Answer
The names of the products when these isomers react with excess $KMnO_{4}$ are:
Propanoic acid ($CH_{3}CH_{2}COOH$) and
Acetone / 2-propanone ($CH_{3}-CO-CH_{3}$)
respectively.
See the below reactions.
Work Step by Step
There are three (03) isomers possible having formula $C_{3}H_{8}O$ which are propanol ($CH_{3}CH_{2}OH$) , 2-propanol ($CH_{3}-CH(OH)-CH_{3}$) and methoxy ethane ($CH_{3}-O-CH_{2}CH_{3}$). Among them, propanol participates in oxidation reaction and produce propanoic acid where $-CH_{2}OH$ group of propanol becomes $-COOH$. 2-propanol also participates in oxidation reaction and produce acetone by converting the $-CH(OH)-$ group of 2-propanol into $-CO-$ group. methoxy methane does not participate in oxidation reaction.
