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Flashcards · 9620 · A Level Topic 28
Aromatic chemistry (A Level) flashcards
11 cards on A Level Topic 28, subtopics 3.3.10.1 to 3.3.10.2. Pick a subtopic or study them all.
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3.3.10.1 Bonding
- Bonding in benzene
- Planar ring; each C forms three σ bonds at 120°. Each C has one p electron; the six p orbitals overlap sideways, so six electrons are delocalised above and below the ring.
- Carbon–carbon bond length in benzene
- All six bonds are equal, with a length between a C–C single bond and a C=C double bond.
- How do enthalpies of hydrogenation show benzene is more stable than cyclohexa-1,3,5-triene?
- Predicted ΔH for the triene = 3 × ΔH(cyclohexene). Benzene’s measured value is less exothermic. Both give cyclohexane, so benzene starts lower in energy: the difference is the delocalisation stabilisation.
- Why does benzene undergo substitution rather than addition?
- Addition would destroy the delocalised ring and lose its extra stability. Substitution keeps the ring intact.
3.3.10.2 Electrophilic substitution
- Nitration of benzene: reagents, electrophile and its formation
- Concentrated HNO₃ and concentrated H₂SO₄. Electrophile: NO₂⁺. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻ Product: nitrobenzene.
- Electrophilic substitution mechanism on benzene: outline
- Arrow from the delocalised ring to the electrophile; intermediate with a horseshoe over five carbons and + inside, the attacked C carrying H and the new group; arrow from the C–H bond into the ring, releasing H⁺.
- Friedel–Crafts acylation: reagents and electrophile
- Acyl chloride with AlCl₃ catalyst, dry conditions. RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻ Product: a phenyl ketone, C₆H₅COR. Catalyst re-formed: H⁺ + AlCl₄⁻ → AlCl₃ + HCl.
- Sulfonation of benzene: reagent and electrophile
- Fuming sulfuric acid, warm (or concentrated H₂SO₄ under reflux). Electrophile: SO₃ (the ring attacks the S). Product: benzenesulfonic acid, C₆H₅SO₃H.
- Nitration of methylbenzene compared with benzene
- Faster, because CH₃ releases electron density into the ring. Substitution mainly at positions 2 and 4: 2-nitromethylbenzene and 4-nitromethylbenzene.
- Chlorine with benzene vs with methylbenzene in UV light
- Benzene: free-radical addition to the ring, giving C₆H₆Cl₆. Methylbenzene: free-radical substitution on the side chain, giving (chloromethyl)benzene, C₆H₅CH₂Cl, and further products.
- Why is chlorobenzene unreactive towards nucleophiles, unlike (chloromethyl)benzene?
- In chlorobenzene a Cl lone pair overlaps with the ring, giving the C–Cl bond partial double-bond character (shorter, stronger), and the delocalised electrons repel nucleophiles.