Paragraph I: An organic compound $P$ with molecular formula $C_5 H_{10} O_2$ decolorizes bromine water and…

Paragraph I: An organic compound $P$ with molecular formula $C_5 H_{10} O_2$ 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: $H=1$, $C=12$, $O=16$] Question: The molecular weight of $U$ is

Solution

$\mathrm{S~} \& \mathrm{~T}$ shows + ve idoform test. $\mathrm{Rmf} \mathrm{C}_5 \mathrm{H}_{10} \mathrm{O}_3$ (M. wt $)_{\mathrm{R}}=70+48=118$ $\mathrm{Q} \mathrm{mf} \mathrm{C}_4 \mathrm{H}_8 \mathrm{O}_3(\mathrm{M} . \mathrm{wt})_{\mathrm{Q}}=56+48=104$ 500 mole $\mathrm{Q}+500$ mole $\mathrm{R} \xrightarrow[\text { polymer }]{\text { Condensation }}$ $\begin{aligned} \mathrm{U} \text { molecular weight } & =500 \times 118+500 \times 104-999 \times 18=500 \times 222-17982 \\ & =111000-17982=93018\end{aligned}$ Ans. is $\Rightarrow 93018$

Asked in: JEE Advanced 2024 (Paper 2)

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