6.1.1 Aromatic Compounds

The Kekulé model has three alternating C–C and C=C bonds. It was an important step historically, but it does not match the bonding in real benzene.

In the delocalised model, all six C–C bonds are equivalent and the six π electrons are spread around the whole ring. On paper, the circle inside the hexagon is the quickest way to show that delocalised π-system.

Each carbon in benzene has a p orbital containing one electron. The six p orbitals overlap sideways all the way around the ring, so the π electrons are not tied to three particular pairs of carbon atoms.

That continuous overlap gives a delocalised π-system above and below the plane of the carbon ring. If you are explaining the model, make the connection explicit: six overlapping p orbitals → six delocalised π electrons.

A Kekulé structure predicts two C–C bond lengths because ordinary single and double bonds have different lengths. Experimentally, all six C–C bonds in benzene have the same length, intermediate between a normal C–C and C=C bond.

So when you use bond lengths as evidence, say all six C–C bonds are equal. Do not say all the bonds in benzene are equal: the C–H bonds have a different length.

Three isolated C=C bonds would be expected to hydrogenate by about −360 kJ mol⁻¹ in total. Benzene gives about −208 kJ mol⁻¹. Its hydrogenation is therefore 152 kJ mol⁻¹ less exothermic than the simple Kekulé prediction.

That tells you benzene starts at a lower energy: the delocalised ring is more stable by about 152 kJ mol⁻¹ in this comparison. You do not need to memorise the numerical values; use the data if the question supplies them.

If benzene really contained three ordinary localised C=C bonds, you would expect much more alkene-like chemistry. The experimental evidence shows a stabilised delocalised π-system, so disrupting that system costs energy.

This is why benzene resists reactions that would add across a C=C bond. Its reactions tend to preserve or restore the aromatic ring, which is the idea you will use again in electrophilic substitution.

For substituted benzene rings, number the ring to give the substituents the lowest possible set of locants, then name the groups in the usual way. OCR can also use familiar aromatic parent names such as methylbenzene and phenol.

Check for equivalent positions when you draw the ring: positions 2 and 6 are the same relative to a single substituent, as are 3 and 5. I’d number the ring lightly before writing the final name if there are several groups.

To nitrate benzene, use concentrated HNO₃ with concentrated H₂SO₄. Sulfuric acid generates the electrophile NO₂⁺ and is regenerated, so it acts as a catalyst.

One H on the benzene ring is replaced by NO₂, giving nitrobenzene. Keep this reaction separate from nitration of phenol later: phenol reacts with dilute nitric acid and does not need concentrated sulfuric acid.

Benzene reacts with a halogen such as Br₂ or Cl₂ in the presence of a halogen carrier. Suitable OCR examples include Fe, FeBr₃, FeCl₃ and aluminium halides such as AlCl₃.

The reaction is substitution: one ring H is replaced by the halogen, giving bromobenzene or chlorobenzene plus HX. Benzene does not rapidly decolourise bromine water on its own in the way an alkene does.

A Friedel–Crafts alkylation joins an alkyl group to the aromatic ring. Use a haloalkane with a halogen carrier; AlCl₃ is the familiar example when the haloalkane is a chloroalkane.

For example, benzene + CH₃Cl gives methylbenzene. The important synthetic result is the new C–C bond between the ring and the alkyl group. If the question uses an unfamiliar haloalkane, keep the carbon skeleton of that alkyl group intact when you attach it.

A Friedel–Crafts acylation uses an acyl chloride with a halogen carrier, commonly AlCl₃. The acyl group, RCO–, becomes attached directly to the aromatic ring and forms an aromatic ketone.

For example, CH₃COCl gives C₆H₅COCH₃. I find it helps to keep the C=O inside the group you attach; losing that carbonyl turns the answer into the wrong reaction.

The useful feature is simple: Friedel–Crafts reactions form a new C–C bond directly to an aromatic ring. That lets you build a larger carbon skeleton from a benzene-type starting material.

When you are planning a synthesis, look for the point where an alkyl or acyl group has appeared on the ring. A haloalkane gives alkylation; an acyl chloride gives acylation. This same C–C bond-forming idea comes back in OCR organic synthesis later in Module 6.

Start by making the electrophile: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. Then draw a curly arrow from the benzene π-system to NO₂⁺. The intermediate has lost full delocalisation and carries a positive charge.

Finally, the C–H bond supplies the electron pair that restores the π-system as H⁺ leaves. HSO₄⁻ removes that proton and H₂SO₄ is regenerated. Check every charge before you finish; a missing + on NO₂⁺ can spoil more than one part of the mechanism.

