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

This guide teaches Topic 18: Organic Chemistry – Arenes of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 18.1 to 18.6. The content sits in Unit 5: Transition Metals and Organic Nitrogen Chemistry, which is an International A2 (IA2) unit. The specification also says that knowledge of the common uses of the organic compounds mentioned in this topic is expected, so uses are flagged as you go.

Course links: the course hub for Edexcel International A Level Chemistry and the printable Edexcel Chemistry checklist. When you finish, turn to the arenes revision notes and test yourself with the arenes practice questions. For the alkene comparison, see the alkenes study guide. A free ten-minute diagnostic shows which topics need the most work.

Topic 18 at a glance

Outcome	What you must be able to do
18.1	Use thermochemical, X-ray diffraction and infrared data as evidence for the structure and stability of the benzene ring
18.2	Explain the delocalised model: p-orbital overlap forming π-bonds
18.3	Explain why benzene resists bromination compared with alkenes
18.4	Know benzene’s reactions with oxygen, bromine (with a catalyst), concentrated nitric and sulfuric acids, fuming sulfuric acid, and halogenoalkanes or acyl chlorides with aluminium chloride
18.5	Give the electrophilic substitution mechanism for halogenation, nitration and Friedel-Crafts reactions, including how the electrophile forms
18.6	Explain the reaction of phenol with bromine water and why it is easier than for benzene

The specification lets you draw benzene either as a hexagon with alternating double bonds (the Kekulé form) or as a hexagon with a circle inside, whichever suits the equation or mechanism.

18.1 Evidence for the structure of benzene

Benzene is C₆H₆. The Kekulé model treats it as cyclohexa-1,3,5-triene: a ring with three C=C bonds and three C–C bonds. Three kinds of data show this model is wrong.

Thermochemical data. If benzene really had three localised C=C bonds, its enthalpy change of formation could be estimated from bond enthalpies. The real value is much less endothermic, so real benzene is lower in energy (more stable) than the Kekulé structure. The same conclusion comes from hydrogenation: adding H₂ to benzene releases far less energy than three times the value for one isolated C=C in cyclohexene.

Worked example 1 (use these values). Enthalpy change of atomisation of carbon = +717 kJ mol⁻¹. Bond enthalpies in kJ mol⁻¹: H–H 436, C–C 347, C=C 612, C–H 413. The experimental enthalpy change of formation of gaseous benzene is +83 kJ mol⁻¹. Estimate the enthalpy change of formation of a Kekulé benzene and comment.

1. Equation: 6C(s) + 3H₂(g) → C₆H₆(g).
2. Energy in: atomise 6 C = 6 × 717 = 4302; break 3 H–H = 3 × 436 = 1308. Total = 5610 kJ mol⁻¹.
3. Energy out: form 3 C–C, 3 C=C and 6 C–H = 1041 + 1836 + 2478 = 5355 kJ mol⁻¹.
4. ΔH(Kekulé) = 5610 − 5355 = +255 kJ mol⁻¹.
5. Real benzene is +83 kJ mol⁻¹, so it is 172 kJ mol⁻¹ more stable than the Kekulé structure on these data. The extra stability comes from delocalisation.

Mean bond enthalpies are averages, so the size of the gap depends on the data used. The direction does not change.

X-ray diffraction data. X-ray diffraction maps electron density, so it shows where the carbon atoms sit. In benzene all six carbon–carbon distances are the same, about 0.140 nm. That lies between a typical C–C single bond (about 0.154 nm) and a C=C double bond (about 0.134 nm). A Kekulé ring would show alternating long and short bonds, and the ring would not be a regular hexagon.

Infrared data. An alkene C=C stretch absorbs at about 1650 cm⁻¹. Benzene does not show that alkene band. Instead it shows absorptions in the 1450–1600 cm⁻¹ region typical of an aromatic ring, plus C–H stretching just above 3000 cm⁻¹. So the ring bonds are not ordinary C=C bonds.

18.2 The delocalised model

Each carbon in benzene uses three of its outer electrons to form σ-bonds: two to neighbouring carbons and one to hydrogen. These σ-bonds lie in one plane at 120°, so the molecule is a flat regular hexagon.

