Animated Solution for Chemistry - Organic Chemistry: Out of the following, the alkene that exhibits optical isomerism is

An Introduction to Optical Isomerism

Imagine looking into a mirror. You raise your right hand, and your reflection raises its left hand. No matter how you twist or turn, you can never perfectly superimpose your right hand onto your left hand. They are non-superimposable mirror images.

In the fascinating world of organic chemistry, molecules can exhibit this exact same property! We call this optical isomerism. When a molecule cannot be superimposed on its mirror image, it is said to be chiral. These chiral molecules have the unique ability to rotate the plane of polarized light, a property that has profound implications in biology and medicine. But how do we spot a chiral molecule just by looking at its structural formula?

What Makes a Carbon Special?

The secret to identifying optical isomerism in simple organic molecules lies in finding a chiral center.

A chiral center is typically a carbon atom that meets two strict conditions. First, it must be $s{p}^3$ hybridized, meaning it forms four single bonds with a tetrahedral geometry. Second, and most importantly, it must be bonded to four completely different groups or atoms. If even two of the attached groups are identical, the molecule will have a plane of symmetry, rendering it achiral (optically inactive).

Therefore, our mission in this problem is straightforward: we need to draw the structures of the given alkenes and hunt for that elusive chiral carbon.

The Process of Elimination

Let's systematically evaluate the options provided in the question.

Option (a): 3-methyl-2-pentene The structure is $C{H}_3-CH=C(C{H}_3)-C{H}_2-C{H}_3$. The carbons involved in the double bond (C2 and C3) are $s{p}^2$ hybridized, so they cannot be chiral. Carbon-4 is a $-C{H}_2-$ group, meaning it has two identical hydrogen atoms. Carbon-1 and Carbon-5 are $-C{H}_3$ groups, possessing three identical hydrogens. No chiral center here!

Option (b): 4-methyl-1-pentene The structure is $C{H}_2=CH-C{H}_2-CH(C{H}_3{)}_2$. Again, the double-bonded carbons are out. Carbon-3 has two hydrogens. Carbon-4 is bonded to a hydrogen, an allyl group, and two identical methyl groups. Because of those two identical methyl groups, it fails the chiral check.

Option (d): 2-methyl-2-pentene The structure is $C{H}_3-C(C{H}_3)=CH-C{H}_2-C{H}_3$. The $s{p}^3$ carbons are either $-C{H}_3$ or $-C{H}_2-$ groups. None of them are bonded to four different groups.

Unveiling 3-methyl-1-pentene

Now, let's turn our attention to Option (c): 3-methyl-1-pentene.

The structural formula is $C{H}_2=CH-CH(C{H}_3)-C{H}_2-C{H}_3$. Let's zoom in and carefully examine Carbon-3. What exactly is attached to this specific carbon atom?

1. Pointing upwards, we have a simple hydrogen atom ($-H$). 2. Pointing downwards, there is a methyl group ($-C{H}_3$). 3. To the left, it is attached to a vinyl group ($-CH=C{H}_2$). 4. To the right, it is bonded to an ethyl group ($-C{H}_{2}C{H}_3$).

Take a moment to visualize this. Hydrogen, methyl, vinyl, and ethyl. All four groups are entirely distinct!

Because Carbon-3 is an $s{p}^3$ hybridized atom bonded to four completely different groups, it perfectly satisfies the condition for chirality. We denote this chiral center with an asterisk ($C^{\ast }$).

Since the molecule possesses a chiral center and lacks any internal plane of symmetry, it will exist as two non-superimposable mirror images (enantiomers). Therefore, 3-methyl-1-pentene is the alkene that exhibits optical isomerism.

By mastering the art of visualizing molecular structures and systematically checking for chiral centers, you can confidently conquer any stereochemistry problem that comes your way!