Chapter 7: Delocalization and Conjugation

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Have you ever looked at, say, a bright red tomato or maybe felt the fabric of your clothes and wondered what actually makes them that way?

Red.

Strong.

It often seems quite abstract.

It does, but often the secret is hidden deep down at the molecular level.

It's about how electrons

arrange themselves.

Exactly.

And that's what we're diving into today.

Welcome to the deep dive.

We're taking a focused look into a really fascinating area of organic chemistry,

delocalization, and conjugation.

Key concepts.

Yeah.

We're pulling insights.

Some surprising facts too, straight from chapter seven of Clayton, Greaves, and Warren's Organic Chemistry, the second edition.

We want to unpack how electrons being shared around profoundly changes things.

Things like stability, how molecules react, even their color, as you mentioned.

Right.

Our goal here is to make these, let's face it, sometimes dense ideas clear and hopefully compelling.

So you understand not just what they are, but really why they matter in the world around us.

Okay.

Where should we start?

Okay.

Let's unpack this by starting simple.

Ethene.

C2H4.

Probably the most basic unsaturated compound.

Right.

Plantar molecule.

Flat.

Flat.

The carbons are CEPI2 hybridized.

You've got that strong sigma bond between them.

But the really interesting part for today is that pi bond.

Formed by the sideways overlap of the pi orbitals.

Yeah.

Above and below the plane.

Exactly.

And that gives rise to the bonding pi orbital and the antibonding pi star orbital, these different energy levels the electrons can be in.

Understanding those is kind of the foundation here.

Absolutely key.

So, okay, that's one double bond.

What happens when you start linking them up?

Especially in a ring.

Which brings us, naturally, to benzene.

Ah, benzene.

C6H6.

An icon.

An icon, but also a real headache for chemists for a long time after it was discovered back in, what, 1825.

What was the big puzzle?

Well, what's fascinating here is that Kekaly came up with this idea of alternating single and double bonds in a six -membered ring.

A brilliant proposal for the time.

But it didn't quite fit, did it?

No.

Because if it really flipped back and forth and resonated between two structures like he drew them, you'd expect, for example, two different versions of 142 -dibromodenzene, depending on whether the bonds between the bromines was single or double in that snapshot.

And they only ever found one.

Only one.

Yeah.

Ever.

So that oscillation idea couldn't be right.

The structure had to be something else.

So molecular orbital theory to the rescue?

Pretty much.

The six p orbitals from the six carbons, they don't just form three separate pi bonds.

They all mix together, combine to form six completely new molecular orbitals that spread over the entire ring.

So the electrons aren't localized in specific double bonds anymore.

Exactly.

They're delocalized, spread out over all six carbons.

You get this sort of doughnut of electron density above and below the ring plane.

Benzy doesn't switch structures.

It is a single average structure with these delocalized electrons all the time.

It's like a constant state, not a flicker.

And is there solid proof for that?

Can we like see it experimentally?

Oh, absolutely.

Electron diffraction studies are really clear.

Benzene is a perfect regular hexagon.

All the carbon bonds are exactly the same length.

What is that length?

It's 139 .5 picometers, which is really interesting because a typical single cc bond is longer, about 154 p .m., and the typical double cc bond is shorter, around 134 p .m.

So benzene is right in the middle.

Perfectly intermediate.

That single average bond length is direct physical proof that the electrons are smeared out evenly.

Right.

And yet, when we draw reaction mechanisms, we often still draw those alternating bonds in benzene, don't we?

We do, yeah.

It's a practical way to keep track of electrons when figuring out reactions.

But we should really use that special double headed arrow, the delocalization arrow, to remind ourselves it's not a real reaction or movement.

It's just showing resonance contributors to the true averaged structure.

Okay, so that helps distinguish things.

Conjugation is about how those double bonds are linked up, the structure, the sequence.

Right, like beads on a string, needing that alternating pattern.

And delocalization is about the electrons themselves being spread across that conjugated system.

Precisely.

The electron behavior.

So this doesn't just happen in rings.

What about straight chains,

like hexatrine?

Good example.

Hexatrine.

Six carbons, three double bonds, no ring.

It's also essentially planar, which is important for overlap.

But it's not quite benzene, is it?

No.

Unlike benzene, its bond lengths aren't all identical.

The double bonds are slightly different.

The single bonds have some double bond character, but they aren't averaged out in the same perfect way as benzene.

You still see evidence of that p orbital overlap across the whole chain, though.

