Aromatic Chemistry

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

• Athree localised C=C double bonds alternating with three longer C–C single bonds, fixed in position around the ring
• Bsix separate carbon atoms held together by hydrogen bonds between the attached hydrogen atoms
• 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 puckered ring of tetrahedral carbon atoms, each one sp³ hybridised

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 at room temperature, in just the same way as an alkene containing three localised double bonds would
• BBenzene has the molecular formula C₆H₆
• CBenzene is a colourless liquid at room temperature, whereas cyclohexa-1,3,5-triene (the Kekulé structure) would be expected to be a gas under the same conditions
• 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 152 kJ mol⁻¹ less stable than the hypothetical Kekulé structure, because delocalisation of the π electrons raises the energy of the ring
• CThe experimental value must be in error, because hydrogenating three C=C bonds in benzene should release exactly the same energy as three separate alkenes
• DBenzene contains fewer than three C=C double bonds, since hydrogenating it releases about 208 kJ mol⁻¹ instead of the predicted 360 kJ mol⁻¹ (3 × 120)

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⁻ (the nitrite ion formed from HNO₃)
• BNO₂⁺ (the nitronium ion)✓
• CNO₃⁻ (the nitrate ion from the nitric acid)
• DHNO₃ (the undissociated nitric acid molecule)

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 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 the same carbon, giving a disubstituted product that is once again aromatic
• BAddition of a hydride ion (H⁻)
• CLoss of an electron pair, leaving a stable 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✓
• BThe π electrons of benzene are locked into three localised double bonds, and these are too tightly held to be attacked by an electrophile such as bromine
• CBenzene molecules are planar and too bulky for two bromine atoms to add across the ring, whereas a single small substituent can be fitted in without strain
• DBenzene behaves as a saturated hydrocarbon (like the alkanes), and saturated compounds of that kind undergo substitution reactions rather than addition reactions

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 itself the electrophile that attacks the delocalised ring directly
• BFriedel–Crafts alkylation; AlCl₃ is a halogen carrier that generates the CH₃CH₂⁺ electrophile✓
• CNitration; AlCl₃ acts as the inert solvent in which the chloroethane and the benzene are mixed together
• DAddition; AlCl₃ reduces the ring so that the ethyl group can add across a carbon–carbon double bond

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 unusually weak, so it breaks first and leaves a highly reactive phenoxide ring that adds bromine across its carbon–carbon double bonds without needing any halogen carrier
• Bphenol is a stronger acid than benzene, and stronger acids react faster with bromine
• Cphenol contains an extra C=C double bond in the ring, which bromine adds across
• 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.)

Why is benzene more stable than the Kekulé structure predicts?

• AIts pi electrons are localised between alternate carbons
• BIts pi electrons are delocalised over the whole ring✓
• CIt contains three genuine carbon-carbon double bonds
• DIt has stronger sigma bonds than other hydrocarbons

Answer: Enthalpy of hydrogenation is about 152 kJ mol⁻¹ less exothermic than expected for three isolated double bonds, and every C–C bond is the same length. Delocalisation of the six pi electrons accounts for both.

Why does benzene undergo substitution rather than addition?

• AAddition also preserves the stable delocalised ring
• BBenzene has no electrons at all available for addition
• CAddition products of benzene cannot exist at all
• DSubstitution preserves the stable delocalised ring✓

Answer: Addition would break the delocalised system and lose the stabilisation, so benzene reacts in ways that keep the ring intact. That is the single most important difference from alkene chemistry.

What is the electrophile in the nitration of benzene?

• AThe nitrate ion, NO₃⁻
• BThe nitronium ion, NO₂⁺✓
• CThe nitrite ion, NO₂⁻
• DNitric acid itself, HNO₃

Answer: Concentrated sulfuric acid protonates nitric acid, which then loses water to give NO₂⁺. Being positively charged, it attacks the electron-rich ring; the negative ions could not.

What is the role of the halogen carrier in Friedel-Crafts acylation?

• AGenerating the nucleophile by donating a halide ion to it
• BGenerating the electrophile by accepting a halide ion✓
• CProtecting the benzene ring from being oxidised during reaction
• DNeutralising the acid produced during the reaction

Answer: Aluminium chloride pulls a chloride from the acyl chloride, leaving an acylium ion strong enough to attack benzene. It is regenerated afterwards and so acts catalytically.

Why is phenol more reactive than benzene towards electrophiles?

• AThe oxygen lone pair donates into the ring, raising electron density✓
• BThe oxygen lone pair withdraws from the ring, lowering electron density
• CPhenol contains an additional carbon-carbon double bond
• DThe hydroxyl group makes phenol a much stronger acid

Answer: A lone pair on oxygen overlaps with the delocalised system, activating the ring so much that phenol brominates with bromine water alone, without a halogen carrier. Nitrobenzene is deactivated for the opposite reason.

Why is phenol more acidic than ethanol?

• AThe ethoxide ion is stabilised by delocalisation
• BPhenol contains more hydroxyl groups than ethanol
• CThe phenoxide ion is stabilised by delocalisation✓
• DPhenol is far more soluble in water than ethanol

Answer: The negative charge on phenoxide spreads into the ring, making the ion more stable and the loss of the proton more favourable. Ethoxide has no such delocalisation, so ethanol is a much weaker acid.

What is formed when benzene reacts with a chloroalkane and aluminium chloride?

• AA halogenobenzene
• BA nitrobenzene
• CA phenol
• DAn alkylbenzene✓

Answer: Friedel-Crafts alkylation attaches the alkyl group to the ring, releasing hydrogen chloride. Acylation with an acyl chloride is often preferred because alkylation tends to substitute more than once.

Why is the reduction of nitrobenzene to phenylamine industrially important?

• APhenylamine is the starting point for making esters
• BPhenylamine is the starting point for making azo dyes✓
• CNitrobenzene cannot be stored safely for long periods
• DThe reaction produces hydrogen as a valuable by-product

Answer: Tin and concentrated hydrochloric acid reduce the nitro group to an amine. Diazotisation of phenylamine below 10 °C then gives a diazonium salt which couples with phenols to form intensely coloured azo dyes.

What conditions are used for the nitration of benzene?

• AConcentrated nitric and sulfuric acids at about 50 °C✓
• BDilute nitric and sulfuric acids together at about 50 °C
• CConcentrated nitric acid alone at ordinary room temperature
• DConcentrated sulfuric acid alone at about 200 °C

Answer: Sulfuric acid acts as catalyst, generating the nitronium ion from nitric acid. Keeping the temperature near 50 °C limits further substitution to dinitrobenzene.