| Cue Column | Note-Taking Column |
|---|---|
| What is the C-C bond length in benzene? | 1.39 Å — intermediate between C-C single (1.54 Å) and C=C double (1.34 Å). All six bonds are identical. |
| Why are all bonds equal? | Pi electrons are fully delocalized over the entire ring. No alternating single-double bonds exist in reality despite Kekule's proposal. |
| What does the Kekule structure represent? | A resonance contributor only — not the actual structure. The true benzene is the resonance hybrid. |
| What geometry does benzene adopt? | Planar hexagonal ring; bond angles = 120°; all C and H atoms are coplanar. |
| SMILES of benzene? | c1ccccc1 (aromatic notation) or C1=CC=CC=C1 (Kekule notation) |
Wikimedia Diagram: Benzene — Hybridization and π Orbital System
C C C C C C H H H H H H $sp^{2}$ hybridized C 120° bond angles All C–C bonds equal: 1.40 Å (between single 1.54 Å and double 1.34 Å) · Delocalized π cloud (6 electrons)Summary (bottom): Benzene's six C-C bonds are all 1.39 Å because its 6 pi electrons are completely delocalized over the ring in a continuous pi-electron cloud above and below the plane. The Kekule structures are resonance contributors, not reality. This delocalization is the source of benzene's exceptional thermodynamic stability (resonance energy ≈ 36 kcal/mol).