Aldehydes, Ketones and Carboxylic Acids Part 3

Concept:The iodoform test (NaOI) is given by compounds with a $\mathrm{CH}_3\mathrm{CO}$– group or a $\mathrm{CH}_3\mathrm{CHOH}$– group, producing a yellow precipitate of $\mathrm{CHI}_3$.Explanation:Statement I: Compound $\mathrm{Z}$ must contain a methyl ketone or a secondary alcohol with a $\mathrm{CH}_3\mathrm{CHOH}$– unit to give a yellow precipitate with NaOI.From the reaction sequence (shown in the image), $\mathrm{Z}$ is formed after steps that introduce such a group, so Statement I is true.Statement II: Compound $\mathrm{Q}$ is reported to have two types of hydrogen atoms: aromatic and aliphatic.The ratio of the number of these hydrogen atoms is given as $1:3$.For typical aromatic compounds like acetophenone or toluene, the ratio is $5:3$ (aromatic:aliphatic).Even for symmetrical compounds like 1,3,5‑trimethylbenzene, the ratio is $3:9 = 1:3$ only if all aromatic H are equivalent and all aliphatic H are equivalent.In the given reaction sequence, the structure of $\mathrm{Q}$ does not satisfy this ratio, making Statement II false.Answer:Statement I is true but Statement II is false, so the correct option is A.