Summary — What This Chapter Covers

1. Benzene — Structure & Stability

The Kekulé Problem

Back in the 1800s, Kekulé proposed that benzene (C₆H₆) was a hexagonal ring of carbon atoms with alternating single and double bonds — basically three "mini alkenes" fused into a ring. It's a neat idea, and it does explain the molecular formula. The problem is: if that were true, benzene should behave like an alkene. Alkenes react eagerly with things like bromine water via electrophilic addition, because their C=C double bond is a small, localised pocket of high electron density that practically begs an electrophile to attack it.

But benzene doesn't decolourise bromine water on its own. It just sits there. So either the Kekulé structure is wrong, or something else is going on. Spoiler: the structure is wrong (or at least incomplete).

The Real (Delocalised) Structure

Here's the modern picture. Every carbon in the ring is sp² hybridised: it forms three σ (single) bonds — two to neighbouring carbons and one to a hydrogen — all lying flat in the same plane, with bond angles of 120°. That accounts for the "hexagon" shape you always draw.

But each carbon also has one leftover electron sitting in a p orbital, sticking up above and below the plane of the ring like a little dumbbell. Because all six of these p orbitals are right next to each other, they don't stay as six separate, isolated electrons — they overlap sideways with their neighbours and merge into one continuous cloud of electron density that stretches all the way around the ring, forming a "doughnut" above the plane and a matching one below it.

This is called a delocalised π system. Instead of three fixed double bonds sitting in three fixed places, you have six electrons smeared evenly around all six carbons. That's why we draw modern benzene as a hexagon with a circle inside it — the circle represents this delocalised ring of electron density, not any specific bond.

The Four Pieces of Evidence

You need to know four separate experimental clues that all point to delocalisation. Examiners love asking you to explain why a piece of data supports the delocalised model over Kekulé's structure — so understand the logic, not just the numbers.

① Enthalpy of hydrogenation. Hydrogenating one C=C double bond (like in cyclohexene) releases −120 kJ mol⁻¹. If Kekulé's benzene really had three independent double bonds, hydrogenating all three should release roughly three times that:

But the actual measured value for benzene is only −208 kJ mol⁻¹ — about 152 kJ mol⁻¹ less exothermic than predicted. Less energy released means the starting molecule (benzene) was already in a lower energy, more stable state than three isolated double bonds would be. That "extra stability" is exactly what delocalisation provides — this difference is often called the "delocalisation energy" or "resonance energy."

② Carbon–carbon bond lengths. A single C–C bond is 154 pm long; a double C=C bond is shorter, at 134 pm (more shared electron density pulls the atoms closer together). If Kekulé were right, X-ray diffraction of benzene should reveal two different bond lengths alternating around the ring. Instead, every single C–C bond in benzene measures 140 pm — right in between the single and double bond lengths, and all identical. That's only possible if the bonding is uniform all the way round, i.e. delocalised.

③ Resistance to bromine water (saturation test). Cyclohexene decolourises bromine water instantly via electrophilic addition, because its localised π bond can polarise the Br₂ molecule. Kekulé's benzene, with three double bonds, should do this even more readily. But real benzene does not decolourise bromine water at all — because the delocalised ring has no small pocket of concentrated charge to attract and polarise a bromine molecule the way a genuine C=C bond does.

④ Infrared spectroscopy. A genuine C=C double bond (as in cyclohexene) gives an IR absorption peak around 1650 cm⁻¹. If benzene had three real double bonds, you'd expect to see that same peak. Instead, benzene shows peaks around 1450, 1500, and 1580 cm⁻¹ — a completely different pattern, characteristic of the delocalised C···C bonds found specifically in aromatic rings.

2. Benzene Reactions

Combustion

Like any hydrocarbon, benzene burns in oxygen to give CO₂ and H₂O:

Because benzene has a high carbon-to-hydrogen ratio and needs a lot of oxygen to burn completely, it's easy for combustion to be incomplete — leftover unburnt benzene and soot particles produce a characteristic smoky, yellow, sooty flame.

Halogenation

Because there's no isolated, exposed double bond for a halogen to polarise, benzene cannot undergo electrophilic addition the way alkenes do. To react with a halogen, you need a halogen carrier catalyst — such as AlCl₃, AlBr₃, or FeBr₃ — which generates a strong enough electrophile (X⁺) to actually attack the delocalised ring.

Nitration

Reacting benzene with a mix of concentrated nitric acid and concentrated sulfuric acid at 25–60 °C substitutes a nitro group (–NO₂) for a hydrogen atom, producing nitrobenzene.

