Propanoic acid CH3CH2COOH

Propanol can have 2 different isomers, propan-1-ol which is CH3CH2CH2OH and propan-2-ol which is CH3CH(OH)CH3.

If propan-1-ol is oxidised by hot acidified KMnO4, propanoic acid, CH3CH2COOH is obtained.

If propan-2-ol is oxidised by hot acidified KMnO4, propanone, CH3COCH3 is obtained.

Propanoic Acid as propanol will undergo oxidation by the acidified KMnO4. 

There are two different propanol: 1-propanol and 2-propanol.

1-propanol is also called n-propanol or just simply propanol. It has a structure of CH3-CH2-CH2-OH. It is a primary alcohol (the -OH group is attached to a CH2 group, with a carbon having 2 hydrogen atoms attached). Oxidation of primary alcohol (by acidified KMnO4) forms carboxylic acid –> simply change -CH2-OH group to -COOH group, with no change of the other parts of the molecule. Hence the answer is CH3-CH2-COOH or simply CH3CH2COOH.

2-propanol is also called isopropanol or i-propanol. It has a structure of CH3-CH(OH)-CH3. Note here the -OH group is bonded to the center carbon. This is a secondary alcohol. Oxidation of secondary alcohol produces a ketone, in this case propanone (CH3-CO-CH3).

Not sure “half help” can help… I recall organic alcohols oxidises to organic acid when left in air

(Beer will turn sour when left in air, being oxidised by O2) – a concept that always sticks in my brain…

So propanol should be oxidised to propanoic acid (CH3CH2COOH)

Exact reduction changes to potassium maganate (VII) (purple potassium permaganate), I’m not sure.

Maybe expert can give equation.