WAEC Chemistry 2024 Paper

Concept:Oxidation of a primary alcohol such as ethanol produces a carboxylic acid after passing through an aldehyde intermediate.Explanation:Ethanol is $\mathrm{CH_3CH_2OH}$.When a strong oxidizing agent like acidified $\mathrm{K_2Cr_2O_7}$ or alkaline $\mathrm{KMnO_4}$ is used, ethanol first loses hydrogen from the carbon atom attached to the $-$OH group.This gives ethanal (acetaldehyde), $\mathrm{CH_3CHO}$.With further oxidation, an oxygen atom is added to the aldehyde carbon, converting ethanal into ethanoic acid (acetic acid), written as $\mathrm{CH_3COOH}$ or $\mathrm{CH_3CO_2H}$.The carboxylic acid retains the two-carbon chain of ethanol.Looking at the options, $\mathrm{C_2H_4CO_2H}$ does not match this structure, $\mathrm{C_2H_5OH}$ is the starting alcohol, and $\mathrm{CH_3OH}$ is a different alcohol.Only $\mathrm{CH_3CO_2H}$ represents ethanoic acid, the correct oxidation product of ethanol.Answer:Option D: $\mathrm{CH_3CO_2H}$ (ethanoic acid).