And that overlap even makes it harder to rotate around the single bonds.

It does.

There's some partial double bond character there, restricting rotation, keeping it planar for better conjugation.

So what's the absolute rule for conjugation?

How close do the double bonds need to be?

The rule is very specific.

The double bonds have to be separated by exactly one single bond.

That's the arrangement that allows the p orbitals on adjacent carbons to overlap side by side continuously.

What if they're further apart, like in, say, some fatty acids?

Right, like a acid.

It's called polyunsaturated.

But the double bonds are separated by sp3 carbons involved only in single bonds.

Those sp3 carbons don't have p orbitals available for this kind of pi overlap, so they act like insulators.

Breaks the conjugation.

Okay.

And what about something like allene, two double bonds right next to each other?

Allene is CCC.

You'd think maybe they conjugate, but the two pi bonds are actually perpendicular to each other.

Imagine one going up and down, the other front to back relative to the central carbon.

So they can't overlap sideways?

No way.

Geometry prevents it.

So no conjugation there either, despite having adjacent double bonds.

It needs that single bond spacer and the right alignment.

Okay.

Here's where it gets really interesting.

Let's dive deeper into the molecular orbitals themselves using butadiene as our guide.

Simplest conjugated dynane.

Four carbons, two double bonds separated by a single bond.

CCC.

So it has four atomic p orbitals involved in the pi system.

What happens when they combine?

Those four atomic p orbitals mix to form four new molecular orbitals.

We usually call them CI1, PS2, PSI3, and PSI4 in order of increasing energy.

And how do these new orbitals affect butadiene's properties?

Let's start with stability.

Okay.

Stability.

But tadagene has four pi electrons.

They fill the two lowest energy MOs, TiO1 and SiO2, which are both bonding orbitals.

The overall energy of the electrons in these orbitals is lower, meaning more stable than if you just had two separate ethene molecules.

So linking them up through conjugation actually makes the whole system more stable thermodynamically.

Correct.

It's an energetically favorable arrangement, like getting a discount for buying in bulk, sort of.

Huh.

Okay.

Now, what about reactivity?

This is where things get a bit weird, right?

Yeah.

This is often counterintuitive at first, if we connect this to the bigger picture.

Butadiene's highest occupied molecular orbital, the HOMO, is CI2.

This HOMO is actually higher in energy than the HOMO of ethene.

Higher energy means the electrons are easier to remove or donate?

Exactly.

So butadiene is more reactive towards electrophiles, things that want electrons, compared to ethene.

Okay.

That part makes some sense.

More available electrons.

Yeah.

But what about the other side?

Right.

Now, at the lowest unoccupied molecular orbital, the LUMO, which is SI3.

Butadiene's LUMO is lower in energy than the LUMO of ethene.

Lower energy LUMO means it's easier to accept electrons.

Precisely.

So butadiene is also more reactive towards nucleophiles, things that donate electrons.

Whoa, hold on.

So it's more reactive towards electron seekers and more reactive towards electron donors.

How does that work?

It seems paradoxical.

Right.

But it's a direct consequence of how conjugation spreads out the energy levels.

The highest occupied level goes up, making it a better donor.

The lowest unoccupied level comes down, making it a better acceptor.

Delocalization enhances both aspects of its reactivity.

It's a fascinating duality.

That is wild.

Okay.

So this energy gap between the HOMO and LUMO, you mentioned stability and reactivity.

Does it affect anything else like light?

Absolutely critical for light absorption.

The size of that energy gap between the HOMO and LUMO determines the wavelength of light a molecule can absorb to promote an electron from the HOMO to the LUMO.

Smaller gap means?

Smaller energy gap means it takes less energy to excite the electron, which corresponds to absorbing light with a longer wavelength.

Okay.

Give me an example.

Ethene absorbs UV light way down at about 185 nanometers.

Short wavelength, high energy.

Butadiene, with its smaller HOMO -LUMO gap, thanks to conjugation, absorbs at 215 nanometers.

Still UV, but significantly longer wavelength.

I see.

And if you keep adding more conjugated double bonds?

The HOMO -LUMO gap gets progressively smaller and smaller.

The absorption wavelength keeps shifting longer and longer.

Until it shifts right into the visible spectrum.

Exactly.

That's the origin of color in so many organic molecules.

If you have a long enough conjugated system, the HOMO -LUMO gap becomes small enough to absorb visible light.

Like the red tomato example.

Perfect example.

Lycopene, the pigment in tomatoes, has 11 conjugated double bonds.

