Chemistry - Hydrocarbon Question with Solution | TestHub

An organic compound A has molecular formula ${\mathrm{C}}_{10}{\mathrm{H}}_{17}\mathrm{Br}$ and it is non-resolvable. A does not decolourize brown colour of bromine water solution. A on treatement with ${\left({\mathrm{CH}}_3\right)}_2\mathrm{COK}/{\left({\mathrm{CH}}_3\right)}_3\mathrm{COH}$ yields B as major product. B on treatment with ${\mathrm{H}}_2/\mathrm{Pt}$ yields $\left({\mathrm{C}}_{10}{\mathrm{H}}_{18}\right)$ which on treatment with ${\mathrm{Cl}}_2/\mathrm{hv}$ yields three monochloro derivative. Also B on boiling with acidic permanganate solution yields $\mathrm{C}\left({\mathrm{C}}_{10}{\mathrm{H}}_{16}{\mathrm{O}}_3\right)$. C on heating with sodalime yields $\mathrm{D}\left({\mathrm{C}}_9{\mathrm{H}}_{16}\mathrm{O}\right)$. D on reducing with ${\mathrm{LiAlH}}_4$ followed by heating the product with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ yields $\mathrm{E}\left({\mathrm{C}}_9{\mathrm{H}}_{16}\right)$ as major product. E on treatment with ozone followed by work-up with $\mathrm{Zn}-{\mathrm{H}}_2\mathrm{O}$ yields 6-Ketononanal.