Question

Explain why lidocaine does not hydrolyze at roomtemperature in aqueous environment ph = 7.2?

Answer 1

The degradation of lidocaine in aqueous solution obeys the expression kobs = (kH+[H +] + ko ) [H+]/([H + ] + Ka + k′oKa([H + ] + Ka) where kH+ is the rate constant for hydronium ion catalysis, and ko and k′oare the rate constants for the spontaneous (or water-catalyzed) reactions of protonated and free-base lidocaine. At 80°C, the rate constants for these processes are 1.31 × 10−7M−l sec−1, 1.37 × 10−9 sec−1, and 7.02 × 10−9sec−1; the corresponding activation energies are 30.5, 33.8, and 26.3 kcal mol−1, respectively. It was found that the room temperature pH of maximum stability is ∼3–6 and that lidocaine is more reactive in the presence of metal ions such as Fe2+and Cu2+. The dissociation constant, Ka, for lidocaine at 25–80°C was also measured at 0.1 M ionic strength and a plot of pKa versus 1/T gave a slope of (1.88 ± 0.05) × 103 K−1 and intercept 1.56 ± 0.16.