That extensive system pushes the absorption well into the visible range around 470 nanometers, which is blue -green light.

And because it absorbs blue -green, we see the leftover light, which is red.

Precisely.

Chlorophyll in plants, beta -carotene in carrots, they all rely on highly conjugated systems.

Generally you need about 8 or more conjugated double bonds before the absorption really creeps into the visible range, and the compound appears colored.

So this is mostly electrons jumping from a pi -bonding orbital to a pi -star anti -bonding orbital.

A p to pi -star transition.

That's the main transition type for these polyanes.

Yes, pi to pi -star.

But there's another important one, especially when you have atoms with lone pairs, like oxygen or nitrogen,

involved in or near the conjugated system.

What's that called?

Those are n to pi -star transitions, n for non -bonding.

An electron from a lone pair gets promoted into an anti -bonding pi -star orbital.

These transitions often require even less energy, so they can also contribute significantly to color.

Any common examples?

Think about indigo dye,

the molecule responsible for the color of blue genes.

It has nitrogen atoms with lone pairs that are part of the conjugated system.

Those n to pi -star transitions help it absorb yellow light, making it appear that characteristic deep blue or indigo color.

Okay, so we've seen conjugation in big systems like benzene and long chains like lycopene.

What about smaller scale, like just three atoms involved?

Ah, yes, the allyl system.

Incredibly common and important motif.

It's basically a three -atom pi system.

Let's start with the allyl anion.

How do you make that?

You can take propene, just a simple three -carbon alkene, and react it with a really strong base.

The base plucks off a proton from the methyl group, the CH3 end.

And that carbon becomes negatively charged.

Crucially, it re -hybridizes from sp3 to sp2, so it now has a p orbital available to hold that negative charge, that lone pair.

This new p orbital lines up perfectly with the existing pi bond of the propene.

Now you have three adjacent p orbitals interacting.

Exactly.

Three atomic p orbitals combine to form three molecular orbitals.

PSI -1 bonding, CSI -2 non -bonding, and CI -3 antibonding.

In the allyl anion, you have the two electrons from the original pi bond plus the two from the lone pair, four electrons total.

Where do they go?

They fill the two lowest energy orbitals,

the bonding PSI -1 and the non -bonding PSI -2.

Now, what's really neat about CSI -2 is that the math shows it has a node, zero electron density,

right on the central carbon atom.

So the extra electron density, the negative charge, isn't on the middle carbon.

Nope.

It's delocalized, but it's primarily concentrated on the two end carbons of the atom system.

The charge is shared between C1 and C3.

That sounds more stable than having it all stuck on one carbon.

Can we actually see this charge distribution?

We can, using carbon -13 NMR.

You look at allyl lithium, which is essentially the allyl anion with a lithium counterion, the NMR spectrum is very revealing.

The middle carbon shows up around 147 ppm.

Pretty typical for double bond carbon.

And the end ones.

The two end carbons are identical by symmetry, and they resonate way upfield at about 51 ppm.

That low number means they're highly shielded.

More shielded means more electron density around them.

Exactly.

It's a direct experimental confirmation that the negative charge density is piled up on those end carbons, just like the MO picture predicts.

Cool.

Okay, what if we go the other way?

Instead of adding electrons, we make an allylocation positive charge.

Same principle, just fewer electrons.

You could start with,

say, allyl bromide.

If the bromide leaves, taking its electrons with it, it leaves behind a positive charge on that carbon.

And that carbon also has an empty p orbital now.

Yes, it becomes spec 2 hybridized with an empty p orbital.

So again, you have three interacting p orbitals forming the same psi i1, psi i2, psi i3 molecular orbitals.

But this time, you only have the two electrons from the original double bond.

So they just go into the lowest bonding orbital, psi 1.

Correct.

Psi 1 is filled, psi 2 and psi r3 are empty.

And just like the anion, the molecular orbitals dictate that the positive charge, the lack of electrons, is delocalized, primarily over the two end carbons.

And I bet NMR shows that too.

It does.

For allylocations, the end carbons show up at extremely low field, maybe around 224 ppm, very deshielded, indicating a strong positive charge character on those end atoms.

So this three atom delocalization pattern is pretty common then?

Extremely common.

Think about a carboxylate anion like acetate,

or COO-, when a carboxylic acid loses its proton.

That looks like C double bond O single bond O minus.

Right.

But that negative charge isn't stuck on one oxygen.

It's delocalized across the OCO system.

