Selectivity of catechol O-methyltransferase has been examined for the three ring-fluorinated norepinephrines to elucidate the role of acidity of the phenolic groups in their methylation. Substitution of fluorine at the 5-position of norepinephrine reverses the selectivity of catechol O-methyltransferase so that p-O-methylation predominates. The 5-fluoro substituent also causes the pKa of the p-hydroxyl group to decrease substantially. In contrast, 2- and 6-fluoronorepinephrines are methylated predominantly at the m-hydroxyl group. These results suggest that acidity of a phenolic group can play an important role in its ability to be methylated by catechol O-methyltransferase. Percentages of p-O-methylation of norepinephrine and its fluorinated derivatives increase with pH. This relative increase in p-O-methylation appears to accompany ionization of a group with pKa of 8.6, 7.7, 7.9, and 8.4 for norepinephrine and its 2-, 5-, and 6-fluoro derivative, respectively. These pKa values are the same as or similar to the pKa values of a phenolic hydroxyl group of these substrates. 3,4-Dihydroxybenzyl alcohol and its 5-fluoro derivative are O-methylated by catechol O-methyltransferase to form p- and m-O-methyl products in approximately 1:1 and 4:1 ratios, respectively, at all pH values. Based on the above results, a catechol-binding site model for catechol O-methyltransferase is proposed in which the two phenolic hydroxyl groups of catechol substrates are postulated to be approximately equally spaced from the methyl group of the cosubstrate S-adenosylmethionine.