IB Chemistry HLTopic 2 — Models of Bonding & StructurePaper 1 & 2Higher level~10 min read
Benzene
Benzene is C6H6, so on paper it should be packed with double bonds and desperate to react. In practice it shrugs off bromine water and refuses to behave like an alkene at all. Four separate experiments say the same thing, and together they make benzene the best evidence for delocalisation on the syllabus.
📘 What you need to know
Benzene is a planar, regular hexagonal ring of six carbons with the formula C6H6.
Kekulé proposed alternating single and double bonds. That model predicts alkene behaviour, which benzene does not show.
Every carbon is sp2 hybridised and forms three σ bonds: two to neighbouring carbons and one to a hydrogen.
Each carbon keeps one unhybridised p orbital perpendicular to the ring.
Those six p orbitals overlap sideways into a delocalised π system above and below the ring, holding six shared electrons.
All six C–C bonds are equal at 140 pm, between a single (154) and a double (134). Bond angles are 120°.
Delocalisation makes benzene about 152 kJ mol⁻¹ more stable than the Kekulé model predicts.
Where the Kekulé model goes wrong
Kekulé’s ring of alternating double and single bonds was a brilliant guess, and it is still how benzene is drawn in most organic mechanisms. But it makes four predictions that all turn out to be false:
It predicts two different bond lengths round the ring. There is only one.
It predicts benzene will decolourise bromine water like any alkene. It does not.
It predicts a hydrogenation enthalpy of about three times that of cyclohexene. The real value is far smaller.
It predicts a strong C=C infrared absorption. Benzene shows nothing in that region.
The bonding, properly
Each carbon uses three of its four outer electrons in σ bonds. That leaves one electron each, sitting in a p orbital that points straight up and down, at right angles to the flat ring. Six p orbitals side by side overlap continuously all the way round, and the six electrons in them are shared by every carbon at once.
From six p orbitals to one ring of electrons
The circle inside the hexagon is not decoration. It is a statement that the six π electrons belong to the whole ring rather than to three fixed double bonds.
🤔 Why does benzene have to be flat?
Sideways overlap between p orbitals only works properly when they are parallel. Bend the ring and neighbouring p orbitals tilt away from each other, the overlap weakens, and the delocalisation is lost. Staying planar with 120° angles keeps all six orbitals lined up, which is worth far more energy than any strain it costs.
Evidence 1: enthalpies of hydrogenation
This is the cleanest experiment on the topic, and it is worth understanding rather than memorising.
Cyclohexene has one real C=C. Hydrogenating it releases 120 kJ mol⁻¹. If benzene really had three of those, hydrogenating it should release 3 × 120 = 360 kJ mol⁻¹. Measure it and you get only 208. Benzene has released 152 kJ mol⁻¹ less than expected, which means it started out 152 kJ mol⁻¹ lower in energy than the Kekulé model assumed. That gap is the delocalisation energy.
The energy that delocalisation saves
Read the diagram from the bottom up: everything ends at cyclohexane, so the shorter the drop, the more stable the starting material was.
Evidence 2, 3 and 4
Evidence
Kekulé predicts
Actually observed
Conclusion
Hydrogenation
−360 kJ mol⁻¹
−208 kJ mol⁻¹
152 kJ mol⁻¹ more stable
X-ray bond lengths
134 pm and 154 pm
all six at 140 pm
one kind of bond only
Bromine water
decolourises
no reaction
no isolated C=C present
Infrared spectrum
C=C peak at 1620–1680 cm⁻¹
weak peaks near 1450–1580 cm⁻¹
a delocalised system, not double bonds
The bond length number worth remembering. A C–C single bond is 154 pm and a C=C double bond is 134 pm. Every bond in benzene measures 140 pm — not one of each, but six identical bonds sitting between the two, exactly as a bond order of 1.5 predicts.
In organic mechanisms you will still draw benzene with alternating double bonds, because it makes electron movement easier to follow. That is fine as a working tool. Just do not claim in a written answer that those double bonds are real.
Worked examples
WE 1
Cyclohexene hydrogenates with ΔH = −120 kJ mol⁻¹. Benzene gives −208 kJ mol⁻¹. Calculate the delocalisation energy.
Step 1: predict from the Kekulé model
Three isolated C=C bonds would give
3 × (−120) = −360 kJ mol⁻¹Step 2: compare with the real value−208 − (−360) = +152 kJ mol⁻¹Step 3: interpret the sign
Less energy released means benzene started lower in energy than predicted.
Delocalisation energy = 152 kJ mol⁻¹quote it as a positive stabilisation, and say benzene is more stable
WE 2
Explain why cyclohexene decolourises bromine water but benzene does not [3]
Mark 1: what cyclohexene has
A real, localised C=C double bond with high electron density that attracts electrophiles.
Mark 2: what happens
Bromine adds across the double bond by electrophilic addition, so the orange colour disappears.
Mark 3: why benzene resists
Benzene has no isolated double bond. Its π electrons are delocalised, and adding across the ring would destroy that stability.
No localised C=C, so no addition and no colour changethis is also why benzene substitutes rather than adds — substitution keeps the ring intact
WE 3
All six C–C bonds in benzene are 140 pm. Explain what this shows about the structure.
Step 1: what would Kekulé predict?
Alternating bonds would give three at 134 pm and three at 154 pm.
Step 2: compare with the measurement
Only one value, 140 pm, and it sits between the two.
Step 3: state the conclusion
All six bonds are identical, so the π electrons must be delocalised evenly round the ring.
Six identical bonds of order 1.5, not alternating singles and doubles“intermediate between a single and a double bond” is the exact phrase to use
💡 Exam tips
Learn the four pieces of evidence as a set. Six-mark questions often ask for two or three of them.
