Edexcel International A Level Chemistry: Organic chemistry: arenes (YCH11) – Revision Notes

These notes condense Topic 18: Organic Chemistry – Arenes from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 18.1 to 18.6. The topic is part of Unit 5: Transition Metals and Organic Nitrogen Chemistry, an International A2 (IA2) unit. For full explanations and worked examples, use the arenes study guide; to practise, go to the arenes practice questions.

Course pages: Edexcel International A Level Chemistry hub and the printable checklist. Unsure what to revise first? A free diagnostic takes about ten minutes.

Key terms

• Arene: an aromatic hydrocarbon, built on one or more benzene rings (benzene, methylbenzene).
• Delocalised electrons: bonding electrons not tied to one pair of atoms; benzene’s six π-electrons belong to the whole ring.
• Electrophile: an electron-pair acceptor; in the mechanisms you must know it is a cation (Br⁺, NO₂⁺, R⁺, RCO⁺).
• Electrophilic substitution: E⁺ takes the place of one ring hydrogen, which leaves as H⁺, and the delocalised system survives.
• Halogen carrier: a catalyst such as FeBr₃ or AlCl₃ that pulls a halide ion away to create the electrophile.
• Nitronium ion: NO₂⁺, made when concentrated sulfuric acid protonates nitric acid.
• Acylium ion: RCO⁺, made from an acyl chloride and AlCl₃; the carbonyl carbon carries the charge.
• Friedel-Crafts reaction: substitution of an alkyl or acyl group onto the ring using a halogenoalkane or acyl chloride with aluminium chloride.
• Fuming sulfuric acid: concentrated sulfuric acid containing dissolved sulfur trioxide.

18.1 Evidence against the Kekulé structure

The Kekulé model is cyclohexa-1,3,5-triene: three localised C=C and three C–C bonds. Three lines of evidence disagree with it.

Evidence	Kekulé prediction	What is observed	Conclusion
Thermochemical (formation from bond enthalpies, or hydrogenation)	Energy values for three isolated C=C bonds	Benzene is lower in energy than predicted	Extra stability from delocalisation
X-ray diffraction	Alternating long C–C (about 0.154 nm) and short C=C (about 0.134 nm) bonds	All six bonds equal, about 0.140 nm; regular hexagon	Bonds are intermediate, not alternating
Infrared	Alkene C=C stretch near 1650 cm⁻¹	No alkene C=C band; aromatic ring bands at 1450–1600 cm⁻¹	Ring bonds are not alkene-type

Method: using thermochemical data

1. Work out the value the Kekulé model predicts (for hydrogenation, three times the cyclohexene value; for formation, use bond enthalpies).
2. Compare it with the experimental value for benzene.
3. State that benzene is more stable (less energy released on hydrogenation, or less endothermic formation) by the difference.
4. Link the difference to delocalisation of π-electrons.

You can draw benzene as the Kekulé hexagon or as a hexagon with a circle, as appropriate in equations and mechanisms.

18.2 The delocalised model in four lines

• Each carbon forms three σ-bonds (two C–C, one C–H) in a plane at 120°.
• Every carbon keeps one electron in a p-orbital perpendicular to the ring plane.
• Each p-orbital overlaps sideways with both neighbours.
• The six π-electrons spread round the whole ring, with density on both faces of the molecule.

18.3 Benzene versus alkenes with bromine

	Alkene (e.g. cyclohexene)	Benzene
π-electron density	Localised between two carbons, high	Delocalised over six carbons, lower between any two
Effect on Br₂	Induces a dipole in Br₂	Cannot polarise Br₂ enough
Conditions	Room temperature, no catalyst	Needs a halogen carrier such as FeBr₃
Reaction type	Electrophilic addition	Electrophilic substitution
Why that type	—	Substitution keeps the stable delocalised ring

18.4 Reaction summary

Reagent	Conditions	Product	Use worth knowing
O₂ (air)	Burn	CO₂ + H₂O; very smoky flame	—
Br₂	FeBr₃ (or Fe) catalyst, room temperature	Bromobenzene + HBr	Intermediate in synthesis
Conc. HNO₃ + conc. H₂SO₄	At or below about 55 °C	Nitrobenzene + H₂O	Reduced to phenylamine for dyes
Fuming H₂SO₄ (SO₃ in H₂SO₄)	Warm	Benzenesulfonic acid	Detergents and dyes
RCl + AlCl₃	Anhydrous, warm	C₆H₅R + HCl	E.g. ethylbenzene → phenylethene → poly(phenylethene)
RCOCl + AlCl₃	Anhydrous, warm under reflux	C₆H₅COR (a ketone) + HCl	Aromatic ketones for synthesis

Smoky flame: high C : H ratio → incomplete combustion → soot. Complete combustion: C₆H₆ + 7½O₂ → 6CO₂ + 3H₂O.

Sulfonation: C₆H₆ + SO₃ → C₆H₅SO₃H. Mechanism not required.

