During the free radical halogenation of 2-methylbutane in the presence of ultraviolet light, what is — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Bromination is highly selective compared to chlorination. The free radical substitution mechanism relies on the stability of the intermediate alkyl radical. 2-methylbutane forms primary, secondary, and tertiary radicals. The tertiary free radical at carbon-2 is highly stable due to hyperconjugation. The highly selective bromine atom will predominantly attack this tertiary position, making 2-bromo-2-methylbutane the major product (>90%).