Aromatic Chemistry

The modern (delocalised) model of benzene describes the bonding as:

• Athree localised C=C double bonds alternating with three C–C single bonds
• Bsix separate carbon atoms held together only by hydrogen bonds
• Ca ring of six carbon atoms with the p-electrons delocalised into a ring of electron density above and below the plane of the carbon atoms✓
• Da tetrahedral arrangement of carbons with sp³ hybridisation

Answer: In benzene each carbon contributes one electron from a p orbital to a delocalised π system spread over all six carbons, forming rings of electron density above and below the planar ring. All six C–C bonds are identical, intermediate between single and double bonds.

Which piece of evidence supports the delocalised model of benzene rather than the Kekulé (alternating double-bond) structure?

• ABenzene readily decolourises bromine water
• BBenzene has the molecular formula C₆H₆
• CBenzene is a liquid at room temperature
• DAll six carbon–carbon bond lengths are equal (~0.139 nm), between a single (0.154 nm) and a double (0.134 nm) bond✓

Answer: X-ray diffraction shows all C–C bonds in benzene are identical (~0.139 nm), intermediate between single and double bonds. The Kekulé structure predicts alternating long and short bonds, so equal bond lengths support delocalisation.

The enthalpy of hydrogenation of cyclohexene (one C=C) is −120 kJ mol⁻¹. The Kekulé structure of benzene (three C=C) would therefore be expected to release 360 kJ mol⁻¹, but the measured value for benzene is only −208 kJ mol⁻¹. What does this difference indicate?

• ABenzene is more stable (lower in energy) than the hypothetical Kekulé structure by about 152 kJ mol⁻¹, owing to delocalisation✓
• BBenzene is less stable than the Kekulé model predicts
• CThe experiment must be in error because the values should be equal
• DBenzene contains four C=C double bonds

Answer: Real benzene releases less energy on hydrogenation (208 kJ mol⁻¹) than the predicted 3 × 120 = 360 kJ mol⁻¹. Releasing less energy means benzene started at a lower energy (more stable) than the Kekulé model by about 360 − 208 = 152 kJ mol⁻¹. This extra stability is the delocalisation (resonance) energy.

Benzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid at about 50 °C to form nitrobenzene. What is the electrophile in this nitration reaction?

• ANO₂⁻
• BNO₂⁺ (the nitronium ion)✓
• CNO₃⁻
• DHNO₃

Answer: Concentrated sulfuric acid protonates nitric acid, which then loses water to generate the nitronium ion: HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺. The electrophile NO₂⁺ attacks the delocalised π system of benzene.

In the nitration of benzene, sulfuric acid is described as a catalyst. Which equations together justify this description?

• AH₂SO₄ is consumed and not regenerated
• BH₂SO₄ only acts as a solvent and takes no part in the reaction
• CH₂SO₄ + HNO₃ → NO₂⁺ + HSO₄⁻ + H₂O, then H⁺ + HSO₄⁻ → H₂SO₄✓
• DH₂SO₄ → SO₃ + H₂O, then SO₃ attacks benzene

Answer: Sulfuric acid first helps generate the NO₂⁺ electrophile (forming HSO₄⁻), and is then regenerated at the end of the mechanism when HSO₄⁻ removes the H⁺ released from the intermediate (H⁺ + HSO₄⁻ → H₂SO₄). Because it is regenerated, H₂SO₄ acts as a catalyst.

In the general mechanism of electrophilic substitution of benzene, after the electrophile E⁺ adds to the ring, an unstable intermediate forms. What is the final step that restores aromaticity?

• AAddition of a second electrophile to give a disubstituted product
• BAddition of a hydride ion (H⁻)
• CLoss of an electron pair to form a carbocation
• DLoss of H⁺ from the carbon bonded to E, regenerating the delocalised ring✓

Answer: The electrophile adds to form a positively charged intermediate in which delocalisation is partly broken. A proton (H⁺) is then lost from the carbon now bonded to E, returning the two electrons to the π system and restoring the stable, fully delocalised aromatic ring.

