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

In the late 1920s in Germany, the Nazi military used Enigma machines – basically typewriters with exchangeable gears– to send and receive secret messages. They encrypted messages by varying the gears used, their starting position, and the location of wires on an attached plugboard. With trillions of possible encryptions, the Nazis believed they had a truly unbreakable code.

However, flaws in the encryption led Polish cryptologists to reverse engineer the components and construct their own Enigma machines. Building on this work, the British fully decrypted Nazi messages by 1940 and likely shortened World War II by several years. We can borrow some cryptological techniques too and use reverse engineering to figure out the steps to make a complicated organic molecule! [Theme Music].

Organic synthesis is the construction of more complex carbon-based structures (like medicines) from simpler, easy-to-get starting materials. In synthesis practice problems, we’re typically given a starting material and a target compound to make in the fewest steps possible. From there, we have to puzzle out what reagents, reaction conditions, and intermediate compounds will get us from one to the other.

We don’t usually show all the mechanism arrow pushing, but it's important to keep in mind because it determines the stereochemistry and regiochemistry of reactions. The best way to practice is by working through examples, so let's try a synthesis problem! The first thing I do is compare the starting material and target compound, like those "spot the difference" pictures.

I start by numbering carbon atoms, because it’s really easy to miss a change in the number of carbons. Here, I can see that we’re introducing one extra carbon on the sulfur atom, but the main chain stays the same with eight carbons. It's also really important to look at the stereochemistry of our starting material and product.

Both molecules have one chiral carbon, and they're both S-enantiomers, which we know by ranking priority of the surrounding groups like we did in episode 8. But one group on the chiral carbon does change from a hydroxyl group to a sulfur. By noticing that change (and what was unchanged, like the carbon chain number and stereochemistry), we have all the puzzle pieces.

For this synthesis problem, we need to think about reactions of alcohols, and how we might replace a hydroxyl group with sulfur. Sulfur is a good nucleophile, one of the key ingredients in substitution reactions. But hydroxide is a poor leaving group, so we need to make it a good one.

In episode 24, we talked about a couple ways to form good leaving groups from alcohols: turning them into sulfonates or halides. So let’s try turning our alcohol into a tosylate. Now, with a good leaving group, we can add in a sulfur nucleophile and do an SN2 reaction.

And... oops, we’ve inverted our stereochemistry, making the R-enantiomer. So let’s try something else: if we perform two sequential SN2 reactions, we’ll invert the stereochemistry twice, and end up with the S-enantiomer product we want. We still need to make the alcohol a better leaving group and we've learned a couple halide-forming reactions that invert the stereochemistry of an alcohol!

Like bromide formation using phosphorus tribromide. Now we’ve got a good leaving group and inverted stereochemistry, so our second SN2 reaction with the sulfur nucleophile inverts it again and gives us the S-enantiomer product. We've solved this puzzle!

With longer puzzles, it’s often easier to reverse engineer the target compound with retrosynthesis – we start with the product and think about bonds we can break to give simpler starting fragments. And we use a new kind of arrow. This retrosynthesis arrow tells us that our product is the more complex molecule on the left, and we’ll make it by forming a bond between the two simpler chemicals on the right.

We have one chiral carbon in our molecule, but the stereochemistry isn't specifically shown as R or S. So we can assume we'll have a racemic mixture of enantiomers as our product. We can start retrosynthesis problems like synthesis problems, by numbering our carbon atoms.

In this case, our target compound has 11, while our starting materials only have 10, so we’ll have to add one carbon during our synthesis. We’ve got to start somewhere, and thinking about carbon atoms we need to add can help. Our target molecule also has an amide on the added carbon… but we haven’t learned a reaction in this series that adds an amide plus a carbon atom to a molecule.

But we know this is a carboxylic acid derivative, so we may get some clues thinking about reactions of these related carboxylic acids. When we change a functional group in retrosynthesis, we call that a functional group interconversion. We can just write FGI over the retrosynthesis arrow as a placeholder for this step, and come back after we've reverse-engineered the whole reaction.

Now, we can remember some reactions that form carboxylic acids. One way we've made them is by oxidizing primary alcohols or aldehydes. Another way involves an organometallic reagent reacting with carbon dioxide, which also adds a carbon atom.

We need one of those because the carbon chain got longer! So, let’s plan to form our carboxylic acid from a halide. A disconnection is a retrosynthetic step that forms a new bond and isn't just changing a functional group.

We show this by putting a squiggly line over that bond! We’re adding a carbon atom in this step, so this will be our first disconnection. And we'll put the halide that our organometallic reagent is made from, plus carbon dioxide, on the other side of the retrosynthesis arrow.

Remember, our goal is to work back to the two starting compounds, and the molecule we have right now is getting closer. We need to eventually break a bond to show the simpler pieces we need, and it might help to think about the forward reaction a bit so we don't get way off track. So let’s leave a little gap in our retrosynthesis while we consider how the two starting materials could react.

Thinking back to Episode 27, carbonyl groups contain a dipole, where carbon is electrophilic. And if we want to form carbon-carbon bonds with aldehydes or ketones, we can use another organometallic reagent, and turn our bromide into a nucleophile. Organometallic reagents react with aldehydes and ketones to produce alcohols, so let’s add that last structure to our retrosynthetic analysis – labeling a second FGI and a second disconnection to finish it up.

