Animated Solution for Chemistry - Organic Chemistry: Which compound (s) out of the following is/are not aromatic?

Welcome to the fascinating world of aromaticity! When we look at cyclic organic compounds, some possess an extraordinary level of stability. We call these compounds aromatic. But how do we distinguish the truly aromatic compounds from the imposters?

To determine if a compound is aromatic, it must pass a strict three-part test: 1. Cyclic and Planar: The molecule must form a closed ring and lie flat in a single plane. 2. Complete Conjugation: Every atom in the ring must have an unhybridized p-orbital, allowing a continuous, unbroken loop of delocalized $\pi$ electrons. 3. Huckel's Rule: The total number of $\pi$ electrons in the conjugated system must equal $4n+2$, where $n$ is an integer ($0,1,2,\dots$).

If a compound is cyclic, planar, and fully conjugated but has $4n$ $\pi$ electrons, it is highly unstable and is termed anti-aromatic. If it fails the cyclic, planar, or conjugation test altogether, it is simply non-aromatic.

Let's put our four suspects to the test.

The Cyclopropenyl Cation

Imagine a three-membered ring with one double bond and a positive charge on the third carbon. This is the cyclopropenyl cation.

The double bond contributes $2$ $\pi$ electrons. What about the positive charge? A carbocation means there is an empty p-orbital. This empty orbital acts as a bridge, allowing the $\pi$ electrons to delocalize completely around the ring.

Let's check Huckel's rule:

Since $n$ is an integer ($0$), this compound perfectly satisfies all conditions. Compound A is aromatic.

The Cyclopentadienyl Cation

Next, we have a five-membered ring with two double bonds and a positive charge. This is the cyclopentadienyl cation.

The two double bonds give us $4$ $\pi$ electrons. The positive charge again provides an empty p-orbital, ensuring complete conjugation around the ring.

However, when we apply Huckel's rule:

Since $n$ is not an integer, it fails Huckel's rule. Instead, it fits the $4n$ rule ($4\times 1=4$). Because it is fully conjugated but has $4n$ $\pi$ electrons, Compound B is anti-aromatic.

Cyclooctatetraene

Now look at the eight-membered ring with four alternating double bonds. At first glance, it looks like a perfect conjugated system with $8$ $\pi$ electrons.

If it were planar, it would fit the $4n$ rule ($4\times 2=8$) and be anti-aromatic. However, anti-aromaticity brings severe instability. To escape this fate, cyclooctatetraene twists itself out of a flat plane into a "tub" shape. By breaking its own planarity, it breaks the continuous overlap of p-orbitals.

Because it is not planar, Compound C is non-aromatic.

Cycloheptatriene

Finally, we examine a seven-membered ring with three double bonds.

The three double bonds provide $6$ $\pi$ electrons, which looks promising for Huckel's rule ($4n+2=6⟹n=1$). But there is a fatal flaw. The top carbon atom in the ring is bonded to two hydrogens (not explicitly drawn, but implied). It has only single bonds, meaning it is $s{p}^3$ hybridized.

An $s{p}^3$ hybridized carbon does not have an unhybridized p-orbital. It acts as a roadblock, completely shattering the continuous loop of conjugation. Without complete conjugation, aromaticity is impossible.

Therefore, Compound D is non-aromatic.

Our investigation reveals that only Compound A possesses the magical stability of aromaticity. Compounds B, C, and D fail the test for various reasons (anti-aromaticity, lack of planarity, and broken conjugation).

Thus, the compounds that are not aromatic are (B), (C), and (D).