The shape of the lesson
By the end of the lesson, learners can
- Orbitals in the ring: Toggle between the Kekulé structure and the delocalised ring in the simulator, and compare the sideways overlap in ethene with the continuous overlap round a benzene ring.
- The evidence against Kekulé: Three independent observations all point the same way.
- Bonding: sp2, 120°, planar, p orbitals overlap sideways.
- Lengths: 0.154 (C–C), 0.139 (benzene), 0.134 nm (C=C).
- Hydrogenation: −360 predicted, −208 actual, 152 more stable.
- Reactivity: substitution, not addition.
One picture every idea returns to
Each carbon in benzene is sp2 hybridised: it forms three sigma bonds (two to C, one to H) in a plane at 120°, so the ring is planar. The remaining p orbital on each carbon overlaps sideways with both neighbours, forming a delocalised pi system of six electrons in rings above and below the plane. The evidence against the Kekulé structure is threefold: all six C–C bonds are 0.139 nm, between C–C (0.154 nm) and C=C (0.134 nm); the enthalpy of hydrogenation is −208 kJ mol−1, not the −360 kJ mol−1 predicted from three C=C bonds, so benzene is about 152 kJ mol−1 more stable; and benzene resists addition, preferring substitution that keeps the delocalised ring.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Six electrons shared by six carbons.. Kekulé drew benzene with three alternating double bonds. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Orbitals in the ring | Toggle between the Kekulé structure and the delocalised ring in the simulator, and compare the sideways overlap in ethene with the continuous overlap round a benzene ring. Project tab 3 of the lab (Organic chemistry (A Level introduction) lab, benzene in 3D) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: The evidence against Kekulé | Three independent observations all point the same way. Project tab 3 of the lab (Organic chemistry (A Level introduction) lab, benzene in 3D) and let a student drive it while the class predicts. | Corners |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 16 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in The evidence against Kekulé, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Corners
Put one answer or viewpoint in each corner of the room. Students move to the corner they agree with, justify the choice to a partner there, then hear one voice from each corner before the reveal.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The circle does not mean benzene has no double bond character: every C–C bond has some. | Stabilisation is 152 kJ mol −1, the gap between −360 and −208, not 208. The pi electrons are above and below the plane of the ring, not in it; the sigma framework lies in the plane. |
Structure of benzene: worksheet
6 questions, 16 marks. Written in the style of the papers; no past paper question is reproduced.
Describe the bonding in a benzene molecule in terms of orbital overlap.
The enthalpy of hydrogenation of cyclohexene is −120 kJ mol−1 and of benzene is −208 kJ mol−1. Calculate the stabilisation energy of benzene.
Explain how C–C bond length data support the delocalised model of benzene.
Cyclohexene decolourises bromine water at room temperature but benzene does not. Explain.
State the shape of a benzene molecule and the H–C–C bond angle.
Cyclohexa-1,3-diene has two C=C bonds and a measured enthalpy of hydrogenation of −232 kJ mol−1. Using −120 kJ mol−1 for one C=C, calculate its stabilisation energy and compare it with that of benzene.
Answer key and marking notes
- Each carbon is sp2 hybridised and forms three sigma bonds: two to C and one to H, at 120° in a plane.
- Each carbon has one remaining p orbital, at right angles to the ring.
- The p orbitals overlap sideways with both neighbours all round the ring.
- This gives a delocalised pi system of six electrons above and below the plane.
- Kekulé prediction = 3 × (−120) = −360 kJ mol−1.
- Stabilisation = −208 − (−360) = 152 kJ mol−1; benzene is more stable by this amount.
- Kekulé predicts three short C=C (0.134 nm) and three long C–C (0.154 nm) bonds.
- X-ray diffraction shows all six bonds are 0.139 nm.
- Equal, intermediate lengths mean the pi electrons are shared equally round the ring.
- Cyclohexene has a localised C=C with a high electron density, so it attacks Br2 and undergoes addition.
- Benzene's pi electrons are delocalised (lower density) and addition would destroy the stable ring, so no reaction without a catalyst.
- Planar (a flat regular hexagon, all 12 atoms in one plane).
- 120°.
- Predicted value for the diene = 2 × (−120) = −240 kJ mol−1, so its stabilisation is only 8 kJ mol−1.
- Benzene: predicted −360, measured −208, stabilisation 152 kJ mol−1.
- The diene's two C=C bonds overlap only slightly; benzene's six p electrons are fully delocalised round a closed ring, giving far greater stability.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
What is the hybridisation of each carbon atom in benzene, and the C–C–C bond angle?
- Asp2, 120°
- Bsp3, 109.5°
- Csp, 180°
- Dsp2, 109.5°
Typical C–C and C=C bond lengths are 0.154 nm and 0.134 nm. What is observed for the carbon-carbon bonds in benzene?
