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Organic chemistry is a great workout for your brain, and to keep its energy up, your brain needs glucose. To maintain blood glucose levels, our bodies go through a process called gluconeogenesis, which involves the important type of organic reaction we’re getting into today: aldol reactions! In this episode of Crash Course Organic Chemistry, we’ll learn what an aldol is, as well as how aldol reactions work and what they do. Plus we’ll learn about a similar reaction, Claisen condensation.

Episode Sources:
Claisen, L., 1887. Ueber die einführung von säureradicalen in ketone. Berichte der deutschen chemischen Gesellschaft, 20(1), pp.655-657.

Series Penicillin References:
Nicolaou, K. C., & Sorensen, E. J. (1996). Classics in total synthesis: targets, strategies, methods. John Wiley & Sons.
Sheehan, J. C. (1982). The enchanted ring: the untold story of penicillin.
Primary literature for Sheehan’s penicillin synthesis: Sheehan, J.C. & Izzo, P.T. J. Am. Chem. Soc. 1948, 70, 1985; Sheehan, J.C. & Izzo, P.T. J. Am. Chem. Soc. 1949, 71, 4059; Sheehan, J.C. & Bose A.K. J. Am. Chem. Soc. 1950, 72, 5158; Sheehan, J.C., Buhle, E.L, Corey E.J., Laubach, G.D. & Ryan J.J. J. Am. Chem. Soc. 1950, 72, 3828; Sheehan, J.C. & Laubach, G.D. J. Am. Chem. Soc. 1951, 73, 4376; Sheehan, J.C. & Hoff, D.R. J. Am. Chem. Soc. 1957, 79, 237; Sheehan, J.C. & Corey E.J. J. Am. Chem. Soc. 1951, 73, 4756

Series Sources:
Brown, W. H., Iverson, B. L., Ansyln, E. V., Foote, C., Organic Chemistry; 8th ed.; Cengage Learning, Boston, 2018.
Bruice, P. Y., Organic Chemistry, 7th ed.; Pearson Education, Inc., United States, 2014.
Clayden, J., Greeves, N., Warren., S., Organic Chemistry, 2nd ed.; Oxford University Press, New York, 2012.
Jones Jr., M.; Fleming, S. A., Organic Chemistry, 5th ed.; W. W. Norton & Company, New York, 2014.
Klein., D., Organic Chemistry; 1st ed.; John Wiley & Sons, United States, 2012.
Louden M., Organic Chemistry; 5th ed.; Roberts and Company Publishers, Colorado, 2009.
McMurry, J., Organic Chemistry, 9th ed.; Cengage Learning, Boston, 2016.
Smith, J. G., Organic chemistry; 6th ed.; McGraw-Hill Education, New York, 2020.
Wade., L. G., Organic Chemistry; 8th ed.; Pearson Education, Inc., United States, 2013.

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You can review content from 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!

Whew, organic chemistry sure gives your brain a good workout! And glucose is the primary energy source used by our brains.

So without it, we’d be in big, big trouble. Thankfully, our brains can get all the glucose they need through chemistry!

One particularly important process is called gluconeogenesis, which helps us maintain the necessary levels of blood glucose when we haven’t just eaten a meal.

The whole process of gluconeogenesis is pretty interesting if you're a biochemistry nerd, but a key step that's relevant to this course is an aldol reaction.

In this aldol reaction, an aldehyde and a ketone – with three carbons each – join to make a six-carbon sugar molecule.

So they're forming tasty carbon-carbon bonds! And we'll be exploring more of these kinds of reactions in this episode.

[Theme Music]

An aldol is a ketone or aldehyde with a beta hydroxy group. In other words, it's a molecule with a carbonyl functional group a couple of carbons away from an alcohol group.

Aldols are the product of aldol reactions, which are another way of forming carbon-carbon bonds in organic chemistry.

Essentially, we combine two simpler carbonyl compounds into a bigger molecule. This reaction was discovered independently by the French chemist Charles-Adolphe Wurtz and the Russian chemist and composer Alexander Borodin in the late 1800s.

The "ald" part of the name comes from the fact that the reaction involved two aldehydes (before we figured out ketones could be involved too).

And the "ol" part comes from the alcohol, because the product is a beta-hydroxy carbonyl compound.

As always, we want to understand the mechanisms at work – not just memorize reactions!

Aldol reactions can be acid-catalyzed or base-catalyzed.

 (02:00) to (04:00)


And they begin by converting an aldehyde (or ketone) to an enol in acid, or enolate in base – like we did in episode 43.

Next, the enol or enolate attacks another aldehyde (or ketone) in an aldol addition step. And then, to finish it off, water is removed in a step called – as you might guess – dehydration.

