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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: sushique on December 29, 2004, 05:03:42 PM

Title: dehydrogenation of alcohols
Post by: sushique 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 H2SO4
Sample two: 1-butanol with KMnO4 and H2SO4
I 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
could sample one's equation be:
CH3-CH-CH3 + KMnO4   CH3-C-CH3 + H2O + KMnO2
         |                               ||
        OH                             O
But no that wouldn't balance, is water even a product?  SOS
Title: Re:dehydrogenation of alcohols
Post by: AWK on December 30, 2004, 03:41:25 AM
Quote
Sample one: 2-propanol with KMnO4 and H2SO4
CH3-CHOH-CH3 => CH3-CO-CH3
C3H8O + KMnO4 + H2SO4 = C3H6O + K2SO4 + MnSO4 + H2O
Balance above equation as for any redox one.
Using oxidation numbers do not afraid to use fractional oxidation numbers for carbon when needed
Quote
Sample two: 1-butanol with KMnO4 and H2SO4
C3H7CH2OH => C3H7COOH
C4H10O + KMnO4 + H2SO4 = C4H8O2 + MnSO4 + K2SO4 + H2O