One looking at the puddle, sees in her dirt, and another – reflected in her star Immanuel Kant Peopl — Physical Chemistry — Thermodynamics Chemistry Question
What Time is it in Baku or Cheating the Death
One looking at the puddle, sees in her dirt, and another – reflected in her star
Immanuel Kant
People use acronyms for convenience sake. Still, sometimes the meanings are not absolutely unambiguous. Thus, AZT generally standing for “Azerbaijan time” has acquired an alternative meaning in pharmacology denoting one of the most usable anti-HIV drugs.
Nucleoside A1, a major component of nucleic acids, was used as a starting material in AZT synthesis. Under complete combustion with subsequent condensation of water A1 gives a gas undergoing an 11-fold volume decrease when bubbled through the 1 M KOH solution.
What are the building blocks of nucleosides?
Model Answer
Nucleosides are glycosylamines with a nucleic acid base linked to a suitable sugar.
Deduce the structures of A1 to A4. Note that:
* when arranged in any order the numbers of atoms in A1 are not described by either arithmetic or geometric progression,
* all the synthetic steps are SN-type reactions,
* A4 is a binary compound (85.8 % N by mass),
* O is the heaviest element in both A3 and AZT.
Model Answer
Since the wanted nucleoside is a major component of nucleic acids, all combinations with ribose or deoxyribose linked via glycosidic bond to one of the five common nitrogenous bases are to be considered. In any case, A1 is composed of atoms of four elements, thus the combustion equation can be written as:
CxHyNzOt + (2x+y/2–t)/2 O2 = x CO2 + y/2 H2O + z/2 N2
An 11-fold reduction of the gas amount when being passed through the alkali is due to absorption of carbon dioxide. Hence, CO2 : N2 = 10 : 1 or C : N = x : z = 5 : 1.
This result excludes purine bases from further consideration. Both sugars (ribose and deoxyribose) contain 5 carbon atoms each. Thus, the nucleic base should also contain 5 carbon atoms to meet the integer relationship, and Thymine is the only option. The choice between deoxythymidine (C10H14N2O5) and methyluridine (C10H14N2O6) is decided in favor of the former, as no progression of the number of atoms in the molecule is possible. А4 includes a light element. If one represents its composition as MaNb, then 14.0×b / (a*M + 14.0*b) = 0.858, and M_r(M) = 2.32 * b/a g mol–1. LiN3 is the only possible solution.
Note that the synthesis is a sequence of nucleophilic substitution steps. Deoxythymidine (A1) is transformed into A2 (do not ignore the relative ease of substitution). A3 is free of sulfur, which manifests in favor of intramolecular cyclization. AZT synthesis is completed by the azide formation and deprotection.
Structures:
A1: Deoxythymidine
A2: 3'-O-mesyl-5'-O-tritylthymidine
A3: 2,3'-anhydrothymidine
A4: LiN3
How many stereoisomers of AZT do exist?
Model Answer
AZT molecule having 3 stereocenters, 8 isomers are theoretically possible.
Give the full wording for the pharmaceutical AZT acronym.
Model Answer
AZT is used to denote azidothymidine. (International Nonproprietary Name – zidovudine.)
Propose the mechanism of AZT antiviral action.
Model Answer
In the infected cells, DNA is actively produced on the virus RNA used as matrix. AZT inhibits the key enzyme of this process, HIV reverse transcriptase. The mechanism of termination of the DNA elongation is as follows. AZT, being a structural analogue of thymidine, is enzymatically converted to the corresponding 5`-triphosphate at the first stage. The latter is erroneously incorporated into the growing nucleic acid chain leading to termination due to the absence of a hydroxyl group in the 3`-position.
Metabolism of AZT is diverse and includes several pathways (X, Y, and Z) shown in the hereunder scheme [VISUAL] (all balanced equations). X, the major of these, leads to a higher soluble product B1 excreted with urine. Y provides the cytotoxic product B2 with the same number of atoms as that in AZT. Z is found in patients infected with a specific bacteria, the product B4 retaining the AZT-type activity.
Draw the structures of B1-B4.
Model Answer
The pathway Х is a conjugation with glucose (glucuronic acid), a classical way of xenobiotic hydrophilization for its subsequent excretion with urine or bile. Counting the number of atoms allows deciphering the pathway Y. B3 turns out to be a diatomic gas (N2), thus suggesting reduction of the azide to the amine (yielding B2, which is 3'-amino-3'-deoxythymidine). Finally, the pathway Z is a mono-oxygenase reaction. Since the product retains its biological activity, the nitrous moiety important for the enzyme recognition is not affected. Thus, the product is a ribose derivative.
