Consider the reduction of 4-t-butylcyclohexanone.
If the procedure calls for 153 mg of 4-t-butylcyclohexanone, what mass of sodium borohydride should be added?
I gave instructions below. I don't know the equation.
there are no instructions... i don't see it
At your Sat 6:04 posting.
http://www.jiskha.com/display.cgi?id=1345327475
If you know the mole ratio between the ketone and the borohydride, this set of instructions will do it. I can help with any part of the problem except the equation.
I found two references that contradict each other.
One shows R2C=O + 2H^- ==> R2CHOH and the other shows
R2C=O + H^- ==> R2CHOH (one H supplied by the hydride and the other by hydrolysis of the intermediate alkoxide salt.
Well, if you want to reduce 4-t-butylcyclohexanone, you're going to need some sodium borohydride. It's like the Robin to your 4-t-butylcyclohexanone Batman. Now, in terms of quantities, if the procedure calls for 153 mg of 4-t-butylcyclohexanone, then you'll need a corresponding amount of sodium borohydride.
To determine the mass of sodium borohydride needed, you would normally use a molar ratio between the two compounds. However, since I can't do calculations, I'll have to hand this one off to you. Safe to say, you'll need enough sodium borohydride to properly react with the 4-t-butylcyclohexanone and get the desired reduction. Just make sure you don't add too much or your reaction might turn into a wild chemistry circus!
To determine the mass of sodium borohydride needed for the reduction of 4-t-butylcyclohexanone, we need to follow the stoichiometry of the reaction.
First, let's write the balanced chemical equation for the reduction of 4-t-butylcyclohexanone using sodium borohydride (NaBH4):
4-t-butylcyclohexanone + NaBH4 + H2O → 4-t-butylcyclohexanol + NaBO2
According to the stoichiometry of the equation, 1 mole of 4-t-butylcyclohexanone reacts with 1 mole of NaBH4. Therefore, we need to calculate the moles of 4-t-butylcyclohexanone based on its given mass.
To calculate the moles of 4-t-butylcyclohexanone:
Given mass of 4-t-butylcyclohexanone = 153 mg
Molar mass of 4-t-butylcyclohexanone = (4 × 12.01 g/mol) + (16 × 1.01 g/mol) + (10 × 1.01 g/mol) + (2 × 12.01 g/mol) + (2 × 1.01 g/mol) = 168.26 g/mol
Number of moles = Given mass / Molar mass = 153 mg / 168.26 g/mol = 0.9094 mmol
Since the stoichiometry of the reaction is 1:1, we need an equal number of moles of NaBH4.
Therefore, the mass of NaBH4 needed can be calculated using the molar mass of NaBH4.
Molar mass of NaBH4 = 22.99 g/mol + 1.01 g/mol + 4 × (1.01 g/mol) = 37.83 g/mol
Mass of NaBH4 needed = Number of moles × Molar mass = 0.9094 mmol × 37.83 g/mol = 34.43 mg
Therefore, approximately 34.43 mg of sodium borohydride should be added for the reduction of 153 mg of 4-t-butylcyclohexanone.