a) If a sample of NaBH_4 is 70% pure by mass (the remaining 30% is incapable of

Question

a) If a sample of NaBH_4 is 70% pure by mass (the remaining 30% is incapable of reducing carbonyls or reacting with the reactant or product), What is the minimum mass of the NaBH_4 sample required to fully reduce 8.26 g of the triketone above? b) How many stereoisomers of the product triol would be formed? How many pairs of enantiomers would there be? How many pairs of diastereomers would there be? c) If there is no molecular symmetry/meso issues (as is the case above), then if there are n stereoisomers, there are n/2 pairs of enantiomers. Derive a formula that is a function of n that will provide the number of pairs of diastereomers for all stereoisomers if there are no symmetry/meso issues.

Explanation / Answer

Post lab

a) moles of triketone = 8.26 g/142 g/mol = 0.0582 mol

4 mole of triketone would get reduced with 3 mole of NaBH4

so moles of NaBH4 needed = 0.0582 mol x 3/4 = 0.04365 mol

with 70% purity of NaBH4 we have in reality = 0.04365 x 0.7 = 0.03055 mol only

we would need = 0.04365 x 0.04365/0.03055 = 0.0624 mol of NaBH4

mass of NaBH4 needed = 0.0624 mol x 37.83 g/mol = 2.36 g

So we would need a minimum of 2.36 g of NaBH4 to reduce triketone.

b) Total number of stereoisomers possible = 6

pairs of enantiomers = 3

pairs of diastereomers = 6 pairs

c) total pair of distereomers = n with no meso symmetry

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