Predicting octanol/water partition coefficients and pKa for the SAMPL7 challenge using the SM12, SM8 and SMD solvation models

Abstract

Blind predictions of octanol/water partition coefficients and pKa at 298.15 K for 22 drug-like compounds were made for the SAMPL7 challenge. Octanol/water partition coefficients were predicted from solvation free energies computed using electronic structure calculations with the SM12, SM8 and SMD solvation models. Within these calculations we compared the use of gas- and solution-phase optimized geometries of the solute. Based on these calculations we found that in general the use of solution phase-optimized geometries increases the affinity of the solutes for water as compared to octanol, with the use of gas-phase optimized geometries resulting in the better agreement with experiment. The pKa is computed using the direct approach, scaled solvent-accessible surface model, and the inclusion of an explicit water molecule, where the latter two methods have previously been shown to offer improved predictions as compared to the direct approach. We find that the use of an explicit water molecule provides superior predictions, and that the predicted macroscopic pKa is sensitive to the employed microstates.

Appendix 1: Relating the distribution coefficient to relative free energies

Let’s assume we have our neutral reference molecule plus five additional states:

• 0. H$_2$A, our neutral reference molecule

• 1. HA⁻, deprotonated species with a formal charge of −1, corresponding to a loss of one hydrogen as compared to our neutral reference molecule

• 2. H₃A⁺, protonated species with a formal charge of +1, corresponding to the gain of one hydrogen as compared to our neutral reference molecule

• 3. AH₂, tautomer of our neutral reference molecule with a formal charge of 0

• 4. A₂, deprotonated species with a formal charge of −2, corresponding to a loss of two hydrogens as compared to our neutral reference molecule

• 5. H₄A⁺², protonated species with a formal charge of +2, corresponding to the gain of two hydrogens as compared to our neutral reference molecule

This leads to:

$$\begin{aligned}&\log_{10} D_1^{\rm o/w}\left( T,P\right) \\&\quad = \log_{10}\left( \frac{c_{1-0,\mathrm {o}}}{c_{1-0,\mathrm {w}}+c_{1-1,\mathrm {w}}+c_{1-2,\mathrm {w}}+c_{1-3,\mathrm {w}} +c_{1-4,\mathrm {w}}+c_{1-5,\mathrm {w}}}\right) \end{aligned}$$

(15)

We can readily compute $\log_{10} P_1^{\rm o/w}$ from the transfer free energy of the neutral solute between the two phases. Let us therefore take the difference between $\log_{10} D_1^{\rm o/w}$ and $\log_{10} P_1^{\rm o/w}$ to obtain an expression that may be used to “correct” our predicted value of $\log_{10} P_1^{\rm o/w}.$ Taking the difference and simplifying we have:

$$\begin{aligned} \begin{aligned}&\log_{10} D_1^{\rm o/w}\left( T,P\right) - \log_{10} P_1^{\rm o/w}\left( T,P\right) \\&\quad = \log_{10} \left( \frac{c_{1-0,\mathrm {o}}}{c_{1-0,\mathrm {w}}+c_{1-1,\mathrm {w}}+c_{1-2,\mathrm {w}}+c_{1-3,\mathrm {w}} +c_{1-4,\mathrm {w}}+c_{1-5,\mathrm {w}}}\right) - \log_{10} \left( \frac{c_{1-0,\mathrm {o}}}{c_{1-0,\mathrm {w}}}\right) \\&\quad = \log_{10} \left( \frac{c_{1-0,\mathrm {o}}}{c_{1-0,\mathrm {w}}+c_{1-1,\mathrm {w}}+c_{1-2,\mathrm {w}}+c_{1-3,\mathrm {w}} +c_{1-4,\mathrm {w}}+c_{1-5,\mathrm {w}}} \cdot \frac{c_{1-0,\mathrm {w}}}{c_{1-0,\mathrm {o}}}\right) \\&\quad = \log_{10} \left( \frac{c_{1-0,\mathrm {w}}}{c_{1-0,\mathrm {w}}+c_{1-1,\mathrm {w}}+c_{1-2,\mathrm {w}}+c_{1-3,\mathrm {w}} +c_{1-4,\mathrm {w}}+c_{1-5,\mathrm {w}}}\right) \\&\quad = \log_{10} \left( \frac{1}{1+\frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} +\frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}}}\right) \\&\quad =-\log_{10} \left( 1+\frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} +\frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}}\right) \end{aligned} \end{aligned}$$

(16)

Next, we will work out expressions for the relative concentrations in terms of the free energies of reaction computed in the present study. Starting with state 1, we have:

$$\begin{aligned} \mathrm {H}_2\text{A} \rightleftharpoons \mathrm {H}^+ + \mathrm {HA}^- \end{aligned}$$

(17)

$$\begin{aligned} K\mathrm {a}_1 = \frac{\left[ \mathrm {H}^+\right] \left[ \mathrm {HA}^-\right] }{\left[ \mathrm {H}_2\text{A}\right] } = \frac{c_{\mathrm {H}^+,\mathrm {w}}c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned}$$

