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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2019 · Paper II

Q1(a). Is [14]annulene aromatic?

  • 10 marks
  • Answered
Question

Is [14] annulene aromatic according to Hückel rule? Justify your answer.

Answer in brief

Yes. [14]Annulene has 4n+2 = 14 π electrons (n=3) in a cyclic conjugated ring that is nearly planar. The inner-H repulsion twists it a little, so the aromatic character is weaker than in benzene, but the NMR ring current (inner H near δ -0.6) and equalised bond lengths show that it is aromatic.

Key points

  • Electron count. 14 π electrons fill the seven bonding π MOs of the ring completely (closed shell), as Hückel's rule requires for n=3.
  • Geometry. A 14-membered ring can take bond angles near 120° only with some trans double bonds. In the stable form four H atoms point into the ring (C-3, C-6, C-10, C-13).
  • Strain. These inner H atoms crowd each other (transannular H/H repulsion), so the ring is twisted slightly out of plane.
  • Diamagnetic ring current in the 1H NMR spectrum: the ring current shields the four inner protons, which appear far upfield (δ about -0.6), while the ten outer protons are deshielded and appear in the aromatic region. A…

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. Hückel rule: a cyclic, planar, fully conjugated ring with (4n+2) π electrons is aromatic. [14]Annulene, C₁₄H₁₄ (cyclotetradecaheptaene), has seven C=C bonds and 14 π electrons, which is 4n+2 with n=3. On electron count it is aromatic; the question is whether the ring can be flat enough.

  1. Electron count. 14 π electrons fill the seven bonding π MOs of the ring completely (closed shell), as Hückel's rule requires for n=3.
  2. Geometry. A 14-membered ring can take bond angles near 120° only with some trans double bonds. In the stable form four H atoms point into the ring (C-3, C-6, C-10, C-13).
  3. Strain. These inner H atoms crowd each other (transannular H/H repulsion), so the ring is twisted slightly out of plane.

Figure: [14]Annulene drawn on the pyrene outline: four H point into the ring.

Evidence that it is still aromatic

  • Diamagnetic ring current in the 1H NMR spectrum: the ring current shields the four inner protons, which appear far upfield (δ about -0.6), while the ten outer protons are deshielded and appear in the aromatic region. A non-aromatic polyene would show all its CH signals near δ 5–6.
  • No bond alternation: X-ray bond lengths lie between 1.35 and 1.41 Å, close to the benzene value, not alternating 1.34/1.46 Å as in a localised polyene.
  • The distortion from planarity is small and is concentrated at the four inner-H carbons, so the p orbitals still overlap round the ring.

The steric strain lowers the aromatic stabilisation: [14]annulene is less stable than benzene, it isomerises between two geometric forms on the NMR time scale, and it is weakly aromatic. When the inner hydrogens are replaced by a bridge (e.g. trans-15,16-dimethyl-15,16-dihydropyrene, a [14]annulene built round a saturated core), the periphery becomes almost planar and the system is strongly aromatic: equal bond lengths and typical substitution reactions.

Conclusion. Yes. [14]Annulene has 4n+2 = 14 π electrons (n=3) in a cyclic conjugated ring that is nearly planar. The inner-H repulsion twists it a little, so the aromatic character is weaker than in benzene, but the NMR ring current (inner H near δ -0.6) and equalised bond lengths show that it is aromatic.

Application. Ring-current shielding of protons inside or above an aromatic ring is used in NMR to recognise aromatic macrocycles such as porphyrins (inner N–H far upfield) and bridged annulenes.

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Model answer to Q1(a), 2019 Paper II: solution page 2
Solution page 2 of 23