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Q9.12(i):

Explain why the following systems are not aromatic?

(i) 

Solution :

Analysis of Aromaticity Requirements

To determine the aromaticity of a cyclic system, we apply the Hückel's Rule, which mandates that a molecule must satisfy four specific criteria simultaneously:

  • Cyclic Structure: The molecule must contain at least one ring.
  • Planarity: All atoms in the ring must be $sp^2$ or $sp$ hybridized, allowing for the alignment of $p$-orbitals.
  • Conjugation: There must be a continuous cycle of overlapping $p$-orbitals throughout the ring.
  • Hückel’s Criterion: The $\pi$-electron system must contain $(4n + 2)$ electrons, where $n = 0, 1, 2, \dots$ (i.e., $2, 6, 10, \dots$ electrons).

Step 1: Structural Evaluation of the Provided System

The provided system is 1,3,5,7-cyclooctatetraene ($C_8H_8$).

Cyclooctatetraene

Step 2: Verification of Criteria

Criterion Observation for Cyclooctatetraene
Cyclic Yes, it is an 8-membered ring.
$\pi$-Electron Count The molecule has 4 double bonds, equating to $4 \times 2 = 8$ $\pi$-electrons. [Per Hückel's Rule, $4n + 2 \neq 8$].
Planarity The molecule adopts a "tub" shape conformation to minimize steric strain between hydrogen atoms, rendering it non-planar.

Step 3: Conclusion on Aromaticity

The system fails the requirements for aromaticity based on two primary factors:

1. Electron Count: With 8 $\pi$-electrons, the system follows the $4n$ rule ($n=2$), which characterizes anti-aromatic character if the molecule were planar. However, it fails the $4n+2$ requirement for aromaticity. [By Hückel's Molecular Orbital Theory].

2. Structural Distortion: Because a planar 8-membered ring with $sp^2$ carbons would experience significant angle strain and transannular repulsion, the molecule undergoes puckering. By exiting the planar geometry, the continuous overlap of $p$-orbitals is disrupted, effectively breaking the conjugation cycle. Consequently, the molecule is classified as non-aromatic rather than anti-aromatic.

Final Solution: The system is non-aromatic because it contains 8 $\pi$-electrons (violating the $4n+2$ rule) and adopts a non-planar 'tub' conformation to relieve steric strain, thereby preventing the continuous cyclic conjugation of $p$-orbitals.


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