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Question 6
Data about the hydrogenation of cyclohexene and of benzene are given. H2 ΔHº = -120 kJ mol⁻¹ C6H6 + 3H2 ΔHº = -208 kJ mol⁻¹ 0 6 . 1 Explain the bonding in and t... show full transcript
Step 1
Answer
Benzene is composed of six carbon atoms, which are arranged in a planar hexagonal ring. Each carbon atom is sp² hybridized, forming three sigma (σ) bonds: one with an adjacent carbon atom and two with hydrogen atoms. The p orbitals on each carbon overlap to form a delocalized π system, which contributes to the stability of the molecule.
The shape of the benzene molecule is planar with bond angles of approximately 120°, resulting from the sp² hybridization. This geometry allows for optimal overlap of the orbitals, providing stability through resonance.
Comparatively, cyclohexa-1,3,5-triene is less stable due to the presence of alternating double bonds, which do not allow for the same level of resonance stabilization as in benzene. The total enthalpy of hydrogenation for benzene is notably less negative (-208 kJ mol⁻¹) than expected, demonstrating the stability provided by resonance. In contrast, cyclohexene (ΔHº = -120 kJ mol⁻¹) indicates that the additional stability from resonance is significant.
Step 2
Answer
The enthalpy of hydrogenation for cyclohexene is -120 kJ mol⁻¹. If we consider cyclohexa-1,3-diene to have two double bonds, we might initially expect its enthalpy of hydrogenation to be around -240 kJ mol⁻¹. However, the value is not exactly double due to the additional resonance stabilization in the diene.
A suggested value for the enthalpy of hydrogenation of cyclohexa-1,3-diene could be around -210 kJ mol⁻¹. This accounts for the presence of conjugation, which allows for some stability, thereby resulting in a less exothermic reaction than doubling the enthalpy of cyclohexene's hydrogenation. This resonance between the double bonds contributes to reducing the overall enthalpy change.
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