Chapter 21, Problem 74E

Interpretation Introduction

Interpretation:

The equation for the reaction between acetic acid and $1\text{-propanol}$ is to be written and the ester formed in this reaction is to be named. Whether the same reaction occurs between acetic acid and $2\text{-propanol}$ is to be determined.

Concept introduction:

Carboxylic acid reacts with alcohol to form an ester. This reaction is known as an esterification reaction or condensation reaction. In this reaction, a water molecule is lost. The general structure of an ester is as follows:

Here, R and ${\text{R}}^{\text{'}}$ represents alkyl groups. In an ester, both $\text{R}$ groups may be similar or different. In an ester, the carboxyl bonded alkyl group is derived from the carboxylic acid and the alkyl group that is directly bonded to the oxygen $\left(\text{O}\right)$ atom is derived from the alcohol.

The steps to write the IUPAC name of an ester are as follows:

Step 1: Identify the parent chain. The alkyl group from the parent acid represents the parent chain.

Step 2: Number the carbon atoms. Since the parent chain is derived from carboxylic acid, therefore; the name of the main acid is alkanoic acid. Drop the letters $\text{-ic acid}$ and add $\text{-ate}$ as a suffix at the end of main acid.

Step 3: Number the carbon atoms of the alcohol part from the carbon which is directly bonded to the oxygen atom. The alcohol part is named as ‘alkyl’.

Step 4: In the case of ester, an alkyl group of the alcohol is written first followed by the anion of carboxylic acid. The format to name an ester is as follows:

$\text{Alcohol alkyl group}+\text{acid anion}$