The big idea: Benzene, C6H6, is the simplest arene (aromatic ring). It is a flat, regular hexagon of six carbon atoms, each bonded to one hydrogen.
Every carbon is sp² hybridised, so it forms three σ bonds in the plane (to two carbons and one hydrogen) at 120°.
That leaves one electron in a p-orbital on each carbon, perpendicular to the ring. These six p-orbitals overlap sideways to form a delocalised π system — two ring-shaped clouds of electrons, one above and one below the plane of the ring.
Key terms: - sp² hybridised — three hybrid orbitals in a plane at 120°, leaving one unhybridised p-orbital. - Delocalised electrons — electrons shared over several atoms, not fixed between two (here, spread around the whole ring). - π (pi) system — the cloud formed by the sideways overlap of p-orbitals, sitting above and below the σ framework. - Aromatic — describes the special stability of a ring with a continuous, delocalised π system.
All six C–C bonds are equal: Because the π electrons are spread evenly around the ring, every C–C bond is identical — each one is intermediate between a single (C–C) and a double (C=C) bond, at about 140 pm.
There are no alternating long-and-short bonds: benzene has one kind of carbon–carbon bond, not two.
A planar hexagon of six sp² carbons; the inner circle represents the delocalised π electrons spread evenly above and below the ring, so all six C–C bonds are equal (~140 pm).
Interactive diagram
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How do we know the π electrons are delocalised rather than fixed in three double bonds? Two classic pieces of experimental evidence point the same way.
| Evidence | What is observed | What it proves |
|---|---|---|
| Equal C–C bond lengths | All six C–C bonds in benzene are the same length (~140 pm), between a single (~154 pm) and a double bond (~134 pm). | There are no separate single and double bonds — the electrons are delocalised over the whole ring. |
| Enthalpy of hydrogenation | Adding H2 to benzene releases less energy than expected from three isolated C=C bonds (≈ −208 kJ mol⁻¹ vs a predicted ≈ −360 kJ mol⁻¹). | Real benzene is lower in energy (more stable) than the hypothetical Kekulé molecule — the difference is the delocalisation (resonance) stabilisation. |
The hydrogenation argument, step by step: - Cyclohexene (one C=C) releases about −120 kJ mol⁻¹ when hydrogenated. - A Kekulé benzene (three separate C=C) would therefore be expected to release 3 × (−120) ≈ −360 kJ mol⁻¹. - Real benzene releases only about −208 kJ mol⁻¹. - The shortfall (≈ 150 kJ mol⁻¹ less energy released) means real benzene started lower in energy — it is more stable than the imaginary Kekulé molecule by that amount. This is the delocalisation (resonance) stabilisation.
Chemists therefore draw benzene two ways. The Kekulé structure (alternating double bonds) is handy for counting electrons; the delocalised structure (a circle inside the hexagon) is the accepted picture because it matches the evidence.
Kekulé structure (older model)
- A hexagon with three alternating C=C double bonds and three C–C single bonds.
- Predicts two different C–C bond lengths (short C=C, long C–C).
- Suggests benzene should behave like an alkene (rapid addition).
- Useful for counting electrons, but it is not the real picture.
Delocalised structure (accepted model)
- A hexagon with a circle inside — the delocalised π electrons.
- All six C–C bonds are identical, intermediate between single and double.
- Explains the extra stability and the resistance to addition.
- The six p-orbitals overlap to form π clouds above and below the ring.
The Kekulé picture shows three alternating C=C double bonds — this predicts two different C–C bond lengths and alkene-like reactivity, which the evidence above disproves. Contrast it with the delocalised ring (inner circle) shown earlier.
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IB-style question — explain the bonding and stability
Benzene, C6H6, does not behave like a typical alkene. Explain, in terms of bonding, why all six carbon–carbon bonds in benzene are the same length and why benzene is unusually stable. [4]
How to score the marks
- sp² carbons: each carbon is sp² hybridised, forming three σ bonds in a plane and leaving one electron in a p-orbital perpendicular to the ring.
- Delocalised π system: the six p-orbitals overlap sideways to form a delocalised π cloud above and below the ring (electrons spread over all six carbons).
- Equal bonds: because the π electrons are spread evenly, every C–C bond is identical — intermediate between a single and a double bond (≈ 140 pm), not alternating.
- Extra stability: delocalisation lowers the energy of the molecule (shown by the less-than-expected enthalpy of hydrogenation), giving the resonance/delocalisation stabilisation.
Final answer
sp² carbons each contribute a p-electron to a delocalised π system; the even spread makes all six C–C bonds equal and lowers the energy, making benzene more stable than a Kekulé structure (evidenced by its enthalpy of hydrogenation).
