0 votes
18 views
ago in Organic Chemistry by (2.4k points)
What is the mechanism of formation of N-methylacetamide from N-methylamine and acetic anhydride?

1 Answer

0 votes
ago by (2.3k points)
The formation of N-methylacetamide from N-methylamine and acetic anhydride is an example of a nucleophilic acyl substitution reaction. Here's the step-by-step mechanism for this reaction:1. Nucleophilic attack: The nitrogen atom in N-methylamine  CH3NH2  acts as a nucleophile, attacking the electrophilic carbonyl carbon in acetic anhydride   CH3CO 2O . This results in the formation of a tetrahedral intermediate and breaking of the carbonyl double bond   bond .2. Proton transfer: The negatively charged oxygen atom in the tetrahedral intermediate abstracts a proton from the nitrogen atom in the N-methylamine moiety, forming a hydroxyl group  -OH  and restoring the positive charge on the nitrogen atom.3. Leaving group departure: The hydroxyl group  -OH  in the tetrahedral intermediate acts as a leaving group, and the carbonyl double bond   bond  is reformed. This results in the expulsion of acetate ion  CH3COO-  as a leaving group.4. Proton transfer: The positively charged nitrogen atom in the N-methylacetamide  CH3NHCOCH3  abstracts a proton from the acetate ion  CH3COO- , forming the final N-methylacetamide product and an acetic acid molecule  CH3COOH .Overall, the reaction can be summarized as follows:N-methylamine  CH3NH2  + Acetic anhydride   CH3CO 2O   N-methylacetamide  CH3NHCOCH3  + Acetic acid  CH3COOH
Welcome to Sarvan Science Q&A, where you can ask questions and receive answers from other members of the community.
...