunusual, they really need heat to make them go, which we’ll talk about more next episode. That gave us a little bit of practice withelimination reactions, but let's do some rapid fire problems to reallymake sure we understand what's going on. We’re going to put three elimination problemsonscreen and predict the likelymechanism and the products. Then, we'll work through the answers,.

So pause right after the question if you wantto 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 expectthis to be an E2 reaction. And there’s only one place the double bondcan go to form a product.

Okay! Here's problem number two. This time we have a tertiary alcohol and phosphoricacid. 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, thetertiary carbocation that forms onthe 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 doublebond 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 slightlyless substituted, but that doesn't clearly tell us if it’s E1 or E2. But sodium methoxide is a strong base, soI'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 hydrogenthat is also axial and antiperiplanar to the chloride.

And when we eliminate it, we get the lesssubstituted double bond. That's the non-Zaitsev product! Super tricky. So a rule can sometimes be broken, and it'sreally important to not take shortcuts when predicting products. With E2 elimination products,we have to check out our Newmanprojections and chair conformations tomake 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, witheverything happening in one step The antiperiplanar transition state is superimportant in E2 reactions, and The position of the double bond can usuallybe predicted with Zaitsev’s rule, but not always! In the next episode we’ll look more closelyat elimination vs. substitution, and make sense of these competing mechanisms. Until then, thanks for watching this episodeof Crash Course Organic Chemistry. If you want to help keep allCrash Coursefree for everybody,forever, you can join our community on Patreon.

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