Structure of Benzene
Kekulé Model (1865)
August Kekulé proposed alternating single and double bonds in a hexagonal ring. Predicts two different C–C bond lengths and that benzene should decolourise Br₂ rapidly (like alkenes). Both predictions are wrong.
Kekulé later suggested rapid equilibrium between two structures — still inadequate.
Delocalised (Resonance) Model
Each C atom is sp² hybridised with a p-orbital perpendicular to the ring plane. All six p-electrons overlap sideways to form a continuous π-cloud above and below the ring. Electrons are delocalised over all 6 carbons — no alternating single/double bonds.
This explains equal bond lengths and stability. Represented by a hexagon with an inscribed circle.
Evidence for the Delocalised Model
| Evidence | What Kekulé Predicts | What is Observed |
|---|---|---|
| C–C bond lengths | Two alternating lengths (134 pm, 154 pm) | All 139 pm (intermediate) ✓ delocalised |
| Reaction with Br₂ | Rapid decolourisation (addition) | No reaction with Br₂(aq); needs catalyst for substitution |
| Thermochemical stability | Enthalpy of hydrogenation ~360 kJ/mol | Only 208 kJ/mol — 152 kJ/mol more stable (delocalisation energy) |
| X-ray crystallography | Two types of bonds | Regular hexagon with equal bonds |
Hückel's Rule & Aromaticity
n=0: 2e; n=1: 6e; n=2: 10e; n=3: 14e…
Criteria for Aromaticity
A molecule is aromatic if it is: (1) Cyclic — closed ring; (2) Planar — all atoms in one plane; (3) Fully conjugated — each atom contributes a p-orbital; (4) Has (4n+2) π-electrons (Hückel).
| Molecule | π-electrons | n | Aromatic? |
|---|---|---|---|
| Benzene C₆H₆ | 6 | 1 | Yes ✓ |
| Naphthalene C₁₀H₈ | 10 | 2 | Yes ✓ |
| Cyclobutadiene C₄H₄ | 4 | — | No — antiaromatic (4n) |
| Cyclopentadienyl anion C₅H₅⁻ | 6 | 1 | Yes ✓ |
| Pyridine C₅H₅N | 6 | 1 | Yes ✓ |
Electrophilic Aromatic Substitution (EAS)
Nitration
Reagents: concentrated HNO₃ + concentrated H₂SO₄ (nitrating mixture), 50–60°C.
Why concentrated H₂SO₄: protonates HNO₃ to generate NO₂⁺; H₂SO₄ is the stronger acid.
Halogenation
Reagents: Cl₂ or Br₂ + Lewis acid catalyst (AlCl₃ or FeBr₃), room temperature, anhydrous.
Friedel-Crafts Alkylation
Reagents: haloalkane RCl + AlCl₃ catalyst, anhydrous.
Friedel-Crafts Acylation
Reagents: acyl chloride RCOCl + AlCl₃, anhydrous.
Sulfonation
Reagents: fuming H₂SO₄ (oleum), heat.
Nomenclature & Positional Isomerism
Naming Benzene Derivatives
When benzene has one substituent: name = substituent + benzene (e.g. chlorobenzene, nitrobenzene) or use retained names (toluene = methylbenzene, aniline = aminobenzene, phenol = hydroxybenzene).
When benzene has two substituents: use numbering or ortho (1,2), meta (1,3), para (1,4) prefixes.
When benzene has three or more substituents: number the ring, choosing the lowest set of locants. Substituents listed alphabetically.
| Position | Prefix | Relationship | Example |
|---|---|---|---|
| 1,2- | ortho (o-) | Adjacent carbons | o-dichlorobenzene |
| 1,3- | meta (m-) | One carbon apart | m-nitrotoluene |
| 1,4- | para (p-) | Opposite sides | p-aminophenol |
Worked Example: How many dichlorobenzene isomers exist?
Three: 1,2- (ortho) — Cl atoms adjacent; 1,3- (meta) — one C apart; 1,4- (para) — Cl atoms opposite. All have formula C₆H₄Cl₂ but different physical properties (e.g. different melting points).
