Question

What is the theoretical yield of 3,3-dimethyl-2-butanol that would be formed if you react 1.5 mL of 3,3-dimethyl-butanone with 155 mg of NaBH4?

Answer 1

The reaction of 3,3dimethylbutanone with NaBH4 will be

4 X 3,3dimethylbutanone + NaBH4 ---> 4 X 3,3-dimethyl-2-butanol

So as per sotichiometry of reaction , 4 moles of ketone will react with 1 mole of NaBH4 to give 4 moles of alcohol

Moles of NaBH4 = Mass / Molecular weight = 155mg / 37.83 = 4.09 millimoles

Density of 3,3-dimethyl-butanone = 0.801 g/mL

So mass of 3,3-dimethyl-butanone reacted = Volume X density = 1.5 X 0.801 = 1.201 grams

Moles of 3,3-dimethyl-butanone = 1.201 / 100 = 12.01 millimoles

For each mole of 3,3-dimethyl-butanone we need 0.25 moles of NaBH4

so for 12.01 millimoles we will need = 12.01 X 0.25 = 3 millimoles of NaBH4

So the limiting reagent is the ketone

1 millimoles of ketone will give 1 millimoles of alcohol

So 12.01 millimoles will give 12.01 millimoles of alochol

So theoretical yield = 12.01 millimoles

Molecular weight of 3,3-dimethyl-2-butanol = 102

So grams of 3,3-dimethyl-2-butanol produced = millimoles X molecular weight = 12.01 X 102 = 1.225 grams