Now, we can use this retrosynthesis to fill in the reaction details going forward! One of our starting materials, bromobenzene, can react with magnesium metal to form a Grignard reagent. That nucleophilically attacks our other starting material, the aldehyde, making an alkoxide.

And an aqueous solution of hydrochloric acid protonates the alkoxide, forming our alcohol. To make our alcohol into a bromide we can use P-Br-3, like we did in the first short synthesis problem. Now, it’s time to make another Grignard reagent and react it with carbon dioxide, adding our carbon atom.

Once we add some aqueous hydrochloric acid, we form our carboxylic acid. Finally, we need to make the carboxylic acid into something more reactive to form the amide– an acid chloride or ester will work. So let’s use the Fisher esterification reaction, and then react that ester with dimethylamine to make our amide.

And we’ve done it! Phew! We don’t always show solvents in these reaction schemes, but they're really important.

Say we put an inorganic salt, like table salt, into a fairly nonpolar organic solvent, like oil. The salt won't dissolve, because an organic phase will only dissolve organic chemicals. But an inorganic table salt will dissolve in an aqueous phase, which is just water.

Let's focus on solvents in one part of the synthesis we just did – where we made a carboxylic acid from a bromide. Diethyl ether is an organic phase, dissolving organic reagents like the bromide so the reaction can happen. And the hydrochloric acid solution is an aqueous phase, dissolving inorganic magnesium salts hanging around (and protonating our carboxylic acid).

Many organic solvents are immiscible with an aqueous solvent. They form separate layers, like the oil and vinegar in salad dressing, or this mixture of oil and water. After I shake it up, it'll eventually separate into two layers again, with the oil floating on top.

So if we did this carboxylic acid synthesis in a lab, we could separate our two immiscible solvents through liquid-liquid extraction. The carboxylic acid, an organic compound, is dissolved in the diethyl ether, while inorganic stuff is dissolved in the aqueous phase. But that's just a start – we have to remove any side products too!

We didn't show this earlier, but any Grignard reagent that didn’t react with carbon dioxide can also react with water to form butylbenzene as a side product. And we can remove this with extraction. To find out how, let’s put on our safety glasses and head to the thought bubble lab:.

To separate the aqueous and organic phases through liquid-liquid extraction, we’ll use a separatory funnel. It has an opening at the top to pour in liquid, and a valve or stopcock at the bottom that lets us open and close the funnel. When we’re done running our reaction, we can transfer our mixture of organic phase, aqueous phase, and various wanted and unwanted compounds into a separatory funnel suspended by a ring stand.

Next, we pop a stopper in the top, pick up the funnel, and flip it upside-down. Sometimes the mixing liquids builds up some pressure, so – with the funnel still inverted – we’ll open the stopcock briefly to let gas out, or vent the funnel. Then, we’ll shake the funnel to make sure everything gets dissolved in the correct phase, venting frequently for safety.

And we'll put it back in the ring stand and wait. Once there's a clear separation between the organic and aqueous phases, we can drain off the more dense, bottom aqueous layer to remove the magnesium salts with the HCl solution. We're left with two organic compounds in the ether solution, but we only want the carboxylic acid.

To separate them, we can add a sodium hydroxide solution to the funnel, which deprotonates the carboxylic acid, making a polar salt. As long as this salt doesn't contain too many carbon atoms it’ll be water-soluble, so by shaking the funnel again, we can force it into the aqueous phase. And the butylbenzene just sits around in the organic layer, since it’s a neutral compound.

So, finally, we can drain off the aqueous layer with our precious carboxylic acid salts, and protonate them. If they're solid, we can just filter them out of the water. Otherwise, we can re-extract our hard earned carboxylic acid into the ether again.

Then, we’ll dry and evaporate off our organic solvent. Thanks, lab bubble! With our purified carboxylic acid in hand, we can move onto the next step of our synthesis reaction.

Synthesis and retrosynthesis road maps on paper show key landmarks, but not the extra purification steps we do in the lab. That's true for our Mold Medicine Map, too. So far, we've been filling in pieces of Dr.

Sheehan’s forward synthesis of penicillin V, but we can look at the whole process with the retrosynthesis skills we learned today. When we’re building a molecule from nature, we’re not given the starting materials, which makes reverse engineering so useful. Whether we’re building an Enigma machine or making complex natural products, the last part we install should be the most sensitive.

You don’t want to be hammering together the sides of the Enigma machine with the delicate wiring of the plugboard already in place! Penicillin was so difficult to isolate because the strained, four-membered beta lactam ring is super sensitive to hydrolysis – it broke open when chemists tried to purify it from the fungus. So, in our retrosynthesis map, that's the first disconnection and the last bond we’ll make in our synthesis!

The other amide in the structure is next. Splitting a carboxylic acid derivative down the middle into two fragments is a common disconnection. A complex, but very clever disconnection comes next.

This reaction uses an iminium ion to form two bonds in one synthetic step from this aldehyde and this amine. And finally, we trace our synthesis back to a compound that we can easily purchase from a company, the amino acid valine. We’ll look at more detailed forward reactions of this penicillin synthesis in future episodes, but retrosynthesis is a good way to picture where Dr.

Sheehan started and where he ended up. In this episode, we:. Practiced multistep synthesis problems.

Applied retrosynthesis to make more complex molecules and. Applied liquid-liquid extraction to a purification step. In the next episode, we’ll take a break from products made by nature to explore some of the most important synthetic polymers humans have created.

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.