Q7(a). Aromatic, anti-aromatic or non-aromatic
Classify the following examples as aromatic, anti-aromatic or non-aromatic. Write their structures also:
(i) Tetrahydrofuran (ii) Cyclooctatrienyl dianion (iii) Azulene (iv) Tropolone
Answer in brief
(i) THF: non-aromatic. (ii) C₈H₈²⁻: aromatic (10π, planar). (iii) Azulene: aromatic (10π). (iv) Tropolone: aromatic (6π tropylium form). None of the four is anti-aromatic.
Key points
- all four carbons are sp³; there is no π system round the ring. Non-aromatic (an ordinary cyclic ether).
- neutral cyclooctatetraene (8π) is tub-shaped and non-aromatic; with potassium it takes two electrons and becomes a planar regular octagon with 10π electrons (n = 2) spread over all eight carbons (equal C–C bonds, one ¹H NMR…
- a planar, fully conjugated 5–7 fused system with 10π electrons round the periphery (n=2). Its dipolar form, a tropylium (6π) ring fused to a cyclopentadienide (6π) ring, explains its dipole moment and its blue colour. Aromatic…
- Tropolone (2-hydroxycyclohepta-2,4,6-trien-1-one): the polar C=O gives a dipolar form, a hydroxytropylium ring with 6π electrons on seven sp² carbons. Nearly equal C–C bonds, pKₐ about 7 (more acidic than phenol, 10),…
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Reading of the question: “cyclooctatrienyl dianion” is the C₈H₈²⁻ ion (cyclooctatetraene dianion), written as a triene with two carbanion centres.
Introduction. Test for each species: is it cyclic, planar and fully conjugated (a p orbital on every ring atom)? If yes, count the π electrons: 4n+2 = aromatic, 4n = anti-aromatic. If the ring is not fully conjugated or not planar, it is non-aromatic.
Scheme: Structures; the circle in (ii) stands for 10 delocalised π electrons.
- Tetrahydrofuran (C₄H₈O, oxolane): all four carbons are sp³; there is no π system round the ring. Non-aromatic (an ordinary cyclic ether).
- Cyclooctatetraene dianion (C₈H₈²⁻): neutral cyclooctatetraene (8π) is tub-shaped and non-aromatic; with potassium it takes two electrons and becomes a planar regular octagon with 10π electrons (n = 2) spread over all eight carbons (equal C–C bonds, one ¹H NMR signal). Aromatic.
- Azulene (C₁₀H₈): a planar, fully conjugated 5–7 fused system with 10π electrons round the periphery (n=2). Its dipolar form, a tropylium (6π) ring fused to a cyclopentadienide (6π) ring, explains its dipole moment and its blue colour. Aromatic (substitution at C-1 and C-3, not addition).
- Tropolone (2-hydroxycyclohepta-2,4,6-trien-1-one): the polar C=O gives a dipolar form, a hydroxytropylium ring with 6π electrons on seven sp² carbons. Nearly equal C–C bonds, pKₐ about 7 (more acidic than phenol, 10), substitution rather than addition: aromatic.
Conclusion. (i) THF: non-aromatic. (ii) C₈H₈²⁻: aromatic (10π, planar). (iii) Azulene: aromatic (10π). (iv) Tropolone: aromatic (6π tropylium form). None of the four is anti-aromatic.
Application. The C₈H₈²⁻ ion is the ligand in uranocene, U(C₈H₈)₂; the azulene chromophore gives the blue of chamomile oil (chamazulene), and the tropolone ring occurs in colchicine.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram


All parts of Question 7
- 7(a) Aromatic, anti-aromatic or non-aromatic10 marks
- 7(b) OsO₄/pyridine hydroxylation of maleic and fumaric acids10 marks
- 7(c) Birch reduction: mechanism and regiochemistry15 marks
- 7(d) Radical chain polymerisation of vinyl chloride with benzoyl peroxide15 marks