Animated Solution for Chemistry - Organic Chemistry: Which of the following molecules is least resonance stabilised?

In the fascinating world of organic chemistry, resonance acts as a molecular safety net. The more a molecule can delocalise its $\pi$ electrons, the more stable it becomes. The ultimate pinnacle of this stability is a phenomenon known as aromaticity.

When a molecule is aromatic, its $\pi$ electrons flow freely in a continuous, closed loop, creating a highly stable electron cloud. Conversely, if a molecule fails to meet the criteria for aromaticity, it is considered non-aromatic and lacks this exceptional resonance stabilization. Our mission in this problem is to identify the least resonance stabilised molecule among four candidates, which simply means finding the one that is non-aromatic.

The Ultimate Stabilizer

To determine if a molecule is aromatic, we rely on Hückel's Rule. A molecule must satisfy three strict conditions: 1. It must be cyclic and planar. 2. It must be fully conjugated, meaning every atom in the ring must possess an unhybridized p-orbital (typically $s{p}^2$ or $sp$ hybridized) to allow continuous electron flow. 3. It must contain exactly $(4n+2)\pi$ electrons, where $n$ is an integer ($0,1,2,\dots$).

Let's put our four contenders to the test.

Option (a): Benzene Benzene is the poster child of aromaticity. It is a perfectly planar hexagonal ring with alternating double and single bonds. This arrangement provides exactly $6\pi$ electrons ($n=1$). Every carbon is $s{p}^2$ hybridized, allowing flawless conjugation. Benzene is highly resonance stabilised.

Option (b): Furan Furan is a five-membered heterocyclic ring containing an oxygen atom. It has two double bonds, which contribute $4\pi$ electrons. The oxygen atom possesses two lone pairs. Crucially, one of these lone pairs resides in an unhybridized p-orbital that aligns perfectly with the ring's $\pi$ system. This brings the total to $6\pi$ electrons. Furan is aromatic and highly stable.

Option (c): Pyridine Pyridine resembles benzene but replaces one carbon with a nitrogen atom. The three double bonds within the ring provide the necessary $6\pi$ electrons. The nitrogen atom has a lone pair, but because nitrogen is already using its p-orbital for the double bond, the lone pair is forced into an $s{p}^2$ orbital that lies in the plane of the ring. It is orthogonal to the $\pi$ system and does not participate in resonance. Nevertheless, the ring itself has $6\pi$ electrons, making pyridine aromatic.

Breaking the Conjugation

Option (d): Cyclohexa-2,5-dien-1-one At first glance, this molecule has a ring and some double bonds. However, we must look closely at the carbon atom at the very top of the ring (opposite the carbonyl group).

This carbon is bonded to two adjacent ring carbons via single bonds. To satisfy its valency of four, it must also be bonded to two hidden hydrogen atoms. With four single (sigma) bonds and zero pi bonds, this carbon is $s{p}^3$ hybridized.

An $s{p}^3$ hybridized carbon lacks an unhybridized p-orbital. It acts as a physical roadblock, completely breaking the continuous loop of conjugation required for aromaticity. Because the $\pi$ electrons cannot delocalise around the entire ring, the molecule is non-aromatic.

Without the profound stabilizing effect of aromaticity, cyclohexa-2,5-dien-1-one is the least resonance stabilised molecule among the choices. Therefore, option (d) is the correct answer.