Hi!

I’m Deboki Chakravarti and welcome to Crash Course Organic Chemistry! Have you ever played an adventure game?

The sort of thing where you start with a character, get some equipment, and go on a quest? We can imagine organic chemicals as these kinds of characters! Our adventuring protagonist is our substrate.

Different substrates can be sorted into classes, which can undergo different transformations. Specifically, our heroes can take on incredible new forms through the substitution and elimination reactions we’ve been learning for the past few episodes. In this episode, we’ll learn how to beat this adventure game: how to take any substrate, combined with any nucleophile, and predict what kind of new adventurer the transformation will produce! [Theme Music].

First of all, let’s set up our “world” to reveal the rules that our substitution and elimination reactions must follow. In terms of starting characters, we have a few fundamental classes: methyl, primary, secondary, and tertiary substrates. Our substrate is always an sp3 hybridized carbon, and the classes refer to how many carbon substituents they have.

Primary is one, secondary is two, and tertiary is three. So a tertiary substrate is our bulkiest character option. Our substrates encounter different nucleophiles, which change the substrate through a chemical reaction.

We can think of nucleophiles as different magic potions. Some are stronger, some are weaker, and the transformative potion effects depend on what the substrate is, and the surrounding reaction conditions. In our world, if a character is in a desert kingdom, a potion could work differently than it would on a character in an icy cavern!

And there are four major transformations in this world -- types of substitution and elimination reactions. If you need more of a refresher, rewatch episodes 20 through 22! In both SN1 and E1 mechanisms, a carbocation intermediate forms.

In the SN1 substitution reaction, the nucleophile can attack the carbocation from either side, so we get a mixture of stereoisomers at a chiral carbon. And in the E1 elimination reaction, the nucleophile acts as a base, deprotonating the substrate, and we get an alkene. On the other hand, both SN2 and E2 mechanisms happen in one, single step.

In the SN2 substitution reaction, the nucleophile's backside attack on the substrate inverts the stereochemistry at a chiral carbon. And in the E2 elimination reaction, we need an antiperiplanar leaving group and beta hydrogen for the nucleophile to deprotonate the substrate, creating an alkene. Okay!

Let’s look at how different potions—or nucleophiles— transform our characters—or substrates. We’ll start with the smallest character class: methyls. Since they only have one carbon, there are no beta hydrogens, so they can't do an E2 transformation.

They also can’t form stable carbocations, which means they're resistant to any SN1 and E1 reactions. That leaves us with one option: methyl substrates only do SN2 reactions, no matter what nucleophile they encounter. And, in fact, with poor nucleophiles, methyl substrates either don’t do anything, or react so slowly that it’s just not worth it.

I mean, come on, we need to take our potion and get on with this adventure! Let’s take a look at a specific methyl substrate, bromomethane, and its reactions with three different nucleophiles. In each case, our starting character is transformed by an SN2 reaction.

There’s only one carbon, so there’s only one place each nucleophile can go, and one option for our character's final form, the reaction product. Easy! Next, let's look at primary substrates.

These characters have more carbons than methyl substrates, but the leaving group is at the end of the chain. So primary substrates also don't form very stable carbocations, which rules out SN1 and E1 reactions. And if they encounter a poor nucleophile, not much will happen.

Mostly, we see SN2 substitution reactions. For example, there's the Williamson etherification, named after its discoverer,. English chemist Alexander William Williamson.

Our character here is a primary alkyl halide, and our potion is a strongly nucleophilic alkoxide. After an SN2 transformation, our hero becomes an ether! But primary substrates can also undergo E2 elimination reactions when there's a certain type of potion... or, nucleophile.

Specifically, when primary substrates encounter a strong, bulky base. Steric hindrance makes it tricky for the big base to displace the leaving group with a backside attack, so it just grabs one of the beta hydrogens and runs. That's elimination, not substitution!

Now, primary substrates also have some special subclasses: primary allylic and primary benzylic substrates. These substrates have double bonds or benzene rings that stabilize carbocations by resonance. Also, polar, protic conditions “slow down” the effects of our nucleophile potions, so the nucleophile doesn’t attack immediately, there's time for a carbocation to form, and we’ll get SN1 reactions.

For example, here's a reaction with a primary benzylic substrate that forms an ether. We can see that hydrobromic acid is a product, which is another hint that we have an SN1 reaction. Remember from episode 21: “when we see acid as a reactant or product, think SN1." If we increase the strength of our nucleophile, or switch from a more sluggish polar, protic environment to an aprotic one, these special substrates act like other primary substrates and favor SN2 reactions.

Time for some secondary characters -- and no, I don't mean sidekicks! Secondary substrates include some of the most important potions and reaction conditions to learn, because they have the most competing mechanisms. Secondary substrates have all the traits of less bulky character classes that allow for SN2 and E2 reactions.

But they also have some branching, which means more stable carbocations, and that SN1 and E1 reactions are possible too. To start, let's look at an SN1 transformation example, with a secondary substrate and acetic acid as our nucleophile -- that hint from episode 21 about acids and SN1 applies again! And these are polar, protic conditions that slow down our nucleophiles and give carbocations time to form.

In SN1 reactions, the nucleophile can attack either face of the carbocation. So if our substrate has a chiral carbon (and in this case, it has two) the stereochemistry gets scrambled. Here, we get a mixture of diastereomers as products.

Remember from episode 9, these can occur when a compound has more than one chiral center and, as a result, has stereoisomers that aren't mirror images of each other. The non-reacting chiral center from the starting material keeps its configuration in both products, but the reacting chiral center ends up R in one diastereomer, and S in the other. But here's why we have to be so careful: when our secondary character drinks this potion, we're not guaranteed an SN1 transformation.

