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

 To communicate with each other, insects often use organic chemistry! Instead of using words, they secrete compounds that can mean "hey, you're cute" or "follow this path for food!" or even "back off, you're gonna get hurt if you come any closer." Soldier termites, for example, defend their nests by secreting a toxic chemical. That wards off invaders just fine but, when they use these chemical weapons, they have to make sure they don’t also hurt their own nest-mates. Turns out, the worker termites in the nest have an enzyme that allows them to reduce the chemical from its toxic form to another, harmless, form. And in this reaction, only the double bond that is conjugated with the carbonyl is reduced. Specifically, the soldier termite’s toxin has a functional group called an enone. And the worker termite's enzyme selectively removes the alkene from that functional group. Enones can undergo a reaction called conjugate addition that simple alkenes can't. And this reaction makes the enone toxic to invaders. Let’s see why, and explore some more reactions and properties of enolates!

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 In the previous episode, we learned about aldol reactions, where two simple carbonyl compounds combine into a larger one. Specifically, we explored what happened when the two reactant molecules were the same. But if we join two different carbonyl compounds – like an enol or enolate and a different aldehyde or ketone – it's called a crossed aldol reaction. With crossed aldol reactions, it’s very important to choose the carbonyls carefully, otherwise we end up with a horrible mess of different products. Generally, it’s a good idea to use an enolizable ketone – with alpha-hydrogens – and a non-enolizable aldehyde. This works well because, first, only the ketone can form an enol (or enolate) and become the nucleophile in the reaction. And second, aldehydes are more reactive than ketones, because they're usually less hindered at the carbonyl carbon and less stabilized by inductive effects. So we minimize the chances of the ketone reacting with itself. It also helps to have the aldehyde in excess, because that way there’s much less chance of a ketone molecule bumping into another ketone molecule at all. The mechanism for a crossed aldol reaction looks like what we've been doing in the last two episodes!

We start with deprotonation of the enolizable ketone. Assuming we’re working under basic conditions, the next step is for the enolate anion to do its nucleophile thing and attack the carbonyl carbon of the aldehyde. Then, we pull a hydrogen off some nearby water to form the hydroxyl, and we have the aldol product. Remember these protons in between the alcohol and the carbonyl are pretty acidic, and we still have basic conditions. So, off goes a proton, and now we have the enolate of the aldol product. Finally, the hydroxide is eliminated through an E-1-c-B elimination, and we’re left with an unsaturated carbonyl compound.

 Honestly, there’s nothing really new here, except we used two different reactants. So let's throw something else into the mix: making an enolate from methyl isopropyl ketone. It has more than one alpha hydrogen, and when we mix it with base, we could form two different enolates. So… which one forms? Or is it both?

To figure it out, let's take a moment to review Zaitsev’s rule: in elimination reactions that form alkenes, we tend to form the alkene product that we get by removing a hydrogen from the carbon with the fewest hydrogens. Or, to put it another way, we mostly get the more substituted alkene product. We can use similar logic here to figure out the enolate product. If we remove a proton from the more substituted alpha-carbon in the methyl isopropyl ketone, we get an enolate with a more substituted carbon-carbon double bond. This is a lower-energy product, and therefore more thermodynamically stable, so we call it the thermodynamic product. If we remove a proton from the less substituted alpha-carbon in the methyl isopropyl ketone, we get an enolate with a less substituted carbon-carbon double bond.

 This product forms more quickly, because the hydrogen being removed is more sterically accessible (there's less stuff around it). So we call it the kinetic product. So we have one enolate that’s lower in energy, but formed more slowly, and one that’s higher in energy, but formed more quickly. Which one we get depends on the reaction conditions. At low temperatures, we don’t have enough energy to get over the larger energy barrier. But we can get over the smaller energy barrier for the quick-forming, less-stable, higher-energy enolate – the kinetic product. At higher temperatures, there’s enough energy to get over both energy barriers and to reverse the reaction. So, over time, the slow-forming, more-stable, lower-energy enolate forms – the thermodynamic product. We also need to take into account what base we're using to form the enolate. Say we have a bigger, stronger, sterically-hindered base, such as lithium diisopropylamide, or LDA, and the deprotonation step is rapid and irreversible. In this case, the kinetic product will be favored. However, with weaker bases, like alkoxides,the pKa of the base and the enolate are close. This means that the enolate can interconvert between the two forms, so over time the more stable product wins and we end up with more of the thermodynamic product.

So to recap: a strong base and a low temperature gives us the kinetic enolate. A weaker base and higher temperature gives us the thermodynamic enolate. Okay, now that we know how to puzzle through tricky enolate formation, let's try a crossed aldol condensation with enolizable hydrogens on both carbonyl substrates. I know earlier I said this could be a horrible mess of different products, but there are some tricks we can use to keep things simple. Let's take it step by step.

We mentioned LDA a moment ago, and said it forms the enolate completely. We’re starting with methyl isopropyl ketone again, and here’s that LDA, nabbing one of the alpha protons to form the kinetic product. Then we add in an aldehyde, and the enolate attacks the carbonyl carbon. We’re still under basic conditions, which are non-aqueous so we can use a strong base like LDA. To finish up the reaction, we toss some acid into the mix at the end to protonate the oxygen and cause elimination. Now, there’s another way to look at these aldol reactions: backwards. Like we learned in Episode 34, using retrosynthesis to figure out how to make something is a lot of what we do in organic chemistry. Retrosynthesis is a bit like cutting up a beautiful cake to figure out the decorations, filling, and ingredients so you can make the same cake yourself from scratch. Just… don’t try to taste your reactions…

 So, for example, here's a cool compound: we notice this structure contains an enone, where an alkene is conjugated with a carbonyl, like a ketone (or aldehyde). This is a dehydrated product that can come from an aldol condensation. We can use retrosynthesis to reverse-engineer it. Let’s imagine cutting that carbon-carbon double bond with magic chemical scissors. The part containing the carbonyl is the enolate we need to form, and we just tack an oxygen onto the other part to figure out the carbonyl compound we need to attack. Ta da!

