Phthalic anhydride is treated with: (i) $\mathrm{CH}{}_3\mathrm{OH}$ (1 equiv), room temperature; (ii) $\mathrm{SOCl}{}_2$; (iii) $\mathrm{CH}{}_3\mathrm{NH}{}_2$ (2 equiv), 0\,^\circ\mathrm{C}. The final product is

(i) ring-opening by methanol gives 2-(methoxycarbonyl)benzoic acid. (ii) converts the free $\mathrm{COOH}$ to $\mathrm{COCl}$ (ester untouched). (iii) The amine converts $\mathrm{COCl}$ to the N-methylamide (second equivalent takes up HCl). The ester remains, so the product is methyl 2-(methylcarbamoyl)benzoate.