[spritemonst3r]

There's a picture of Compound B. It's C9H12O3, cyclic, with three carbonyl groups at 1, 4, and 7 positions. I dunno what its called, but i would guess something like cyclonon-1,4,7-trione?

Anyway the question says:

"Compound B was reduced with sodium borohydride (NaBH4). The student used 1.1015g of the molecule. She prepared the sodium borohydride solution by dissolving 1.0085g of sodium borohydride in sufficient solvent to prepare 17.00mL of solution. Calculate the volume of the sodium borohydride solution required to reduce Compound B."



I don't know where to go with this question....All I've done so far is gotten the molar mass of compound B and the number of moles used of everything. And I know that to reduce the compound, you would only need one molecule of NaBH4 per molecule of compound B because its (NaBH4) got 4 H's and there are only 3 carbonyl groups on the compound.


My calculations:

Molar mass of comp. B: 168.186 g/mol
Molar mass of NaBH4: 37.83 g/mol

Moles used of comp. B: 0.00655 moles
Moles used of NaBH4: 0.0267 moles

There is about 4X more NaBH4 than Comp. B.

Where do I go with this??? And how to I incorporate the volume (17mL) to find the answer?


Thanks

[movies]

I really don't like this question because there are an awful lot of variables.  Depending on the solvent, NaBH₄ can deliver different numbers of hydrides (up to 4).  Without knowing the solvent that is used, it is pretty difficult to say what they are after.  In reality, sodium borohydride reductions are usually carried out with a fair excess of the reducing agent as well.  Generally I would only count on one equivalent of reducing ability from NaBH₄ in a typical solvent like methanol.  From there it should be easy for you to figure out how many equivalents of the reagent you would need.  From there you will need to relate moles and volume.  There is a very common unit for this, but you are probably more used to seeing it with respect to amounts of acid or base in solvent.  Ringing any bells?