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Chapter 18: Problem 36

Write a mechanism for the acid-catalyzed formation of ethyl hexanoate from ethanol and hexanoic acid.

Short Answer

Expert verified
Ethyl hexanoate is formed through the acid-catalyzed reaction of ethanol and hexanoic acid, involving protonation, nucleophilic attack, proton transfer, water elimination, and deprotonation.

Step by step solution

01

Protonation of the Carboxyl Group

The reaction begins with the protonation of the carbonyl oxygen of hexanoic acid by an acid catalyst, such as sulfuric acid. This makes the carbonyl carbon more electrophilic.
02

Nucleophilic Attack by Ethanol

Ethanol, acting as a nucleophile, attacks the carbonyl carbon of the protonated hexanoic acid, forming a tetrahedral intermediate.
03

Proton Transfer

A proton transfer occurs within the intermediate, making the hydroxyl group a better leaving group by converting it into water.
04

Formation of Water and Elimination

Water is eliminated from the intermediate, leading to the formation of an oxonium ion.
05

Deprotonation

Finally, the oxonium ion loses a proton, forming ethyl hexanoate.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Protonation
The acid-catalyzed esterification mechanism starts with protonation. Here, the carbonyl oxygen of the hexanoic acid molecule gets protonated by the acid catalyst (often sulfuric acid).
When protonation occurs, the lone pairs on the carbonyl oxygen grab a proton (H⁺) from the acid catalyst.
This process makes the carbonyl carbon much more electrophilic.
Being more electrophilic means that the carbonyl carbon is now more prone to react with nucleophiles because it has a stronger positive charge.
Nucleophilic Attack
Ethanol acts as a nucleophile in the next step.
A nucleophile is a chemical species that donates an electron pair to form a chemical bond.
Ethanol attacks the carbonyl carbon of the protonated hexanoic acid.
This nucleophilic attack changes the shape of the molecule and leads to the formation of a tetrahedral intermediate.
This is because the carbon-oxygen double bond (C=O) opens up, allowing the ethanol to attach.
Proton Transfer
In the intermediate formed in the previous step, a proton transfer takes place.
This transfer turns the hydroxyl (-OH) group into a much better leaving group.
The proton from the hydroxyl group moves to another area within the intermediate, often to an oxygen atom connected to the incoming parts.
This transfer is crucial because it changes the -OH group into water (H₂O), which is more likely to leave the molecule.
Elimination
After the proton transfer, water is eliminated from the tetrahedral intermediate.
The bond between the carbon and the hydroxyl group breaks, freeing water from the structure.
This elimination process leads to the formation of an oxonium ion.
It is called an oxonium ion because it has an oxygen atom with a positive charge, usually represented as R-O-H₂⁺.
Deprotonation
The final step in the mechanism is deprotonation.
The oxonium ion loses a proton (H⁺), which neutralizes its charge.
This deprotonation results in the formation of the ester, ethyl hexanoate.
The final product is stable and has no extra positive charge.
At the end of this process, ethyl hexanoate and water are formed, completing the acid-catalyzed esterification.

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Most popular questions from this chapter

Devise syntheses for the following molecules. You may start with benzene, methyl alcohol, sodium methoxide, ethyl alcohol, sodium ethoxide, butyl alcohol (BuOH), phosgene, and propanoic acid as your sources of carbon. You may also use any inorganic reagent and solvents as needed.

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