 Solution

First, proton transfer from acetic acid to the carbon of diazomethane gives methyldiazonium and an acetate ion:Draw an electron-pair arrow from carbon, using the carbon-negative resonance contributor, to the acidic hydrogen, and another from the acid O–H bond to oxygen. Second, an acetate oxygen attacks the diazonium methyl carbon by an SN2 reaction; an arrow from the C–N bond to nitrogen expels neutral ${\mathrm{N}}_2$:The acid's hydroxyl hydrogen has become the third hydrogen of the newly installed methyl group. This is a methyl-transfer sequence driven by formation of molecular nitrogen, without forming a free methyl carbocation or invoking Fischer esterification.