(18)

where:

$$\begin{aligned} \begin{aligned} \mathrm {pH}&= -\log_{10} c_{\mathrm {H}^+,\mathrm {w}} \\ \ln c_{\mathrm {H}^+,\mathrm {w}}&= -\mathrm {pH} / \log_{10} e \end{aligned} \end{aligned}$$

(19)

and then:

$$\begin{aligned} & -\frac{\varDelta G_{1-0}^{\rm rxn}\left( T,P\right) }{RT}= \ln K\mathrm {a}_1 = \ln c_{\mathrm {H}^+,\mathrm {w}} + \ln \frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ &-\frac{\varDelta G_{1-0}^{\rm rxn}\left( T,P\right) }{RT}= -\frac{\mathrm {pH}}{\log_{10} e} + \ln \frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ &-\frac{\varDelta G_{1-0}^{\rm rxn}\left( T,P\right) }{RT}+ \frac{\mathrm {pH}}{\log_{10} e} = \ln \frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned}$$

(20)

where $\varDelta G_{1-0}^{\rm rxn}$ is the corresponding free energy of reaction computed in the present study. This leads to the desired result:

$$\begin{aligned} \frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = \exp \left[ -\frac{\varDelta G_{1-0}^{\rm rxn}\left( T,P\right) }{RT} + \frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned}$$

(21)

Next, we move on to state 2:

$$\begin{aligned}&\mathrm {H}_2\text{A} + \mathrm {H}^+ \rightleftharpoons \mathrm {H}_3\text{A}^+ \end{aligned}$$

(22)

$$\begin{aligned}&\quad K\mathrm {a}_2 = \frac{\left[ \mathrm {H}_3\text{A}^+\right] }{\left[ \mathrm {H}_2\text{A}\right] \left[ \mathrm {H}^+\right] } = \frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}c_{\mathrm {H}^+,\mathrm {w}}} \end{aligned}$$

(23)

and then:

$$\begin{aligned} \begin{aligned} -\frac{\varDelta G_{2-0}^{\rm rxn}\left( T,P\right) }{RT}&= \ln K\mathrm {a}_2 = -\ln c_{\mathrm {H}^+,\mathrm {w}} + \ln \frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{2-0}^{\rm rxn}\left( T,P\right) }{RT}&= \frac{\mathrm {pH}}{\log_{10} e} + \ln \frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{2-0}^{\rm rxn}\left( T,P\right) }{RT}&- \frac{\mathrm {pH}}{\log_{10} e} = \ln \frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned} \end{aligned}$$

(24)

where $\varDelta G_{2-0}^{\rm rxn}$ is the corresponding free energy of reaction computed in the present study. This leads to the desired result:

$$\begin{aligned} \frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = \exp \left[ -\frac{\varDelta G_{2-0}^{\rm rxn}\left( T,P\right) }{RT} - \frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned}$$

(25)

Next, we move on to state 3, a tautomer of the neutral reference molecule:

$$\begin{aligned} \mathrm {H}_2\text{A} \rightleftharpoons \mathrm {AH}_2 \end{aligned}$$

(26)

$$\begin{aligned} K\mathrm {a}_3 = \frac{\left[ \mathrm {AH}_2\right] }{\left[ \mathrm {H}_2\text{A}\right] } = \frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned}$$

(27)

and then:

$$\begin{aligned} -\frac{\varDelta G_{3-0}^{\rm rxn}\left( T,P\right) }{RT} = \ln K\mathrm {a}_3 = \ln \frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned}$$

(28)

where $\varDelta G_{3-0}^{\rm rxn}$ is the corresponding free energy of reaction computed in the present study. This leads to the desired result:

$$\begin{aligned} \frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = \exp \left[ -\frac{\varDelta G_{3-0}^{\rm rxn}\left( T,P\right) }{RT}\right] \end{aligned}$$

(29)

Next, we move on to state 4:

$$\begin{aligned} \mathrm {H}_2\text{A} \rightleftharpoons 2\mathrm {H}^+ + \text{A}^{-2} \end{aligned}$$

(30)

$$\begin{aligned} K\mathrm {a}_4 = \frac{\left[ \mathrm {H}^+\right] ^{2}\left[ \text{A}^{-2}\right] }{\left[ \mathrm {H}_2\text{A}\right] } = \frac{c_{\mathrm {H}^+,\mathrm {w}}^{2}c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned}$$

(31)

and then:

$$\begin{aligned} \begin{aligned} -\frac{\varDelta G_{4-0}^{\rm rxn}\left( T,P\right) }{RT}&= \ln K\mathrm {a}_4 = 2 \ln c_{\mathrm {H}^+,\mathrm {w}} + \ln \frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{4-0}^{\rm rxn}\left( T,P\right) }{RT}&= -2\frac{\mathrm {pH}}{\log_{10} e} + \ln \frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{4-0}^{\rm rxn}\left( T,P\right) }{RT}&+ 2 \frac{\mathrm {pH}}{\log_{10} e} = \ln \frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned} \end{aligned}$$