It's perfectly analogous to the allyl anion.

Three atoms,

four pi electrons, from the CO pi bond and the O lone pair.

And does that affect the bond lengths?

It sure does.

In the carboxylate ion, both carbon -oxygen bonds are identical in length, about 136 picometers,

perfectly intermediate between a typical CO double bond and a CO single bond.

Proof of delocalization?

Any other examples?

The nitro group, NO2.

It's isoelectronic with carboxylate, same number of electrons, similar arrangement, but it's neutral overall.

Still has significant delocalization.

Just have to remember nitrogen usually forms three bonds, not five like the resonance structures might imply.

Okay.

And what about amides?

You see those everywhere, especially in biology.

Amides are crucial.

Proteins are held together by amide bonds.

Nylon is a polyamide.

In an amide, you have a nitrogen atom attached to a carbonyl group.

And the nitrogen has a lone pair.

Yes.

And that lone pair can participate in conjugation with the carbonyl pi bond.

For this to happen effectively, the nitrogen atom needs to be species 2 hybridized and planar, aligning its p orbital containing the lone pair with the p orbitals of the carbonyl group.

So what are the consequences of this amide delocalization?

It sounds important.

Oh, profoundly important.

First, it gives the carbon -nitrogen bond significant partial double bond character.

Meaning it's harder to rotate around that CN bond.

Much harder.

Way more than a typical CN single bond.

Take NN dimethylformamide, DMF.

It takes about 88 kilojoules per mole of energy to force rotation around that CN bond.

That's a substantial barrier.

Can you see that effect?

You can.

In the 13CNMR of DMF at room temperature, the two methyl groups attached to the nitrogen actually show up as two separate signals.

Because the rotation around the CN bond is low on the NMR time scale, they're in different average environments.

Wow.

What else does amide delocalization do?

It makes the carbonyl oxygen more electron rich than it would be otherwise, and the nitrogen less electron rich, or less basic.

It also significantly stabilizes the entire amide group.

So what does this all mean for how amides react?

Several things.

The carbonyl carbon in an amide is less electrophilic, less reactive towards nucleophiles, compared to say a ketone or an ester.

And because the nitrogen's lone pair is tied up in delocalization, the nitrogen is much less basic than a typical amine nitrogen.

So if you try to protonate an amide, it doesn't go for the nitrogen easily.

Often, protonation will actually occur on the carbonyl oxygen because it's become more electron rich.

Due to that resonance contribution from the nitrogen, it flips the expected basicity site.

It's a really key aspect of aminamide chemistry.

Okay, this theme of delocalization leading to extra stability.

Let's loop back to benzene one more time, because its stability is really off the charts, isn't it?

It really is.

We talked about the structure, but its reactivity is the dead giveaway.

Normal alkenes, like cyclohexene, react easily with bromine, by addition the double bond breaks.

Bromines add across it.

But benzene doesn't do that.

No.

Benzene needs a catalyst, like iron bromide, to react with bromine.

And even then, it undergoes substitution, not addition.

A hydrogen atom gets replaced by a bromine atom, but that stable ring system stays completely intact.

It fights hard to keep its special electron arrangement.

It has that shield we mentioned.

Pretty much.

Yeah.

Compare that to cyclooctatrain, COT.

It's cyclic.

It has alternating double bonds, C8H8.

So you might think it's like a bigger benzene.

You might, but it's completely different.

First, it's not planar.

It adopts this floppy, tub -like shape.

The key orbitals can't overlap properly all the way around.

So no continuous conjugation.

Nope.

And chemically, it behaves just like a normal alkene.

Reacts readily with bromine via addition, breaking the double bonds.

No special stability.

No aromatic shield.

Is there a way to quantify benzene's extra stability?

Yes.

Heats of hydrogenation are perfect for this.

When you add hydrogen across a double bond, it releases energy.

For cyclohexene, one double bond, it releases about 120 kilojoule.

So for benzene with three double bonds, you'd expect three times 120, which is 360 kilojoule.

That's what you'd predict if they were just three normal, isolated double bonds.

But when you actually hydrogenate benzene, it only releases about 208 kilojoule.

Wow.

That's way less.

About 150 kilojoule less stable than expected, or rather, 150 kilojoule more stable.

Exactly.

It's 150 kilojoule more stable than predicted.

That difference is called the aromatic stabilization energy, or resonance energy.

It's a huge amount of extra stability.

And cyclooctatrain, does it show any extra stability?

None at all.

