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:
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They have no true nucleus as the DNA is not contained within a membrane or separated from the rest of the cell, but is coiled up in a region of the cytoplasm called the nucleoid. Prokaryotic organisms have varying cell shapes.
Explanation:
Answer:
1200 mL
Explanation:
Given data
- Initial pressure (P₁): 600.0 mmHg
- Initial volume (V₁): 400.0 mL
- Final pressure (P₂): 200.0 mmHg
For a gaseous sample, there is an inverse relationship between the pressure and the volume. If we consider the gas as an ideal gas, we can find the final volume using Boyle's law.

Answer:
A base that forms K⁺ and OH⁻ ions.
Explanation:
The KOH is an Arrhenius base.
A is <em>wrong</em>. A base does not form H⁺ ions.
B is <em>wrong</em>. A metal hydroxide forms K⁺ ions, not KO⁻ ions.
D is <em>wrong</em>. The metal forms K⁺ ions, KO⁻ ions.