The fourth outer electron of every carbon occupies a p-orbital pointing straight up and down through the ring plane. Every p-orbital overlaps sideways with the p-orbitals on both of its neighbours, not just one. The overlap builds two doughnut-shaped regions of π-electron density, one on each face of the ring. Those six π-electrons are delocalised: they belong to the ring as a whole, not to three separate double bonds. This explains the equal bond lengths and the extra stability.

18.3 Why benzene resists bromination

An alkene such as cyclohexene decolourises bromine quickly at room temperature with no catalyst. Benzene does not react with bromine under those conditions.

• In an alkene, the two π-electrons are localised between two carbon atoms. This region of high electron density can induce a dipole in an approaching Br₂ molecule, and the δ+ bromine is attacked.
• In benzene, the π-electrons are spread over six carbons, so each carbon–carbon region carries less π-electron density. That is not enough to induce a large enough dipole in Br₂ to start a reaction.
• Addition across the ring would also destroy the delocalised system and lose its extra stability. So benzene needs a stronger electrophile, and it then undergoes substitution, which keeps the ring intact, rather than addition.

18.4 Reactions of benzene

Reagent and conditions	Organic product	Type
Oxygen in air, burning	CO₂ and H₂O (with soot)	Combustion
Bromine, with FeBr₃ or iron as catalyst, room temperature	Bromobenzene, C₆H₅Br, + HBr	Electrophilic substitution
Concentrated HNO₃ and concentrated H₂SO₄, below about 55 °C	Nitrobenzene, C₆H₅NO₂, + H₂O	Electrophilic substitution (nitration)
Fuming sulfuric acid (H₂SO₄ containing dissolved SO₃), warm	Benzenesulfonic acid, C₆H₅SO₃H	Sulfonation
Halogenoalkane with anhydrous AlCl₃, warm	Alkylbenzene + HCl	Friedel-Crafts alkylation
Acyl chloride with anhydrous AlCl₃, warm under reflux	Aromatic ketone + HCl	Friedel-Crafts acylation

Combustion. Benzene burns with a very smoky flame. Its carbon to hydrogen ratio is high, so there is not enough oxygen at the flame for complete combustion and unburnt carbon forms soot. Complete combustion: C₆H₆ + 7½O₂ → 6CO₂ + 3H₂O.

Worked example 2. Compare the percentage of carbon by mass in benzene and hexane (C = 12.0, H = 1.0).

• Benzene: 72.0 ÷ 78.0 × 100 = 92.3%.
• Hexane, C₆H₁₄: 72.0 ÷ 86.0 × 100 = 83.7%.

The higher carbon content of benzene is why its flame is much sootier.

Nitration. Keep the temperature at or below about 55 °C, otherwise more than one nitro group is substituted. Use: nitrobenzene is reduced to phenylamine, which is used to make dyes (you meet this in Topic 19).

Sulfonation. Equation: C₆H₆ + SO₃ → C₆H₅SO₃H. Sulfonic acids of alkylbenzenes are used to make detergents, and sulfonation is also used in making dyes. The mechanism of sulfonation is not required.

Friedel-Crafts reactions make a new C–C bond to the ring. Alkylation with chloromethane gives methylbenzene, a common solvent; alkylation with chloroethane leads to ethylbenzene, which is converted to phenylethene for making poly(phenylethene). The AlCl₃ must be dry, because water reacts with it and destroys the catalyst.

Worked example 3. Benzene reacts with benzoyl chloride, C₆H₅COCl, and AlCl₃ to give diphenylmethanone, (C₆H₅)₂CO. A student uses 7.80 g of benzene with excess benzoyl chloride and obtains 12.4 g of product. Calculate the percentage yield.

1. Equation: C₆H₆ + C₆H₅COCl → C₆H₅COC₆H₅ + HCl (1 : 1).
2. n(benzene) = 7.80 ÷ 78.0 = 0.100 mol.
3. Mr(C₁₃H₁₀O) = 156.0 + 10.0 + 16.0 = 182.0, so the theoretical mass = 0.100 × 182.0 = 18.2 g.
4. Yield = 12.4 ÷ 18.2 × 100 = 68.1%.