The overall process of two carbonyls coming together with loss of water is called aldol condensation.

In this episode, we're only going to join two identical aldehydes or ketones. It’s possible to join two different carbonyl molecules, but we’re saving that for next time!

With that being said, let’s take a look at some simple examples. First, an aldol reaction under basic conditions.

As we saw in the last episode, a base steals a proton from the alpha carbon of the carbonyl. This gives us an enolate ion with a resonance-stabilized negative charge.

This enolate is only formed in small amounts.

So next, an enolate molecule attacks the carbonyl of one of the ketones in the sea of acetone molecules, which are acting as electrophiles.

And once this addition step is done, we’ve formed a carbon-carbon bond!

The now-negative oxygen of the alkoxide swipes a proton from some nearby water. Now we’ve got an aldol product with a carbonyl and an alcohol!

But we’re not quite done. At this point, the aldol product will often undergo dehydration.

And remember, this is a base-catalyzed reaction, which means the base is reformed at the end. This dehydration is a little tricky, so let’s look at it carefully.

To start, the base nabs one of the alpha hydrogens in between the carbonyl and alcohol groups, and we form an enolate again.

But now it’s the enolate of the aldol product. Then, hydroxide is eliminated, leaving us with an unsaturated carbonyl compound.

Wait a second. Hydroxide is a bad leaving group, right? Or at least Deboki-of-the-past said so. So how are we getting away with this?

Well, it’s what’s known as an E-1-c-B elimination, which happens in multiple steps.

 (04:00) to (06:00)


As a quick reminder, we’ve already met a couple of elimination mechanisms: E1 and E2. The E2 mechanism occurs in one step.

And the E1 mechanism has two steps: loss of a leaving group followed by deprotonation, and the rate depends on the concentration of the substrate.

E-1-c-B joins this family as a special type of E1 mechanism, so the hydroxide leaving doesn't affect the reaction rate.

This mechanism typically happens when we can make a stabilized anion on carbon, like adjacent to a carbonyl, and have a poor leaving group on the carbon next to the anion, like hydroxide.

First, the base grabs one of the acidic alpha hydrogens and forms a new enolate – in this stabilized anion there’s a build-up of negative charge on the molecule.

Then, the lone pair of electrons moves to the neighboring atom, kicking out the leaving group and forming a double bond.

In general, E-1-c-B mechanisms can be drawn like this. EWG and X are the electron stabilizing groups, and LG stands for leaving group.

So E-1-c-B is the cherry-on-top of a base-catalyzed aldol reaction. We don't have anything more to say about those for now – time for the acid-catalyzed version!

As we saw in the last episode (again):  the carbonyl oxygen is protonated, then the alpha carbon is deprotonated by the conjugate base of our acid, and we get an enol.

Now another protonated ketone enters the mix, with its electropositive carbon vulnerable to nucleophilic attack.

The electrons push down from the oxygen, kicking out the double bond of our nucleophilic enol, which makes a carbon-carbon bond with our protonated ketone.

That's our aldol addition step!

Then, the protonated carbonyl group is deprotonated and, voilà, we have a neutral aldol product! But, again, we're not quite done.

There's also a potential elimination step here, especially if the aldol product is heated in these acidic conditions.

There’s a few possible mechanisms for the elimination. For example, the hydroxyl group could get protonated first. Then, water might leave on its own

 (06:00) to (08:00)


in an E1 mechanism, followed by a deprotonation. It’s also possible to eliminate in one step, through the E2 mechanism.

A third possibility is that the reaction mechanism goes through a more complex mechanism involving an enol.

This beta-elimination uses a little push from the carbonyl kinda like the E-1-c-B.

But this time we’re going through an enol instead of an enolate.

Speaking of which, it's been a little while since we looked at our penicillin V Mold Medicine Map, which we started in episode 30.

It's the perfect time to revisit that synthesis because it has an E-1-c-B step, too.

If we compare the starting material with the product we’d like to make, we can see that we have to get rid of two functional groups: a hydroxyl from our acid and a chloride.

We also want to join the two oxygens up to give us an ester-type linkage, and add some double bonds.

This is one of the most complicated sequences we’ll look at in this series, but we can puzzle through some of it using the rules we’ve learned.

So, buckle up!

The acetic anhydride reactant has a built-in leaving group – acetate – and it can even react with weak nucleophiles.

Looking at our molecule, let’s think about the most nucleophilic spot... it’s the carboxylic acid.

So the acetic anhydride reacts with our carboxylic acid, forming a new anhydride in the first step.

Now, we’ve put a good leaving group on our carboxylic acid, which sets us up for an intramolecular reaction.

