Question

The IUPAC name of the following compound is$\text{1-Chloro-3-methyl-but-3-en-2-ol}$.

A) True
B) False

Answer 1

Hint: The IUPAC nomenclature rules are used to name an organic compound. An alcohol$\text{OH}$ act functional group. Alcohol can be named by the replacement of ‘e’ in the parent chain by the ‘ol’.Here, the compound has the double bond therefore the primary suffix changes to ‘ene’ followed by the secondary suffix which is –ol.

Complete step by step answer:
The complete IUPAC nomenclature of organic compound is represented as:
$\text{Prefix+Word root+Primary Suffix+Secondary suffix}$
According to the IUPAC nomenclature, alcohols are named by replacing the ‘e’ in the name of parent alcohol by ‘ol’, $\text{Alkane-e+ol=Alkanol}$

In this system, the following rule is followed to name alcohol:
(i) The longest continuous chain containing the carbon atom bonded to $\text{OH}$ the group is selected as the parent chain. The $\text{OH}$ group is considered as the functional group in the molecule. It is added as the primary suffix. For alcohol the primary suffix is$\text{-Ol}$.
(ii) The carbon atoms in the chain are numbered in such a way that the carbon atom carrying the hydroxyl group ($\text{(-OH)}$ gets the lowest number.
(iii) The position of the substituents is indicated by suitable numbers.
We are given with the following organic molecule,

The compound contains both the functional group and a double bond.
1) Find out the word root: The word root represents the number of carbon atoms in the compound. First, select the longest continuous chain of carbon atoms. This is called the parent chain. Here, the longest chain is as numbered from ${{\text{C}}_{\text{1}}}\text{-}{{\text{C}}_{\text{4}}}$. The methyl group and chloro group are not a part of the parent chain thus they act as a substituent. Therefore, the word root is $\text{butane}$.

2) Here, the numbering can be done in two ways, one starting from the carbon next to the choro group or the other from the double-bonded carbon-carbon atom.
                        

Let us apply the lowest locant rule. According to which carbon carrying the substituents gets the lower number position. Thus the (II) set of locations is preferable.

3) The compound contains the carbon atom which bears the functional group, $\text{(-OH)}$.Thus the numbering of the carbon chain is done in such a way that the carbon with the functional group gets the lowest number. Here, the $\text{(-OH)}$ group can get a position at ${{\text{C}}_{\text{2}}}$ or ${{\text{C}}_{3}}$. Therefore we will prefer the (II) set over the (I).
4) The double bond is at the 4 positions. Therefore the primary suffix is modified as ‘ene’ along with the position of the double bond. Thus primary suffix can be written as $\text{3-ene}$
5) The secondary suffix is modified $\text{-Ol}$ along with the position of alcohol.
6) In the given compound, chlorine and methyl group are treated as the substituents. Thus they are always written before the word root along with their positions on the chain.
Thus, the chlorine group is written as $\text{1-Chloro}$ and the methyl as $\text{3-methyl}$
The substituents are always arranged in the alphabetical order.

Thus on combining we get the IUPAC name of the compound. We get,
$\begin{align}
  & \text{Prefix+Word root+Primary Suffix+Secondary suffix} \\
 & \text{1-Chloro-3-methylbut + 3-en + 2-ol} \\
\end{align}$

Thus IUPAC name of the compound is, $\text{1-Chloro-3-methyl-but-3-en-2-ol}$

The given nomenclature is true. So, the correct answer is “Option A”.

Note: Remember that the double bond position is taken as the carbon on which the double exists and has the lowest number on the chain. Note that extra ‘a’ is dropped from the word root if the secondary suffix starts with a consonant (a, e, i, o, u). Remember the following trend if you encounter a compound with a functional group, double triple bond, and substituents.
$\text{Functional group }\rangle \text{ Double bond }\rangle \text{ Triple bond }\rangle \text{ Substituents/side chains}$