Animated Solution for Chemistry - Organic Chemistry: Which of the following compound will exhibit geometrical isomerism?

Geometrical isomerism, often referred to as cis-trans isomerism, is a fascinating spatial property of certain molecules. But what exactly makes a molecule capable of showing this behavior? It boils down to two non-negotiable conditions.

First, the molecule must possess a site of restricted rotation. In organic chemistry, this is most commonly provided by a carbon-carbon double bond ($C=C$). Unlike single bonds, which allow atoms to spin freely like a wheel on an axle, double bonds lock the atoms into a rigid, planar geometry.

Second, and this is the condition where most students stumble, each of the two carbon atoms involved in the double bond must be attached to two different groups. Mathematically, if we represent the double bond as $abC=Ccd$, the condition demands that $a eq b$ and $c eq d$. If even one of the double-bonded carbons is attached to two identical groups (like two hydrogen atoms), the molecule cannot exhibit geometrical isomerism.

When tackling multiple-choice questions in competitive exams, time is your most valuable asset. Let's look at option (d): 1,1-diphenyl-1-propane.

Notice the suffix "-ane". This immediately tells us that the molecule is an alkane. It is a completely saturated hydrocarbon consisting entirely of single bonds. Because there is no double bond, there is no restricted rotation. Without restricted rotation, geometrical isomerism is fundamentally impossible. We can confidently eliminate option (d) without even drawing its structure.

Now, let's evaluate the remaining options. Option (b) is 3-phenyl-1-butene and option (c) is 2-phenyl-1-butene.

Both of these compounds share a common structural feature: they end in "-1-butene". This indicates that they are terminal alkenes, meaning the double bond is located at the very end of the carbon chain, specifically between $C_1$ and $C_2$.

Let's focus on $C_1$. In any terminal alkene, the terminal carbon atom is bonded to two identical hydrogen atoms ($=C{H}_2$). Because these two groups are the same, the crucial second condition for geometrical isomerism is violated. Therefore, neither option (b) nor option (c) can exist as cis or trans isomers.

Finally, we arrive at option (a): 1-phenyl-2-butene.

Here, the double bond is internal, situated between $C_2$ and $C_3$. Let's meticulously examine the groups attached to each of these carbons:

Carbon-2 ($C_2$) is attached to a hydrogen atom ($-H$) and a benzyl group ($-C{H}_{2}Ph$). These are two different groups. Carbon-3 ($C_3$) is attached to a hydrogen atom ($-H$) and a methyl group ($-C{H}_3$). These are also two different groups.

Since both carbons of the double bond are attached to two distinct groups, this molecule perfectly satisfies all conditions. It can exist in two distinct spatial arrangements: the cis isomer (where the bulky groups are on the same side) and the trans isomer (where they are on opposite sides).