JEE Advanced 2024 Paper 2 · Q14 · Aldehydes & Ketones
PARAGRAPH I: An organic compound P with molecular formula $C_9H_{18}O_2$ decolorizes bromine water and also shows positive iodoform test. P on ozonolysis followed by treatment with $H_2O_2$ gives Q and R. While compound Q shows positive iodoform test, compound R does not give positive iodoform test. Q and R on oxidation with pyridinium chlorochromate (PCC) followed by heating give S and T, respectively. Both S and T show positive iodoform test. Complete copolymerization of 500 moles of Q and 500 moles of R gives one mole of a single acyclic copolymer U. [Given, atomic mass: H = 1, C = 12, O = 16]
Sum of number of oxygen atoms in S and T is _____.
Reveal answer + step-by-step solution
Correct answer:2, 4
Solution
P has formula $C_9H_{18}O_2$ with one degree of unsaturation, decolourises bromine water (so has C=C) and gives iodoform (so has CH$_3$CO$-$ or CH$_3$CH(OH)$-$). Ozonolysis with H$_2$O$_2$ gives carboxylic acids Q and R. Standard mainstream interpretation (FIITJEE): P = 8-hydroxy-non-5-en-2-ol structure giving Q = 4-hydroxy-pentanoic acid (gives iodoform from the CH$_3$CH(OH) end) and R = 3-hydroxy-pentanoic acid (no iodoform). PCC oxidation followed by heating gives S = acetone (CH$_3$COCH$_3$, 1 O) and T = butan-2-one (CH$_3$COC$_2$H$_5$, 1 O). Sum of O atoms = 1 + 1 = 2. Alternative valid interpretation in the official solution gives S = a methyl-ketone bearing an additional carboxylic acid (3 O) and T = formaldehyde-type after $-CO_2$ loss (1 O), totalling 4. Hence the official key accepts "2 OR 4".
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