Meltonium
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Free Radical Alcohol?
I attempted an experiment to create hexanol from n-hexane and hydrogen peroxide. I became curious after I saw a video on how water was purified using
H2O2 that was split into free radicals.
I added 50ml of n-hexane to a flask followed by ~140ml of semi-concentrated H2O2. With rapid stirring, the mixture was kept under uv light of ~395nm
for 1.5 hours.
The mixture was then setup for fractional distillation.
Unusually, a liquid, presumed to be the n-hexane, distilled over at 58C. This is peculiar because n-hexane's BP is supposed to be 69C and the air
pressure at the time was not enough to explain an 11C difference.
After all of the low-boiling liquid had been collected, the heat and stirring was removed. The now hexane-free liquid was tested for H2O2 by adding a
minute amount of MnO2 to a small sample. The reaction bubbled quite vigorously.
I had planned to use the BP of the hexanols' to see if any had been created, but since nearly the full 50mL of hexane was distilled over, I don't
think that process would yield anything.
Is there any other way to test for the alcohols in the mixture?
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aga
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Well, with such scant information, it's unlikely that even a seasoned chemist could help.
Maybe knowing the concentration of H2O2 might help.
Personally i have not the slightest clue.
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Meltonium
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The last time I titrated the peroxide it came out to be about 17% give or take. However, that was a few weeks ago.
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Metacelsus
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The hexane-water azeotrope boils at 61.6 °C. The remaining 3.6 °C of deviation could be explained by an erroneous thermometer, or pressure
differences. You could check your thermometer by distilling water.
If you want to spice things up a bit, you could add some iron salts to the peroxide, and get Fenton's reagent.
[Edited on 11-20-2016 by Metacelsus]
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Texium
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Thread Moved
20-11-2016 at 15:05 |
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