The amide oxygen is pretty electron-rich because of amide resonance, so that attacks the acid anhydride carbon.

Our leaving group – acetate again – is kicked out.

And we get the five-membered ring we see in our product, more or less – except we need to lose H-Cl.

We still have a basic molecule in here, the acetate.

Hmm, that hydrogen next to an electronegative nitrogen plus a carbonyl is pretty acidic. And deprotonation even makes a bonus aromatic ring intermediate!

So... we make an enolate!

Then the chlorine is eliminated in an E-1-c-B step – it’s just one extra double bond away from what we saw before.

 (08:00) to (10:00)


More substituted double bonds are more stable, so a kind of brain-bendy isomerization occurs, and all our double and single bonds are where we want them.

Hooray!

Now, as exciting as penicillin is, there's another reaction we still have to cover – as you may have guessed from the title of this episode!

We've got the basics of aldol reactions down, but there’s a similar reaction called the Claisen Condensation.

It’s named after the German chemist Rainer Ludwig Claisen.

He was born in 1851, began studying chemistry in 1869, was a military nurse, and then returned to academia and worked in Kekulé's lab for a while – yes, Kekulé the benzene guy!

Anyway, in the Claisen condensation, an ester joins with either another ester or another carbonyl compound, in the presence of a strong base.

Once again, we’re making a carbon-carbon bond!

The ultimate product of this Claisen reaction is a beta-keto ester.

We don’t want to end up with a horrible mess, so we need to use a base that won’t get involved in some sort of nucleophilic substitution or addition reaction with a carbonyl carbon.

The alkoxide salt of the alcohol that’s kicked out at the end of the reaction is a good option.

So in this example, we're using sodium ethoxide as our strong base, because the alcohol that gets kicked out is ethanol.

First, the oxygen of the ethoxide grabs one of the alpha hydrogens from the ester, which gives us an enolate.

The next bit looks similar to other reactions we’ve seen: the enolate acts as a nucleophile and attacks the carbonyl carbon of another ester molecule.

The ethoxide group is kicked out, and promptly picks up an alpha hydrogen, to form the alcohol.

Now we have a resonance-stabilized carbanion that gets protonated to form the neutral product, when some acid is added.

And that's a Claisen condensation in a nutshell!

Unlike in the aldol reaction, where the base was catalytic and remade at the end, Claisen condensation needs a stoichiometric amount of base.

The deprotonation of the beta-keto ester at the end is effectively pulling the whole reaction along,

 (10:00) to (12:00)


so we need enough base for that to happen.

We also need our enolizable partner to have a minimum of two protons to start, so the deprotonation can happen.

The reaction we just did involved two identical esters, but we can also do a crossed Claisen condensation, where we have two different starting molecules.

In this case, we need to deliberately choose one molecule that can form an enolate, and one that can’t – to avoid a messy mixture of four different products.

Basically, we need to make sure that one of the substrates doesn’t have alpha-hydrogens.

Also, we set this up with an excess of the non-enolizable reactant.

That way, there’s a better chance that the enolate will react with the non-enolizable substrate – which is what we want – rather than react with another molecule of itself – which we don’t.

Overall, the mechanism of a crossed Claisen condensation is the same as a Claisen condensation – and we end up with a carbonyl group beta to our ester.

In this particular example, we’ve made a diester.

Now that we have this reaction in our toolkit, we’re heading back to our Mold Medicine Map for another loop of the rollercoaster – so check your safety harness!

To do a crossed Claisen condensation on this penicillin precursor, let's take a peek at the alcohol that’s kicked out at the end.

So for our strong base, we'll use the alkoxide salt: sodium tert-butoxide.

And our other reactant is tert-butyl formate.

In the mechanism, sodium tert-butoxide plucks the acidic hydrogen off the alpha carbon in between the nitrogen and the ester.

Next, the nucleophilic enolate we’ve just formed attacks the carbonyl carbon in the tert-butyl formate.

The tetrahedral intermediate collapses, and the tert-butoxide is kicked out.

The final deprotonation makes another enolate, and once we protonate that, we're left with a neutral product and a shiny, new carbon-carbon bond!

Two new reactions, and two new steps in our penicillin V synthesis! Not bad for one episode.

In this episode we learned:

An aldol product is a ketone or aldehyde with a beta hydroxy group,

Aldol reactions join together two aldehydes or ketones, and can be acid or base catalyzed

 (12:00) to (12:35)


The aldol condensation involves an elimination step, which, in base, is an E1CB reaction, and

The Claisen condensation is used to join up esters.

There are even more enols and enolates coming up in the next episode, as we learn crossed aldol reaction and conjugate addition.

Until next time, thanks for watching this episode of Crash Course Organic Chemistry.

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