A chemist needs to add water across an alkene that has a tertiary carbon adjacent to the double bond — a substrate prone to carbocation rearrangement. Which method should they choose, and why?
AAcid-catalyzed hydration, because it forms the most stable carbocation intermediate
BOxymercuration-demercuration, because the mercurinium ion intermediate prevents skeletal rearrangements
COxymercuration-demercuration, because it produces anti-Markovnikov products that avoid the rearrangement site
DAcid-catalyzed hydration, because strong acid suppresses rearrangement pathways
Oxymercuration is preferred precisely because it avoids a free carbocation. The mercury bridges both carbons of the double bond as a mercurinium ion, preventing the hydride and methyl shifts that plagued the acid-catalyzed route. It still delivers Markovnikov selectivity — not anti-Markovnikov — because the more substituted carbon bears more positive character within the bridged ring.
Question 2 Multiple Choice
In oxymercuration, the nucleophilic attack of water on the mercurinium ion gives anti addition geometry. What structural feature of the mercurinium ion causes this?
AThe positive charge on mercury repels incoming nucleophiles to the opposite face
BMercury sits on one face of the ring, so the nucleophile must attack the opposite face
CThe carbonyl character of mercury blocks syn attack by steric bulk
DAnti addition is enforced by the NaBH₄ reduction step, not the mercurinium ring opening
The mercurinium ion is a three-membered ring with mercury bonded to both carbons on one face of what was the double bond. This bridge physically blocks approach from that face, so the nucleophile (water or alcohol) must attack from the opposite face — yielding anti addition. The NaBH₄ step that follows is not stereospecific and does not enforce the anti geometry.
Question 3 True / False
Oxymercuration produces Markovnikov products because a free carbocation forms on the more substituted carbon, just as in acid-catalyzed hydration.
TTrue
FFalse
Answer: False
This is the central misconception to avoid. Oxymercuration does NOT form a free carbocation — that is the whole point of the reaction. Instead, mercury bridges both carbons as a mercurinium ion. Markovnikov selectivity is preserved because the more substituted carbon bears more partial positive character within the bridged ring, directing nucleophilic attack there. The advantage over acid-catalyzed hydration is precisely that the skeleton stays intact.
Question 4 True / False
The NaBH₄ demercuration step in oxymercuration-demercuration is stereospecific, proceeding with inversion of configuration at the carbon that bore the mercury.
TTrue
FFalse
Answer: False
The demercuration step with NaBH₄ is NOT stereospecific. The C–Hg bond is replaced by C–H via a radical-like mechanism that proceeds without strict retention or inversion. The overall stereochemical outcome of the sequence is governed by the anti addition geometry established in the mercurinium ring-opening step, not by the demercuration.
Question 5 Short Answer
Why does oxymercuration-demercuration reduce the likelihood of carbocation rearrangements compared to acid-catalyzed hydration, and what intermediate is responsible for this advantage?
Think about your answer, then reveal below.
Model answer: Acid-catalyzed hydration generates a free carbocation, which can undergo hydride or methyl shifts to produce rearranged products. Oxymercuration avoids this by forming a mercurinium ion — a three-membered ring in which mercury bridges across both carbons of the former double bond. Because the carbon skeleton is held rigidly in the bridged ring, the skeletal rearrangements that require a fully open carbocation are largely prevented.
The key insight is that the mercurinium ion distributes the positive charge without fully exposing either carbon as a naked carbenium. Some rearrangement is still theoretically possible if a highly stabilized carbocation can form transiently, but the bridged structure substantially suppresses this pathway. Whenever a substrate's structure would lead to rearrangement under acidic conditions, oxymercuration is the standard synthetic solution.