Animated Solution for Chemistry - Organic Chemistry: Among the following, the number of aromatic compound (s) is-

Welcome to one of the most fascinating puzzles in organic chemistry! We are presented with nine different cyclic structures and asked a simple yet profound question: How many of these are aromatic?

To solve this, we must rely on our trusty guide, Hückel's Rule. For a compound to be crowned aromatic, it must pass four strict tests: it must be cyclic, it must be planar, it must have a fully conjugated continuous loop of p-orbitals, and it must possess exactly $4n+2$ $\pi$ electrons (where $n$ is an integer like 0, 1, 2, etc.). If a molecule is cyclic, planar, and fully conjugated but has $4n$ $\pi$ electrons, it falls into the dreaded category of anti-aromaticity—a state so unstable that molecules will often twist out of shape just to avoid it.

Let's put on our detective hats and examine each suspect one by one.

Our first suspect is cyclooctatetraene. At first glance, it looks like a larger version of benzene. It has four double bonds, which means $8$ $\pi$ electrons. Since $8$ is a multiple of $4$ ($4n$ system), this molecule is staring down the barrel of anti-aromaticity. To escape this highly unstable fate, it cleverly bends out of its flat plane into a 3D tub shape. Because it is no longer planar, the conjugation is broken, and it becomes simply non-aromatic.

Next, we look at the three-membered rings. The cyclopropenyl anion has a double bond ($2$ $\pi$ electrons) and a negative charge (a lone pair contributing $2$ more $\pi$ electrons). That's a total of $4$ $\pi$ electrons in a planar, continuous loop. Unfortunately, this makes it strictly anti-aromatic.

But what about its sibling, the cyclopropenyl cation? The double bond gives $2$ $\pi$ electrons, and the positive charge provides an empty p-orbital. This empty orbital perfectly completes the conjugation circuit without adding any extra electrons. With exactly $2$ $\pi$ electrons, it perfectly satisfies Hückel's rule ($4(0)+2=2$). We have found our first aromatic compound!

Moving on to 1,4-cyclohexadiene. Look closely at the top and bottom carbon atoms. They are $s{p}^3$ hybridized, meaning they don't have unhybridized p-orbitals to share. This completely shatters the continuous loop required for delocalization. Without full conjugation, it is non-aromatic.

Now, behold the tropylium ion, a majestic seven-membered ring. It boasts three double bonds ($6$ $\pi$ electrons), and the positive charge acts as a bridge to complete the circuit. Six is a classic Hückel number ($4(1)+2=6$). This is our second aromatic compound.

Let's examine the five-membered rings. The cyclopentadienyl cation has two double bonds ($4$ $\pi$ electrons) and an empty p-orbital. Being planar and fully conjugated with $4$ $\pi$ electrons makes it highly unstable and anti-aromatic.

Conversely, the cyclopentadienyl anion has the same two double bonds, but its negative charge contributes a lone pair to the $\pi$ system. This brings the total to $6$ $\pi$ electrons. It is fully conjugated, planar, and satisfies Hückel's rule. This is our third aromatic compound.

Finally, we enter the realm of fused polycyclic systems.

Naphthalene consists of two fused benzene-like rings. If we count the double bonds across the entire conjugated system, we find five of them. Five double bonds mean $10$ $\pi$ electrons. Since $10$ fits the $4n+2$ rule perfectly ($n=2$), naphthalene is our fourth aromatic compound.

Our last suspect is phenanthrene, a beautiful tricyclic system. Counting its double bonds reveals a total of seven. Seven double bonds yield $14$ $\pi$ electrons. Does $14$ fit Hückel's rule? Yes, it does! ($4(3)+2=14$). Therefore, phenanthrene is our fifth aromatic compound.

We have meticulously interrogated all nine structures. We found that the cyclopropenyl cation, the tropylium ion, the cyclopentadienyl anion, naphthalene, and phenanthrene all possess the magical stability of aromaticity.

Counting them up, we have exactly $5$ aromatic compounds. The mystery is solved!