The two-stage Kober reaction with oestriol, oestrone and oestradiol-17β has been investigated. The factors involved in the maximum production of colour are summarized. Reducing agents were important in both the first and second stages of the Kober reaction. Oestriol in small amounts did not form the red Kober colour when reducing agents were not present in the first stage. In the absence of water, sulphuric acid did not give optimal results in the first stage. Oestradiol-17β failed to give the Kober colour reaction in the presence of high concentrations of sulphuric acid. Optimal sulphuric acid-water ratios differed for the three oestrogens and were 76 % sulphuric acid for oestriol, 66 % acid for oestrone and 60 % acid for oestradiol-17β. The development of fluorescence was similarly affected by sulphuric acid concentration. The phenol-sulphuric acid reagent used by many workers appears to owe its efficacy to the action of the phenol, partly as a reducing agent and partly as a diluent of the sulphuric acid. Oestriol reacted in the first stage at a slower rate than oestrone and oestradiol. Further water and heating was usually required after the first stage (i.e. for the second stage of the Kober reaction) for complete formation of the red colour. Maximum intensity and stability of the red colour in the second stage was obtained in the presence of 50–60 % sulphuric acid and water. Under these optimum conditions the second heating time was not critical. In the second stage, concentrations of sulphuric acid lower than 50 % caused instability of the colour and fading during heating. In the second stage of the Kober reaction, reducing agents with oxidation-reduction potentials of the order of the hydroquinone-quinhydrone couple were also required for the production of maximum density and stability of the red colour. A colour method based on these findings is presented.