Daily answer writing · Sat, 31 Oct, 2026
Hückel's rule and Frost circles: why cyclopentadiene is acidic and tropylium bromide is ionic
State Hückel's $(4n+2)$ rule of aromaticity. Using Frost circle (polygon-in-circle) MO diagrams, explain why cyclopentadiene ($\text{p}K_a \approx 16$) is far more acidic than cycloheptatriene ($\text{p}K_a \approx 36$), and why cycloheptatrienyl bromide is an ionic solid. Also comment on the aromaticity of cyclooctatetraene and its dianion.
Syllabus: Common to CSE Paper II, IFoS Paper II and Geo-Scientist Mains Paper III (organic), with MO theory from Paper I: aromaticity, Hückel's rule, annulenes and aromatic ions.
Model answer
Hückel's rule
A cyclic, planar, fully conjugated system (a p orbital on every ring atom) is aromatic if it has $(4n+2)$ π electrons (2, 6, 10 …) and antiaromatic if it has $4n$. Non-planar rings are non-aromatic.
Frost circles
Hückel energies for an N-ring: $E_k = \alpha + 2\beta\cos\dfrac{2\pi k}{N}$ (β < 0). For N = 5: $\alpha+2\beta$, $\alpha+0.618\beta$ (×2), $\alpha-1.618\beta$ (×2). For N = 7: $\alpha+2\beta$, $\alpha+1.247\beta$ (×2), $\alpha-0.445\beta$ (×2), $\alpha-1.802\beta$ (×2).
Acidity
$\ce{C5H6 -> C5H5^- + H+}$: the anion has 6π electrons filling all three bonding MOs, a closed shell (n = 1) → aromatic, strongly stabilised, so $\text{p}K_a \approx 16$, close to water.
Loss of H+ from cycloheptatriene gives $\ce{C7H7^-}$ with 8π: two electrons enter the antibonding pair at $\alpha-0.445\beta$ (4n, antiaromatic if planar). No stabilisation, so it is about $10^{20}$ times weaker as an acid.
Tropylium bromide
Loss of $\ce{Br^-}$ gives $\ce{C7H7^+}$: 6π electrons fill the three bonding MOs → aromatic cation. So $\ce{C7H7Br}$ exists as $\ce{[C7H7]^+Br^-}$: high-melting, water-soluble, gives AgBr at once.
Cyclooctatetraene
8π (4n). If planar, two electrons would occupy two non-bonding MOs (at α) → antiaromatic diradical. So COT folds into a tub, with alternating bonds (1.34 and 1.47 Å): non-aromatic, reacts like a polyene. $\ce{C8H8^2-}$ (from K) has 10π, is planar and aromatic.
Conclusion. Filling all bonding MOs with (4n + 2) electrons, as the Frost circle shows, is what makes C₅H₅⁻ and C₇H₇⁺ unusually stable.
Where the marks are
- Hückel's rule with the planarity and conjugation conditions; antiaromatic and non-aromatic defined (2 marks).
- Frost circles for C5 and C7 rings with correct filling (2 marks). A diagram is expected.
- Acidity of cyclopentadiene vs cycloheptatriene from 6π anion vs 8π anion (2 marks).
- Ionic tropylium bromide from the 6π cation, with evidence (2 marks).
- COT tub-shaped non-aromatic; dianion aromatic (2 marks).