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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.
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