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.
Two German chemists, Justus von Liebig (1803–1873) and Friedrich Wöhler (1800–1882), were responsible for the emergence of organic chemistry in the early nineteenth century. Their quantitative analytical methods helped establish the constitution of newly isolated and synthesized carbon compounds.
The balanced chemical reaction for the described reaction above is,
Na2CO3 + 2HCl ---> 2NaCl + H2CO3
From the reaction, 1 mole of Na2CO3 is needed to produce 2 moles of NaCl. In terms of mass, 106 grams of Na2CO3 are needed to produce 116.9 grams of NaCl. From this,
(23.4 g NaCl) x (106 g Na2CO3 / 116.9 NaCl = 21.22 g Na2CO3
Thus, approximately 21.22 g Na2CO3 is needed for the desired reaction.
There is no reaction.
<em>Molecular equation
:</em>
K₂CO₃(aq) + 2NH₄Cl(aq) ⟶ 2KCl(aq) + (NH₄)₂CO₃(aq)
<em>Ionic equation
:</em>
2K⁺(aq) + CO₃²⁻(aq) + 2NH₄⁺(aq) +2Cl⁻(aq) ⟶ 2K⁺(aq) + 2Cl⁻(aq) + 2NH₄⁺(aq) + CO₃²⁻(aq)
<em>Net ionic equation
:</em>
Cancel all ions that appear on both sides of the reaction arrow (underlined).
<u>2K⁺(aq)</u> + <u>CO₃²⁻(aq)</u> + <u>2NH₄⁺(aq</u>) +<u>2Cl⁻(aq)</u> ⟶ <u>2K⁺(aq)</u> + <u>2Cl⁻(aq</u>) + <u>2NH₄⁺(aq)</u> + <u>CO₃²⁻(aq)</u>
<em>All ions cancel</em>. There is no net ionic equation.