JEE Advanced 2024 — Paper 2 — CHEM-IV solution

JEE Advanced2024Paper 2CHEM-IV

PARAGRAPH I

An organic compound $P$ with molecular formula $C_{9} H_{18} O_{2}$ decolorizes bromine water and also shows positive iodoform test. $P$ on ozonolysis followed by treatment with $H_{2} O_{2}$ 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]

Q1. Sum of number of oxygen atoms in $S$ and $T$ is _____.

Your answer

Solution

Decoding the structure of $P$ . Formula $C_{9} H_{18} O_{2}$ has one degree of unsaturation, so $P$ contains one $C = C$ double bond (consistent with decolorising bromine water). The positive iodoform test indicates a $C H_{3} C H (O H) -$ or $C H_{3} C O -$ moiety. Oxidative ozonolysis ( $O_{3}$ then $H_{2} O_{2}$ ) cleaves the double bond to give two oxidised fragments: $Q$ as a hydroxy-carboxylic acid (still iodoform-positive, so it retains a $C H_{3} C H (O H) -$ group) and $R$ as a hydroxy-carboxylic acid (iodoform-negative, so the $- O H$ on $R$ is not on a $C H_{3} C H (O H) -$ carbon).

A consistent assignment is:

$P$ : $C H_{3} C H (O H) - C H = C H - C H_{2} - C H (O H) - C H_{2} C H_{2} C H_{3}$ (a $C_{9}$ unsaturated diol — total $C_{9} H_{18} O_{2}$ , one $C = C$ ).

Oxidative ozonolysis cleaves the central $C = C$ :

$Q$ : $C H_{3} C H (O H) - C O O H$ (lactic acid analogue, iodoform-positive).

$R$ : $H O O C - C H_{2} - C H (O H) - C H_{2} C H_{2} C H_{3}$ (a $β$ -hydroxy carboxylic acid; the $- O H$ is on a carbon flanked by $- C H_{2} -$ and propyl, not by $- C H_{3}$ , so iodoform-negative).

Q and R $\to$ S and T (PCC oxidation, then heat). PCC oxidises each secondary $- O H$ to a $C = O$ without touching the $- C O O H$ . Heating then promotes decarboxylation of the resulting $β$ -keto acid.

For $Q$ : $C H_{3} C H (O H) - C O O H P C C C H_{3} C O - C O O H Δ C H_{3} C H O + C O_{2}$ (after decarboxylation, $S$ = acetaldehyde, $C H_{3} C H O$ ; iodoform-positive). $S$ has $1$ oxygen.

For $R$ : $H O O C - C H_{2} - C H (O H) - C_{3} H_{7} P C C H O O C - C H_{2} - C O - C_{3} H_{7} Δ C H_{3} - C O - C_{3} H_{7} + C O_{2}$ ( $T$ = pentan-2-one, $C H_{3} C O C_{3} H_{7}$ ; iodoform-positive because of the methyl ketone). $T$ has $1$ oxygen.

$∴$ Sum of oxygen atoms in $S$ and $T$ = $1 + 1 = 2.$

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