Chapter 17, Problem 1PP

Practice Problem 17.1

Give an IUPAC systematic name for each of the following:

(a)

(b)

(c)

(d)

(e)

Interpretation Introduction

Interpretation:

The IUPAC name for each given carboxylic compound is to be identified.

Concept introduction:

The carboxylic group $\left(-{\text{CO}}_{\text{2}}\text{H}\right)$ or $\left(-\text{COOH}\right)$ is one of the most widely occurring functional groups in the study of organic chemistry. It is the parent group of a large number of acyl compounds and carboxylic acids. Most of the carboxylic acids have common names that have been used since a long time. But they are also given the names derived using IUPAC system along with their microscopic structure, and are used by the science world.

Answer to Problem 1PP

Solution:

$2-\text{Methyl butanoic acid}$

$\left(Z\right)-\text{3}-\text{Pentenoic acid}$

$\text{Sodium}-4-\text{Bromobutanoate}$

$5-\text{Phenylpentanoic acid}$

$\left(E\right)-3-\text{Ethyl}-3-\text{pentenoic acid}$

Explanation of Solution

a)

Start numbering the chain beginning from the carbon of the carboxylic group.

This chain contains four carbons and so the base name is butanoic acid. In the position-2, it has a methyl group attached. Therefore, the compound is $2-\text{Methyl butanoic acid}$.

b)

The numbering of the chain begins from the carbon of the carboxylic group.

This chain contains five carbons. So, the compound belongs to pentanoic acid. Here, in position-3, it has a double bond. Thus, by considering this double bond and its stereochemistry, the compound identified is $\left(Z\right)-\text{3}-\text{Pentenoic acid}$.

c)

The numbering of the chain begins from the carbon of the carboxylic group.

This chain contains four carbons, and so, the name of the compound must be butane. Here, the hydrogen of the carboxylic group is replaced by sodium ion $\left(\text{Na}\right)$, and thus the compound is a carboxylic salt. It also contains bromide ion at the end of the chain at position-5. Therefore, by considering all these conditions, the name of the compound is $\text{Sodium}-4-\text{Bromobutanoate}$.

d)

The numbering of the chain begins from the carbon of the carboxylic group.

This chain contains five carbons, and so, it is pentanoic acid. It has a phenyl compound at position-5. Thus, the name of the compound is $5-\text{Phenylpentanoic acid}$

e)

The numbering of the chain begins from the carbon of the carboxylic group.

This chain contains five carbons. So, the name is pentanoic acid. There is a double bond and an ethyl group in position-3. Therefore, by considering the above conditions and the stereochemistry involved, the name of the compound is $\left(E\right)-3-\text{Ethyl}-3-\text{pentenoic acid}$.