Animated Solution for Chemistry - Organic Chemistry: Among the following compounds, geometrical isomerism is exhibited by

Geometrical isomerism is a fascinating phenomenon in organic chemistry where molecules with the exact same connectivity of atoms differ in their spatial arrangement. This occurs primarily due to restricted rotation around a bond, most commonly a carbon-carbon double bond ($C=C$) or within a rigid ring structure.

For a molecule to exhibit geometrical isomerism across a double bond, it must satisfy a strict mathematical and physical condition: each carbon atom of the double bond must be attached to two completely different groups. If we represent the double bond as $abC=Ccd$, the condition dictates that $a eq b$ and $c eq d$. If either carbon is attached to two identical groups, the molecule possesses a plane of symmetry that renders cis-trans distinctions impossible.

In this problem, we are presented with cyclohexane rings featuring an exocyclic double bond—a double bond where one carbon is part of the ring, and the other is outside it. Let's systematically evaluate the options based on our core condition.

Starting with Option (a), we look at the terminal carbon of the double bond (the one outside the ring). It is bonded to two identical hydrogen atoms ($H,H$). Because these two groups are the same, the molecule immediately fails the condition for geometrical isomerism. No matter how we arrange the rest of the molecule, swapping the two hydrogens produces the exact same structure.

Clockwise vs Anti-clockwise

When dealing with the carbon atom that is part of the ring, determining whether its two attached "groups" are different requires a unique approach. We must treat the two directions around the ring—clockwise and anti-clockwise—as the two separate groups.

Let's apply this to Option (c). The terminal carbon has a hydrogen atom and a chlorine atom, which are different. Great! Now, let's trace the ring from the double-bonded ring carbon. If we travel clockwise, we encounter a $-C{H}_2-$ group, then another $-C{H}_2-$ group, and finally the carbon bearing the methyl group ($-CH(C{H}_3)-$ at position 4). If we travel anti-clockwise, we encounter the exact same sequence: $-C{H}_2-$, then $-C{H}_2-$, and then the $-CH(C{H}_3)-$ group. Because both paths are perfectly identical, the ring carbon is effectively attached to two identical groups. Thus, Option (c) cannot show geometrical isomerism.

Now, let's examine Option (b). The terminal carbon again has different groups ($H$ and $Cl$). We trace the ring paths once more.

Traveling clockwise, we immediately hit a $-C{H}_2-$ group, and right after that, we arrive at the carbon with the methyl group ($-CH(C{H}_3)-$ at position 3). However, if we travel anti-clockwise, we must pass through three consecutive $-C{H}_2-$ groups before finally reaching that same methyl-bearing carbon.

The clockwise path and the anti-clockwise path in Option (b) are fundamentally different. This means the ring carbon is attached to two distinct "groups". Since both carbons of the double bond satisfy the condition ($a eq b$ and $c eq d$), the molecule in Option (b) will exhibit geometrical isomerism, existing as distinct cis and trans (or E and Z) isomers. This elegant interplay of ring symmetry and double bond rigidity is a classic concept in organic stereochemistry.