Benzene and electrophilic substitution

Benzene, C₆H₆, is planar and regular hexagonal, with all six C–C bonds the same length (0.139 nm) — between a single and a double bond. Each carbon contributes one electron from a p orbital to a delocalised π system above and below the ring.

The thermochemical proof of delocalisation: the enthalpy of hydrogenation of cyclohexene is −120 kJ mol⁻¹, so a hypothetical cyclohexa-1,3,5-triene (three isolated C=C) should give 3 × −120 = −360 kJ mol⁻¹. The measured value for benzene is only −208 kJ mol⁻¹. Benzene is therefore about 152 kJ mol⁻¹ more stable than the Kekulé structure predicts — that gap is the delocalisation (resonance) energy.

Because the delocalised system is stable and the electron density is lower than in an isolated C=C, benzene undergoes substitution, not addition — it preserves the ring. It does not decolourise bromine water.

• Nitration — conc. HNO₃ + conc. H₂SO₄, 50–60 °C. The electrophile is the nitronium ion, NO₂⁺, generated by: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O.
• Halogenation — Br₂ with a halogen carrier (AlBr₃ or FeBr₃), which polarises Br₂ to generate Br⁺.
• Friedel–Crafts acylation — RCOCl with AlCl₃ → a phenylketone.

Phenol is more reactive than benzene: a lone pair on the oxygen is partially delocalised into the ring, raising the electron density. Phenol therefore reacts with bromine water (no catalyst needed) to give a white precipitate of 2,4,6-tribromophenol.

Think it through

Aldehydes, ketones and nucleophilic addition

The C=O bond is strongly polar (C^δ+=O^δ−), so unlike C=C it is attacked by nucleophiles, not electrophiles. The mechanism is nucleophilic addition.

• Reduction with NaBH₄ (the nucleophile is the hydride ion, H⁻): aldehyde → primary alcohol; ketone → secondary alcohol.
• Addition of HCN (with KCN, the nucleophile is CN⁻) → a hydroxynitrile, which lengthens the carbon chain by one.

Racemic mixtures — a classic exam question. When CN⁻ attacks propanone, the planar C=O can be attacked with equal probability from either face. Two enantiomers therefore form in equal amounts, giving a racemic mixture with no net optical activity.

Distinguishing tests:

• 2,4-DNPH (Brady's reagent) — an orange precipitate confirms a carbonyl (aldehyde or ketone). Purified, its melting point identifies the exact compound.
• Tollens' reagent — a silver mirror confirms an aldehyde (which is oxidised to a carboxylic acid). A ketone gives nothing.

Think it through

Optical isomerism

A carbon atom bonded to four different groups is a chiral centre (an asymmetric carbon). A molecule with one chiral centre exists as two enantiomers — non-superimposable mirror images, like left and right hands.

• Enantiomers have identical physical and chemical properties in an achiral environment.
• They differ in one way: they rotate the plane of plane-polarised light by equal amounts in opposite directions.
• A 50 : 50 mixture is a racemic mixture and is optically inactive — the rotations cancel.

Why the pharmaceutical industry cares: biological receptors are themselves chiral, so the two enantiomers of a drug can behave completely differently — one active, the other inactive or harmful. This is why single-enantiomer synthesis matters so much.

Your turn — calculation 1

Carboxylic acids, esters and acyl chlorides

Carboxylic acids are weak acids, but strong enough to react with carbonates — which is the test that distinguishes them from phenols and alcohols:

• Esterification: carboxylic acid + alcohol, with a conc. H₂SO₄ catalyst → ester + water. It is reversible and the yield is modest.
• Acid hydrolysis of an ester (dilute acid, reflux) → carboxylic acid + alcohol — reversible and incomplete.
• Alkaline hydrolysis (NaOH, reflux) → the carboxylate salt + alcohol. It goes to completion, so the yield is better.
• Acyl chlorides (RCOCl) are far more reactive: with water → carboxylic acid; with an alcohol → ester (irreversible, high yield); with ammonia → primary amide; with a primary amine → N-substituted amide. All release HCl — steamy white fumes.

Your turn — calculation 2

Amines, amino acids and condensation polymers

Amines are bases — the lone pair on nitrogen accepts a proton. Their strength depends on how available that lone pair is:

• Ethylamine > ammonia — the alkyl group is electron-releasing (positive inductive effect), so it pushes electron density onto N and makes the lone pair more available.
• Phenylamine < ammonia — the nitrogen lone pair is delocalised into the benzene ring, so it is much less available to accept a proton. Phenylamine is a very weak base.

Amino acids contain both –NH₂ and –COOH, so they are amphoteric. At intermediate pH they exist as a zwitterion (⁺H₃N–CHR–COO⁻), which is why they are high-melting ionic-like solids. The pH at which the zwitterion is the dominant form is the isoelectric point.

Condensation polymers lose a small molecule (H₂O or HCl) each time a link is made:

• Polyester — diol + dicarboxylic acid → ester links, losing water.
• Polyamide (nylon, and proteins) — diamine + dicarboxylic acid (or diacyl chloride) → amide links.

Why condensation polymers are biodegradable and addition polymers are not: the ester and amide links can be hydrolysed back to the monomers. A poly(alkene) backbone is a chain of inert, non-polar C–C bonds with nothing for water or enzymes to attack.

Think it through

Your turn — calculation 3

Addition, condensation, or not a polymerisation?

Tap a reaction, then tap what kind of process it is.

Chromatography and NMR spectroscopy

Chromatography separates a mixture between a stationary phase and a mobile phase. The more strongly a component is adsorbed onto the stationary phase, the slower it moves.

Gas chromatography identifies a component by its retention time against known standards, and the peak area gives the amount present.

NMR — the single most powerful structural tool at A-level. A nucleus (¹H or ¹³C) in a magnetic field absorbs radio-frequency radiation. TMS is the reference standard, defined as δ = 0.

• ¹³C NMR — the number of peaks = the number of different carbon environments. Simple and decisive.
• ¹H NMR — three separate pieces of information: chemical shift (δ) tells you the environment; integration gives the ratio of hydrogens in each environment; splitting follows the n + 1 rule, where n is the number of hydrogens on the adjacent carbon(s).
• O–H and N–H protons give a broad singlet and do not split. Shaking the sample with D₂O makes their peak disappear — a neat way of identifying them.

Worked splitting: in ethanol, CH₃CH₂OH — the CH₃ sees the 2 H of the CH₂, so it is a triplet (2 + 1). The CH₂ sees the 3 H of the CH₃, so it is a quartet (3 + 1). The OH is a singlet. Integration ratio 3 : 2 : 1.

Your turn — calculation 4

Your turn — calculation 5

Your turn — calculation 6

Match the evidence to the conclusion

Tap a piece of spectroscopic evidence on the left, then what it tells you on the right.

Think it through

Think it through

The big ideas to know

That is Module 6 — and with it, the whole of OCR Chemistry A. Press Finish to see your score.