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An ordinary alkene has a localised C=C double bond that is a reactive, electron-rich target, so alkenes undergo rapid addition (e.g. with bromine, the orange colour vanishes at once).
Benzene protects its ring: Benzene's stability comes from its delocalised π system. An addition reaction would force two of those π electrons into a new σ bond and break up the delocalisation — destroying the very stability that makes benzene special.
So benzene strongly resists addition. Instead it tends to undergo substitution: a hydrogen atom on the ring is swapped for another atom or group, and the delocalised ring stays intact.
Alkene — addition
- Has a localised C=C double bond.
- Adds across the double bond (e.g. Br2, HBr, H2O).
- Decolourises bromine water rapidly with no catalyst.
- The product loses the C=C double bond.
Benzene — substitution
- Has a delocalised π system (extra stable).
- Substitutes a ring H (e.g. nitration, halogenation) — usually needs a catalyst.
- Reacts only slowly and does not readily decolourise bromine without a catalyst.
- The product keeps the delocalised ring.
Set-up only — the mechanism is separate: Here we only explain why benzene substitutes (to preserve the stable ring). The detailed electrophilic substitution mechanism (the curly arrows for nitration / halogenation of benzene) is taught in micro 6.4.3.
IB-style question — addition vs substitution
Cyclohexene reacts rapidly with bromine, Br2, but benzene does not under the same conditions. Explain this difference and state the type of reaction benzene undergoes with bromine instead. [3]
How to score the marks
- Cyclohexene: has a localised, electron-rich C=C double bond, so it readily undergoes addition with Br2 (the bromine is decolourised).
- Benzene: its π electrons are delocalised; addition would disrupt the delocalisation and lose the stabilisation, so benzene resists addition.
- Type of reaction: benzene instead undergoes substitution (with a catalyst), replacing a ring H and keeping the delocalised ring intact.
Final answer
Cyclohexene's localised C=C adds Br2; benzene's delocalised ring resists addition (which would cost the stabilisation), so benzene undergoes substitution instead.
How this is tested: At HL, organic chemistry is assessed as reaction pathways: you are given a starting material and a target, and asked to deduce the route — the reagents, conditions and intermediate(s) for a two- or three-step functional-group interconversion.
- Paper 2 typically asks you to deduce a product, state the reagents/conditions for each step, or construct the overall scheme. - The skill is recognising what each functional group can be converted into, then chaining the steps in the right order.
A toolkit of common functional-group interconversions you can chain together:
| Conversion | Reagent / conditions | What changes |
|---|---|---|
| alkene → halogenoalkane | HBr (or HCl) | C=C + H–X across the double bond (addition) |
| halogenoalkane → alcohol | warm aqueous NaOH (or KOH) | –X replaced by –OH (substitution) |
| alcohol → carboxylic acid | acidified KMnO4 / K2Cr2O7, heat under reflux | primary alcohol oxidised to –COOH |
| alcohol → halogenoalkane | HX (or PCl3, SOCl2) | –OH replaced by –X |
Planning a route: (1) Identify the functional group in the start and the target. (2) List the conversions that bridge them. (3) Give the reagent and conditions for each step (a step with no reagent earns no mark). (4) Check each intermediate is sensible before moving on.
IB-style question — deduce the conversion route
Suggest a two-step synthesis of propan-1-ol, CH3CH2CH2OH, starting from propene, CH3CH=CH2. State the reagent and conditions for each step and identify the intermediate. [3]
How to score the marks
- Step 1 (addition): react propene with HBr → 1-bromopropane, CH_{3}CH_{2}CH_{2}Br (H–Br adds across the C=C). [The intermediate is the halogenoalkane.]
- Step 2 (substitution): warm 1-bromopropane with aqueous NaOH (or KOH) → propan-1-ol, as –Br is replaced by –OH.
- Identify the intermediate: 1-bromopropane, CH3CH2CH2Br (the bromine must add to the end carbon to give propan-1-ol).
Final answer
Propene + HBr → 1-bromopropane (CH3CH2CH2Br); then warm aqueous NaOH → propan-1-ol. Intermediate = 1-bromopropane.
IB-style question — deduce the product
Ethanol is heated under reflux with acidified potassium dichromate(VI), K2Cr2O7/H2SO4, until no further reaction occurs. Deduce the organic product and explain the colour change of the dichromate. [2]
How to score the marks
- Product: with excess oxidising agent under reflux, the primary alcohol is fully oxidised to a carboxylic acid — ethanoic acid, CH_{3}COOH.
- Colour change: the dichromate is reduced from orange (Cr2O72-) to green (Cr3+).
Final answer
Ethanoic acid, CH3COOH; the orange dichromate turns green as it is reduced to Cr3+.