Molecular Orbital Treatment of Benzene: MDCAT Chemistry notes
Molecular Orbital Treatment of Benzene for MDCAT: sp2 carbons, delocalized pi cloud, planar hexagon, 1.39 Å bond length and sigma to pi ratios.
Building benzene from orbitals
Benzene has the formula $\mathrm{C_6H_6}$. Its structure is explained in three steps.
- Hybridization. Each of the six carbons is $sp^2$ hybridized. Its three $sp^2$ orbitals form three σ bonds: two C–C and one C–H. Because $sp^2$ orbitals lie at 120° in one plane, the molecule is a regular, flat (planar) hexagon and every bond angle is 120°.
- Unhybridized p orbitals. Each carbon keeps one p orbital, standing perpendicular to the plane of the ring.
- Delocalization. The six parallel p orbitals overlap sideways with both neighbours, not just one. This forms a continuous π molecular orbital. The six π electrons are delocalized over all six carbons, as ring-shaped (doughnut) clouds above and below the plane of the ring.
The "three alternate double bonds" of the Kekulé formula therefore do not exist as fixed bonds. They are delocalized bonds, often drawn as a circle inside the hexagon.
Consequences of delocalization
- Equal bond lengths: every C–C bond is 1.39 Å (139 pm). This lies between a C–C single bond (1.54 Å) and a C=C double bond (1.34 Å). The C–C bond order is about 1.5.
- Extra stability (resonance energy): benzene is more stable than a hypothetical cyclohexatriene with three fixed double bonds.
- Substitution rather than addition: addition would destroy the stable delocalized system. So benzene prefers electrophilic substitution, which keeps the ring intact. It does not decolourize bromine water like an alkene does.
- The electron-rich π cloud is what attracts electrophiles.
Counting bonds and electrons
| Quantity | Value |
|---|---|
| σ bonds | 12 (6 C–C + 6 C–H) |
| π bonds | 3 |
| σ : π bonds | 4 : 1 (12 : 3) |
| σ electrons | 24 |
| π electrons | 6 |
| σ : π electrons | 4 : 1 (24 : 6) |
Both ratios come out as 4 : 1, because each bond holds two electrons.
Comparison with other hydrocarbons
| Compound | Carbon hybridization | C–C bond length |
|---|---|---|
| Ethane | $sp^3$ | 1.54 Å |
| Benzene | $sp^2$ | 1.39 Å |
| Ethene | $sp^2$ | 1.34 Å |
| Ethyne | $sp$ | 1.20 Å |
Common MDCAT traps
- Benzene is flat. It is not a 3-D hexagon, and its angles are 120°, not 109.5°.
- 1.34 Å belongs to ethene, 1.54 Å to ethane and 1.20 Å to ethyne. Benzene is 1.39 Å.
- The π cloud lies above and below the ring plane, not between the C–C bonds in the plane.
- Do not forget the six C–H σ bonds when counting. The total is 12 σ, not 6.
- Its bonds are delocalized, not fixed, coordinate or ionic.
Quick revision
- All six carbons are $sp^2$, and the ring is a planar regular hexagon.
- There are six delocalized π electrons above and below the ring.
- All C–C bonds are equal at 1.39 Å.
- The σ : π ratio is 4 : 1 for both bonds and electrons.
- Delocalization makes benzene favour substitution.