Kekule vs Delocalised Benzene Model, Nitration & Friedel-Crafts Acylation

1

Bonding in Benzene vs Kekulé's Cyclohexa-1,3,5-triene Model

Benzene (C₆H₆) is a flat hexagonal ring of six carbon atoms with 120° bond angles. Each carbon forms three planar covalent σ-bonds (to two adjacent carbons and one hydrogen), leaving one unhybridised p-orbital perpendicular to the ring containing 1 electron. The six p-orbitals overlap sideways above and below the ring to form a continuous delocalised π-electron cloud (6 delocalised electrons shared over all 6 carbons), giving benzene high thermodynamic stability.

Save

2

Three Pieces of Evidence Disproving Kekulé's Model

1. Bond Lengths (X-ray diffraction): Kekulé's alternating single and double bond model predicts three long C-C bonds (0.154 nm) and three shorter C=C bonds (0.134 nm); in reality, all six carbon-carbon bonds in benzene are the exact same intermediate length (0.140 nm). 2. Thermochemical Hydrogenation Data: Hydrogenation of cyclohexene (1 C=C bond) has ΔH = -120 kJ mol⁻¹, so theoretical cyclohexa-1,3,5-triene (3 C=C bonds) should release 3 × -120 = -360 kJ mol⁻¹; actual hydrogenation of benzene is only -208 kJ mol⁻¹, proving benzene is 152 kJ mol⁻¹ more stable due to delocalisation energy. 3. Chemical Reactivity: Unlike alkenes, benzene does not decolourise bromine water by electrophilic addition; instead, it undergoes electrophilic substitution to retain the stable delocalised π-ring.

Save

3

Nitration of Benzene: Generating the NO₂⁺ Electrophile & Uses

Reagents and conditions: Concentrated nitric acid (HNO₃) and concentrated sulfuric acid (H₂SO₄ catalyst) heated at 50°C (above 50°C causes multiple substitutions to form 1,3-dinitrobenzene). Because H₂SO₄ is a stronger acid than HNO₃, it protonates nitric acid to generate the powerful nitronium ion electrophile (NO₂⁺): HNO₃ + 2H₂SO₄ → NO₂⁺ + 2HSO₄⁻ + H₃O⁺ (or simplified: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O). Used to manufacture explosives (TNT, 2,4,6-trinitromethylbenzene) and reduced with Sn/conc HCl to form aromatic amines (phenylamine) for azo dyes.

Save

4

Electrophilic Substitution Mechanism (Horseshoe Wheland Intermediate)

For any electrophile E⁺ (where E⁺ = NO₂⁺ or R-C⁺=O): 1. Step 1: Draw a curly arrow starting from the inside of the delocalised benzene ring circle to the positively charged electrophile E⁺ (4 of the 6 delocalised electrons remain in the partial ring). 2. Draw the unstable cyclohexadienyl ('horseshoe') intermediate: a broken dashed horseshoe ring covering the 5 sp² carbons (not extending past the two carbons adjacent to the tetrahedral sp³ carbon bonded to both H and E), with a + charge inside the horseshoe. 3. Step 2: Draw a curly arrow from the middle of the C-H covalent bond back into the inside of the horseshoe ring to restore the 6-electron delocalised aromatic ring and release an H⁺ ion (which regenerates the catalyst: H⁺ + HSO₄⁻ → H₂SO₄).

Save

5

Friedel-Crafts Acylation of Benzene (AlCl₃ Catalyst)

Reagents and conditions: An acyl chloride (e.g. ethanoyl chloride, CH₃COCl) or acid anhydride in the presence of an anhydrous aluminium chloride (AlCl₃) halogen-carrier catalyst under reflux in a non-aqueous solvent (anhydrous conditions required because AlCl₃ and RCOCl both hydrolyse vigorously with water). 1. Generation of the acylium ion electrophile: CH₃COCl + AlCl₃ → CH₃C⁺=O + AlCl₄⁻. 2. Electrophilic substitution onto benzene forms phenylethanone (C₆H₅COCH₃) + H⁺. 3. Catalyst regeneration: H⁺ + AlCl₄⁻ → AlCl₃ + HCl(g) (steamy acidic fumes).

Save

6

Reduction of Nitrobenzene to Phenylamine

Nitrobenzene (C₆H₅NO₂) is reduced to phenylamine (C₆H₅NH₂) in a two-stage laboratory process: 1. Heat nitrobenzene under reflux with tin (Sn) and concentrated hydrochloric acid (HCl) reducing agent: C₆H₅NO₂ + 6[H] + HCl → C₆H₅NH₃⁺Cl⁻ + 2H₂O (forms the soluble phenylammonium chloride salt because the basic amine reacts with the acid). 2. Add excess aqueous sodium hydroxide (NaOH) to deprotonate the salt and liberate free phenylamine: C₆H₅NH₃⁺Cl⁻ + NaOH → C₆H₅NH₂ + NaCl + H₂O, followed by steam distillation.

