Write the important resonance structures which contribute to the resonance hybrid of diazomethane and show how these can be used to rationalize the formation of methyl acetate from diazomethane and acetic acid.

First of all, we need to draw the diazomethane molecule. Diazomethane has the molecular formula CH2N2, which we can represent as CH2=N+=N·.

Now, we will draw the two important resonance structures of diazomethane: 1. CH2=N+=N· 2. CH2-+≡N-N· These structures show the delocalization of electrons in the diazomethane molecule.

Next, we need to draw the acetic acid molecule. Acetic acid has the molecular formula CH3COOH, which can be represented as CH3-C(=O)-OH.

To explain the formation of methyl acetate from diazomethane and acetic acid, let's follow these steps: 1. Proton transfer: In the first step, the unshared electron pair on the nitrogen atom in the diazomethane molecule attacks the hydrogen of the acetic acid molecule. This results in the formation of a \(CH_3C(=O)O^-\) ion (acetate ion) and a protonated diazomethane molecule. CH2=N+=N· + CH3-C(=O)-OH → CH2=N+=NH^+ + CH3-C(=O)O^- 2. Nucleophilic attack: The negatively charged oxygen atom of the acetate ion attacks the positively charged carbon atom of the protonated diazomethane molecule, forming a new carbon-oxygen bond. CH2=N+=NH^+ + CH3-C(=O)O^- → CH2-N+=NH - CH3-C(=O)-O 3. Proton transfer: Finally, the protonated nitrogen atom donates a proton to the negatively charged oxygen atom. CH2-N+=NH - CH3-C(=O)-O → CH2=N+=N· + CH3-C(=O)-OH The overall result is that diazomethane and acetic acid have reacted to form a new molecule, methyl acetate (CH3-C(=O)-OCH3) and nitrogen gas (N2). CH2=N+=N· + CH3-C(=O)-OH → CH3-C(=O)-OCH3 + N2 The resonance structures of diazomethane facilitate the nucleophilic attack of the acetate ion on the positively charged carbon atom in diazomethane, which leads to the formation of methyl acetate.