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

The German chemist August Kekulé is one of those people who’s often considered a founder of modern organic chemistry. In a speech at a symposium in 1890, he told the  audience that, thirty years earlier, he was dozing near a fire in Ghent, Belgium when the image  of a snake devouring its own tail came to him. This dream, he said, was what led him  to the chemical structure of benzene: a flat ring of six carbon  atoms with the formula C6H6.

This story has been repeated widely,   but some scientists have questioned it. In 1854, for example, an article was published in a French journal that  showed benzene as a hexagonal ring. The author was Auguste Laurent,   but the work was actually published after  his death by some of his colleagues.

Laurent is probably most famous for  his work developing the functional group nomenclature that we still use today – that whole business of alkanes ending  in “ane”, alkenes in “ene,” and so on. Other scientists have speculated that Kekulé chose to describe his “revelation” as a dream to avoid sharing credit with his colleagues! Perhaps that’s true, but for now it’s  his name that we find in textbooks.

In any case, I'm not here  to debate science history. So let’s take a look at what makes benzene, and other aromatic compounds, so special. [Theme music]. Benzene is drawn in a couple different ways.

We can think of it with alternating  single and double bonds,   or we can imagine the pi bonds  “smeared” across the whole molecule,   and represent that resonance hybrid with  a dotted line or a circle in the middle. As the statistician George Box famously said,  “all models are wrong, but some are useful.” And that’s definitely the case here! Neither diagram is completely right or  completely wrong, but both have their uses.

The “alternating bonds” version is  helpful when we’re trying to show which way electrons move in mechanisms. It’s also better if we’re trying to  draw polycyclic aromatic hydrocarbons – molecules with joined-up aromatic rings. On the other hand, the "smeared"  version more accurately represents the electron distribution in the benzene ring.

You see, carbon-carbon double bonds  are typically 0.134 nanometers long,   while single carbon bonds are slightly longer. So, if you drew out Kekulé’s model,  you’d expect a sort of distorted hexagon. But in a famous paper published in 1929,   the English crystallographer Kathleen  Lonsdale showed that benzene rings are flat,   symmetrical around a central point, and  the bonds were all the same length.

Plus, in chemical reactions, benzene doesn't react like an alkene with some double and single carbon bonds. For example, alkenes react  easily with molecular bromine – the bromines add with anti-addition  across the double bond. This doesn’t happen with benzene.

We need harsher conditions or a catalyst to  make anything happen, and even when it does,   only one bromine atom adds  and we keep the pi bonds. It turns out that the electron distribution in benzene actually makes it less nucleophilic than a simple alkene, thanks to conjugation. Essentially, in a molecule with alternating  double and single carbon-carbon bonds,   the p orbitals overlap, giving all of  the bonds partial double-bond character.

And because benzene is a ring, the p  orbitals on each carbon atom align. We get something kind of like a hamburger – the pi bonds are the “bread” and  the sigma bonds are the “meat." There are six pi electrons,  one for each p orbital,   and they’re evenly distributed around the ring. So we say the pi bonds in benzene are delocalized.

Overall, a pi bond in benzene  has lower electron density than in an alkene, where electrons are  localized between two carbon atoms. And this explains the lower reactivity – with delocalized electrons, benzene is less  effective at polarizing other molecules! Benzene is part of a family of aromatic compounds.

The name comes from the fact that many of these  substances were first extracted from smelly stuff. For example, the smell of wintergreen  leaves comes from methyl salicylate,   while the smell of cinnamon  comes from cinnamaldehyde. Besides their scents, aromatic compounds  all share four key characteristics.

First, aromatics are cyclic – they all have a ring structure. Second, their rings are planar,  which means they're flat. Third, they have conjugation  throughout the entire ring – the "smeared", continuous stretch of delocalized  pi electrons we’ve been talking about.

And finally, they follow Hückel’s rule. This rule is named after German  chemist and physicist Erich Hückel,   who noticed that if the number of pi  electrons in a compound equals 4n+2,   where n is an integer, then  the compound is aromatic. For example, let's say n equals 1 – so, 4 times 1, plus 2, equals 6 pi electrons.

Benzene has 6 pi electrons, and it’s aromatic! And if n equals 2, we get 4 times  2, plus 2, equals 10 pi electrons. That works for naphthalene,  which is also aromatic!

By the way, if we drew naphthalene with  circles in the middle of each ring,   it sort of implies two pi bonds in the  middle where the rings are joined – which isn't the case. So this is why it's better to use the "alternating  bonds" model when we're drawing polycyclics! Now, not all aromatic compounds  are perfect 6-carbon hexagons.

We see aromaticity in certain  ions, too, like a tropylium cation: a 7-membered ring with a positive charge. It has an empty p orbital, and since we can draw resonance structures that delocalize the positive charge and the six pi electrons around the ring, the pi electron system is continuous. So it's cyclic, planar, conjugated, and  with those 6 pi electrons, like benzene,   it follows Hückel’s rule.

It's aromatic! And tropylium salts, such as tropylium  bromide, are remarkably stable.   The cyclopentadienyl anion  is an aromatic ion, too. A strong base can deprotonate cyclopentadiene,  putting a pair of electrons into a p orbital.

The anion that results is cyclic,   planar, has a continuous series of  p orbitals, and has 6 pi electrons! We also have heterocyclic  compounds with aromaticity,   which are rings that include  an atom other than carbon. For example, pyridine is C5H5N.

It's cyclic, planar, and has  6 delocalized pi electrons,   one of which is coming from  the nitrogen, so it’s aromatic! Nitrogen also has a lone pair in an sp2 orbital,  but it’s in the same plane as the sigma bonds. You know, the meat part of the molecule “burger”!

