The kinetics of the hydrolysis of 1,8-bis(trifluoroacetylamino)naphthalene to 1-amino-8-trifluoroacetylaminonaphthalene have been studied in 70%, 80% and 90%(v/v) Me2SO–H2O in the presence of hydroxide ion. Under these conditions the product of the reaction is present as the amide anion and does not undergo further reaction. Approximate values of pK1 5.7 and pK2 15.9 have been measured in 70%(v/v) Me2SO–H2O for dissociation of the amide protons of 1,8-bis(trifluoroacetylamino)naphthalene. The hydrolysis involves reaction of the amide monoanion because negligible concentrations of the undissociated amide are present and the dianion is unreactive. The dependence of the first-order rate coefficient (kobs) on [OH–] in 70% and 80%(v/v) Me2SO–H2O is fitted by the expression kobs=(k0+k1[OH–]+k2[OH–]2)/(1 +K″[OH–]) where K″ is the equilibrium constant for proton removal by hydroxide ion from the monoanion of 1,8-bis(trifluoroacetylamino)naphthalene to give the dianion. The magnitude of k0 in 70%(v/v) Me2SO–H2O suggests that the attack of water on the amide carbonyl at the 1-position is assisted by intramolecular catalysis involving the amide anion at the 8-position. The data in 90%(v/v) Me2SO–H2O are fitted by the expression kobs=(k2/K″)[OH–]. Under these conditions the amide is practically fully dissociated into the dianion and the k0 and k1[OH–] terms are negligibly small in comparison with the k2[OH–]2 term.