Reduction of 9-fluorenone by sodium borohydride will result in the formation of

Question

Reduction of 9-fluorenone by sodium borohydride will result in the formation of 9-fluorenol. How to calculate the percent yield of 9-fluorenol and % unreacted 9-fluorenone from the NMR spectrum?

the formula below can be used for the calculations

Relative amount of 9-fluorenol =
8 x (integration of methine H) / total integration of aromatic peaks
Relative amount of 9-fluorenone =
(total integration of aromatic peaks) – (8 x integration of methine H) / total integration of
aromatic peaks
of course you will convert these to percentages by multiplying by 100.

Explanation / Answer

I feel that there are some errors in the formulae you have mentioned. So, I may be wrong, but I'm giving my best to solve your problem. Please check out the following calculations.

Integration of alcohol H + methine H = 1.08 + 0.87 = 1.95

Total integration of all the peaks = 3.93+2.12+1.08+0.87 = 8

Now, the relative amount of 9-fluorenone = (Integration of alcohol H + methine H)/total integration = 1.95/8 = 0.24375

The percentage of 9-fluorenone = 0.24375*100 = 24.375 %

And the relative amount of 9-fluorenol = 100-percentage of 9-fluorenone = 100-24.375 = 75.625 %

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