Dissolve 200mg (1.11mmol) of 9-fluorenone in 2.5 ml ethanol and introduce 50 mg
ID: 622861 • Letter: D
Question
Dissolve 200mg (1.11mmol) of 9-fluorenone in 2.5 ml ethanol and introduce 50 mg (1.32mmol) of NABH4. What is the limiting reagent and the theoretical yield?Explanation / Answer
To obtain a balanced equation you need to be somewhat familiar with the mechanism of the reaction. It can be broken down into 3 steps: 1) Nucleophilic addition of hydride (by borohydride) to the carbonyl carbon of fluorenone C13H8O + BH4- ==> C13H9O- + BH3 2) Complexation of BH3 with ethanol BH3 + CH3CH2OH ==> CH3CH2OHBH3 Note that in a Lewis structure the product should appear with a formal positive charge on O and a formal negative charge on B. 3) Formation of product (proton transfer) C13H9O- + CH3CH2OHBH3 ==> C13H9OH + CH3CH2OBH3- Next, the ethoxyborohydride formed here can react with 3 additional equivalents of fluorenone, thereby picking up additional ethoxy groups in the process. Adding these steps together and canceling like terms, we arrive at: 4C13H8O + BH4- + 4CH3CH2OH ==> 4C13H9OH + B(OCH2CH3)4- You may also write the equation with counterions (Na+) as appropriate. We see that there is a 4:1 molar ratio between fluorenone and borohydride in the reactants. That is, 1.32 mmol of borohydride require 4 x 1.32 mmol = 5.28 mmol of fluorenone to react completely. However, there are only 1.11 mmol of fluorenone present. Therefore, fluorenone is the limiting reagent.