Animated Solution for Chemistry - Organic Chemistry: The correct combination of names for isomeric alcohols with molecular formula ${\text{C}}_4{\text{H}}_{10}\text{O}$ is/are-

When we are given a molecular formula like ${\text{C}}_4{\text{H}}_{10}\text{O}$, our first instinct should always be to calculate the Degree of Unsaturation (DU). This simple calculation tells us if we are dealing with any rings or double bonds.

The formula for DU is $C+1-\frac{H}{2}+\frac{N}{2}-\frac{X}{2}$. Plugging in our values, we get $4+1-\frac{10}{2}=0$.

A DU of zero is fantastic news! It means our molecule is completely saturated and acyclic. Given the single oxygen atom, the possible functional groups are either alcohols or ethers. Since the question specifically asks for "isomeric alcohols," we can confidently ignore the ethers and focus entirely on drawing all possible alcohol structures for a four-carbon skeleton.

Let's start with the simplest skeleton: a straight chain of four carbon atoms.

If we attach the hydroxyl ($-\text{OH}$) group to the terminal carbon (carbon-1), we get ${\text{CH}}_3-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$. The common name for this is n-butyl alcohol (where 'n' stands for normal, indicating a straight chain). According to IUPAC rules, this is named butan-1-ol, but it is also frequently referred to as n-butanol. Looking at our choices, Option (C) pairs "n-butanol and butan-1-ol", which is a perfect match.

What if we move the $-\text{OH}$ group to the second carbon? We get ${\text{CH}}_3-{\text{CH}}_2-\text{CH}(\text{OH})-{\text{CH}}_3$. This is sec-butyl alcohol (or secondary butyl alcohol), and its IUPAC name is butan-2-ol. While this is a valid isomer, it doesn't appear in our options, so we move on.

Now, let's introduce branching. We can create a three-carbon main chain with a methyl group attached to the central carbon.

If we place the $-\text{OH}$ group on one of the terminal carbons, the structure is ${\text{CH}}_3-\text{CH}({\text{CH}}_3)-{\text{CH}}_2-\text{OH}$. The common name for this specific arrangement is isobutyl alcohol. To find its IUPAC name, we number the longest chain containing the $-\text{OH}$ group, giving the $-\text{OH}$ the lowest possible number. This makes it 2-methylpropan-1-ol. Option (D) pairs "isobutyl alcohol and 2-methylpropan-1-ol", which is absolutely correct.

Finally, what if we attach the $-\text{OH}$ group to the central carbon of our branched skeleton? The structure becomes $({\text{CH}}_3{)}_3\text{C}-\text{OH}$. The carbon holding the $-\text{OH}$ is attached to three other carbons, making it a tertiary alcohol. Hence, its common name is tert-butyl alcohol or tert-butanol.

Naming tert-butanol using IUPAC rules presents an interesting scenario. The standard approach is to find the longest continuous carbon chain containing the $-\text{OH}$ group, which is three carbons long. Numbering it to give the $-\text{OH}$ the lowest number, we get 2-methylpropan-2-ol. This makes Option (A) correct.

However, there is an older, alternative IUPAC convention that is still technically valid. In this system, you can name complex alcohols as derivatives of simpler ones. If we consider the parent chain to be just two carbons long (ethane), the first carbon holds the $-\text{OH}$ group and two methyl groups. This leads to the name 1,1-dimethylethan-1-ol. Because this is a recognized systematic name, Option (B) is also correct.

In JEE Advanced, "Multiple Correct" questions are designed to test the absolute depth of your knowledge. We have systematically proven that every single option provided is a valid pair of names for one of the isomers of ${\text{C}}_4{\text{H}}_{10}\text{O}$. Therefore, the correct answer includes all four options: (A), (B), (C), and (D).