The structure of benzene

Benzene is C₆H₆: a planar, regular hexagon of carbon atoms with bond angles of 120°. Each carbon forms three σ bonds (two C–C, one C–H) and has one electron left in a p orbital, perpendicular to the ring plane.

Those six p orbitals overlap sideways, all the way round, producing a delocalised π system — a ring of electron density above and below the plane. The six π electrons belong to the whole ring, not to any pair of carbons.

You must be able to quote the evidence — that is the next screen, and it is worth several marks.

Three pieces of evidence for delocalisation

1. Bond lengths. X-ray diffraction shows all six C–C bonds are identical, 0.139 nm — between a C–C single bond (0.154 nm) and a C=C double bond (0.134 nm). The Kekulé structure predicts alternating long and short bonds. It doesn't have them.

2. Enthalpy of hydrogenation. Cyclohexene (one C=C) + H₂ gives ΔH = −120 kJ mol⁻¹. If benzene really had three isolated C=C bonds, you would predict 3 × (−120) = −360 kJ mol⁻¹. The measured value is only −208 kJ mol⁻¹ — far less exothermic than predicted. Benzene is therefore more stable (lower in energy) than Kekulé benzene would be, by about 152 kJ mol⁻¹: the delocalisation (resonance) energy.

3. Lack of reactivity. Benzene does not decolourise bromine water at room temperature, whereas an alkene does instantly. The delocalised π system has a lower electron density in any one place than a localised C=C, so it cannot polarise a Br₂ molecule on its own.

Your turn — delocalisation energy

Reading the hydrogenation data

Electrophilic substitution — nitration

Benzene's π cloud is electron-rich, so it attracts electrophiles. But it always substitutes (swaps an H) rather than adds — addition would break up the delocalised system and lose 152 kJ mol⁻¹ of stability.

Conditions: concentrated HNO₃ + concentrated H₂SO₄ (catalyst), warmed to about 50 °C (above 55 °C you start to get dinitrobenzene).

The three steps:

• Two of the delocalised π electrons attack NO₂⁺ (curly arrow from inside the ring to the N).
• An unstable intermediate forms — draw it as a horseshoe of partial delocalisation, a + in the middle, and the carbon bearing both the H and the NO₂.
• HSO₄⁻ removes that H⁺, the C–H bonding pair drops back into the ring and the delocalised system is restored. H⁺ + HSO₄⁻ → H₂SO₄, so the catalyst is regenerated.

The nitrating electrophile

Halogenation and Friedel–Crafts

Benzene's π cloud is too spread out to polarise a halogen molecule by itself, so you must add a halogen carrier (a Lewis acid such as AlCl₃, FeCl₃, FeBr₃ or AlBr₃) to make a strong enough electrophile.

• Halogenation: AlCl₃ + Cl₂ → AlCl₄⁻ + Cl⁺ (the electrophile is the full cation Cl⁺, not a δ+ atom). Then C₆H₆ + Cl₂ → C₆H₅Cl + HCl. Chlorobenzene, at room temperature.
• Friedel–Crafts alkylation: AlCl₃ + CH₃Cl → AlCl₄⁻ + CH₃⁺. Then C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl. You have made methylbenzene — a new C–C bond.
• Friedel–Crafts acylation: AlCl₃ + CH₃COCl → AlCl₄⁻ + CH₃CO⁺ (the acylium ion). Then C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl. The product, phenylethanone, is an aromatic ketone.

Why Friedel–Crafts matters: it is the standard way of joining an alkyl or acyl group onto an aromatic ring — the first step in making dyes, drugs and detergents. The mechanism is the same three steps as nitration; only the electrophile changes.

Friedel–Crafts product

Your turn — yield of nitrobenzene

Phenol — an activated ring

Phenol is C₆H₅OH: an –OH group bonded directly to the benzene ring. One lone pair on the oxygen sits in a p orbital that overlaps with the ring's π system, so it is partially delocalised into the ring. That raises the electron density of the ring — phenol is activated and reacts far more readily with electrophiles than benzene does.

• Bromination: phenol decolourises bromine water at room temperature with no catalyst, giving a white precipitate of 2,4,6-tribromophenol (antiseptic smell). Benzene needs Br₂ + FeBr₃ and gives only mono-substitution.
• Nitration: phenol reacts with dilute HNO₃ at room temperature (benzene needs conc. HNO₃ / conc. H₂SO₄ at 50 °C).
• Directing effect: the –OH group directs incoming groups to the 2, 4 and 6 positions.

Phenol is also weakly acidic. The phenoxide ion C₆H₅O⁻ is stabilised by delocalisation into the ring, so phenol reacts with NaOH to give sodium phenoxide + water. But it is a weaker acid than a carboxylic acid, so — unlike a carboxylic acid — it does not fizz with sodium carbonate.

Why is phenol more reactive?

