Parsing substituent pKa values and determining charge states at pH 2 and 7.Mapping the three pKa values to charge states at pH 2 and 7.4.Extending the pKa mapping to net charge at pH 2 and 7.4 with Henderson-Hasselbalch.Calculating percent ionization and net charge for three groups at pH 2 and 7.4 using Henderson-Hasselbalch.Extending the per-group ionization analysis to net charge and solubility implications at pH 2 and 7.4.Calculating group-wise charge states and net charge at pH 2 and 7.4 with pI estimate. Always write every group as an acid: ``` HA <=> A- + H+ pH = pKa + log [A-]/[HA] ``` Even for a base. For a base the "HA" is the conjugate acid BH+. Do not use pKb. Rule: pH < pKa : predominantly protonated HA form pH > pKa : predominantly deprotonated A- form +/- 1 unit = 90:10, +/- 2 units = 99:1 Protonated does NOT always mean +. * Neutral acid, e.g. R-O-SO3H, enolizable beta-dicarbonyl: HA neutral 0 <=> A- anionic -1 + H+ * Cationic acid, e.g. aniline: BH+ Ar-NH3+ cationic +1 <=> B Ar-NH2 neutral 0 + H+ You have, as average pKa of the conjugate acid: 1. sulfate / sulfonate: R-SO3H <=> R-SO3- + H+ , pKa ~1.5 neutral to -1 2. aniline: Ar-NH3+ <=> Ar-NH2 + H+ , pKa ~3 +1 to neutral 3. beta-dicarbonyl enol: keto-enol-H <=> enolate- + H+ , pKa ~7 neutral to -1, negative delocalized O-C-C-C-O ### At pH 2 stomach acid **Sulfate pKa 1.5:** pH > pKa, deprotonated. pH = pKa + log[A-]/[HA] 2.0 = 1.5 + log ratio, ratio = 10^0.5 ~3.2 ~76% R-SO3-, ~24% R-SO3H. In integers: **-1** negative delocalized over 3 O's. **Aniline pKa 3:** pH < pKa, protonated. 2.0 = 3.0 + log[B]/[BH+], [B]/[BH+] = 10^-1 = 0.1 ~91% Ar-NH3+, ~9% Ar-NH2. In integers: **+1** localized on N. **Beta-dicarbonyl pKa 7:** pH << pKa, protonated. % ionized = 100/(1+10^(pKa-pH)) = 100/(1+10^5) ~0% In integers: **0** neutral keto / enol. In integers, as separate benzenes: dicarbonyl-benzene 0, aniline-benzene +1, sulfate-benzene -1. If all 3 on the SAME benzene: -1 +1 +0 = net 0, a zwitterion. This is close to the isoelectric point (1.5+3)/2 = 2.25. ### At pH 7.4 plasma **Sulfate pKa 1.5:** pH >> pKa, 100% deprotonated. **-1** **Aniline pKa 3:** pH >> pKa, 100% deprotonated to free base. % BH+ = 100/(1+10^(pH-pKa)) = 100/(1+10^4.4) ~0% **0** neutral Ar-NH2 **Beta-dicarbonyl pKa 7:** pH > pKa, mostly deprotonated. 7.4 = 7.0 + log ratio, ratio = 10^0.4 ~2.5 ~72% enolate-, ~28% neutral. In integers: **-1** negative shared by the two carbonyl O's. In integers, as separate benzenes: dicarbonyl-benzene -1, aniline-benzene 0, sulfate-benzene -1. If all 3 on the SAME benzene: -1 +0 -1 = net -2.

Medicinal Chem