Animated Solution for Chemistry - Organic Chemistry: The IUPAC name of the compound is

Imagine you are an explorer mapping out a newly discovered island. In the world of organic chemistry, molecules are these islands, and the IUPAC nomenclature system is our universal map-making guide. It ensures that no matter where you are in the world, a specific chemical name points to one, and only one, molecular structure. Today, we are going to decode the IUPAC name of a cyclic organic compound step by step.

When we look at the skeletal structure provided in the problem, the first thing that stands out is the central ring. It is a closed loop made of six carbon atoms with no double bonds inside it. This forms the foundation of our molecule, known as a cyclohexane ring.

However, this ring isn't bare. It has attachments, or 'substituents', branching off from it. To name this molecule correctly, we must identify these attachments and determine their exact addresses on the ring.

The golden rule of IUPAC nomenclature is to establish a hierarchy. Not all attachments are created equal. We must scan the molecule for the principal functional group, as this group will dictate both the suffix of our name and the starting point for our numbering.

In our molecule, we see an $-OH$ group attached to one of the ring carbons. This is a hydroxyl group, which classifies our compound as an alcohol. Because alcohols sit higher on the IUPAC priority list than simple carbon branches (alkyl groups), the $-OH$ group takes the crown as the principal functional group.

Consequently, the carbon atom of the cyclohexane ring that is directly bonded to this $-OH$ group is granted the VIP status of being Carbon-1 ($C_1$).

Now that we have our starting point, let's survey the rest of the ring. Two carbons away from our hydroxyl group, we notice a carbon atom with two straight lines branching outward.

In skeletal structures, the end of a line represents a methyl group ($-C{H}_3$). Since there are two lines originating from the exact same carbon, we have a gem-dimethyl group. These are our secondary substituents, and we need to figure out their exact locant (number) on the ring.

Lowest Locant Rule

We are standing at Carbon-1 and need to number the rest of the ring. We have two choices: walk clockwise or walk counter-clockwise. Which path is the correct one?

This is where the Lowest Locant Rule comes into play. This rule dictates that we must number the parent chain in the direction that assigns the lowest possible numbers to the substituents.

Let's test both paths:

Path A (Clockwise): If we move clockwise, the next carbon is $C_2$, and the carbon bearing the two methyl groups becomes $C_3$. This gives our methyl substituents the locants $3,3$.

Path B (Counter-Clockwise): If we move counter-clockwise, we have to travel much further around the ring. The carbon with the methyl groups would end up being $C_5$. This gives our substituents the locants $5,5$.

Comparing the two sets, $3$ is clearly lower than $5$. Therefore, the clockwise path is the undisputed winner. Our methyl groups are officially located at Carbon-3.

Like building a puzzle, we now have all the pieces required to construct the final IUPAC name. We build it in three parts: Prefix, Parent, and Suffix.

1. The Prefix: This part lists our substituents. We have two methyl groups at position 3. To indicate there are two identical groups, we use the multiplier 'di'. Thus, our prefix is $3,3$-dimethyl.

2. The Parent: Our main structural framework is a six-membered saturated ring. This gives us the parent name cyclohexane.

3. The Suffix: Because our principal functional group is an alcohol, we drop the terminal 'e' from cyclohexane and add the suffix -ol.

When we seamlessly stitch these components together, we get:

The '$1$' is placed right before the parent name to explicitly state the position of the hydroxyl group. While in simple cyclic alcohols the '$1$' is sometimes omitted because it is understood by default, including it leaves absolutely zero room for ambiguity.

Looking at our options, this perfectly matches option (c). By systematically applying the rules of priority and lowest locants, we've successfully navigated the map of this molecule!