Indian Forest Service (Mains), Chemistry optional · 2023 · Paper II
Q1(a). Aromaticity of heterocycles
Question
Classify the five printed heterocycles as aromatic, antiaromatic or non-aromatic. [structures of the five heterocycles (an N-methyl azirine, a borole, a silabenzene, an N-methylazepine and a borepine) given in the paper]
Answer in brief
Formal Hückel classification: I and II antiaromatic; III and V aromatic; IV non-aromatic in its non-planar ground-state structure.
Key points
- I: the N-methyl three-membered ring has one C=C bond and a nitrogen lone pair: four pi electrons if planar and conjugated, the antiaromatic count.
- II: the borole has two double bonds and an empty boron p orbital: four pi electrons in the conjugated circuit, hence antiaromatic.
- III: the silabenzene has three alternating pi bonds, six pi electrons, and the aromatic 4n + 2 count.
- IV: the N-methylazepine would have eight pi electrons if planar and fully conjugated. It avoids that arrangement by non-planarity and is non-aromatic.
- V: the borepine has three double bonds and an empty boron p orbital: six pi electrons, formally aromatic.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
CoversaromaticityantiaromaticityHückel ruleborolesilabenzeneazepine
All parts of Question 1
- 1(a) Aromaticity of heterocycles5 marks
- 1(b) Evidence for a tetrahedral intermediate5 marks
- 1(c) Benzyne capture by tetracyclone5 marks
- 1(d) Para-alkyl effects on benzyl solvolysis5 marks
- 1(e) Dehydrobromination through an ion pair5 marks
- 1(f) Why nucleophilic shifts are common5 marks
- 1(g) A double oxy-Cope rearrangement5 marks
- 1(h) Substituent effects in Cannizzaro reaction5 marks