aimnova.
DashboardMy LearningPaper MasteryStudy Plan

Stay in the loop

Study tips, product updates, and early access to new features.

aimnova.

AI-powered IB study platform with personalised plans, instant feedback, and examiner-style marking.

IB Subjects
  • All IB Subjects
  • IB Diploma
  • IB ESS
  • IB Economics
  • IB Business Management
  • IB Math AI
  • IB Math AA
  • IB Physics
  • IB Biology
  • IB Chemistry
  • IB Geography
  • IB Spanish B
  • IB German B
  • IB French B
  • IB English B
Question Banks
  • ESS Question Bank
  • Economics Question Bank
  • Business Management Question Bank
  • Math AI Question Bank
  • Math AA Question Bank
  • Physics Question Bank
  • Biology Question Bank
  • Chemistry Question Bank
  • Geography Question Bank
  • Spanish B Question Bank
  • German B Question Bank
  • French B Question Bank
  • English B Question Bank
Predicted Topics 2026
  • ESS Predictions 2026
  • Economics Predictions 2026
  • Business Management Predictions 2026
  • Math AI Predictions 2026
  • Math AA Predictions 2026
  • Physics Predictions 2026
  • Biology Predictions 2026
  • Chemistry Predictions 2026
  • Geography Predictions 2026
  • Spanish B Predictions 2026
  • German B Predictions 2026
  • French B Predictions 2026
  • English B Predictions 2026

Study Resources

  • Free Study Notes
  • Mock Exams
  • Revision Guide
  • Flashcards
  • Exam Skills
  • Command Terms
  • Past Paper Feedback
  • Grade Calculator
  • Exam Timetable 2026

Company

  • Features
  • Pricing
  • About Us
  • Blog
  • Contact
  • Terms
  • Privacy
  • Cookies

© 2026 Aimnova. All rights reserved.

Made with 💜 for IB students worldwide

v0.1.1455
NotesChemistry HLTopic 3.2Benzene and extended organic reactions (HL)
Back to Chemistry HL Topics
3.2.64 min read

Benzene and extended organic reactions (HL)

IB Chemistry • Unit 3

IB exam ready

Study like the top scorers do

Access a smart study planner, AI tutor, and exam vault — everything you need to hit your target grade.

Start Free Trial

Contents

  • The structure and bonding of benzene
  • Evidence for delocalisation
  • Why benzene substitutes rather than adds
  • Exam-style question — multi-step pathways
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

Explore the labelled diagram, charts and maps for this topic in full study mode.

Unlock free for 7 days

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.

EvidenceWhat is observedWhat it proves
Equal C–C bond lengthsAll 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 hydrogenationAdding 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.

Interactive diagram

Explore the labelled diagram, charts and maps for this topic in full study mode.

Unlock free for 7 days

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

  1. 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.
  2. 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).
  3. 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.
  4. 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).

Stop wasting time on topics you know

Our AI identifies your weak areas and focuses your study time where it matters. No more overstudying easy topics.

Try Smart Study Free7-day free trial • No card required

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

  1. Cyclohexene: has a localised, electron-rich C=C double bond, so it readily undergoes addition with Br2 (the bromine is decolourised).
  2. Benzene: its π electrons are delocalised; addition would disrupt the delocalisation and lose the stabilisation, so benzene resists addition.
  3. 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:

ConversionReagent / conditionsWhat changes
alkene → halogenoalkaneHBr (or HCl)C=C + H–X across the double bond (addition)
halogenoalkane → alcoholwarm aqueous NaOH (or KOH)–X replaced by –OH (substitution)
alcohol → carboxylic acidacidified KMnO4 / K2Cr2O7, heat under refluxprimary alcohol oxidised to –COOH
alcohol → halogenoalkaneHX (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

  1. 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.]
  2. Step 2 (substitution): warm 1-bromopropane with aqueous NaOH (or KOH) → propan-1-ol, as –Br is replaced by –OH.
  3. 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

  1. Product: with excess oxidising agent under reflux, the primary alcohol is fully oxidised to a carboxylic acid — ethanoic acid, CH_{3}COOH.
  2. 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+.

IB Exam Questions on Benzene and extended organic reactions (HL)

Practice with IB-style questions filtered to Topic 3.2.6. Get instant AI feedback on every answer.

Practice Topic 3.2.6 QuestionsBrowse All Chemistry HL Topics

How Benzene and extended organic reactions (HL) Appears in IB Exams

Examiners use specific command terms when asking about this topic. Here's what to expect:

Define

Give the precise meaning of key terms related to Benzene and extended organic reactions (HL).

AO1
Describe

Give a detailed account of processes or features in Benzene and extended organic reactions (HL).

AO2
Explain

Give reasons WHY — cause and effect within Benzene and extended organic reactions (HL).

AO3
Evaluate

Weigh strengths AND limitations of approaches in Benzene and extended organic reactions (HL).

AO3
Discuss

Present arguments FOR and AGAINST with a balanced conclusion.

AO3

See the full IB Command Terms guide →

Related Chemistry HL Topics

Continue learning with these related topics from the same unit:

3.1.1The periodic table: periods, groups and blocks
3.1.2Periodic trends in atomic properties
3.1.3Chemical trends of groups and period 3
3.1.4Transition elements and oxidation states (HL)
View all Chemistry HL topics

Improve your exam technique

Command terms, paper structure, and mark-scheme tips for Chemistry HL

Previous
3.2.5Isomerism: structural and stereoisomers (HL)
Next
Exothermic and endothermic reactions4.1.1

2 questions to test your understanding

Reading is just the start. Students who tested themselves scored 82% on average — try IB-style questions with AI feedback.

Start Free TrialView All Chemistry HL Topics