[BROe]

I recently learned about organic reactions and thought I'd try my hand at the oxidation of isopropyl alcohol to acetone. The oxidizer I'm using is acidified (with H₂SO₄) potassium permanganate simply because it is the most available to me. I did a preliminary run of the reaction to acquaint myself with how it works and looks (did not use stoichiometric amounts) and it appeared to work well, I observed the deep purple color of the permanganate go from clear to opaque, then to brown, and finally colorless. I believe that the brown color came from MnO₂ and then the colorless solution being solvated Mn²⁺ (probably in the form of MnSO₄).

Now that I know the reaction works, I would like to attempt it using stoichiometric amounts. This is where I get a bit fuzzy on the reaction stoich as I never formally learned organic redox and was only taught to memorize that secondary alcohols will react with oxidizers to form ketones (obviously this only takes me so far). I looked around on the internet about how to balance these types of redox reactions and I think I may have it right. I would greatly appreciate it if anyone with more experience than myself would make sure I'm on the right track and possibly point out anything that I may be missing.

The website I used to learn how to balance organic redox:
http://web.pdx.edu/~wamserc/C335W02/redox.htm

My Work: (shows IPA to acetone)

Oxidation: CH₃CHOHCH₃  CH₃COCH₃ + 2H

Reduction: MnO₄^- + 3H⁺ + 5H  Mn²⁺ + 4H₂O

I then combined the two reactions such that the arbitrary H atoms would cancel ("multiplying" oxidation by 5 and reduction by 2), and I was left with the final equation
5CH₃CHOHCH₃ + 2MnO₄^- + 6H⁺ 2Mn²⁺ + 8H₂O + 5CH₃COCH₃

[AWK]

Reaction is correctly balanced (this may be checked by mass and charge balance).

For more complicated organic structures you can simplify work using only part of structuce that is oxidized or reduced, eg:
CHOH  CO + H₂O (your case)

COOH  CH₃ + H₂O

CH=CH  2CHO or 2COOH
and so on.
In organic Chemistry most oxidation with KMnO₄ undergo without acid (MnO₂ is formed) with better yield of needed compound.

http://chemwiki.ucdavis.edu/Core/Organic_Chemistry/Reactions/Oxidation_and_Reduction_Reactions/Oxidation_of_Organic_Molecules_by_KMnO4

[BROe]

Thanks for reading through my huge block of text, I know it was alot.

According to the link posted, MnO₄^- will oxidize an alcohol under essentially any conditions, be they alkaline, acidic, or neutral. So out of the three is a neutral pH then the best way to oxidize IPA with permanganate to get the highest yield? Also does this hold true for other organic oxidations or is there some way of determining what general pH will give a better yield?

[AWK]

For high yield specific oxidation of organic compounds more oxidizer are used (see additional reading in UCDAVIS link above), search also for oxidation with osmium tetroxide (OsO₄), peracids oxidation or catalytic oxidation.

[BROe]

Thanks for all the help, I ran the reaction a few weeks ago both with and without using acid and using just the oxidizer (MnO₄^-) and using no acid really did seem to improve the yield.
A few things that I noticed was that the reaction required a large amount of oxidizer for the amount of product produced, however that just seems to be due to the high molar mass of the oxidizer compared to the low molar mass of the desired product. Also, in the version of the oxidation that doesn't use acid, aqueous KOH is produced. If I run the reaction at the boiling point of acetone (≈60C) would I have to worry about damage to my flask over time due to the corrosive effect of KOH on glass?

*Edit*: Just thought of this, one of the byproducts of the reaction without acid is MnO₂, which can used as an oxidizing agent, so could this compound be used as an oxidizer in further organic syntheses if I were to recover it?