SolutionsBenzene's Puzzle
1. Heats of formation
Calculate the heats of formation of cyclohexene, 1,4-cyclohexadiene, 1,3-cyclohexadiene and benzene from their heats of combustion: −896.9, −853.3, −854.6 and −780.8 kcal/mol, respectively (with liquid water as product). What conclusions can you draw?
Data: C(graphite) + O2 → CO2, ΔH° = −94.1 kcal/mol; H2 + ½ O2 → H2O(l), ΔH° = −68.3 kcal/mol. The heat of formation is the difference between the heat released by the combustion of the elements and that released by the combustion of the compound: ΔH°f = ΔH°comb(elements) − ΔH°comb(compound).
| Compound | ΔH°comb | Combustion of the elements | ΔH°f |
|---|---|---|---|
| Cyclohexene (C6H10) | −896.9 | 6(−94.1) + 5(−68.3) = −906.1 | −9.2 |
| 1,4-Cyclohexadiene (C6H8) | −853.3 | 6(−94.1) + 4(−68.3) = −837.8 | +15.5 |
| 1,3-Cyclohexadiene (C6H8) | −854.6 | 6(−94.1) + 4(−68.3) = −837.8 | +16.8 |
| Benzene (C6H6) | −780.8 | 6(−94.1) + 3(−68.3) = −769.5 | +11.3 |
Cyclohexene is more stable than its elements, and each additional double bond raises the heat of formation by about 25 kcal/mol. By extrapolation, a "cyclohexatriene" with three non-interacting double bonds would have ΔH°f ≈ +42 kcal/mol; benzene has +11.3, i.e. it is about 30 kcal/mol more stable than expected.
This is the same conclusion reached with the heats of hydrogenation: benzene is not a cyclohexatriene but an especially stable aromatic system.
2. Kekulé isomers
If benzene were a cyclohexatriene rapidly interconverting between its two Kekulé structures, how many isomers of 1,2-dimethylbenzene should be isolable? How many are known?
With localized double bonds there would be two 1,2-dimethylbenzenes: one with a double bond between the carbons bearing the methyl groups and one with a single bond between them. Only one is known, because the six C–C bonds are identical (1.39 Å): benzene is a resonance hybrid, not an equilibrium between two structures.
3. Hydrogenation
The hydrogenation of cyclohexene releases 28.6 kcal/mol and that of benzene 49.3. Calculate the resonance energy of benzene.
Three isolated double bonds would release 3 × 28.6 = 85.8 kcal/mol. The difference, 85.8 − 49.3 = 36.5 kcal/mol, is the resonance energy (aromatic stabilization) of benzene. If the small conjugation stabilization of a triene (about 3 × 2.3 kcal/mol) is subtracted, about 30 kcal/mol remain that are due to aromaticity.