For OCR, you may assume the electrophile is X⁺ in the halogenation mechanism. Draw the first curly arrow from the benzene π-system to X⁺, then show the positively charged intermediate with the ring only partly delocalised.

The final curly arrow starts from the C–H bond and reforms the π-system as H⁺ leaves. If the question also asks about the halogen carrier, show how it forms the electrophile and is regenerated using the information supplied.

In an alkene, the π electrons are localised between two carbon atoms, so there is a concentrated region of electron density that can polarise Br₂ and attract an electrophile readily.

Benzene spreads its π-electron density around all six carbons. The electron density at any one part of the ring is lower, so benzene is less susceptible to electrophilic attack and needs a halogen carrier for bromination. Make the comparison with the alkene explicit if the question asks for relative reactivity.

With an unfamiliar aromatic substitution, first find the electron-poor species that will act as the electrophile. Then replace one ring H with that group unless the question gives evidence for a different substitution pattern.

Use any reagents, charges or directing information the question supplies. You do not need to have seen the exact reaction before: the ring is the electron-pair donor, the electrophile is attacked, and the aromatic π-system is restored at the end.

The same three mechanism steps work with different electrophiles. The aromatic π-system attacks E⁺, you draw the positively charged intermediate, then loss of H⁺ restores the delocalised ring.

The part you may have to work out is how the reagents make E⁺ or which species removes H⁺. Use the information in the question for that. I’d get the electron movement on the ring right first, then add the reagent-specific chemistry around it.

Phenol is a weak acid, but it is acidic enough to react with aqueous NaOH. The O–H proton is removed and you form sodium phenoxide plus water.

A simple equation is C₆H₅OH + NaOH → C₆H₅ONa + H₂O. If you are comparing acids, keep the scale in mind: phenol is more acidic than an alcohol, but much less acidic than a carboxylic acid.

Phenol is not acidic enough to react with carbonate ions, so adding aqueous carbonate gives no CO₂ effervescence from a phenol group.

A carboxylic acid does react with carbonate and releases CO₂. That makes carbonate a useful way to tell the two apart when both molecules contain an O–H group. In a molecule containing both groups, it is the carboxylic acid group that reacts with carbonate.

Phenol reacts readily with bromine water without a halogen carrier. Three ring hydrogens are substituted, giving 2,4,6-tribromophenol.

The orange/brown bromine colour disappears and a white precipitate forms. For the equation, three Br₂ molecules are needed and three HBr molecules are produced: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr.

Phenol is activated enough to react with dilute HNO₃; you do not need the concentrated HNO₃/H₂SO₄ mixture used for benzene.

For monosubstitution, NO₂ goes mainly to the 2- and 4-positions relative to –OH, giving 2-nitrophenol and 4-nitrophenol. If you draw both products, check that they are genuinely different positions and that the –OH group is still carbon 1.

One lone pair on the oxygen in phenol can occupy a p orbital and donate electron density into the benzene π-system. The ring therefore has higher electron density than benzene and attracts electrophiles more strongly.

That is why phenol reacts with bromine water without a halogen carrier and with dilute nitric acid. For OCR, keep the explanation at this level: electron-pair donation makes the ring more susceptible to attack; you do not need to discuss the stability of the intermediate.

For OCR, learn –OH and –NH₂ as 2- and 4-directing groups. Their electron donation increases electron density most usefully at positions 2, 4 and 6 around the ring, so a new electrophile is directed mainly to 2 or 4.

When you number a monosubstituted ring, positions 2 and 6 are equivalent, so you do not need to draw both as separate products. The same applies to positions 3 and 5.

The –NO₂ group is electron-withdrawing and 3-directing in OCR aromatic chemistry. So if a nitro group is already on the ring, a further electrophilic substitution is favoured at the 3-position relative to it.

On a monosubstituted ring, positions 3 and 5 are equivalent. Draw one correct 3-substituted structure and use the lowest locant in the name; rotating the same molecule does not create another isomer.

Start with the group already attached to the ring and ask what directing information you have. For OCR you know –OH and –NH₂ are 2,4-directing and –NO₂ is 3-directing; any extra directing data you need will be supplied.

Then mark the favoured ring positions before you draw the product. This small step prevents a lot of numbering mistakes, especially when two positions are equivalent or the ring already has more than one substituent.

In a multi-step aromatic synthesis, the order of substitution can decide which isomer you make. Look at the directing effect of the group introduced in step 1, then see whether it sends the next electrophile to the position you need.

If it does not, try the steps in the other order or choose a route that introduces a different directing group first. OCR can give you unfamiliar directing data, so the skill is applying the information to the target structure, not memorising a huge list of groups.