Aldehydes, Ketones and Carboxylic Acids Part 3

Concept:The iodoform test detects the $\mathrm{CH}_3\mathrm{CO}-$ group (methyl ketone) or the $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})-$ group (secondary alcohol oxidizable to methyl ketone).A positive test gives a yellow precipitate of $\mathrm{CHI}_3$.Explanation:Pair A: Methanol ($\mathrm{CH}_3\mathrm{OH}$) has no $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})-$ group, so it gives a negative test.Ethanol ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OH}$) oxidizes to acetaldehyde ($\mathrm{CH}_3\mathrm{CHO}$), which contains a $\mathrm{CH}_3\mathrm{CO}-$ group, so it gives a positive test.Thus, they can be differentiated.Pair B: Both acetic acid ($\mathrm{CH}_3\mathrm{COOH}$) and propanoic acid ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COOH}$) are carboxylic acids and do not give the iodoform test.Both give negative tests, so they cannot be differentiated.Pair C: Cyclohexene is an alkene with no required group, giving a negative test.Cyclohexanone is a ketone but lacks a $\mathrm{CH}_3$ group attached to the carbonyl, so it also gives a negative test.Thus, they cannot be differentiated.Pair D: Diethyl ether ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{OCH}_2\mathrm{CH}_3$) has no $\mathrm{CH}_3\mathrm{CO}-$ or $\mathrm{CH}_3\mathrm{CH}(\mathrm{OH})-$ group, giving a negative test.Pentan-3-one ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{COCH}_2\mathrm{CH}_3$) is a ketone but not a methyl ketone, so it gives a negative test.Thus, they cannot be differentiated.Pair E: Anisole ($\mathrm{C}_6\mathrm{H}_5\mathrm{OCH}_3$) is an ether with no required group, giving a negative test.Acetone ($\mathrm{CH}_3\mathrm{COCH}_3$) is a methyl ketone, giving a positive test.Thus, they can be differentiated.Answer:Only pairs A and E can be differentiated using the iodoform test.The correct option is C: A & E Only.