An emerging strategy to improve the pharmacokinetics (PK) of drugs takes advantage of the kinetic is — Physical Chemistry — Kinetics Chemistry Question
Zoniporide
An emerging strategy to improve the pharmacokinetics (PK) of drugs takes advantage of the kinetic isotope effect. Molecules containing non-radioactive heavy isotopes in metabolically relevant positions may be cleared more slowly from the body. Zoniporide, a cardioprotective inhibitor of the Na+/H+ antiporter 1 protein, was considered a candidate for improved PK upon deuteration, since the major metabolic pathway of zoniporide involves the oxidation by aldehyde oxidase in position 2 of the quinoline core.
2H-Zoniporide, deuterated in position 2 of the quinoline core is synthesized from ester 1 by the following sequence of reactions:
[VISUAL]
Draw the structures of intermediates A through C and reagent D.
Model Answer
The structures of the intermediates A through C and reagent D are:
- A: ethyl 5-cyclopropyl-1-(1-oxidoquinolin-5-yl-1H-pyrazole-4-carboxylate
- B: 5-cyclopropyl-1-(2-deuterio-1-oxidoquinolin-5-yl-1H-pyrazole-4-carboxylic acid
- C: 5-cyclopropyl-1-(2-deuterioquinolin-5-yl-1H-pyrazole-4-carboxylic acid
- D: guanidine (or guanidinium salt)
[VISUAL]
Ammonium formate decomposes upon mild heating in the presence of palladium on charcoal (transformation B → C) into three gaseous products, one of which is the reducing agent required for the aforementioned transformation. Draw the structures of these compounds.
Model Answer
Ammonia (NH3), carbon dioxide (CO2) and hydrogen (H2)
The active site of the aldehyde oxidase enzyme contains a molybdenum(VI) cofactor chelated by pyranopterin dithiolate (PD). Two mechanisms have been proposed for the oxidation of the drug in position 2 of the quinoline core. Mechanism 1 involves three major individual steps: formation of a molybdate ester, a hydride transfer and a final hydrolytic step.
Draw the intermediate active site structure E involved in the proposed Mechanism 1 of zoniporide oxidation.
[VISUAL]
Model Answer
Structure E represents the molybdate ester intermediate in which the oxygen of the molybdenum cofactor is covalently bound to the position 2 of the quinoline core. The nitrogen of the quinoline core is protonated by the lysine residue, and the molybdenum is in the (VI) oxidation state.
[VISUAL]
Give the oxidation state of molybdenum in each of the intermediate structures E, 3 and 4.
Model Answer
The oxidation states of molybdenum are:
- Structure E: Molybdenum(VI)
- Structure 3: Molybdenum(IV)
- Structure 4: Molybdenum(IV)
On the other hand, another proposed mechanism, Mechanism 2, involves a concerted substitution step yielding intermediate 3, which is further hydrolyzed to 4 in the same fashion as in Mechanism 1.
Draw the transition state structure F for the concerted substitution step 2 → 3. Use a dotted line for bonds which are being formed and cleaved.
[VISUAL]
Model Answer
The transition state structure F shows a concerted process where the C2–O bond is forming (oxygen of molybdenum attacking C2 of quinoline) and the C2–H bond is cleaving (hydride transferring to Mo or sulfur), while a proton is transferred from Lys to the quinoline nitrogen. Dotted lines are used to represent these forming and breaking bonds.
[VISUAL]
The following experimental evidence was gathered to determine whether the transformation 2 → 3 in the mechanism of the oxidation of zoniporide (and related nitrogen heterocycles) by aldehyde oxidase is stepwise (Mechanism 1) or concerted (Mechanism 2):
a) The kinetic isotope effect, kH / kD, for zoniporide (deuterated in quinoline position 2) oxidation by aldehyde oxidase was 5.8 at 37 °C.
b) The introduction of electron withdrawing groups on the heterocycle core led to an increase in the reaction rate and a slight decrease in kH / kD.
Which of the two mechanisms (Mechanism 1 or Mechanism 2) of quinoline oxidation by aldehyde oxidase is more plausible based on the aforementioned experimental evidence? Rationalize your answer.
Model Answer
Mechanism 2
Rationalization:
From a) the KIE is >> 1 which indicates that the C2–H bond is being cleaved during the rate determining step (RDS). For Mechanism 1 the RDS would have to be E → 3, for Mechanism 2 the RDS would be the concerted 2 → 3 transformation.
