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Chapter 18: Problem 12

Esters can also react with amines to give amides. Write a mechanism for the reaction of methyl acetate (CH3COOCH3) and ammonia to form acetamide (CH3CONH2).

Short Answer

Expert verified
Ammonia attacks the carbonyl carbon, forms a tetrahedral intermediate, ejects methanol, and tautomerizes to form acetamide.

Step by step solution

01

- Nucleophilic Attack

Ammonia (NH3) acts as a nucleophile and attacks the carbonyl carbon of methyl acetate (CH3COOCH3). This results in the formation of a tetrahedral intermediate.
02

- Proton Transfer

The oxygen atom in the tetrahedral intermediate gets protonated by a proton transfer from the incoming ammonia molecule, generating a positively charged tetrahedral intermediate.
03

- Elimination of Methanol

The tetrahedral intermediate collapses, ejecting methanol (CH3OH) and forming an imidic acid intermediate (CH3C(OH)=NH).
04

- Tautomerization

The imidic acid tautomerizes to form acetamide (CH3CONH2) by shifting a hydrogen atom from the nitrogen to the oxygen, creating a more stable amide structure.

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Key Concepts

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

Nucleophilic Attack
The first step in forming an amide from an ester and ammonia involves a nucleophilic attack. Here, ammonia (NH3) serves as the nucleophile because it donates a pair of electrons to another atom. Specifically, ammonia targets the carbonyl carbon of methyl acetate (CH3COOCH3). The carbonyl carbon is electrophilic, meaning it's an attractive site for the electron-rich ammonia to attack. This forms a four-membered, unstable intermediate known as a tetrahedral intermediate.
Proton Transfer
Following the nucleophilic attack, the tetrahedral intermediate that forms isn't stable. To stabilize it, a proton transfer occurs. In this step, an ammonia molecule donates a proton (H+) to the oxygen atom within the tetrahedral intermediate. This conversion makes the intermediate positively charged. Proton transfers like this one help make the molecules more stable energetically, which is crucial for the reaction to proceed effectively.
Tetrahedral Intermediate
The tetrahedral intermediate plays a vital role in the mechanism. When ammonia attacks the carbonyl carbon, the structure of the molecule becomes more complex but less stable. Initially, the carbon has four different groups attached to it, resembling a tetrahedral geometry—hence the name. The intermediate must stabilize temporarily to allow the next steps in the reaction. Though brief, this intermediate is essential for progressing from reactants to products.
Methanol Elimination
In the next phase, the tetrahedral intermediate undergoes a significant change through the elimination of methanol (CH3OH). This means the intermediate collapses and expels a methanol molecule. The result is an imidic acid intermediate (CH3C(OH)=NH). This elimination is key because it removes unnecessary components, simplifying the structure and making subsequent steps possible. Methanol elimination is a crucial step for achieving the final amide product.
Tautomerization
The final step in forming acetamide (CH3CONH2) from methyl acetate and ammonia involves tautomerization. Tautomerization is a process where a hydrogen atom shifts from one position to another within a molecule, altering the positions of double bonds. Starting from the imidic acid (CH3C(OH)=NH), a hydrogen moves from the nitrogen to the oxygen. This shift makes the structure more stable, converting it into the final amide product. This step solidifies the transformation into acetamide, ensuring the molecule is in its most stable form.

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