Its heat of hydrogenation is pretty much exactly what you'd expect for four isolated double bonds.

Confirms it's not aromatic.

This raises an important question.

Why?

Why is benzene so special?

Why does having six pi electrons in that ring confer so much stability?

Is it just the number six?

It's largely the number six in this case.

It comes down to the pattern of the molecular orbitals.

Benzene's six pi electrons completely fill all of its available bonding molecular orbitals, with none left over for nonbonding or antibonding orbitals.

This creates a stable closed shell electron configuration.

Very favorable.

Like noble gases having filled electron shells in atoms.

That's a great analogy.

It's like a noble gas configuration for the pi system.

Now, COT has eight pi electrons.

If it were planar, those eight electrons wouldn't create a stable closed shell.

Two electrons would end up in nonbonding orbitals, which isn't nearly as stable.

That's partly why it avoids being planar.

But you mentioned COT can become aromatic.

Yes.

If you add two electrons to COT to make the dianion, ten pi electrons, or remove two electrons to make the cation, six pi electrons, then these ions are planar and show aromatic stability.

Because six and ten fit the magic number rule.

Okay, so what is the magic number rule?

It's Huckel's rule.

It states that for a system to be aromatic, it needs to be one, cyclic, two, planar, three, fully conjugated all the way around the ring, and four must have four n plus two pi electrons, where n is zero, or any positive integer, zero one two three.

So four n plus two gives you two n zero, six n one, ten n two, fourteen n three, and so on.

Those are the aromatic numbers.

Those are the Huckel numbers for aromaticity.

Benzene has six n one.

The Cot dianion has ten n two two.

The Cot dication is six n one.

They fit.

And what about systems with four n pi electrons, like four eight twelve?

Those are predicted by the theory to be anti -aromatic if they stay planar.

Anti -aromatic systems are generally unstable, often highly reactive, or they distort from planarity to avoid that unfavorable electron configuration, like neutral Cot does.

Does this apply to bigger rings, too, these anulines?

It does.

For example, ten anuline has ten pi electrons that fits four n plus two n two.

But the molecule is too strained to be perfectly planar.

The hydrogens inside the ring bump into each other.

Since it's not planar, it's not aromatic.

Ah, so planarity is a strict requirement.

Absolutely.

But then you look at eighteen anuline.

It has eighteen pi electrons, n four, which fits the rule.

And this larger ring is big enough to be essentially planar.

And indeed, eighteen anuline shows aromatic properties.

Fascinating.

And this extends beyond simple hydrocarbon rings.

Oh, yes.

Fused rings like naphthalene to benzene rings stuck together are aromatic, though slightly less stable per ring than benzene itself.

And very importantly, heterocyclic compounds rings containing atoms other than carbon, like nitrogen or oxygen or sulfur.

Like pyridine.

Pyridine is a classic example.

It's like benzene, but with one CH replaced by nitrogen.

It has six pi electrons, fits Huckel's rule, planar, aromatic.

The nitrogen's lone pair sits in an sp2 orbital in the plane of the ring, not part of the aromatic pi system.

What about pyrrole?

Another nitrogen ring, but five -membered.

Pyrrole is also aromatic with six pi electrons.

But here, the nitrogen atom does contribute its lone pair into the pi system to reach that magic number six.

So the lone pair is delocalized as part of the aromatic cloud.

So wrapping this up then, it feels like this idea of electron delocalization, electrons being shared across multiple atoms in these pi systems.

It's not just some minor detail.

Not at all.

It's absolutely fundamental.

It dictates so much about a molecule structure, its stability, how it reacts, even as we saw its color.

It really explains a huge range of properties we observe, from why benzene is so weirdly unreactive in some ways, to why tomatoes are red, to why amides behave the way they do in proteins.

Exactly.

And understanding these molecular orbital interactions, how conjugation and aromaticity work, gives chemists incredible power.

We can predict how molecules will behave, and even start to design new molecules with specific properties in mind.

It's amazing that something so seemingly abstract, the way electrons dance around in orbitals, can have such concrete effects we see every day.

The stability of benzene, the function of a protein, the color of dye.

It connects the microscopic quantum world to the macroscopic properties we experience.

So here's a final thought.

If understanding electron delocalization allows us to explain the color of a tomato or design a stable drug molecule,

what other everyday mysteries might be unlocked by looking at these invisible electron dances?

What phenomena are out there just waiting for us to understand the underlying molecular choreography?

Makes you think, doesn't it?