Friedel–Crafts Reactions

Because benzene's delocalised ring makes it fairly unreactive, chemists need a way to "activate" it as a starting material for building more complex molecules. Friedel–Crafts reactions do exactly that, substituting either an alkyl group (alkylation) or an acyl group (acylation, containing a C=O) onto the ring — both catalysed by AlCl₃ under reflux.

Sulfonation

Warming benzene with fuming sulfuric acid (sulfur trioxide dissolved in concentrated H₂SO₄) at 40 °C for around 30 minutes substitutes a sulfonyl group (–SO₃H) onto the ring, giving benzenesulfonic acid.

3. Electrophilic Substitution — The Mechanism

This is the part examiners test hardest, because you're drawing curly-arrow mechanisms and every single detail matters for the marks. The brilliant thing is: every arene substitution reaction follows the exact same three-step skeleton. Learn the skeleton once, and you can adapt it to nitration, halogenation, or Friedel–Crafts with just a different electrophile.

Mechanism: Nitration of Benzene

Step 1 — Generate the electrophile. Concentrated nitric acid reacts with concentrated sulfuric acid (which acts as a catalyst/acid here). Sulfuric acid protonates nitric acid, which then loses water to generate the electrophile: the nitronium ion, NO₂⁺.

Step 2 — Electrophilic attack. A pair of electrons from the delocalised ring is donated to NO₂⁺, forming a new covalent C–N bond. This breaks the full delocalisation, leaving a positively charged intermediate where only 5 carbons still share delocalised electrons (the horseshoe).

Step 3 — Restore aromaticity. The C–H bond on the carbon that now holds both the NO₂ group and an H atom breaks heterolytically. The bonding pair of electrons swings back into the ring, restoring full delocalisation over all six carbons, and H⁺ is released.

Step 4 — Regenerate the catalyst. The released H⁺ reacts with HSO₄⁻ to reform H₂SO₄, so the sulfuric acid is a true catalyst — used up in step 1, regenerated in step 4.

Mechanism: Halogenation of Benzene (Bromination)

Step 1 — Generate the electrophile. Br₂ alone isn't a strong enough electrophile. The halogen carrier (AlBr₃) accepts a lone pair from one bromine atom of Br₂, forming a dative covalent bond and heterolytically breaking the Br–Br bond to generate Br⁺.

Step 2 — Electrophilic attack. Br⁺ is attacked by a pair of ring electrons, forming a C–Br covalent bond and the same horseshoe intermediate (5 delocalised carbons, + charge, one carbon now bonded to both H and Br).

Step 3 — Restore aromaticity. The complex ion [AlBr₄]⁻ removes the H atom as H⁺, the C–H bond breaks heterolytically, and the electron pair restores full ring delocalisation.

Step 4 — Regenerate the catalyst. AlBr₃ is reformed, confirming it acted as a true catalyst throughout.

Mechanism: Friedel–Crafts Acylation

Step 1 — Generate the electrophile. An acyl chloride (e.g. propanoyl chloride, CH₃CH₂COCl) reacts with AlCl₃, which accepts a lone pair from the chlorine atom, generating an acylium ion (e.g. CH₃CH₂CO⁺).

Step 2 — Electrophilic attack. The ring's electrons attack the positively charged carbon of the acylium ion, forming a new C–C bond and the familiar horseshoe intermediate.

Step 3 — Restore aromaticity. [AlCl₄]⁻ removes H⁺ from the intermediate; the C–H bond breaks heterolytically and the ring's full delocalisation is restored.

Step 4 — Regenerate the catalyst. H⁺ reacts with [AlCl₄]⁻ to reform HCl gas and AlCl₃.

4. Phenol Bromination

Why Phenol Is So Much More Reactive

Phenol is just benzene with an –OH group attached — but that one small change makes an enormous difference to reactivity. Oxygen in the –OH group has lone pairs of electrons, and one of these lone pairs sits in a p orbital that lines up perfectly with the ring's p orbitals. That lone pair overlaps and merges into the delocalised π system, feeding extra electron density into the ring.

The result: phenol's ring has a noticeably higher electron density than plain benzene's, which makes it far more attractive to electrophiles. This –OH group is described as activating, and because of the way the extra electron density distributes around the ring, it specifically directs incoming electrophiles to the 2, 4, and 6 positions (i.e. the positions ortho and para to the –OH group).

Bromination of Phenol

This extra reactivity is dramatic enough that phenol reacts with bromine water at room temperature, with no catalyst needed at all — a stark contrast to benzene, which needs a halogen carrier and heat. All three activated positions (2, 4, and 6) get substituted at once.

The orange bromine water is decolourised, and a white precipitate of 2,4,6-tribromophenol forms. This is actually a classic test used to confirm the presence of a phenol group in an unknown compound.

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