 We know which starting materials we need! Besides being fun to make, enones can also be reagents in some interesting chemical reactions. Let’s take a quick look at their structure: they’re often numbered based on the atoms in their conjugated system, starting with the high-priority oxygen atom. Looking at some resonance structures, we can see that the 4 position is electrophilic, which means it can be attacked by nucleophiles! Of course, the carbonyl carbon is also electrophilic– as they've been throughout this whole series…and the known universe!

 So there are two possibilities for an enone reaction. There’s 1,2-nucleophilic addition, where the nucleophile adds to the carbonyl carbon. Alternatively, we could get 1,4-nucleophilic addition, also known as conjugate addition, where it adds at the 4 position. It’s using the whole conjugated system to do the reaction, and that’s where it gets its name. In 1,2-addition, the nucleophile attacks the carbonyl carbon. The pi electrons in that bond are shunted up to the oxygen, which then picks up a proton to form a hydroxyl group. Fairly simple and familiar!

In the conjugate addition mechanism, the nucleophile attacks the 4 position, and the enolate that forms is stabilized by resonance. Because of this, one of two things can happen next – but the product is the same either way! In one possible mechanism, the oxygen is protonated, and tautomerization gives the ketone. In the other, the alpha carbon picks up a proton to form the ketone directly. We have two possibilities again. This time, the nucleophilic attack is kinetically favored at the carbonyl carbon, meaning it's faster. On the other hand, the nucleophilic attack is thermodynamically favored at the alkene, meaning it's more stable. It turns out that conjugate addition reactions aren't unique to enones – conjugated carboxylic acid derivatives can undergo this kind of reaction, under the right conditions. The substrate's structure plays a big role in what reaction happens. For example, a nucleophile attacking an acid chloride is more likely to attack the carbonyl carbon. The strongly electron-withdrawing chlorine really polarizes the carbonyl carbon, making it super-attractive to nucleophiles and favoring 1,2-addition. But a nucleophile attacking an amide is headed for the 4 position. The nitrogen has lone pairs that stabilize the molecule by resonance, which makes the carbonyl less reactive and favors conjugate addition. The nucleophile is even more important, though!

 At this point, it’s helpful to think of them in two categories: hard nucleophiles and soft nucleophiles. Hardness and softness is related to charge and polarizability, which we learned about in Episode 11. Think of hard nucleophiles as kind of like…sharp weapons. They’re aggressive. They’re very electronegative, relatively small, and not very polarizable – think fluorides, alkoxides, organolithiums, and Grignard reagents. Soft nucleophiles, on the other hand, are cuddly. They say, "Hey, let’s not fight. Let’s hug it out instead!" They have lower electronegativity, they’re bigger, more polarizable, and sometimes neutrally charged – think sulfur compounds, iodide ions, and resonance stabilized enolates. Hard and soft nucleophiles do have hard and soft electrophile counterparts… and we’ll just touch on this topic briefly with our enone example. All we need to know for now is that soft nucleophiles are more likely to react with soft electrophiles, and hard nucleophiles are more likely to react with hard electrophiles. I guess cuddlers like to cuddle and fighters like to fight!

So when we have an enone with two electrophilic sites, the carbonyl carbon is a harder electrophile because it’s closer to the electron-pulling oxygen, while the carbon at the 4 position is much softer. And based on that cuddling-versus-fighting pattern, a hard nucleophile is more likely to give us the 1,2-addition product, while a soft nucleophile is more likely to give us the conjugate addition product. In fact, remember our termites from the intro? The toxic chemical weapon of the soldier termites is an enone. And because so many biological groups have thiols, which are soft nucleophiles, the toxin causes biological upset thanks to conjugate addition reactions. The inactivated toxin, which forms after exposure to the worker termites’ enzyme, doesn’t have a double bond conjugated with the carbonyl group. It’s unreactive with soft nucleophiles like thiols, and our worker termites live to build another day!

Not only that, but conjugate addition of thiols are part of our penicillin V synthesis, too! This step of our Mold Medicine Map starts off with the ring compound we made in the last episode. First, we mix methoxide with hydrogen sulfide gas, and the methoxide nabs a proton. The newly-formed bi-sulfide anion – a soft nucleophile – attacks the enone at the beta position to form the conjugate addition product. After that we have a cleavage step. Methoxide – which is a harder nucleophile– attacks the carbonyl carbon. When the electrons move down from the oxygen, the ring cleaves to leave an ester group a tone end of the molecule.

At the other end, we have our leaving group, and the protonation at nitrogen leaves us with an amide. So, overall, Dr. Sheehan used a conjugate addition reaction to add the sulfur atom needed for the penicillin ring! In Episode 31, we saw that the next step was hydrolysis of both the ester and the amide. Comparing that to the 5-membered ring of penicillin, we can see that this early step in the synthesis introduced all but one of these atoms. But that was a real brain bender, so we’ll end this episode here!

 This time we learned that: A crossed aldol reaction involves joining an enolizable ketone and an aldehyde, Kinetic and thermodynamic enolates form under different conditions, Soft nucleophiles are more likely to attack soft electrophiles, and Hard nucleophiles are more likely to attack hard electrophiles For now, we're done with enols and enolates– but not with penicillin! Next time we’ll cover amines, and learn how Dr. Sheehan closed up that 5-membered ring.

Until next time, 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.