Answer:
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Answer:
Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.
Explanation:
Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.
Answer:
3.4 M
Explanation:
M = grams/molar mass = ans./volume(L)
M = 919/180 = ans./1.5
Answer:
0.595
Explanation:
Let's consider the decomposition of hydrogen chloride.
2 HCl(g) ⇄ H₂(g) + Cl₂(g)
The pressure equilibrium constant (Kp) is equal to the product of the partial pressure of the products raised to their stoichiometric coefficients divided by the product of the partial pressure of the reactants raised to their stoichiometric coefficients.
Kp = pH₂ × pCl₂ / pHCl²
Kp = 77.9 × 54.4 / 84.4²
Kp = 0.595