And because of the trigonal planar  geometry, the lone pair of electrons can’t be delocalized into the ring (or  burger bun) and break Hückel’s rule. In fact, this lone pair is  available to accept a proton,   which is why we’ll see pyridine used  as a base in some organic reactions. By contrast, the geometry of this heterocyclic compound pyrrole means that its nitrogen lone pair does contribute to the pi system.

Just like the cyclopentadienyl anion, we  end up with 6 delocalized pi electrons,   and pyrrole is aromatic, too. BUT this lone pair is all tied up with  its buddies, making an aromatic pi system. So pyrrole can't be used as a base like pyridine.

The same goes for furan, only there's  an oxygen instead of a nitrogen. Because of this compound's geometry, only one of the oxygen lone pairs is incorporated into the ring system, so we've still got 6 delocalized pi electrons. And this makes the oxygen atom sp2 hybridized.

So far, we've looked at  lots of aromatic compounds,   but plenty of cyclic compounds  don't quite make the cut. For example, cyclo-deca-penta-ene follows  Hückle’s rule with 10 pi electrons,   but it's not aromatic. If it were planar, a 10-membered  ring would have 144º angles,   which is much larger than the comfy  120º angles of sp2 hybridized carbons.

So this ring puckers to become non-planar,  breaking the overlap between the p orbitals. No conjugation, no aromaticity – sorry! There are also some cyclic, planar, conjugated  compounds that don’t follow Hückel’s rule.

They look good at first, but then  you realize something is a bit off – like creepy alien clones! These are antiaromatic compounds, and  tend to be very unstable with 4n electrons (where n is still an integer). One antiaromatic example is  cyclobutadiene, with 4 pi electrons.

Or there's pentalene, which looks a little bit  like naphthalene, but it has 8 pi electrons. But while this 8-membered cyclo-octa-tetra-ene gets close to being antiaromatic because it's cyclic and has 8 pi electrons,    it's not planar. Because it’s large enough to bend, it adopts  a “tub” shape to avoid antiaromaticity.

In the lab, when you have some stuff  in a flask,. Nuclear Magnetic Resonance can be a great way to find out  whether a molecule has aromaticity. We met NMR in episode 26, so rewatch  that video if you want a refresher.

But we'll try to ease back into  reading spectra with an example! This is the proton NMR spectrum of 4-chlorophenol,   which is a benzene ring with a Cl  on one side, and an OH at the other. There are 3 peaks here – and remember, each peak represents a  proton, or group of protons, in a molecule. we can see that the peak at 9.5 ppm with an integral of 1 is the one OH proton on the phenol – easy!

Now, the protons attached to a benzene  ring are a little less downfield – they're between 8.5 and 6.5 ppm or so. Because of the way the pi orbital system  interacts with the magnetic field,   these protons are quite deshielded. And there are 2 peaks in this region, each with  an integral of 2, which represents 2 protons.

But looking at the structure of 4-chlorophenol,   we can see there are four  protons on the aromatic ring. We only see two peaks in the NMR spectrum  because this compound is symmetrical – if you put a mirror though the middle of the ring,   the protons are effectively  the same on both sides. In other words, they're chemically  equivalent to each other.

Specifically, the two protons near the  chlorine are equivalent to each other,   and so are the two protons near the hydroxide. The chlorine and hydroxide have slightly  different shielding effects, which is why we see 2 peaks, each with an integral of 2. Now that we've done some practice,   let's try to figure out an aromatic  compound based on the proton NMR spectrum.

We know its formula: C10H12O. And we have an IR spectrum with a carbonyl  stretch, so we know it's an aldehyde or ketone. So let's dig into these NMR peaks and integrals!

By looking at our NMR table, we know that an  aldehyde proton would appear between 9 and 10 ppm. There's no peak there, so it looks  like our mystery compound is a ketone. Next, let’s look at the  huge peak over on the right,   with an integral of 6 – that’s weird, right?

One carbon can't have 6 protons attached! But! We know that chemically equivalent  protons can all combine into 1 peak.

So there has to be some symmetry going on  here, and we just need to puzzle it out. 6 hydrogen atoms can come from the protons  on 2 chemically equivalent methyl groups,   which could be attached to the  same atom – let’s try a carbon! So here we have an isopropyl group  with 6 chemically equivalent protons,   because there’s a line of  symmetry though the middle. If we look a little downfield, there’s a  peak at 3.25 ppm with an integral of 1.

It looks like a CH group, but  it’s split into… well, lots. Zooming in, we can count 7 split peaks. If the n+1 rule gives us 7, then n has to be 6,  and there must be 6 protons on adjacent carbons– which fits with the isopropyl  group we just deduced!

So on one end of our mystery ketone,  we have a CH group with two CH3 groups. And at the other end, there's  a substituted benzene. Time to tackle the benzene ring  peaks between 7.5 and 8 ppm!

The integrals suggest we have 2 sets  of 2 chemically equivalent protons,   and 1 proton that’s a little different. Adding those up, we know there are 5 protons  sticking off the benzene ring of this compound,   so it only has one non-proton  thing substituted onto it. We'll just represent that as R for now!

From our NMR spectrum, we know our puzzle pieces: a benzene ring with 1 substitution,   an isopropyl group, and the ketone  that our IR spectrum told us about. Putting them together, we get... isobutyrophenone. We used what we've learned about aromatic  compounds and identified one using NMR – go us!

In this episode, we saw that:. Aromatic compounds are cyclic, planar,  conjugated, and satisfy Hückel's rule,. The aromatic ring system is less  reactive than simple alkenes, and.

Symmetry leads to chemically  equivalent protons in NMR spectra. We also briefly mentioned that  a halogen like bromine adds to   a benzene ring in a different  way than a plain old alkene. And we’ll dive into how next time, when we look  at electrophilic aromatic substitution reactions.

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.