Structures:
B1: AZT 5'-O-glucuronide
B2: 3'-amino-3'-deoxythymidine
B3: N2
B4: 3'-azido-3'-deoxy-5'-O-beta-D-glucuronide of thymidine, or similar ribose-modified glucuronide derivative.
"Dallas Buyers Club", a movie from 2013, received numerous awards including three Oscars. It is based on a true story describing ambiguous results of the first clinical AZT trials. The protagonist suffering from AIDS lacks a positive effect from AZT and switches over to drugs unauthorized by US FDA. One of the latter is based on the substance D.
D (51.17 % of C by weight) and nucleoside N (a major component of nucleic acids; 47.57 % of C by weight) differ by one atom only. Under complete combustion with subsequent condensation of water, the mixture of D and N of any composition gives gaseous products undergoing a 7-fold decrease of their total volume when bubbled through the 1 M KOH solution. It was found that D imitates AZT with respect to the mechanism of its action.
Calculate the number of C and N atoms in D without deciphering its structure.
Model Answer
The absence of variations in volume ratio of the combustion gases before and after the absorption by the alkali solution with that of compounds D and N unambiguously indicates that the numbers of C and N atoms in these compounds are the same. The number of H atoms may differ in such compounds by an even value only. Hence, molecules of D and N differ by one oxygen atom, with D having less. From the equation (x − the number of C atoms):
12.01 * 100 / (12.01 * x / 0.4757 - 16.00) = 0.5117, one finds x = 9. Using the C:N ratio of 3:1 (derived from 7-fold decrease: CO2 : N2 = 6 : 1, i.e., C : N = 9 : 3), the number of nitrogens (z) in D is z = 3.
Draw the structures of D and N.
Model Answer
The above results (9 carbon and 3 nitrogen atoms) allow establishing the nucleoside N. Only cytidine and deoxycytidine out of the nucleosides formed by the primary bases are in correlation with the requirement. The given mass percent of carbon allows final assignment in favor of deoxycytidine. D has one oxygen atom less. Out of the four oxygens present in N, removal of only the hydroxyl group at the 3'-carbon atom can provide for the mechanism of antiviral action identical to that of AZT (D − 2′,3′-dideoxycytidine, or zalcitabine).
Structures:
N: Deoxycytidine
D: Zalcitabine (ddC)
Do you support the “Dallas Buyers Club” principal character in his belief that D possesses a major clinical advantage over AZT? Comment why physicians used to administer AZT and D in a combination.
Model Answer
Despite AZT and zalcitabine (ddC) having identical mechanisms of action, their clinical effects can vary significantly. For example, AZT can be metabolized to a cytotoxic amine compound, whereas ddC, of course, cannot. This may cause the non-identical safety profiles of drugs as well as the difference in patients` tolerance due to adverse effects. Moreover, mutant variants of reverse transcriptase exhibit different specificities for non-identical substrates (AZT and ddC), which also affects the relative clinical efficiency. Thus, in some cases the appearance of AZT-resistant HIV strains is overcome by ddC administration, since the virus mutant strains retain sensitivity to the latter drug. Therefore, it is no surprise that a combination (AZT + ddC) is applied in real clinical practice. Besides, the absence of any significant clinical benefit from taking AZT by the main character of the movie could be due to the combination of drugs with alcohol.
The compound D turns out to be a prodrug. Its active metabolite M (17.23 % P and 35.62 % O by mass) is produced by a number of manufacturers for research (non-clinical) purposes only.
Draw the structure of M.
Model Answer
The presence of phosphorus in M together with the knowledge received in part 24.5 allows us classifying this metabolite as a nucleotide with the molar ratio of phosphorus and oxygen:
17.23/30.97 : 35.62/16.00 = 1: 4
Any ddC-based nucleotide should have the general formula C9Hn+13N3O3n+3Pn. Hence, the value of n is: (3n+3)/n = 4, which gives n = 3.
The molar mass of M is equal to 539.2 g mol–1, which is much higher than that of C9H16N3O12P3 (451.2 g mol–1). Considering М is a salt helps overcoming the contradiction. Then it contains a cation with the molar mass of:
(539.2 + 4 * 1.008 - 451.2)/4 ≈ 23 a.m.u. Hence M is tetrasodium ddC-triphosphate.
Of course, in some cells (e.g., CD4-lymphocytes) mono-, di-, and trianionic ddC-triphosphates will be found alongside with M. However, the chemical industry manufactures tetrasodium ddC-triphosphate (M) for research objectives.