(32)

where $\varDelta G_{4-0}^{\rm rxn}$ is the corresponding free energy of reaction computed in the present study. This leads to the desired result:

$$\begin{aligned} \frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = \exp \left[ -\frac{\varDelta G_{4-0}^{\rm rxn}\left( T,P\right) }{RT} + 2\frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned}$$

(33)

And last, we move on to state 5:

$$\begin{aligned} \mathrm {H}_2\text{A} + 2\mathrm {H}^+ \rightleftharpoons \mathrm {H}_4\text{A}^{+2} \end{aligned}$$

(34)

$$\begin{aligned} K\mathrm {a}_5 = \frac{\left[ \mathrm {H}_4\text{A}^{+2}\right] }{\left[ \mathrm {H}_2\text{A}\right] \left[ \mathrm {H}^+\right] ^{2}} = \frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}c^{2}_{\mathrm {H}^+,\mathrm {w}}} \end{aligned}$$

(35)

and then:

$$\begin{aligned} \begin{aligned} -\frac{\varDelta G_{5-0}^{\rm rxn}\left( T,P\right) }{RT}&= \ln K\mathrm {a}_5 = -2 \ln c_{\mathrm {H}^+,\mathrm {w}} + \ln \frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{5-0}^{\rm rxn}\left( T,P\right) }{RT}&= 2\frac{\mathrm {pH}}{\log_{10} e} + \ln \frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ -\frac{\varDelta G_{5-0}^{\rm rxn}\left( T,P\right) }{RT}&- 2\frac{\mathrm {pH}}{\log_{10} e} = \ln \frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \end{aligned} \end{aligned}$$

(36)

where $\varDelta G_{5-0}^{\rm rxn}$ is the corresponding free energy of reaction computed in the present study. This leads to the desired result:

$$\begin{aligned} \frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = \exp \left[ -\frac{\varDelta G_{5-0}^{\rm rxn}\left( T,P\right) }{RT} - 2\frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned}$$

(37)

Relating back to original expression (16) we have:

$$\begin{aligned} \begin{aligned}1+&\frac{c_{1-1,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-2,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-3,\mathrm {w}}}{c_{1-0,\mathrm {w}}} +\frac{c_{1-4,\mathrm {w}}}{c_{1-0,\mathrm {w}}}+\frac{c_{1-5,\mathrm {w}}}{c_{1-0,\mathrm {w}}} \\ =&1+ \exp \left[ -\frac{\varDelta G_{1-0}^{\rm rxn}}{RT} + \frac{\mathrm {pH}}{\log_{10} e}\right] + \exp \left[ -\frac{\varDelta G_{2-0}^{\rm rxn}}{RT} - \frac{\mathrm {pH}}{\log_{10} e}\right] \\&+ \exp \left[ -\frac{\varDelta G_{3-0}^{\rm rxn}}{RT}\right]+\exp \left[ -\frac{\varDelta G_{4-0}^{\rm rxn}}{RT} + 2\frac{\mathrm {pH}}{\log_{10} e}\right]\\ &+ \exp \left[ -\frac{\varDelta G_{5-0}^{\rm rxn}}{RT} - 2\frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned} \end{aligned}$$

(38)

We find that the expression may be equivalently expressed as:

$$\begin{aligned} 1+\sum_{i=1}^{5} \frac{c_{1-i,\mathrm {w}}}{c_{1-0,\mathrm {w}}} = 1 + \sum_{i=1}^{5} \exp \left[ -\frac{\varDelta G_{i-0}^{\rm rxn}}{RT} -q_{i} \frac{\mathrm {pH}}{\log_{10} e}\right] \end{aligned}$$

(39)

where $q_i$ is the molecular formal charge of state i.

Using this result, we can write a general expression for N states:

$$\begin{aligned} \begin{aligned}&\log_{10} D_1^{\rm o/w}\left( T,P\right) - \log_{10} P_1^{\rm o/w}\left( T,P\right) \\&\quad = -\log_{10} \left( 1+\sum_{i=1}^{N} \frac{c_{1-i,\mathrm {w}}}{c_{1-0,\mathrm {w}}}\right) \\&\quad = -\log_{10} \left( 1 + \sum_{i=1}^{N} \exp \left[ -\frac{\varDelta G_{i-0}^{\rm rxn}}{RT} -q_{i} \frac{\mathrm {pH}}{\log_{10} e}\right] \right) \end{aligned} \end{aligned}$$

(40)

and then:

$$\begin{aligned}&\log_{10} D_1^{\rm o/w}\left( T,P\right) = -\frac{\log_{10} e}{RT}\left[ \varDelta G_{1-0,\mathrm {o}}^{\mathrm {solv}}\left( T,P\right) -\varDelta G_{1-0,\mathrm {w}}^{\mathrm {solv}}\left( T,P\right) \right] \\&\quad -\log_{10} \left( 1 + \sum_{i=1}^{N} \exp \left[ -\frac{\varDelta G_{i-0}^{\rm rxn}}{RT} -q_{i} \frac{\mathrm {pH}}{\log_{10} e}\right] \right) \end{aligned}$$

(41)