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:bubuvuvuvftctcrchj
Explanation:
To determine the heat dissipated when a substance freezes, we multiply the heat of fusion of the substance to the mass of the substance that freezes. We calculate as follows:
Heat = -3.16 (64/32.06) = - 6.32 kJ
Hope this answers the question.
I think the correct answer would be D. The reaction that involves an acid and a covalent base would be the reaction of sulfuric acid and water or H2SO4 + 2H2O → 2H3O+ + SO42– . The acid would be H2SO4 and the covalent base would be H2O since it is being held by covalent bonds and when in solution it will have equal amounts of OH- and H+ ions.
Answer:
B
Explanation:
CH4 + O2 →H2O + CO2
Left side
C = 1
H = 4
O = 2
Right side
C = 1
H = 2
O = 3
So find common denominator in this case would be 6 O
CH4 + 3O2 →2H2O + CO2
Left side
C = 1
H = 4
O = 6
Right side
C = 1
H = 4
O = 6