Aldehydes, Ketones and Carboxylic Acids Part 3

Concept:P gives positive iodoform test (contains $-\mathrm{COCH}_3$ or $-\mathrm{CH(OH)CH}_3$) but negative Tollen’s test (no free aldehyde).Dilute acid hydrolysis of P produces Q which gives both iodoform and Tollen’s tests, meaning Q has both a methyl ketone (or methyl carbinol) and an aldehyde group.Thus P must contain a masked aldehyde (as an acetal) that is unmasked on hydrolysis.Explanation:The molecular formula of P is $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_3$ with one degree of unsaturation (one carbonyl group).P gives iodoform test → presence of $\mathrm{CH}_3\mathrm{CO}-$ or $\mathrm{CH}_3\mathrm{CH(OH)}-$ group.Negative Tollen’s test confirms no free $-\mathrm{CHO}$ group.Acid hydrolysis of P gives Q. Since Q gives both iodoform and Tollen’s tests, Q must contain both a $\mathrm{CH}_3\mathrm{CO}-$ (or $\mathrm{CH}_3\mathrm{CH(OH)}-$) and an aldehyde group.This implies that in P, the aldehyde group is protected as an acetal ($-\mathrm{CH(OR)}_2$), which on hydrolysis releases a free aldehyde.A plausible structure for P is $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH(OCH}_3)_2$ (4,4-dimethoxybutan-2-one).In P, the $\mathrm{CH}_3\mathrm{CO}-$ group gives iodoform test; the acetal does not react with Tollen’s reagent.Upon hydrolysis: $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH(OCH}_3)_2 \xrightarrow{\mathrm{H}^+/\mathrm{H}_2\mathrm{O}} \mathrm{CH}_3\mathrm{COCH}_2\mathrm{CHO} + 2\mathrm{CH}_3\mathrm{OH}$.The product Q ($\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CHO}$) has both an aldehyde (positive Tollen’s) and a methyl ketone (positive iodoform), satisfying all observations.The other options do not yield such a Q or give inconsistent test results.Answer:The structure of P is $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CH(OCH}_3)_2$, corresponding to option A (assuming the image shows this linear acetal).