JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 3)
Compound $\mathrm{A}, \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}$, given a tetraacetate with $\mathrm{Ac}_{2} \mathrm{O}$ and oxidation of $\mathrm{A}$ with $\mathrm{Br}_{2}-\mathrm{H}_{2} \mathrm{O}$ gives an acid, $\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{6}$. Reduction of $\mathrm{A}$ with $\mathrm{HI}$ gives isopentane. The possible structure of $\mathrm{A}$ is
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Explanation
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Formation of tetraacetate with AcO2 means compound A has four –OH linkage.
Reduction of A with HI gives Isopentane i.e. molecule contains five carbon atoms.
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