Question
Benzene has the molecular formula C₆H₆. Draw the two Kekulé structures of benzene and explain why benzene is more stable than either structure suggests.
Solution — Step by Step
For C₆H₆, the degree of unsaturation = (2×6 + 2 − 6) / 2 = 4. That means a ring plus three double bonds. This is our first hint that benzene isn’t just a simple cycloalkene.
Place six carbons in a regular hexagon. Assign alternating double bonds — say, between C1–C2, C3–C4, and C5–C6. Each carbon gets one hydrogen. This is Kekulé structure I.
Shift all three double bonds by one position — now they sit between C2–C3, C4–C5, and C6–C1. Every carbon still has one H. This is Kekulé structure II.
These two structures are mirror images in terms of double bond placement, and neither one alone correctly represents benzene.
Benzene is the resonance hybrid of both Kekulé structures. We represent this with a double-headed arrow (↔) between them, or by drawing a circle inside the hexagon. The actual molecule has all six C–C bonds equal in length (140 pm — between a single bond at 154 pm and a double bond at 134 pm).
Benzene is 36 kJ/mol more stable than cyclohexatriene would be if it existed. This extra stabilisation is the resonance energy (also called delocalisation energy). It is why benzene undergoes substitution, not addition — breaking that delocalisation would cost too much energy.
Why This Works
The six π electrons in benzene are not locked between specific pairs of carbons. Each carbon contributes one unhybridised p orbital, and all six overlap sideways to form a continuous π cloud above and below the ring. This is delocalisation — the electrons belong to the whole ring, not to any one bond.
When two valid Lewis structures can be written that differ only in electron placement (not atom position), the real molecule is a weighted average of both. Neither Kekulé structure alone can show equal C–C bond lengths, but their hybrid can.
Experimental hydrogenation of benzene gives −208 kJ/mol, while cyclohexene (one double bond) gives −120 kJ/mol. If benzene had three isolated double bonds, we’d expect −360 kJ/mol. The difference of 152 kJ/mol is the resonance stabilisation (NCERT uses 36 kcal/mol ≈ 150 kJ/mol — same idea).
Alternative Method — Molecular Orbital View
Instead of resonance structures, we can use MO theory. The six p orbitals combine to form six π molecular orbitals — three bonding, three antibonding. All six π electrons fill the three bonding MOs completely.
A completely filled set of bonding MOs with no antibonding electrons = maximum stabilisation. This is the MO language for what resonance theory calls “delocalisation energy.” For board exams, the resonance picture is enough. For JEE Advanced or NEET conceptual questions, knowing the MO argument gives you an edge.
The circle-in-hexagon symbol is the shorthand for the resonance hybrid. In organic mechanisms, always use this symbol for benzene — drawing Kekulé structures in reaction schemes can confuse the electron-pushing arrows.
Common Mistake
Writing resonance structures with atoms in different positions. Students sometimes draw one Kekulé structure and then move a carbon or hydrogen when drawing the second. That would make them structural isomers, not resonance structures. In resonance, only electron pairs move — the nuclear skeleton is frozen. If your two structures have atoms in different positions, they are not resonance structures of the same compound.
A related error: calling the two Kekulé structures “isomers of benzene.” They are not — benzene is a single compound with one structure (the hybrid). Kekulé structures are just our imperfect pen-and-paper representation of a molecule that doesn’t fit standard Lewis theory.
Final Answer: Benzene has two Kekulé resonance structures with alternating single and double bonds. The actual molecule is the resonance hybrid, with all C–C bonds equal (140 pm) and a resonance stabilisation energy of approximately 36 kcal/mol (150 kJ/mol). This delocalisation of six π electrons over the ring makes benzene exceptionally stable.