Animated Solution for Chemistry - Organic Chemistry: Which one of the following structures has the IUPAC name 3-ethynyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid ?

A Visual Journey

When faced with a complex IUPAC name and a set of skeletal structures, it is easy to get overwhelmed by the zigzag lines and multiple bonds. The key to solving these problems flawlessly is to break the name down into its fundamental components: the principal functional group, the main carbon chain, and the substituents.

Let's start by looking at the very end of the name: 3-ethynyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid. The suffix is -oic acid. This immediately tells us that the principal functional group is a carboxylic acid ($-COOH$). According to IUPAC nomenclature rules, the carbon atom of the carboxylic acid group is given the highest priority and is designated as Carbon-1.

Next, we look for the root word, which is hex. This indicates that our longest continuous carbon chain contains exactly six carbon atoms.

But this isn't just a simple alkane chain. The name contains 3-en and 5-yn. This means there is a double bond starting at Carbon-3 and a triple bond starting at Carbon-5. Therefore, our main chain skeleton looks like this:

It is crucial to remember that the principal chain must contain the maximum number of multiple bonds, even if a longer chain exists without them. In this case, our 6-carbon chain perfectly accommodates both the double and the triple bond.

Now, we systematically attach the substituents to our main chain based on their locants:

1. 2-hydroxy: An $-OH$ group is attached to Carbon-2. 2. 3-ethynyl: An ethynyl group ($-C\equiv CH$) is attached to Carbon-3. 3. 4-methyl: A methyl group ($-C{H}_3$) is attached to Carbon-4.

With our theoretical structure built, we can now evaluate the given options. We are looking for a structure where the central double bond connects Carbon-3 and Carbon-4, with an ethynyl group on $C_3$ and a methyl group on $C_4$.

If we carefully trace the skeletal structures: - Option (A) has an ethyl group on $C_3$ and two ethynyl groups on $C_4$. Incorrect. - Option (B) has an ethyl group on $C_4$ and a methyl group on $C_3$. Incorrect. - Option (C) has a methyl group on $C_3$ and two ethynyl groups on $C_4$. Incorrect. - Option (D) has an ethynyl group on $C_3$ and a methyl group on $C_4$. This matches our derived connectivity perfectly!

While the given IUPAC name did not specify stereochemistry (like E or Z), it is a great exercise to determine it for our correct structure. In Option (D), if we assign Cahn-Ingold-Prelog priorities to the groups on the double bond: - At $C_3$, the $-CH(OH)COOH$ group has a higher priority than the $-C\equiv CH$ group. - At $C_4$, the $-C\equiv CH$ group (part of the main chain) has a higher priority than the $-C{H}_3$ group.

Since both high-priority groups are on the same side of the double bond, Option (D) represents the (Z)-isomer. Mastering these systematic steps ensures you can confidently tackle any nomenclature problem JEE throws your way.