Common Retained Names
No videos added yet for this unit.
| Compound | IUPAC Name | Retained Name |
|---|---|---|
| C₆H₅CH₃ | Methylbenzene | Toluene |
| C₆H₅OH | Hydroxybenzene | Phenol |
| C₆H₅NH₂ | Aminobenzene | Aniline |
| C₆H₅CHO | Benzenecarboxaldehyde | Benzaldehyde |
| C₆H₅COOH | Benzenecarboxylic acid | Benzoic acid |
| C₆H₅COCH₃ | 1-phenylethan-1-one | Acetophenone |
Exercises
- Give THREE pieces of evidence that benzene has a delocalised electron structure rather than the Kekulé alternating bond structure.1. Equal C–C bond lengths (all 139 pm, intermediate between single 154 pm and double 134 pm) — Kekulé predicts two alternating lengths. 2. Thermochemical stability: enthalpy of hydrogenation is only 208 kJ/mol vs expected ~360 kJ/mol for three isolated double bonds. The 152 kJ/mol difference is the delocalisation (resonance) energy. 3. Resistance to addition: benzene does not decolourise Br₂(aq) — Kekulé structure predicts rapid addition; instead, electrophilic substitution occurs to preserve aromaticity.
- Write the mechanism for the nitration of benzene, showing the formation of the electrophile, the two steps of EAS, and the product.Step 1 — Formation of electrophile: HNO₃ + H₂SO₄ → NO₂⁺ + HSO₄⁻ + H₂O. The nitronium ion NO₂⁺ is the electrophile.
Step 2 — EAS Step 1: NO₂⁺ attacks benzene π-system → σ-complex (arenium ion), a delocalized carbocation (non-aromatic).
Step 3 — EAS Step 2: HSO₄⁻ acts as base, removes H⁺ from sp³ carbon → rearomatisation → nitrobenzene C₆H₅NO₂ + H₂SO₄.
Overall: C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O. - Explain why Friedel-Crafts acylation is preferred over alkylation for introducing a carbon chain onto benzene.1. No polysubstitution: the ketone product (C₆H₅COR) is electron-withdrawing, deactivating the ring. Further acylation is much slower. In alkylation, the alkyl group activates the ring → polyalkylation occurs. 2. No rearrangement: acylium ions (RCO⁺) are stable and do not rearrange. Carbocations in alkylation can rearrange (e.g. 1° → 2° or 3°), giving unexpected products. After acylation, the ketone can be reduced to the alkyl group (Clemmensen: Zn/Hg, HCl; or Wolff-Kishner: N₂H₄, KOH, heat).
- Apply Hückel's rule to determine whether cyclopentadienyl anion (C₅H₅⁻) is aromatic.C₅H₅⁻ has 5 carbons in a ring. The anion has a lone pair on C5 in a p-orbital. All 5 carbons contribute one p-electron, plus the extra electron from the negative charge → total = 6 π-electrons. 4n+2: n=1 gives 6 ✓. The anion is also cyclic and planar. Therefore C₅H₅⁻ is aromatic. This explains the acidity of cyclopentadiene (pKa ~16) — its conjugate base is stabilised by aromaticity.
- Name the following compounds: (a) benzene with Cl at 1 and NO₂ at 4; (b) benzene with CH₃ at 1 and OH at 3; (c) benzene with Br at 1, Cl at 2, and CH₃ at 4.(a) 1-chloro-4-nitrobenzene or p-chloronitrobenzene (para). (b) 3-methylphenol (using phenol as base name with OH as principal group) or m-cresol. (c) Number to give lowest locants: Br=1, Cl=2, CH₃=4 → 1-bromo-2-chloro-4-methylbenzene. List substituents alphabetically.
- In the halogenation of benzene, what is the role of FeBr₃? Why does benzene not react with Br₂ directly?FeBr₃ is a Lewis acid catalyst. It accepts a lone pair from Br₂: Br₂ + FeBr₃ → Br⁺ [FeBr₄]⁻. This polarises/breaks the Br–Br bond, generating the electrophilic Br⁺.
Without a catalyst, Br₂ is not electrophilic enough to attack benzene's stable π-system (delocalisation energy = 152 kJ/mol). The ring is electron-rich but the high activation energy is not overcome by Br₂ alone. FeBr₃ lowers this energy by generating a better electrophile.
Quiz — 25 Questions
Unit 4: Benzene
25 QsThe C–C bond length in benzene (139 pm) is:
The electrophile in the nitration of benzene is:
In Friedel-Crafts alkylation, a Lewis acid catalyst is needed to:
Hückel's rule states an aromatic compound has:
Why does benzene undergo substitution rather than addition reactions?
