Animated Solution for Chemistry - Organic Chemistry: Which of the following compounds shows geometrical isomerism?

Geometrical isomerism (often referred to as cis-trans isomerism) arises in molecules where free rotation around a bond is restricted. The most common scenario in organic chemistry is the carbon-carbon double bond ($C=C$). Because the $\pi$-bond locks the molecule in a planar geometry, the spatial arrangement of the groups attached to the doubly bonded carbons becomes fixed.

However, simply having a double bond is not enough. There is a strict mathematical and physical condition that must be met: Each of the doubly bonded carbon atoms must be attached to two different atoms or groups.

If we represent an alkene as $abC=Ccd$, it will only show geometrical isomerism if $a eq b$ and $c eq d$. If either $a=b$ or $c=d$, rotating the molecule 180 degrees in space will result in the exact same superimposable structure, meaning no isomers exist.

The Terminal Alkene Trap

When faced with a multiple-choice question like this, drawing the structures is your best weapon. Let's systematically break down the given options.

Option (a): 2-methylpent-2-ene If we draw the structure, $C{H}_3-C(C{H}_3)=CH-C{H}_2-C{H}_3$, we immediately notice the left carbon of the double bond. It is attached to two identical methyl ($-C{H}_3$) groups. Because these two groups are the same, swapping them doesn't create a new molecule. Thus, it fails the condition.

Options (c) and (d): 4-methylpent-1-ene and 2-methylpent-1-ene Here is a pro-tip for competitive exams: Terminal alkenes (1-enes) almost never show geometrical isomerism. Why? Because the double bond is at the end of the chain ($C{H}_2=C...$), meaning the terminal carbon is attached to two identical hydrogen atoms. Both options (c) and (d) fall into this trap and can be instantly eliminated.

Option (b): 4-methylpent-2-ene Let's draw its structure carefully: $C{H}_3-CH=CH-CH(C{H}_3{)}_2$.

Let's inspect the doubly bonded carbons: 1. The left carbon (C2): It is attached to a hydrogen atom ($-H$) and a methyl group ($-C{H}_3$). These are different. 2. The right carbon (C3): It is attached to a hydrogen atom ($-H$) and an isopropyl group ($-CH(C{H}_3{)}_2$). These are also different.

Since both carbons independently satisfy the condition of having two different groups, this molecule will exhibit geometrical isomerism. It can exist as a cis isomer (where the two hydrogen atoms are on the same side of the double bond) and a trans isomer (where the hydrogen atoms are on opposite sides).

By systematically applying the fundamental rule of geometrical isomerism and drawing the structures, we can confidently conclude that only 4-methylpent-2-ene has the necessary asymmetry around its double bond to exist as distinct geometrical isomers.