JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Hydrocarbons

Bromine atoms abstract hydrogen in a strongly endothermic step, so the transition state resembles the radical being formed and the reaction is highly selective for the most stable radical. Isobutane offers one tertiary hydrogen and nine primary ones; abstraction of the tertiary hydrogen gives $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}\cdot$, which is stabilised by hyperconjugation from nine $\mathrm{C}-\mathrm{H}$ bonds, and this selectivity far outweighs the numerical advantage of the primary hydrogens. The tertiary radical then takes bromine from $\mathrm{B}{\mathrm{r}}_2$, and the major product is 2-bromo-2-methylpropane, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{CBr}$.