The product of the reaction of benzene with Br₂/FeBr₃ is:
The delocalisation energy (resonance energy) of benzene is approximately:
In the Friedel-Crafts acylation, the electrophile is:
The intermediate in EAS is called a:
Para-dichlorobenzene (1,4-dichlorobenzene) is an isomer of:
The conditions for nitration of benzene are:
What is the IUPAC name of aniline?
In radical halogenation of benzene with Cl₂ in UV light, the product is:
Why is Friedel-Crafts acylation preferred over alkylation for synthesis?
The sulfonation of benzene is reversible. The reverse reaction (desulfonation) occurs when:
The name of the intermediate in EAS formed when NO₂⁺ bonds to benzene is:
How many structural isomers of trichlorobenzene (C₆H₃Cl₃) exist?
The term "ortho" in benzene nomenclature means substituents are at:
Benzene is more stable than expected because:
The reagents for Friedel-Crafts acylation are:
Pyridine (C₅H₅N) is aromatic because:
The product of nitration of benzene at >60°C is most likely:
Which compound is used as the electrophilic species in the bromination of benzene?
In EAS, the second step (loss of H⁺ from the arenium ion) is driven by:
The retained name "toluene" refers to:
Unit 4 Quiz — Benzene (25 Questions)
Select one answer eachThe molecular formula of benzene is:
The Kekulé structure of benzene proposed alternating single and double bonds. This was disproved because:
The delocalisation of electrons in benzene is shown by:
Electrophilic substitution is the characteristic reaction of benzene because:
In the nitration of benzene, the electrophile NO₂⁺ is generated by:
Nitration of benzene is carried out at approximately:
The mechanism of electrophilic aromatic substitution (EAS) proceeds via:
Halogenation of benzene requires a Lewis acid catalyst (FeBr₃ or AlCl₃) because:
Benzene reacts with chlorine in ultraviolet light to give:
Friedel-Crafts alkylation limitation: rearrangement occurs because:
Benzene is oxidised by hot concentrated KMnO₄ to give:
The heat of formation of benzene is less negative than expected for cyclohexatriene because:
Sulfonation of benzene with fuming H₂SO₄ is reversible. This is useful because:
The resonance structures of benzene contribute equally to the actual structure because:
π electrons in benzene occupy:
Hückel's rule states that aromatic compounds have:
Naphthalene (C₁₀H₈) is:
Cyclopentadienyl anion (C₅H₅⁻) is aromatic because:
The first step in EAS (attack of electrophile E⁺ on benzene) gives:
Activated benzene rings undergo EAS:
Deactivated benzene rings (e.g. nitrobenzene) undergo EAS:
Benzene is used as a starting material for aniline via:
The Born-Haber explanation of benzene's stability uses the concept of:
Benzene reacts with CH₃COCl + AlCl₃ to give:
The stability of benzene compared to cyclohexatriene is best demonstrated by:
Unit Test — 50 marks
Section A — Short Answer
30 marksDescribe the Kekulé and delocalised models of benzene. Give three pieces of experimental evidence that support the delocalised model over Kekulé's. [6]
Delocalised: all 6 C are sp² hybridised; p-orbitals overlap to form a π-cloud above and below the ring; 6 π-electrons delocalised over all 6 carbons; represented by hexagon with inscribed circle.
Evidence: (1) All C–C bonds = 139 pm (intermediate, equal) — not two alternating lengths. (2) Benzene resists Br₂(aq) — no addition; requires catalyst for substitution. (3) Enthalpy of hydrogenation = 208 kJ/mol, not 360 kJ/mol — 152 kJ/mol more stable than Kekulé predicts (delocalisation energy). [2 marks each evidence]
Write the mechanism for the nitration of benzene. Include: generation of the electrophile, both steps of the substitution mechanism, and the overall equation. [5]
Compare Friedel-Crafts alkylation and acylation: (a) reagents [2]; (b) electrophile formed [1]; (c) explain why acylation gives a better-defined product [2]. [5]
Draw and name all positional isomers of dichlorobenzene (C₆H₄Cl₂). Predict which isomer has the highest melting point and explain. [4]
Para isomer has highest MP (~53°C vs ~24°C and ~25°C for ortho and meta). Para-dichlorobenzene is the most symmetric — molecules pack efficiently in the crystal lattice (highest crystal symmetry), maximising London dispersion forces and requiring more energy to melt. [2]
Apply Hückel's rule to determine whether the following are aromatic, antiaromatic, or non-aromatic: (a) benzene C₆H₆ [1]; (b) cyclobutadiene C₄H₄ [2]; (c) cyclopentadienyl anion C₅H₅⁻ [2]. [5]
(b) Cyclobutadiene: 4 π-electrons (4n with n=1), cyclic, planar, conjugated. 4n → antiaromatic. This molecule is highly unstable, exists only transiently.
