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Gelneren [198K]
3 years ago
7

10. Isolation of a pure sample of the third product, which has been determined to be an isomer of the major and minor products,

gives the following 13C NMR data. Determine the structure of this product. 2 points

Chemistry
1 answer:
Ymorist [56]3 years ago
7 0

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.

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Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to pre
xenn [34]

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at 25^oC is, 2.79\times 10^{-39}

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L

The chemical reaction will be:

Fe(OH)_3\rightarrow Fe^{3+}+3OH^-

The expression of solubility constant will be:

K_{sp}=[Fe^{3+}]\times [3OH^-]^3

Now put all the given values in this expression, we get the concentration of hydroxide ion.

2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3

[OH^-]=1.5\times 10^{-11}M

Now we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1.5\times 10^{-11})

pOH=10.8

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2

Therefore, the pH will be, 3.2

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