Question 10 of 13: Markovnikov vs. Anti-Markovnikov HBr Addition

Question 10: Markovnikov vs. Anti-Markovnikov HBr Addition (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. The two conditions trigger two completely different mechanisms — ionic (Markovnikov) addition without peroxide, vs. radical-chain (anti-Markovnikov) addition with peroxide — and each mechanism is intercepted by a different rearrangement/regiochemical event, which is why the two products differ not just in orientation but in carbon skeleton.

1. HBr alone — ionic (Markovnikov) mechanism, WITH a hydride shift. Protonation of the terminal alkene carbon (Markovnikov's rule: H adds to the carbon with more H's already) would naively give a secondary carbocation at C2. But C2 is directly adjacent to the fully substituted, quaternary-like C3 (bearing two methyls) — a 1,2-hydride shift from C3 to C2 converts this secondary cation into a far more stable tertiary carbocation centred at C3: $$\mathrm{(CH_3)_3C{-}CH{=}CH_2 \xrightarrow{H^+} (CH_3)_3C{-}\overset{+}{C}HCH_3 \xrightarrow{1,2\text{-hydride shift}} (CH_3)_2\overset{+}{C}{-}CH(CH_3)CH_3}$$ Bromide then captures this rearranged tertiary cation, giving 2-bromo-2,3-dimethylbutane — note the carbon skeleton itself has changed (a hydrogen migrated), which is only possible because a discrete, rearrangement-prone carbocation intermediate exists in the ionic pathway.
Product without peroxide: 2-bromo-2,3-dimethylbutane (rearranged, Markovnikov)
2. HBr + peroxide — radical chain (anti-Markovnikov) mechanism, NO rearrangement. Peroxide initiates a radical chain: it homolyses to generate RO•, which abstracts the H of H–Br to generate a bromine radical (Br•) first — the reverse order from the ionic mechanism. Br• then adds to the terminal (less hindered) alkene carbon, generating the more stable tertiary radical directly at C2 (adjacent to the fully substituted C3), with no need for any rearrangement (radicals, unlike cations, essentially never undergo 1,2-hydride shifts under these conditions because there is no low-energy pathway/driving force for a radical shift the way there is for a cation): $$\mathrm{(CH_3)_3C{-}CH{=}CH_2 + Br^{\bullet} \longrightarrow (CH_3)_3C{-}\overset{\bullet}{C}H{-}CH_2Br}$$ This tertiary carbon radical then abstracts an H atom from another H–Br molecule, propagating the chain and delivering the bromine to the terminal carbon: $$\mathrm{(CH_3)_3C{-}\overset{\bullet}{C}H{-}CH_2Br + HBr \longrightarrow (CH_3)_3C{-}CH_2{-}CH_2Br + Br^{\bullet}}$$ giving 1-bromo-3,3-dimethylbutane — the original, unrearranged carbon skeleton, with Br on the terminal (originally less-substituted) carbon.