Topic 18: Organic Chemistry A2 – Arenes
- Syllabus
- 2017
- Topic
- —
- Level
- A2
| Evidence | Observation | Structural conclusion |
|---|---|---|
| thermochemical | three isolated C=C bonds would hydrogenate by about 3 × -120 = -360 kJ mol^-1, but benzene is about -208 kJ mol^-1 | benzene is about 152 kJ mol^-1 more stable than the localised cyclohexa-1,3,5-triene model |
| X-ray diffraction | all six C-C bonds have the same length, intermediate between typical C-C and C=C | the ring does not contain three fixed single and three fixed double bonds |
| infrared | all ring C-C bonds give the same aromatic stretching pattern, rather than separate fixed C-C/C=C sets | the six carbon-carbon bonds are equivalent |
The evidence is consistent with six π electrons delocalised around a planar six-carbon ring. Benzene may be drawn as a hexagon with a circle or as a Kekulé hexagon when equations and curly-arrow mechanisms require explicit electron movement.
A Kekulé drawing is a representation, not evidence that benzene rapidly switches between two localised structures. The measured molecule has equivalent bonds and additional delocalisation stability.
Each carbon in benzene is trigonal planar and uses three orbitals to make σ bonds: two C-C bonds in the ring and one C-H bond. This leaves one unhybridised p orbital perpendicular to the ring plane on every carbon.
The six parallel p orbitals overlap sideways with both neighbours. Their electron density joins into one continuous delocalised π system above and below the carbon ring, containing six π electrons rather than three isolated electron pairs.
Because the π electrons are shared across all six carbon atoms, every C-C bond has the same order, length and strength, intermediate between a localised single and double bond.
The π system is formed from overlapping p orbitals; there are not six separate 'π orbitals' or three fixed π bonds located on alternating edges.
| Feature | Alkene | Benzene |
|---|---|---|
| π electron density | localised between two carbons | spread around six carbons |
| attraction/polarisation of Br2 | strong enough under normal conditions | weaker; an electrophile must be generated with a catalyst |
| reaction | electrophilic addition, rapidly decolourises bromine | electrophilic substitution, requiring FeBr3/Fe and heat |
| stability cost | local π bond is replaced | high-energy intermediate temporarily loses aromatic delocalisation |
Benzene can react with an electrophile, but formation of the non-aromatic intermediate has a larger activation-energy barrier. Substitution then restores the delocalised ring; addition would destroy its stabilisation in the product.
The delocalised electrons do not repel electrophiles. Benzene is less reactive because its π density is spread out and disrupting aromatic delocalisation creates a kinetic barrier.
| Reaction | Reagent/conditions | Organic product or observation |
|---|---|---|
| combustion | oxygen in air | CO2 and H2O in complete combustion; smoky flame because of high carbon content |
| bromination | Br2 with FeBr3 (or Fe forming catalyst), heat | bromobenzene + HBr |
| nitration | concentrated HNO3 + concentrated H2SO4, warm | nitrobenzene + H2O |
| sulfonation | fuming H2SO4 | benzenesulfonic acid |
| Friedel-Crafts alkylation | halogenoalkane + anhydrous AlCl3 | alkylbenzene + HX |
| Friedel-Crafts acylation | acyl chloride + anhydrous AlCl3 | aryl ketone + HCl |
\ce{C6H6 + CH3COCl ->[AlCl3] C6H5COCH3 + HCl}
The syllabus list is deliberately limited. Keep the catalysts and concentrated/fuming conditions distinct, and do not substitute a carboxylic acid for the acyl chloride in Friedel-Crafts acylation.
| Reaction | Electrophile generation | Electrophile |
|---|---|---|
| bromination | Br2 + FeBr3 → Br+ + FeBr4- | Br+ |
| nitration | HNO3 + H2SO4 → NO2+ + HSO4- + H2O | NO2+ |
| Friedel-Crafts alkylation | RCl + AlCl3 → R+ + AlCl4- | R+ |
| Friedel-Crafts acylation | RCOCl + AlCl3 → RCO+ + AlCl4- | RCO+ |
The catalyst is regenerated: FeBr4- + H+ → HBr + FeBr3, or AlCl4- + H+ → HCl + AlCl3. In nitration, HSO4- accepts H+ to regenerate H2SO4.
Curly arrows begin at electron pairs or bonds, never at a positive charge. The first step disrupts aromaticity; the second must restore the ring rather than produce an addition product.
\ce{C6H5OH + 3Br2 -> 2,4,6-C6H2Br3OH + 3HBr}
| Starting material | Bromine conditions | Result |
|---|---|---|
| benzene | Br2 requires FeBr3/Fe and heat | bromobenzene by substitution |
| phenol | bromine water at room temperature, no catalyst | bromine decolourises and white 2,4,6-tribromophenol precipitate forms |
One lone pair on the phenol oxygen overlaps with the ring π system and donates electron density into it. The ring is therefore more electron-rich, especially at the 2, 4 and 6 positions, so it polarises bromine and undergoes electrophilic substitution much more readily than benzene.
Phenol does not react more readily because the O-H bond is acidic or because phenol is simply 'a nucleophile'. The required explanation is lone-pair overlap with the ring and increased ring electron density.