Save

1. Q1

   Explain with reference to bond lengths, enthalpies of hydrogenation, and chemical reactions why the delocalised model of benzene is preferred over Kekulé's cyclohexa-1,3,5-triene structure. (6 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Bond length evidence: Kekulé's model has alternating C-C single (0.154 nm) and C=C double (0.134 nm) bonds, predicting two different bond lengths [1 mark]. 2. X-ray diffraction shows that all six carbon-carbon bonds in benzene are identical in length (0.140 nm), intermediate between a single and a double bond [1 mark]. 3. Enthalpy of hydrogenation evidence: Hydrogenation of cyclohexene (one C=C bond) has ΔH = -120 kJ mol⁻¹, so Kekulé's structure with three localised C=C bonds should have ΔH = 3 × (-120) = -360 kJ mol⁻¹ [1 mark]. 4. Experimental hydrogenation of benzene is only -208 kJ mol⁻¹ (152 kJ mol⁻¹ less exothermic), showing benzene is 152 kJ mol⁻¹ lower in energy (more stable) due to the delocalised 6-electron π-ring [1 mark]. 5. Reactivity evidence: If benzene had three localised C=C bonds like an alkene, it would rapidly decolourise bromine water at room temperature via electrophilic addition [1 mark]. 6. Instead, benzene does not react with bromine water and requires a halogen-carrier catalyst (FeBr₃) to undergo electrophilic substitution, preserving the stable delocalised ring [1 mark].

2. Q2

   Describe the mechanism for the nitration of benzene to form nitrobenzene, including the equations for the formation of the electrophile and the regeneration of the sulfuric acid catalyst. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Electrophile generation equation: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O (where H₂SO₄ acts as a Brønsted-Lowry acid protonating HNO₃) [1 mark]. 2. Step 1 of electrophilic substitution: Draw a curly arrow starting from the delocalised π-electron circle inside the benzene ring to the nitrogen atom of the NO₂⁺ electrophile [1 mark]. 3. Draw the correct cyclohexadienyl intermediate with both -H and -NO₂ attached to one carbon, a horseshoe covering the other five carbons (gap facing the C(H)(NO₂) carbon), and a + charge inside the horseshoe [1 mark]. 4. Step 2 of mechanism: Draw a curly arrow from the middle of the C-H covalent bond back into the hexagon inside the horseshoe to restore the delocalised π-system and eliminate H⁺ [1 mark]. 5. Catalyst regeneration equation: H⁺ + HSO₄⁻ → H₂SO₄ [1 mark].

3. Q3

   Benzene reacts with ethanoyl chloride (CH₃COCl) in the presence of anhydrous aluminium chloride (AlCl₃) to form an aromatic ketone. Name the mechanism and the organic product, write the equation for the formation of the electrophile, and outline the curly-arrow mechanism. (5 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. Name of mechanism: Electrophilic substitution (Friedel-Crafts acylation); Name of product: Phenylethanone [1 mark]. 2. Equation for formation of the acylium ion electrophile: CH₃COCl + AlCl₃ → CH₃C⁺=O + AlCl₄⁻ [1 mark]. 3. Step 1: Draw a curly arrow from the delocalised electron ring of benzene to the positively charged carbon atom of the CH₃C⁺=O electrophile [1 mark]. 4. Draw the intermediate showing the -H and -COCH₃ groups attached to the same ring carbon, with a correct horseshoe and internal + charge [1 mark]. 5. Step 2: Draw a curly arrow from the C-H bond back into the ring to restore the delocalised aromatic ring, forming C₆H₅COCH₃ and releasing H⁺ (which reacts with AlCl₄⁻ to form AlCl₃ + HCl) [1 mark].

4. Q4

   Explain why cyclohexene reacts readily with bromine in the dark at room temperature without a catalyst, whereas benzene requires an iron(III) bromide (FeBr₃) catalyst to react with bromine. (4 marks)

   Check Mark Scheme

   Click “Check Mark Scheme” to compare your answer, 1. In cyclohexene, the two π-electrons of the C=C double bond are localised between just two carbon atoms, giving a region of very high electron density [1 mark]. 2. In benzene, the six π-electrons are delocalised (spread out) above and below all six carbon atoms in the ring, giving a much lower π-electron density at any single bond [1 mark]. 3. Consequently, cyclohexene has sufficient electron density to polarise an approaching Br₂ molecule (inducing a Br^δ+ - Br^δ- dipole) and undergo electrophilic addition directly [1 mark]. 4. Benzene's lower electron density cannot polarise Br₂ strongly enough on its own, so a Lewis acid halogen carrier (FeBr₃) is needed to react with Br₂ (Br₂ + FeBr₃ → Br⁺ + FeBr₄⁻) and generate a strongly positive Br⁺ electrophile [1 mark].

Related study guides