Hello everyone, kinda need help understanding on this topic.
How do I calculate the percentage yield and theoretical yield of the product (cyclohexanol) ?
The weight of the product is = 1.502g
Volume of cylcohexanone is = 2mL
Using NaBH4 and methanol
Reduction of Cyclohexanone to Cylcohexanol
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- ChenBeier
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Re: Reduction of Cyclohexanone to Cylcohexanol
First develop the chemical reaction.
Then you can see how many mole of the keton correspond to the
Then convert the 2 ml into mole and calculate how much theoretically product can be obtained. Then also convert the obtained mass and compare.
Then you can see how many mole of the keton correspond to the
Then convert the 2 ml into mole and calculate how much theoretically product can be obtained. Then also convert the obtained mass and compare.
Re: Reduction of Cyclohexanone to Cylcohexanol
Okay thank you
So does the equation be something like this?
C6H10O + 2 NaBH4 = C6H11OH + 2 BH3 + 2 Na
- ChenBeier
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Re: Reduction of Cyclohexanone to Cylcohexanol
No,
Ok the keton get to alcohol is correct, but there will be nor sodium neither Borane obtained.
Product is sodiumborate.
But you know now that 1 mol keton gives 1 mol alcohol.
Now procedure the next step.
Ok the keton get to alcohol is correct, but there will be nor sodium neither Borane obtained.
Product is sodiumborate.
But you know now that 1 mol keton gives 1 mol alcohol.
Now procedure the next step.
Re: Reduction of Cyclohexanone to Cylcohexanol
Ohh i see its sodiumborate, thank you. But how do i convert 2ml to mole?
- ChenBeier
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Re: Reduction of Cyclohexanone to Cylcohexanol
First you need specific gravity δ = m/V, convert this to mass, then convert with molar mass to mole.
n =m/M
n =m/M