Question

CRUDE PRODUCT COMPOSITION: (as determined by NMR) Show Calculations for % 9-fluorenol and % unreacted 9-fluorenone:

A05-1.1.fid 2800 2600 2400 2200 2000 1800 1600 1400 1200 -1000 -800 -600 400 200 0 -200 -2 -3 5 4 1 1 8 7 10 11 14 13 12 16 15 f1 (ppm) You will notice that the main difference in the spectra are the presence of non-aromatic absorbances around 5.5 and 2.0 ppm in the NMR of the product alcohol. These of course arise from the methine-H and the O-H The aromatic absorbances (between 7 and 8 ppm) are common to both starting ketone and product alcohol. Also, both compounds contribute 8 H's to the aromatic absorbances. Therefore when we have a mixture of these two molecules, the aromatic peaks will arise from BOTH species present whereas the upfield protons will only come from the product alcohol. For this reason, we can use the integrations to obtain an approximate, RELATIVE ratio of the two species present in the crude product mixture. Remember that the numbers printed on the NMR spectrum are relative and not absolute numbers Below I provide you with a formula for determining the relative amounts of each substance. Make sure that you fully understand how this formula was derived as it would not be surprising to see it appear on an exam! 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. Point Assignments: Correct calculation of % 9-fluorenol Correct calculation of % unreacted 9-fluorenone: NMR attached to report

Answer 1

Relative amount of 9-fluorenol = {(8 × 0.13)/(1.00 + 0.66 + 0.04 + 1.13)} × 100% = 36.75 %

Therefore relative amount of 9-fluorenone = (100-36.75)% = 63.25%