18.5 Electrophilic substitution mechanisms

All three mechanisms share the same two steps once the electrophile E⁺ exists.

1. Attack. A curly arrow goes from the delocalised ring (start it inside the circle) to E⁺. A bond forms between E and one carbon.
2. The intermediate. That carbon now holds both H and E and is no longer part of the delocalised system. Show this with a partial (horseshoe) ring spanning the remaining five carbons and a + sign drawn within it. This ion is less stable than benzene because the full delocalisation has been lost.
3. Loss of H⁺. The electron pair of the C–H bond moves back into the ring (show this with a curly arrow). H⁺ is released, the delocalised ring is restored and the product is C₆H₅E.

What changes is how E⁺ is made.

Halogenation. Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻. After substitution, H⁺ + FeBr₄⁻ → FeBr₃ + HBr, so the catalyst is regenerated. Iron filings work because iron reacts with bromine to form FeBr₃ in the flask.

Nitration. The acids react to form the nitronium ion: HNO₃ + 2H₂SO₄ → NO₂⁺ + H₃O⁺ + 2HSO₄⁻. Sulfuric acid is the stronger acid and protonates nitric acid, which then loses water. At the end, H⁺ + HSO₄⁻ → H₂SO₄, so sulfuric acid acts as a catalyst.

Friedel-Crafts. AlCl₃ accepts a chloride ion:

• alkylation: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻;
• acylation: RCOCl + AlCl₃ → RCO⁺ + AlCl₄⁻ (an acylium ion).

Then H⁺ + AlCl₄⁻ → AlCl₃ + HCl regenerates the catalyst.

Worked example 4. Describe the mechanism for benzene with chloromethane and aluminium chloride.

• Electrophile: CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄⁻.
• Curly arrow from the ring to the carbon of CH₃⁺.
• Intermediate: one ring carbon now bonded to both H and CH₃; the partial ring and positive charge cover the rest.
• The C–H bond on the attacked carbon breaks, its electrons return to the ring (curly arrow) and H⁺ leaves, giving methylbenzene.
• H⁺ + AlCl₄⁻ → AlCl₃ + HCl.

18.6 Phenol and bromine water

Phenol, C₆H₅OH, reacts with bromine water at room temperature with no catalyst. The orange colour disappears and the product, 2,4,6-tribromophenol, comes out of solution as a white precipitate:

C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr

Why it is easier than for benzene. Oxygen in phenol has a lone pair in a p-orbital, and that orbital overlaps with the ring’s π-cloud. Some of this electron density spreads into the ring, so the electron density of the ring rises. The ring can now polarise Br₂ by itself, so no catalyst is needed, and three hydrogens are replaced instead of one. Uses of phenol include making resins and antiseptics.

Worked example 5. 0.600 g of phenol reacts with excess bromine water. Calculate the mass of precipitate (H = 1.0, C = 12.0, O = 16.0, Br = 79.9).

1. n(phenol) = 0.600 ÷ 94.0 = 6.38 × 10⁻³ mol.
2. Mr(C₆H₃Br₃O) = 72.0 + 3.0 + 239.7 + 16.0 = 330.7.
3. Mass = 6.38 × 10⁻³ × 330.7 = 2.11 g.

Practical note. The specification suggests carrying out these reactions in the lab, using methylbenzene or methoxybenzene where appropriate. Benzene itself is carcinogenic, so school labs use these less hazardous arenes instead.

Common errors

• Saying benzene “has no double bonds”. It has delocalised π-bonding, not three localised ones.
• Explaining stability without data. In 18.1 questions, quote the numbers and say which way they point.
• Starting the first curly arrow from a single carbon atom rather than from the delocalised ring.
• Drawing the intermediate horseshoe over all six carbons, or forgetting the + charge.
• Leaving the H on the attacked carbon out of the intermediate.
• Forgetting to show how the electrophile forms, or how the catalyst is regenerated.
• Describing sulfonation conditions as “concentrated sulfuric acid” when the specification names fuming sulfuric acid.
• For phenol, writing that oxygen “donates electrons” with no mention of its lone pair overlapping with the ring π-system.

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.