I recently learned that an oxidizing agent such as KMnO4 will oxidize EtOH to acetic acid in a two-step process:
CH3CH2OH --> CH3CHO --> CH3COOH
But when I tried to draw the reaction out, I ran into a problem in the first reaction. There is an imbalance in the number of electrons on either
side.
What am I misunderstanding? Is the extra electron perhaps being given back to the K cation?tetrahedron - 24-10-2012 at 13:26
hi thelonious
1. there are already threads about this oxidation.
2. your ethanol is missing some hydrogens.
3. on the pruduct side you have formaldehyde instead of acetaldehyde.
4. the Mn in MnO2 is tetravalent, hence it should add "16 electrons" to your count.
5. regarding electron balance: use H+ or e- or OH- (instead of only H2O). the organic
species are uncharged and the Mn species have clearly defined charges.
6. a carbon disappears as well.
7. the stoichiometry is probably wrong too..but try to correct the obvious mistakes first.
[Edited on 24-10-2012 by tetrahedron]kristofvagyok - 24-10-2012 at 13:38
KMnO4 is one of the strongest oxidizers used in the lab. It was used for destructive oxidation of aromatics, with proper rection conditions it will
oxidize even the highly stable aromatic ring to CO2.
So, acetaldehyde, a highly sensitive aldehyde what is usually used as it's acetal what is stable, won't survive the power of the permanganate, even
with relative special reaction conditions. Why? Because the ethanol is oxidized relative hardly, but the aldehyde is almost autooxidative to the
carboxylic acid.
This is why they produce no acetaldehyde industrially from ethanol.
P.S.: Heated copper(II) oxide will do it with small amounts, but forget the KMnO4(:thelonious - 24-10-2012 at 13:58
1. there are already threads about this oxidation.
2. your ethanol is missing some hydrogens.
3. on the pruduct side you have formaldehyde instead of acetaldehyde.
4. the Mn in MnO2 is tetravalent, hence it should add "16 electrons" to your count.
5. regarding electron balance: use H+ or e- or OH- (instead of only H2O). the organic
species are uncharged and the Mn species have clearly defined charges.
6. a carbon disappears as well.
7. the stoichiometry is probably wrong too..but try to correct the obvious mistakes first.
[Edited on 24-10-2012 by tetrahedron]
Oh boy, it seems i screwed up a few things copying it from my notes onto paint.
That aside, could I use the change in oxidation numbers to determine the stoichiometry of the reaction? ie. C on EtOH is oxidized by 2 (+1 to -1), Mn
is reduced by 3 (+7 to +4) therefore 3:2 ratio of MnO4-:EtOH?Adas - 25-10-2012 at 05:44
The best method of acetaldehyde preparation is leading ethanol vapor through red hot copper wire. It will dehydrogenate the ethanol.Waffles SS - 25-10-2012 at 09:39
Trimethylammonium flurochromate (TriMaFC):A convenient new and mild reagent for oxidation of organic
substrates
(I tried this route for making benzaldehyde and i am sure this method is useable for acetaldehyde too)
[Edited on 25-10-2012 by Waffles SS]AndersHoveland - 1-12-2012 at 04:48
I found this: http://en.wikipedia.org/wiki/Parikh-Doering_oxidation
Apparently the adduct of pyridine and SO3, in the presence of trimethylamine which acts as the base, can be used to oxidize alcohols to aldehydes, in
up to 99% yield.
Probably the adduct of trimethylamine would optionally also work instead.
What is interesting about this is that pyridium chlorochromate also has the same sort of selective alcohol-to-aldehyde chemistry.
And then there is 2-iodoxybenzoic acid, which is also a common regent used to selectively oxidize alcohols to aldehydes. The 2-Iodoxybenzoic acid can
then be re-oxidized and recycled after completion of the reaction.
It is a very useful reagent, but the pure form is an explosive, so it is not completely without hazard. 2-Iodoxybenzoic acid can oxidize methanol to
formaldehyde in 94% yield, and can similarly oxidize ethylene glycol (vehicle anti-freeze) to glyoxal (although the solvent dimethyl sulfoxide (DMSO)
can not be used for the latter, as its pressence will cause the ethylene glycol to be oxidized to formaldehyde instead)
2-Iodoxybenzoic acid can be prepared by the slow addition, over a half hour, of potassium bromate (76.0 g, 0.45 mol) to a vigorously stirred sulfuric
acid mixture (0.73 M, 730 mL) containing 2-iodobenzoic acid (85.2 g, 0.34 mol).
Otherwise, it is generally quite difficult to oxidize alcohols to aldehydes without further oxidation to a carboxylic acid; yields are normally low.
Selenium dioxide is another reagent, which is moderately selective, but yields are still only around 60%.
[Edited on 1-12-2012 by AndersHoveland]zed - 6-12-2012 at 15:30
Sounds to me, like a pain in the ass. Acetaldehyde isn't really hard to make. Try Vogel.
Another interesting thought is to distill it out from Cinnamaldehyde, via the reverse aldol.
Two useful products are formed. Benzaldehyde and Acetaldehyde. The trick, is to condense out the Water-Benzaldehyde first, while allowing the
Acetaldehyde vapor to pass to a secondary condenser of colder temperature.
Hey, it could happen!
sparkgap - 6-12-2012 at 16:34
Nobody answered the guy, so...
Quote:
That aside, could I use the change in oxidation numbers to determine the stoichiometry of the reaction?
You certainly can; this is a redox reaction you're dealing with, after all.
sparky (~_~)SM2 - 6-12-2012 at 18:00
I like the Cu wire dehydrogenation. Once Rxn. is ensuued, is it endothermic, or does heat need to be applied the whole time? Also, is 190 proof
everclear OK to use?Adas - 7-12-2012 at 07:39
I like the Cu wire dehydrogenation. Once Rxn. is ensuued, is it endothermic, or does heat need to be applied the whole time? Also, is 190 proof
everclear OK to use?
The purer your alcohol, the better. And this reaction is endothermic, you need to constantly heat the Cu wire. I think the best method would be to
heat the copper wire with induction heating, because just bunsen burner under the vessel will heat the glass, not the copper, which is the catalyst.