(c) C₅H₅⁻: 5 C, each contributes 1 p-electron + extra e⁻ from negative charge = 6 π-electrons. 4n+2, n=1 ✓. Cyclic, planar. → aromatic. Explains high acidity of cyclopentadiene (pKa 16).
Name the following compounds using IUPAC nomenclature or retained names where appropriate: (a) C₆H₅OH; (b) C₆H₅NH₂; (c) 1-bromo-4-chlorobenzene; (d) 1,2-dimethylbenzene; (e) C₆H₅COOH. [5]
Section B — Extended Response
20 marks(a) Describe what is meant by "electrophilic aromatic substitution" and explain why benzene undergoes substitution rather than addition. [3]
(b) For THREE of the following reactions of benzene, give: reagents and conditions, the electrophile formed, and the organic product: (i) nitration; (ii) bromination; (iii) Friedel-Crafts alkylation with CH₃Cl; (iv) sulfonation. [6]
(c) In each EAS reaction, what drives the loss of H⁺ in the second mechanistic step? [1]
(b) (i) Nitration: conc HNO₃/conc H₂SO⃨, 50–60°C; electrophile: NO₂⁺; product: nitrobenzene C₆H₅NO₂. (ii) Bromination: Br₂/FeBr₃, anhydrous, RT; electrophile: Br⁺ [FeBr⃨]⁻; product: bromobenzene C₆H₅Br + HBr. (iii) F-C alkylation: CH₃Cl/anhydrous AlCl₃; electrophile: CH₃⁺; product: methylbenzene (toluene) C₆H₅CH₃ + HCl. (iv) Sulfonation: fuming H₂SO⃨ (oleum), heat; electrophile: SO₃/H₂S₂O₇ (SO₃); product: benzenesulfonic acid C₆H₅SO₃H. [6: 2 per reaction, any three]
(c) Loss of H⁺ is driven by restoration of aromaticity — the arenium ion is non-aromatic and high-energy; loss of H⁺ re-establishes the delocalised π-system and releases ~152 kJ/mol. [1]
(a) What is Hückel's rule? State the four conditions for aromaticity. [3]
(b) Explain how the delocalised model accounts for the following observations: (i) benzene does not decolourise bromine water; (ii) all C–C bonds in benzene are the same length; (iii) benzene is less reactive than alkenes towards electrophilic addition. [4]
(c) Describe the sulfonation of benzene, including the reversibility of the reaction and how this reversibility can be exploited in organic synthesis. [3]
(b) (i) Benzene's π-electrons are delocalised and stabilised by 152 kJ/mol — not as accessible/reactive as alkene π-bonds. Bromine water (without catalyst) cannot provide an electrophile strong enough to overcome this activation energy. No addition across individual C=C bonds because no localised double bonds exist. [~1.5] (ii) All six C atoms are equivalent sp² carbons sharing delocalised electrons equally. Bond order is 1.5 for each C–C, giving uniform 139 pm. No alternating bond lengths because no localised single/double bonds exist. [~1.5] (iii) Alkenes have localised, accessible π-electrons with no resonance stabilisation — react readily with Br₂. Benzene's delocalised electrons are lower in energy and much more stable. Any addition product would be a non-aromatic cyclohexadiene — losing 152 kJ/mol stabilisation. Alkenes gain stability on addition; benzene loses it. [1]
(c) Sulfonation: C₆H₆ + H₂SO⃨ ⇌ C₆H₅SO₃H + H₂O, using fuming H₂SO⃨ (oleum), warm. The reaction is reversible: heating C₆H₅SO₃H with steam at ~200°C regenerates benzene (desulfonation). Synthetic exploitation: the –SO₃H group can be used as a blocking group. Add –SO₃H to block a position on the ring (e.g. para), carry out another EAS at the remaining positions (e.g. nitration goes ortho to –SO₃H), then remove –SO₃H by desulfonation to get a product with a specific substitution pattern. [3]