Animated Solution for Chemistry - Organic Chemistry: Which of the following molecules does not show stereoisomerism ?

Imagine you are a molecular detective. Your job is to figure out if a molecule can exist in multiple 3D spatial arrangements, even though its connectivity remains exactly the same. This phenomenon is known as stereoisomerism.

To crack this case, we need to look for two primary suspects: Geometrical Isomerism and Optical Isomerism.

Geometrical isomerism requires a restricted rotation, typically a carbon-carbon double bond ($C=C$). But that's not enough! Each carbon atom involved in that double bond must be attached to two different groups. If even one carbon has two identical groups, geometrical isomerism is impossible.

Optical isomerism, on the other hand, requires a chiral center. A chiral center is an $s{p}^3$ hybridized atom (usually carbon) that is bonded to four completely different groups. If you find a chiral center, the molecule will show optical isomerism.

Let's bring our first suspect to the interrogation room. The IUPAC name is 3,4-dimethylhex-3-ene.

When we draw its structure, we get $C{H}_3-C{H}_2-C(C{H}_3)=C(C{H}_3)-C{H}_2-C{H}_3$. The double bond is located right in the middle, between Carbon-3 and Carbon-4.

Let's inspect Carbon-3. It is attached to an ethyl group ($-C{H}_{2}C{H}_3$) and a methyl group ($-C{H}_3$). These are different! Now let's look at Carbon-4. It is also attached to an ethyl group and a methyl group. Since both carbons of the double bond satisfy the condition of having two different groups, this molecule will definitely exhibit Geometrical Isomerism (it can exist as cis and trans isomers).

Next up is 3-methylhex-1-ene. Its structure is $C{H}_2=CH-C{H}^{\ast }(C{H}_3)-C{H}_2-C{H}_2-C{H}_3$.

Right away, we notice a terminal double bond ($C{H}_2=$). Because the first carbon is attached to two identical hydrogen atoms, geometrical isomerism is immediately ruled out.

But wait, let's look deeper into the chain. Carbon-3 is an $s{p}^3$ hybridized carbon. What is it attached to? It has a hydrogen atom ($-H$), a methyl group ($-C{H}_3$), a vinyl group ($-CH=C{H}_2$), and a propyl group ($-C{H}_{2}C{H}_{2}C{H}_3$). All four groups are entirely different! This makes Carbon-3 a chiral center, meaning the molecule exhibits Optical Isomerism.

Now we arrive at 3-ethylhex-3-ene. This is where many students fall into a trap.

Let's carefully draw the structure: $C{H}_3-C{H}_2-C(C{H}_{2}C{H}_3)=CH-C{H}_2-C{H}_3$. The double bond is between Carbon-3 and Carbon-4.

Let's interrogate Carbon-3. It is attached to an ethyl group that is part of the main chain. But look at its substituent... it is also an ethyl group! Because Carbon-3 is attached to two identical ethyl groups, it completely fails the test for geometrical isomerism.

Furthermore, if we scan the entire molecule, we will not find a single $s{p}^3$ carbon attached to four different groups. There is no chiral center. Therefore, this molecule shows absolutely no stereoisomerism.

Just to be absolutely thorough, let's check the final option: 4-methylhex-1-ene.

Its structure is $C{H}_2=CH-C{H}_2-C{H}^{\ast }(C{H}_3)-C{H}_2-C{H}_3$. Similar to option (b), the terminal double bond prevents any geometrical isomerism.

However, if we inspect Carbon-4, we find it is attached to a hydrogen atom, a methyl group, an allyl group ($-C{H}_{2}CH=C{H}_2$), and an ethyl group. Four distinct groups! This makes Carbon-4 a chiral center, so the molecule exhibits Optical Isomerism.

After a rigorous investigation of all four molecules, we have our culprit. Molecule (c), 3-ethylhex-3-ene, lacks both the restricted rotation asymmetry required for geometrical isomerism and the chiral center required for optical isomerism.

This perfectly illustrates why you should never rely solely on IUPAC names. Always draw the skeletal structures to visually confirm the presence of identical groups or chiral centers!