Three different organic compounds have the formula \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\). Only two of these isomers react with KMnO_ (a strong oxidizing agent). What are the names of the products when these isomers react with excess \(\mathrm{KMnO}_{4} ?\)

There are three different organic compounds with the molecular formula of C3H8O: methoxyethane, propan-1-ol, and propan-2-ol. 1. Methoxyethane: \(CH_{3} O CH_{2} CH_{3}\) 2. Propan-1-ol: \(CH_{3} CH_{2} CH_{2} OH\) 3. Propan-2-ol: \(CH_{3} CH(OH) CH_{3}\)

Out of the three isomers, only the ones containing an alcohol functional group (propan-1-ol and propan-2-ol) can react with potassium permanganate (KMnO\(_{4}\)). Methoxyethane does not contain an alcohol group and therefore does not react with KMnO\(_4\). 1. Propan-1-ol: \(CH_{3} CH_{2} CH_{2} OH\) 2. Propan-2-ol: \(CH_{3} CH(OH) CH_{3}\)

When the alcohol-containing isomers react with an excess of KMnO\(_{4}\), they will be oxidized. Different products are formed depending on the type of alcohol (primary or secondary) since the reaction involves cleavage of the C-C bonds. 1. Propan-1-ol is a primary alcohol (the -OH group is attached to a carbon atom with one R group). The reaction with excess KMnO\(_{4}\) oxidizes it to propanoic acid: \(CH_{3} CH_{2} COOH\). 2. Propan-2-ol is a secondary alcohol (the -OH group is attached to a carbon atom with two R groups). Oxidation by excess KMnO\(_{4}\) leads to the formation of two compounds: acetone (\(CH_{3} C(=O) CH_{3}\)) and carbon dioxide (CO\(_{2}\)). So, the products of the reactions of propan-1-ol and propan-2-ol with excess KMnO\(_{4}\) are propanoic acid, acetone, and carbon dioxide.