Amines — making them, and their basicity

An amine is NH₃ with one or more H replaced by an alkyl or aryl group: primary RNH₂, secondary R₂NH, tertiary R₃N. The nitrogen keeps its lone pair, so an amine is a Brønsted–Lowry base (it accepts H⁺) and a good nucleophile.

Three preparations you must know:

• From a halogenoalkane: heat with excess ethanolic ammonia under pressure. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br (nucleophilic substitution). The excess NH₃ limits further substitution — otherwise you get a mixture of secondary, tertiary and quaternary ammonium salts.
• By reducing a nitrile: LiAlH₄ in dry ether (or H₂ with a Ni catalyst). CH₃CN + 4[H] → CH₃CH₂NH₂. Since a nitrile is made from a halogenoalkane + KCN, this route lengthens the carbon chain by one and gives a clean primary amine.
• By reducing a nitro compound: nitrobenzene + Sn and conc. HCl, then add NaOH(aq) to liberate the free amine. C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O. This is how you get phenylamine, the starting point for azo dyes.

Explain it by asking how available the nitrogen lone pair is. An alkyl group is electron-donating (positive inductive effect): it pushes electron density onto N, so the lone pair is more available and ethylamine is a stronger base than ammonia. In phenylamine the nitrogen lone pair is partially delocalised into the benzene ring, so it is much less available — phenylamine is a much weaker base than ammonia.

Ranking the bases

Amino acids and zwitterions

A 2-amino acid has the general formula RCH(NH₂)COOH: an acidic –COOH group and a basic –NH₂ group on the same carbon. They react with each other.

• At low pH (acidic) the excess H⁺ protonates the COO⁻: the amino acid exists as a positive cation, HOOC–CH(R)–NH₃⁺.
• At high pH (alkaline) OH⁻ removes a proton from the NH₃⁺: it exists as a negative anion, ⁻OOC–CH(R)–NH₂.
• The isoelectric point (pI) is the pH at which the amino acid exists as the zwitterion, with zero net charge — so in electrophoresis it does not move towards either electrode.

For a simple amino acid with no ionisable side chain, pI = ½(pK_a1 + pK_a2), where pK_a1 is for the –COOH group and pK_a2 is for the –NH₃⁺ group.

Amino acids are chiral (except glycine, R = H, whose central carbon carries two H atoms). Every naturally occurring amino acid in a protein is the same single enantiomer — a link straight back to Topic 17.

Your turn — isoelectric point

Charge at a given pH

Condensation polymers

In addition polymerisation (alkenes) every atom of the monomer ends up in the polymer — 100% atom economy. In condensation polymerisation, each new link expels a small molecule (usually H₂O, or HCl if you use an acyl chloride), so each monomer must have two reactive groups.

• Polyesters — a diol + a dicarboxylic acid (or diacyl chloride), joined by ester links –COO–. Terylene / PET is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.
• Polyamides — a diamine + a dicarboxylic acid (or diacyl chloride), joined by amide links –CONH–. Nylon-6,6 is made from hexanedioic acid and 1,6-diaminohexane; Kevlar uses benzene-1,4-dicarboxylic acid and 1,4-diaminobenzene.

Disposal: ester and amide links can be hydrolysed (by acid, alkali, or enzymes), so condensation polymers are far more biodegradable than poly(alkenes), whose inert C–C backbone is not attacked. That is the standard 3-mark comparison question.

Your turn — mass of a repeat unit

Which link is which?

Multi-step synthesis and functional group interconversion

An A-level synthesis question gives you a start and a finish and asks for the route. Learn the arrows, not the compounds — and remember which ones change the number of carbon atoms.

• Halogenoalkane → nitrile: KCN in ethanol, reflux (+1 C)
• Nitrile → carboxylic acid: reflux with dilute HCl(aq)
• Nitrile → primary amine: LiAlH₄ in dry ether, or H₂/Ni
• Halogenoalkane → amine: excess ethanolic NH₃, heat under pressure
• Halogenoalkane → alcohol: warm with NaOH(aq)
• Alcohol → aldehyde: K₂Cr₂O₇/H₂SO₄, distil off; → carboxylic acid: the same reagents but reflux
• Carbonyl → hydroxynitrile: HCN/KCN (+1 C)
• Carboxylic acid → ester: alcohol + conc. H₂SO₄; or better, go via the acyl chloride (SOCl₂), then add the alcohol
• Benzene → nitrobenzene → phenylamine: conc. HNO₃/conc. H₂SO₄ at 50 °C, then Sn/conc. HCl followed by NaOH(aq)

Planning rule: count the carbons first. If the product has one more carbon than the starting material, your route must contain a KCN or HCN step.

Planning a route

What is each set of conditions for?

Tap a set of reagents, then tap the box it belongs in.

Match the reaction to the product

Tap an item on the left, then its partner on the right.

The big ideas to know

You've now covered Topic 18: Organic Chemistry III of the Edexcel A-level Chemistry (9CH0) specification. Press Finish to see your score.