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sushique

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dehydrogenation of alcohols

« on: December 29, 2004, 05:03:42 pm » I am having a hard time writing the chemical equations for two oxidation reactions that were done on alcohols.Sample one: 2-propanol with KMnO4 and H2SO4Sample two: 1-butanol with KMnO4 and H2SO4I know that sample one produced propanone and a brown sludgey precipitate while sample two produces first the aldehyde butanal then goes on to produce butanoic acid. as well as complete seperation between layers. What are those layers precisely and what is the equation? I can't seem to find any examples of equations that actually show specific reactants and their products, the only thing I seem to find are equations showing

• as the oxidizing agent.

could sample one's equation be:CH3-CH-CH3 + KMnO4 CH3-C-CH3 + H2O + KMnO2 | || OH OBut no that wouldn't balance, is water even a product? SOS « Last Edit: December 29, 2004, 05:19:53 pm by sushique » Logged

AWK

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Re:dehydrogenation of alcohols

« Reply #1 on: December 30, 2004, 03:41:25 am » Quote

Sample one: 2-propanol with KMnO4 and H2SO4

CH3-CHOH-CH3 => CH3-CO-CH3C3H8O + KMnO4 + H2SO4 = C3H6O + K2SO4 + MnSO4 + H2OBalance above equation as for any redox one.Using oxidation numbers do not afraid to use fractional oxidation numbers for carbon when neededQuote

Sample two: 1-butanol with KMnO4 and H2SO4

C3H7CH2OH => C3H7COOHC4H10O + KMnO4 + H2SO4 = C4H8O2 + MnSO4 + K2SO4 + H2O Logged AWK

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