Substitution and elimination can compete! Once a carbocation forms, the nucleophile can deprotonate our substrate, so some E1 products get formed too. Now, let's look at an SN2 transformation, with a tosylated alcohol as our substrate and a weakly basic nucleophile.

DMSO as a solvent gives us polar aprotic conditions, which tend to give the nucleophile a bit of a boost! In other words, polar, aprotic conditions strengthen our potions, so even a weak nucleophile can attack our substrate, favoring an SN2 reaction and inverted stereochemistry in the product. Now, if a secondary substrate encounters a strongly basic nucleophile, we tend to see E2 reactions.

Like I said, lots of competing mechanisms with this character class! Stronger bases are very reactive and quickly steal a beta hydrogen -- so long as it’s antiperiplanar to the leaving group, of course! This Newman projection reminds us why we get the E-alkene as our final form -- it has to do with the antiperiplanar hydrogen and leaving group rotating around the carbon-carbon bond.

Pop back to episode 22 if you need a refresher. This example shows another way that the surrounding reaction conditions affect how our nucleophile potions work. Notice this reaction was heated -- maybe this part of the game is in a desert?

And heat favors elimination reactions. To understand why this is true, we have to remember our Gibbs Free Energy equation from episode 15: ΔG = ΔH − TΔS. Increasing the temperature makes the entropy factor more significant, which means we end up with a more negative delta G.

So, with higher heat, a reaction that increases entropy is more likely to happen spontaneously. Elimination reactions create more entropy than substitution reactions because they give us a greater number of products. We get a whole extra molecule by comparison, because the base runs off with a beta hydrogen.

Now it's time for tertiary substrates -- our final class of characters! With all their branches, tertiary substrates are really good at forming stable carbocations. So SN1 and E1 reactions are possible with weak nucleophiles.

Being the bulkiest character class means that tertiary substrates have a lot of steric hindrance too, and can't undergo SN2 reactions. Nucleophiles can’t get close enough to attack the electrophilic carbon from behind, so our only all-at-once reaction choice is E2 elimination. Substitution and elimination reactions compete, so we have to pay close attention to the nucleophiles that favor one transformation over the other.

For example, here are some tertiary substrates and two different reactants: hydrobromic acid or water. Bromide and water are good nucleophiles, but poor bases -- so they're not so great at grabbing protons. That means they favor substitution over elimination, and our only substitution option is SN1.

To encourage elimination over substitution, we can pick a different reactant: sulfuric acid, which forms the conjugate base bisulfate anion when it loses a proton. Remember from episode 22, a bisulfate ion is a poor nucleophile but it can act as a base. That means it's better at grabbing a proton, and E1 elimination is favored with our tertiary, and benzylic substrate.

Now, let’s check out an example of an E2 elimination reaction. The key here is the strong base, which usually gives us E2... When tertiary or secondary substrates transform through elimination, usually the major product that forms is the most substituted alkene.

That's Zaitsev’s rule, which can be really helpful when predicting our characters' final forms. In fact, we can see it in action here with a tertiary substrate and a sodium ethoxide potion. However, Zaitsev's rule has an exception with a special type of potion: a nucleophile that's a very bulky base.

For example, the bulk of sodium tert-butoxide gets in the way. To form the Zaitsev product, this nucleophile would need to grab a proton from a beta carbon. But as the oxygen from the tert-butoxide ion gets closer, there's a steric clash causing lots of repulsion, which slows the reaction down.

Instead, the tert-butoxide ion just grabs the proton from our less substituted beta carbon, reducing the clash, speeding up the reaction, and forming the less-substituted, non-Zaitsev product. Substitution and elimination reactions can be really tough. Trust me, I've been there.

So here's a table to organize what we've learned. Remember that it'll take more than memorization to solve these puzzles: really imagine our game world, the substrate characters, and how the nucleophile potions might transform them. And, of course, practice helps.

So it's time for some rapid fire problems! We’re going to put four problems on screen that could be SN1, SN2, E1 or E2 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.

Player 1, are you ready to start this game? Here's problem number one. We have a secondary benzylic substrate with methanethiol as our nucleophile.

Looking at the table, negatively charged sulfur groups are weak bases, so we might expect SN2, but here we have a thiol, and since the sulfur is protonated, an “S-H” group, we have acidic or neutral conditions. Using our handy hint, remember acidic conditions favor SN1! Because we have a chiral center, we also get a mixture of enantiomers.

And don't forget that pesky minor product by E1. Here's problem number two. This time the substrate is secondary again, but the nucleophile is an acetylide anion.

Acetylide anions are strong nucleophiles and super strong bases, which means this reaction strongly favors elimination -- specifically E2. And there’s a mixture of products. Following Zaitsev’s rule, we would expect but-2-ene as the major product since it’s more substituted.

Then, looking at that major product, we expect more trans than cis, because that’s the form where we get the most energetically stable arrangement -- when the methyl groups are spaced as far apart as possible. Here's problem number three. We have another secondary substrate with methanol, a poor nucleophile.

We also have polar, protic conditions. Looking at our reactants, we expect to see HBr as a product, which is acidic -- so using that handy hint, we think we're dealing with SN1. But there’s something else to consider.

The intermediate is a secondary carbocation which is less stable than a tertiary carbocation. And if a less stable carbocation can get more stable through the migration of a C-H bond, then a rearrangement can occur. We've got to look out for these tricks!

This migration is called a hydride shift. So we actually end up with a tertiary carbocation, and this substitution product. And lastly, here's problem number four.

Here we have one more secondary substrate, and sodium acetate, which is a weak nucleophile. This is a nice simple setup, and it reacts easily by the SN2 mechanism. That’s it for this episode!

Next time, we’re going to be looking more at alcohols, ethers and epoxides. 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.