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04-BS-12 · December 2016

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

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exam 04-BS-12, Organic Chemistry — December 2016. 3 hours, closed-book examination (no non-communicating calculator restriction beyond the standard aid sheet, 8.5×11", hand-written both sides). Ten questions constitute a complete exam paper (only the first 10 questions as they appear in the answer book are marked, each of equal value) — the source paper in fact prints thirteen questions; all thirteen are answered in full below.

Reference texts: McMurry, Organic Chemistry, 9th ed. (acid/base theory of drugs, SN1/SN2 stereochemistry, carbocation rearrangements, alkyne synthesis via acetylide alkylation, IR/NMR structure elucidation, electrophilic aromatic substitution and synthesis design, amino-acid pKa); Clayden, Organic Chemistry, 2nd ed. (amide resonance and β-lactam reactivity, radical vs. ionic HBr addition mechanisms); a standard biomaterials reference for the poly(ester amide) drug-delivery polymer chemistry of Question 13 (Katsarava-type AABB poly(ester amide)s built from diacids, diols, and protected diamino acids).

Question 10: Markovnikov vs. Anti-Markovnikov HBr Addition: Rearrangement vs. Radical Mechanism

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 completely different mechanisms — ionic (Markovnikov, via a carbocation) without peroxide, and radical (anti-Markovnikov, via the peroxide-initiated chain) with peroxide — and only the cationic intermediate is prone to rearrangement.

3,3-dimethyl-1-butene
  1. HBr alone — ionic (Markovnikov) addition through a secondary cation that rearranges. H+ (from HBr) adds first to the terminal, less-substituted alkene carbon (Markovnikov protonation), generating the more stable of the two possible carbocations: a secondary cation at C2, immediately adjacent to a fully substituted (no H) neopentyl-type carbon (C3, bearing three methyls).
    secondary cation (C2) — adjacent to a quaternary-type carbon
    This secondary cation sits directly next to a carbon bearing a methyl group that can migrate with its bonding electron pair (a 1,2-methyl shift). Migration converts the secondary C2 cation into a tertiary cation at C3 — a substantially more stable carbocation — so the rearrangement is strongly downhill and occurs essentially instantaneously, faster than Br− can capture the original secondary cation. Br− then captures the rearranged, more stable tertiary cation, giving the rearranged, more highly substituted bromide as the major product.
    2-bromo-2,3-dimethylbutane (rearranged, Markovnikov)
  2. HBr + peroxide — radical chain (anti-Markovnikov) addition; no cation, no rearrangement. Peroxide initiates a radical chain: RO• abstracts the H of H–Br to generate Br•, which then adds to the alkene first — and it adds to the less substituted (terminal) carbon, because that leaves the unpaired electron on the more substituted, more stable secondary carbon radical (the same substitution-stability preference as cations, but here it is a radical, not a cation, and it does not undergo skeletal 1,2-shifts the way a carbocation does). This secondary radical then abstracts an H atom from another H–Br to give the product and propagate the chain, with Br ending up on the terminal carbon and no rearrangement possible at any point.
    1-bromo-3,3-dimethylbutane (anti-Markovnikov, unrearranged)
ConditionsMechanismIntermediateRearranges?Product
HBr aloneionic (electrophilic addition)2° carbocation → 3° carbocationyes (1,2-methyl shift)2-bromo-2,3-dimethylbutane
HBr + peroxideradical chain2° carbon radicalno1-bromo-3,3-dimethylbutane