Question

What reagents can be used to prepare the ester \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}\) ?

Short answer

The reagents needed to prepare the ester \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}\) are Propanoic acid (\( \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COOH}\)), Methanol (\( \mathrm{CH}_{3}\mathrm{OH}\)), and the catalyst is concentrated Sulphuric acid (\( \mathrm{H}_{2}\mathrm{SO}_{4}\)).

Step-by-step solution

Identifying the Carboxylic Acid

The ester is formed by esterification process which involves a carboxylic acid and an alcohol. Since the resultant ester is \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}\), we can recognize the two original components by identifying the unique parts of the ester. The \( \mathrm{CO}_{2}\) part indicates the carboxylic acid. On the left side of \( \mathrm{CO}_{2}\), it's \( \mathrm{CH}_{3} \mathrm{CH}_{2}\) which is the simplified structural form of propanoic acid (CH3CH2COOH). So, the carboxylic acid used in the formation of this ester is Propanoic acid.

Identifying the Alcohol

The right side of the \( \mathrm{CO}_{2}\) part provides the clue for the alcohol used. Here it is \( \mathrm{CH}_{3}\), which indicates Methanol (\( \mathrm{CH}_{3} \mathrm{OH}\)). That means, the alcohol used in the formation of this ester is Methanol.

Identifying the Catalyst

Esterification usually requires a catalyst to speed up the reaction. A commonly used catalyst in this reaction is concentrated sulphuric acid \( \mathrm{H}_{2}\mathrm{SO}_{4}\).

Final Reaction

The complete reaction to formulate the ester \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{CH}_{3}\) from propanoic acid and methanol in the presence of \( \mathrm{H}_{2}\mathrm{SO}_{4}\) is: \[ \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{COOH} + \mathrm{CH}_{3}\mathrm{OH} \rightarrow \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CO}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\].

Key concepts

Carboxylic Acid

Carboxylic acids are organic compounds that play a critical role in chemical reactions like esterification. They are defined by the presence of a carboxyl group, denoted as \( -COOH \). This functional group is highly reactive and forms the basis for the carboxylic acid's properties.
In our original exercise, the carboxylic acid involved is Propanoic acid. Propanoic acid is a simple, three-carbon carboxylic acid with the formula \( \text{CH}_3\text{CH}_2\text{COOH} \). It provides the \( \text{CH}_3\text{CH}_2 \) chain in the formation of the ester.
The structure of carboxylic acids enables them to lose a proton (\( H^+ \)), which facilitates their participation in esterification. When they react with alcohols in the presence of an acid catalyst, esters are formed, with water being a byproduct of this reaction.

Methanol

Methanol, known chemically as \( \text{CH}_3\text{OH} \), is a simple form of alcohol used in many chemical processes, including esterification. It is the alcohol component that reacts with the carboxylic acid to form an ester.
In the original problem, Methanol provides the \( \text{CH}_3 \) group that links with Propanoic acid to form the ester\( \text{CH}_3\text{CH}_2\text{CO}_2\text{CH}_3 \). This formation occurs through the reaction of methanol's hydroxyl (-OH) group with the carboxyl (-COOH) group of the acid, resulting in the release of water.
Methanol is a small, single-carbon alcohol which is notable for its ability to dissolve in water, making it a versatile solvent in numerous reactions. However, in the context of esterification, its role is pivotal in contributing the alcohol component that bonds with the acid to create the ester.

Sulphuric Acid Catalyst

A catalyst is a substance that increases the speed of a chemical reaction without being consumed in the process. In esterification reactions, a common catalyst is concentrated sulphuric acid (\( \text{H}_2\text{SO}_4 \)). Its primary function is to protonate the carboxyl group of the acid, enhancing its electrophilicity and making it more reactive to the alcohol.
Sulphuric acid as a strong acid helps in breaking the O-H bond in the carboxylic acid, allowing the alcohol to easily attack and bind. This results in the removal of water and the formation of the ester.
The use of sulphuric acid is crucial because it not only speeds up the reaction but also shifts the equilibrium towards the formation of the ester rather than the reactants. This makes it highly effective in promoting ester formation, ensuring a successful and efficient esterification process.