Assertion (A): The compound cyclooctatetraene has a cyclic structural formula (a puckered, tub-shaped eight-me

NCERT Exemplar · Q44

Q. Assertion (A): The compound cyclooctatetraene has a cyclic structural formula (a puckered, tub-shaped eight-membered ring). It is cyclic and has conjugated 8π-electron system but it is not an aromatic compound. Reason (R): (4n + 2) π electrons rule does not hold good and ring is not planar.

(i) Both A and R are correct and R is the correct explanation of A.

(ii) Both A and R are correct but R is not the correct explanation of A.

(iii) Both A and R are not correct.

(iv) A is not correct but R is correct.

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Concept understanding — Benzene Aromatic Stability

Benzene Aromatic Stability

Imagine you have a ring of six carbon atoms, each holding one hydrogen atom. That's benzene — $C_6H_6$ . Now, if you drew it with alternating single and double bonds (the Kekulé structure), you'd expect it to behave like any other alkene: reactive, ready to add things across those double bonds.

But benzene doesn't behave that way. It's stubbornly unreactive toward addition reactions. It burns with a sooty flame, yes, but it resists the kind of chemistry that typical alkenes love. Why?

The Intuition: A Circle of Electrons

The key is that the double bonds in benzene aren't really "fixed" in place. The six $p$ orbitals (one on each carbon, perpendicular to the ring) overlap sideways to form a continuous ring of electron density — a delocalised $\pi$ system. Picture a doughnut of negative charge above and below the plane of the carbon atoms.

This delocalisation spreads the electrons out, lowering the energy of the molecule. It's like having six people in a room who can all share one big, comfortable sofa instead of being forced into three separate, cramped chairs. The shared arrangement is far more stable.

The Precise Statement

Benzene is aromatic — a term that describes a cyclic, planar molecule with a continuous ring of overlapping $p$ orbitals containing $4n+2$ $\pi$ electrons (Hückel's rule, where $n$ is a whole number). For benzene, $n=1$ , so it has $6$ $\pi$ electrons.

\text{Aromatic stability} = \text{Resonance energy} \approx 150 \text{ kJ/mol}

This resonance energy is the extra stability benzene has compared to a hypothetical "cyclohexatriene" with three fixed double bonds. It's not a small effect — it's about the energy of a strong covalent bond.

Why It Matters

Because benzene is so stable, it doesn't undergo addition reactions (which would break the aromatic ring). Instead, it undergoes electrophilic substitution — a reaction that preserves the aromatic system. This is the single most important reaction in aromatic chemistry. …

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Cyclooctatetraene (not cyclooctane) has 8π electrons in a conjugated system but is non-aromatic because it fails both Hückel criteria: it does not satisfy the $(4n+2)$ rule and adopts a non-planar tub conformation. Both assertion and reason are correct, and the reason explains the assertion.

The question tests your understanding of aromaticity through Hückel's rule. A compound is aromatic only when it satisfies all of the following conditions:

The molecule must be cyclic.
The molecule must be planar (or nearly so) to allow orbital overlap.
The molecule must have a fully conjugated system of p-orbitals.
The molecule must contain (4n + 2) π electrons where $n = 0, 1, 2, 3, \ldots$

When any one of these fails, aromaticity is lost. The compound in question is almost certainly cyclooctatetraene (COT), $\ce{C8H8}$ , not cyclooctane $\ce{C8H16}$ (which has no π system at all). Cyclooctatetraene has alternating single and double bonds around an eight-membered ring, giving it 8π electrons.

Watch out

The assertion says "cyclooctane" but describes a conjugated π system. Cyclooctane is a saturated hydrocarbon with no double bonds. The intended molecule is cyclooctatetraene.

Why cyclooctatetraene is not aromatic

Count the π electrons.

Cyclooctatetraene has four $\ce{C=C}$ double bonds, contributing $4 \times 2 = 8$ π electrons.
Check the $(4n+2)$ rule.

For aromaticity, the number of π electrons must equal $4n+2$ . Setting $8 = 4n+2$ gives $n = 1.5$ , which is not an integer. So cyclooctatetraene has $4n$ electrons ( $n=2$ ), characteristic of antiaromatic systems if planar.
Examine the geometry.

A planar eight-membered ring with 8π electrons would be antiaromatic and highly unstable. To avoid this destabilisation, cyclooctatetraene distorts into a non-planar, tub-shaped (or "puckered") conformation. This breaks the continuous overlap of p-orbitals, eliminating both aromaticity and antiaromaticity. The molecule behaves as a typical polyene with localised double bonds. …

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Q.Assertion (A): The compound cyclooctatetraene has a cyclic structural formula (a puckered, tub-shaped eight-membered ring). It is cyclic and has conjugated 8π-electron system but it is not an aromatic compound. Reason (R): (4n + 2) π electrons rule does not hold good and ring is not planar.

Why cyclooctatetraene is not aromatic

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Q.Assertion (A): The compound cyclooctatetraene has a cyclic structural formula (a puckered, tub-shaped eight-membered ring). It is cyclic and has conjugated 8π-electron system but it is not an aromatic compound.
Reason (R): (4n + 2) π electrons rule does not hold good and ring is not planar.

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