JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Hydrocarbons

${\mathrm{C}}_4{\mathrm{H}}_8$ has one degree of unsaturation, and both reagents named act only on a carbon-carbon double bond, so a saturated ring such as cyclobutane or methylcyclopropane would be recovered unchanged and A must be an alkene. Cleaving the double bond divides the four carbons between two fragments, and B keeps three of them, so the double bond has to lie between a terminal $\mathrm{C}{\mathrm{H}}_2$ group and a carbon bearing two methyl groups. That alkene is 2-methylpropene: ozonolysis cuts it to methanal and propan-2-one, and hot acidic permanganate oxidises the $\mathrm{C}{\mathrm{H}}_2$ end right through to carbon dioxide while leaving the same propan-2-one, ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$, as the isolable product. But-2-ene would instead give two molecules of ethanal. Compound A is 2-methylpropene.