- Aall six are 0.139 nm
- Bthree are 0.154 nm and three are 0.134 nm
- Call six are 0.154 nm
- Dall six are 0.134 nm
The diagram compares the enthalpy changes of hydrogenation of a Kekulé structure (predicted from 3 × −120 kJ mol−1) and of real benzene. What is the size of the gap marked ?, the stabilisation energy of benzene?
- A152 kJ mol−1
- B208 kJ mol−1
- C360 kJ mol−1
- D88 kJ mol−1
Bromine water is shaken with benzene at room temperature, with no catalyst. What is observed?
- Athe orange colour is not removed
- Bthe bromine water is decolourised at once
- Ca white precipitate forms
- Dsteamy white fumes are given off
How many sigma bonds are there in one molecule of benzene, C6H6?
- A12
- B6
- C9
- D15
Which statement about the delocalised electrons in benzene is correct?
- Asix p electrons, one from each carbon, form a pi system above and below the plane of the ring
- Bsix electrons are delocalised in the plane of the ring, between the carbon nuclei
- Ctwelve electrons are delocalised, two from each carbon
- Dthe delocalised electrons alternate between three fixed C=C bonds
How many carbon atoms in methylbenzene, C6H5CH3, are sp2 hybridised and how many are sp3 hybridised?
- Asix sp2 and one sp3
- Bseven sp2
- Cseven sp3
- Dfive sp2 and two sp3
Another data source gives the enthalpy change of hydrogenation of cyclohexene as −119 kJ mol−1 and of benzene as −208 kJ mol−1. What stabilisation energy does this give for benzene?
- A149 kJ mol−1
- B89 kJ mol−1
- C152 kJ mol−1
- D357 kJ mol−1
Quiz key
Three sigma bonding regions around each carbon means sp2 hybridisation, with the three bonds in a plane at 120°; the unhybridised p orbital forms the delocalised pi system. sp3 and 109.5° describe a carbon with four sigma bonds, as in cyclohexane. sp and 180° describe a carbon with two bonding regions, as in ethyne. sp2 with 109.5° mixes the two: sp2 orbitals are always at 120°.
X-ray diffraction shows six identical bonds of 0.139 nm, between single and double, because the pi electrons are delocalised equally round the ring. Three long and three short bonds is the Kekulé prediction, which is not observed. All 0.154 nm would mean no pi bonding at all, as in cyclohexane. All 0.134 nm would need six full double bonds, which carbon's valency does not allow.
The Kekulé level is 360 kJ mol−1 above cyclohexane and benzene is 208 kJ mol−1 above it, so benzene is 360 − 208 = 152 kJ mol−1 lower than the Kekulé structure. 208 and 360 are the individual enthalpy changes, not the gap. 88 compares benzene with one C=C bond only (208 − 120).
Benzene does not undergo addition, because that would destroy the stable delocalised ring, and without a FeBr3 catalyst it does not substitute either; the bromine simply stays orange (it moves into the benzene layer). Rapid decolourisation is the test result for an alkene such as cyclohexene. A white precipitate (2,4,6-tribromophenol) forms with phenol, whose ring is activated by the OH group. Steamy fumes of HBr only appear when substitution occurs with a catalyst.
There are six C–C sigma bonds round the ring and six C–H sigma bonds: 12 in total. 6 counts only one type. 9 counts three single C–C bonds and six C–H bonds, forgetting that every C=C in a Kekulé drawing also contains a sigma bond. 15 counts the three Kekulé pi bonds as well, but pi bonds are not sigma bonds.
Each sp2 carbon has one electron in a p orbital perpendicular to the ring; these overlap sideways all round to give a ring of electron density above and below the plane, holding six electrons. Electron density in the plane between the nuclei belongs to the sigma bonds. Each carbon gives only one electron to the pi system, so there are six, not twelve. Fixed alternating C=C bonds is the Kekulé model, which the equal bond lengths rule out.
The six ring carbons each have three sigma bonds and a p orbital in the delocalised system, so they are sp2; the CH3 carbon has four single bonds, so it is sp3 and tetrahedral. The methyl carbon has no p orbital in the pi system, so seven sp2 is wrong. The ring carbons cannot be sp3, or the ring would not be planar. The ring carbon bonded to CH3 still has three sigma bonds and a p orbital, so it is sp2, not sp3.
Kekulé prediction = 3 × (−119) = −357 kJ mol−1; stabilisation = −208 − (−357) = 149 kJ mol−1. 89 compares benzene with only one C=C (208 − 119). 152 is the value from −120, not from the data given. 357 is the predicted enthalpy change itself, not the difference.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.