Reduction of 9-fluorenone: objective of this lab was to reduce 9-florenone(a ket

Question

Reduction of 9-fluorenone: objective of this lab was to reduce 9-florenone(a ketone), to 9-fluorenol (an alcohol) using sodium borohydride as the reducing agent.

Experimental data:

9-fluorenone: MW 180.20 g/mol mass used 0.6 grams

methanol: MW 32.04 g/mol volume used: 6ml density: 0.792

NaBH4: MW 37.83 g/mol mass used 0.05 grams

9-fluorenol: MW 182.22 g/mol

Balanced equation for the reaction is: 4C13H8O + NaBH4+ 4CH3OH --> 4C13H10O + NaB(OCH3)4

Calculate the potential yield (in grams) if 9-fluorenol for 9-fluorenone, methanol and NaBH4. From these potential yields identify the limiting reagent and theoretical yield.

I tried calculating this on my own and I feel like I did this wrong. Please help

Explanation / Answer

Inspect stoichiometry of the given reaction as per balanced equation as below-

4C13H8O + NaBH4 + 4CH3 OH ------------> 4C 13 H 10 O + NaB(OCH 3 )4

4 moles 1 mole 4 moles 4 moles 1 mole

hence , mole ratio 1 0.25 1 1

given weights for reactants 0.6gms. 0.05 gms 6 x 0.792 x (?)gms.

wts. used in terms of moles 0.6/ 180.20 0.05 /37.83 4.792/ 32.04 x/182.22

or, = 0.0033 0.0013 0.15 ?

Now, out of the three reactants NaBH 4 acts as a reducing agent only and hence the reaction can occur only when NaBH4 is present in relatively much smaller amounts. But the yield of the product ,9-fluorenol , is not governed by its quantity. Theoretically when equimolar amounts of reactants ( 9-fluorenone and methanol ) are taken for reaction in presence of NaBH4 an equivalent quantity of product , 9- fluorinol is obtained , as inspected in above stoichometric relations. But practically we have used lesser number of moles of 9-fluorenone than the theoretically needed , hence in this case it should serve as the limiting reagent.

Note : 9- fluorenone , methanol & NaBH4 are the reactants , while the product 9-fluorenol would be obtained

= 0.0033 moles

or, = 0.0033 x182.22 = 0.60 gms.

this would represent the theoretical yield of the product 9-fluorenone , as we have used excess of methanol thus facilitating the reaction to go to completion.

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