Haloalkanes and Haloarenes Part 2

Concept:In presence of peroxide, HBr addition to alkene follows free radical mechanism (peroxide effect) via more stable radical intermediate.Explanation:For $\mathrm{Ph{-}CH{=}CH_2}$ (styrene) with peroxide, $\mathrm{Br}^{\bullet}$ adds to terminal carbon $\mathrm{CH_2}$, forming benzylic radical $\mathrm{Ph{-}\dot{C}H{-}CH_2Br}$, which is resonance-stabilized.This radical abstracts H from HBr to give $\mathrm{Ph{-}CH_2{-}CH_2Br}$ (1-bromo-2-phenylethane) as major product.Statement A is true: reaction proceeds via more stable radical intermediate.Statement B is false: peroxide does not generate $\mathrm{H}^{\bullet}$; it generates radicals that ultimately produce $\mathrm{Br}^{\bullet}$ (chain-carrying radical).Statement C is true: benzoyl peroxide can form phenyl radicals $\mathrm{Ph}^{\bullet}$, which abstract H from HBr forming benzene ($\mathrm{PhH}$) as byproduct.Statement D is false: 1-bromo-2-phenylethane is the major product, not minor.Statement E is true: in absence of peroxide, HBr adds by electrophilic addition via carbocation intermediate (benzylic carbocation), giving Markovnikov product.Thus, correct statements are A, C, and E.Answer:Option C (A, C & E Only)