Benzene undergoes substitution rather than addition with electrophiles such as bromine (which alkenes add readily). Why does benzene resist addition reactions?

• AAddition would disrupt the stable delocalised π system, losing the delocalisation (stabilisation) energy, whereas substitution preserves it✓
• BBenzene has no π electrons to react
• CBenzene molecules are too large to react
• DBenzene is saturated and so cannot react at all

Answer: The delocalised π system gives benzene extra stability (delocalisation energy). An addition reaction would permanently break this delocalisation, which is energetically unfavourable. Substitution allows the aromatic ring to be reformed, retaining the stabilisation, so benzene reacts by substitution.

Benzene reacts with chloroethane (CH₃CH₂Cl) in the presence of anhydrous aluminium chloride to give ethylbenzene. This is an example of which reaction, and what is the role of AlCl₃?

• AFriedel–Crafts acylation; AlCl₃ is the electrophile
• BFriedel–Crafts alkylation; AlCl₃ is a halogen carrier that generates the CH₃CH₂⁺ electrophile✓
• CNitration; AlCl₃ is the solvent
• DAddition; AlCl₃ reduces the ring

Answer: Reaction of benzene with a haloalkane and AlCl₃ is a Friedel–Crafts alkylation. AlCl₃ acts as a halogen carrier (catalyst): it accepts a chloride from CH₃CH₂Cl to form the carbocation electrophile CH₃CH₂⁺ (and AlCl₄⁻), which then attacks the ring. AlCl₃ is regenerated at the end.

Benzene reacts with ethanoyl chloride (CH₃COCl) and an AlCl₃ catalyst. What type of reaction is this, and what is the organic product?

• AFriedel–Crafts alkylation, giving ethylbenzene (C₆H₅CH₂CH₃)
• BNitration, giving nitrobenzene
• CFriedel–Crafts acylation, giving phenylethanone (C₆H₅COCH₃)✓
• DHalogenation, giving chlorobenzene

Answer: An acyl chloride plus AlCl₃ carries out Friedel–Crafts acylation. AlCl₃ generates the acylium electrophile CH₃CO⁺, which substitutes onto benzene to give the aromatic ketone phenylethanone (acetophenone), C₆H₅COCH₃, plus HCl.

Phenol (C₆H₅OH) reacts with bromine water rapidly at room temperature, decolourising it and forming a white precipitate of 2,4,6-tribromophenol — no catalyst is needed. Benzene, by contrast, does not react with bromine water. The increased reactivity of phenol is because:

• Athe O–H bond in phenol is very weak
• Bphenol is a stronger acid than benzene
• Cphenol contains an extra C=C double bond
• Da lone pair of electrons on the oxygen atom is partially delocalised into the ring, increasing the electron density and making the ring more readily attacked by electrophiles✓

Answer: One of the lone pairs on the oxygen of the –OH group overlaps with (is delocalised into) the benzene ring's π system, raising the electron density of the ring. This activates the ring towards electrophiles, so phenol brominates rapidly with bromine water and without a halogen carrier; the –OH group is 2,4-directing.

Phenol is weakly acidic and reacts with aqueous sodium hydroxide. Which equation correctly represents this reaction?

• AC₆H₅OH + NaOH → C₆H₅ONa + H₂O✓
• BC₆H₅OH + NaOH → C₆H₅ONa + H₂
• CC₆H₅OH + Na₂CO₃ → C₆H₅ONa + CO₂ + H₂O
• DC₆H₅OH + NaOH → C₆H₆ + NaOH + ½O₂

Answer: Phenol is acidic enough to react with the strong base NaOH in a neutralisation, forming sodium phenoxide and water: C₆H₅OH + NaOH → C₆H₅ONa + H₂O. (Phenol is too weak an acid to react with the weaker base sodium carbonate, which distinguishes it from carboxylic acids.)