18.5 Electrophiles and catalyst regeneration

Reaction	Forming E⁺	Regenerating the catalyst
Bromination	Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻	H⁺ + FeBr₄⁻ → FeBr₃ + HBr
Nitration	HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻	H⁺ + HSO₄⁻ → H₂SO₄
Alkylation	RCl + AlCl₃ → R⁺ + AlCl₄⁻	H⁺ + AlCl₄⁻ → AlCl₃ + HCl
Acylation	RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻	H⁺ + AlCl₄⁻ → AlCl₃ + HCl

Method: drawing the mechanism

1. Show E⁺ with its charge on the correct atom (N in NO₂⁺, C in R⁺ and RCO⁺).
2. Curly arrow from inside the ring (the delocalised electrons) to E⁺.
3. Intermediate: the attacked carbon carries H and E; partial ring spanning the remaining five carbons, with + inside it.
4. Arrow from the C–H bond pair back to the ring; H⁺ is released and full delocalisation returns.
5. Write the regeneration equation if the question asks about the catalyst.

Must-know distinctions

• Halogen carrier (FeBr₃, AlCl₃) versus reactant: the carrier is regenerated, the halogen or organic halide is used up.
• Alkylation adds an alkyl group (C₆H₅R); acylation adds an acyl group and gives a ketone (C₆H₅COR).
• Nitration electrophile is NO₂⁺, not NO₂ or HNO₃.
• Fuming sulfuric acid supplies SO₃; it is not the same as concentrated sulfuric acid.
• AlCl₃ must be anhydrous: water reacts with it and the catalyst is lost.

18.6 Phenol with bromine water

• Conditions: room temperature, bromine water, no catalyst.
• Observations: orange solution decolourised; white precipitate.
• Equation: C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr.
• Product: 2,4,6-tribromophenol, three bromines substituted.
• Reason: a lone pair on O, in a p-orbital, overlaps with the ring π-system and is partly delocalised into it. Ring electron density increases, so the ring polarises Br₂ itself.
• Phenol is used to make resins and antiseptics.
• Practical work uses methylbenzene or methoxybenzene in place of benzene, which is carcinogenic.

Quick self-test

1. Give benzene’s C–C–C angle and describe the molecule’s geometry.
2. Give the C–C bond length in benzene and say how it compares with C–C and C=C bonds.
3. Which absorption in the infrared spectrum of cyclohexene is absent from benzene’s spectrum?
4. Name the catalyst for the bromination of benzene and write the equation for forming the electrophile.
5. Write the equation for forming the nitronium ion.
6. Why must the nitration of benzene be kept at or below about 55 °C?
7. Give the reagent needed to make benzenesulfonic acid and calculate its Mr (H = 1.0, C = 12.0, O = 16.0, S = 32.1).
8. Hexanoyl chloride, CH₃(CH₂)₄COCl, reacts with benzene and AlCl₃. Name the ketone formed.
9. How many moles of Br₂ react with 0.0200 mol of phenol?
10. 6.24 g of benzene is fully converted to bromobenzene. What mass of bromobenzene forms (Br = 79.9)?
11. Why is the Friedel-Crafts intermediate less stable than benzene?
12. Calculate the percentage by mass of carbon in benzene, and say how it links to the flame.

Answers

1. 120°; planar regular hexagon.
2. About 0.140 nm; between C–C (about 0.154 nm) and C=C (about 0.134 nm).
3. The alkene C=C stretch at about 1650 cm⁻¹.
4. FeBr₃ (or iron, which forms FeBr₃); Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻.
5. HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻.
6. Higher temperatures cause more than one nitro group to be substituted.
7. Fuming sulfuric acid; Mr(C₆H₆O₃S) = 72.0 + 6.0 + 48.0 + 32.1 = 158.1.
8. 1-phenylhexan-1-one, C₆H₅CO(CH₂)₄CH₃.
9. 3 × 0.0200 = 0.0600 mol.
10. n = 6.24 ÷ 78.0 = 0.0800 mol; Mr(C₆H₅Br) = 156.9; mass = 12.6 g.
11. The attacked carbon is no longer part of the delocalised system, so the full delocalisation of six π-electrons over six carbons is lost.
12. 72.0 ÷ 78.0 × 100 = 92.3%; the high carbon content leads to incomplete combustion and a smoky flame.

Where marks are usually lost

• Writing “all bonds are the same” without giving the data or saying the length is intermediate.
• Quoting thermochemical figures with no statement of which structure is more stable.
• Describing π-bonds as end-on overlap, or overlap with only one neighbour.
• Explaining benzene’s resistance to bromination only by “stability” without mentioning electron density and polarising Br₂.
• Missing the condition “fuming” for sulfonation, or “anhydrous” for Friedel-Crafts.
• Placing the + charge of NO₂⁺ on an oxygen, or aiming the first arrow at the oxygen of the acylium ion instead of its carbon.
• Starting the attacking arrow at one ring carbon; it must begin in the circle of delocalised electrons.
• An intermediate with the horseshoe over six carbons or with no hydrogen on the sp³ carbon.
• Writing 2,4,6-tribromophenol as a monobromo product, or leaving 3HBr out of the equation.
• Giving “orange to colourless” for phenol and forgetting the white precipitate.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 5, Topic 18: Organic Chemistry – Arenes, outcomes 18.1 to 18.6.