From b) we know that electron withdrawing groups (EWGs) on the heterocyclic core speed up the reaction. This indicates that the RDS involves either buildup of negative charge on the quinoline ring (e.g. by a nucleophilic attack) or loss of positive charge from the ring (e.g. by deprotonation). In Mechanism 1, this is true for the 2 → E step (an electron rich nucleophile adds to the quinoline core) but not for the E → 3 step (expulsion of a hydride nucleofuge is disfavoured in the presence of EWGs). This contradiction disproves Mechanism 1; therefore, the correct answer is Mechanism 2.
The molybdenum cofactor further needs to be reoxidized to its original state. The reducing equivalents from one reaction are transferred, via an iron sulfide cluster cofactor and a flavin cofactor, to a single molecule of oxygen as the stoichiometric oxidant.
What small molecule byproduct is formed by the reduction of O2 in this process?
Model Answer
Hydrogen peroxide (H2O2)
Deuterium is not the only heavy isotope of hydrogen. In theory, an even higher kinetic isotope effect would be expected using tritium. The isotope 3H is not used in practice to slow down the metabolism of drugs due to economic and safety reasons but let us at least theoretically look at 3H-zoniporide.
Calculate the theoretical tritium (kH / kT) kinetic isotope effect for the oxidation of zoniporide by aldehyde oxidase at 37 °C. The deuterium kinetic isotope effect for the same reaction is 5.8. Consider the following approximations:
- The harmonic oscillator approximation
- Isotope exchange does not alter the rate determining step transition state structure
- The KIE is solely affected by the 12C−H/D/T stretching vibration mode
- The KIE is solely determined by zero-point vibrational energies (the role of higher vibrational levels is negligible)
m(1H) = 1.0078 amu; m(2H) = 2.0141 amu; m(3H) = 3.0160 amu; m(12C) = 12.0000 amu
Hint: You need to calculate 1) the relevant reduced masses; and 2) the force constant for the C–H/D bond before you get to the final KIE calculation.
Model Answer
The calculation steps are as follows:
1) Calculate the reduced masses of C–H, C–D and C–T bond stretching modes (using m_C = 12.0000 amu):
- mu_H = (m_H * m_C) / (m_H + m_C) = (1.0078 * 12) / (1.0078 + 12) = 0.9297 amu = 1.5438 * 10^-27 kg
- mu_D = (m_D * m_C) / (m_D + m_C) = (2.0141 * 12) / (2.0141 + 12) = 1.7246 amu = 2.8638 * 10^-27 kg
- mu_T = (m_T * m_C) / (m_T + m_C) = (3.0160 * 12) / (3.0160 + 12) = 2.4102 amu = 4.0023 * 10^-27 kg
2) Calculate the force constant (k) using the deuterium KIE of 5.8 at 37 °C (T = 310.15 K):
Using the expression derived from zero-point energy differences:
ln(k_H / k_D) = (hbar * sqrt(k) / (2 * kB * T)) * (1 / sqrt(mu_H) - 1 / sqrt(mu_D))
- Substituting kB = 1.3806 * 10^-23 J/K, hbar = 1.0546 * 10^-34 J s, T = 310.15 K, and ln(5.8) = 1.7579:
- This yields sqrt(k) = 21.105 kg^(1/2) s^-1, corresponding to a force constant k = 445.4 kg s^-2.
3) Calculate the tritium KIE (k_H / k_T):
Using the same force constant:
ln(k_H / k_T) = (hbar * sqrt(k) / (2 * kB * T)) * (1 / sqrt(mu_H) - 1 / sqrt(mu_T))
- Substituting the values yields:
- ln(k_H / k_T) = 2.506
- k_H / k_T = e^(2.506) = 12.3 (or 12.26)
Unfortunately, the kH/kD kinetic isotope effect of 5.8 for the oxidation of zoniporide by aldehyde oxidase does not translate into a more complex system. The degradation rate of 2H-zoniporide in human liver cells is only 1.9× lower than that of 1H-zoniporide. This is because aldehyde oxidase is not the only enzyme involved in zoniporide catabolism. Nonspecific cellular hydrolases, as well as cytochrome P450 enzymes compete with aldehyde oxidase for the degradation of zoniporide.
Draw the two products of zoniporide hydrolysis by nonspecific cellular hydrolases.
Hint: Non-enzymatic aqueous alkaline hydrolysis under mild conditions would result in the same products.
Model Answer
The two products of the amide-like guanidine hydrolysis of zoniporide are:
1. 5-cyclopropyl-1-(quinolin-5-yl-1H-pyrazole-4-carboxylic acid (or its conjugate carboxylate form)
2. Guanidine (or its conjugate acid guanidinium form)
[VISUAL]