Oxymercuration–Demercuration

Introduction

Acid-catalysed hydration gives a Markovnikov alcohol but may rearrange through a free carbocation. Hydroboration–oxidation avoids that cation but gives the opposite OH orientation. Oxymercuration–demercuration combines Markovnikov OH placement with a bridged intermediate that largely avoids ordinary carbocation shifts. It is a two-stage laboratory sequence: mercury(II) acetate and water first, then sodium borohydride to replace mercury with hydrogen.

Core explanation

In the first stage, Hg(OAc)₂ supplies an electrophilic mercury centre to the alkene in a solvent such as aqueous THF. The alkene pi bond interacts with mercury, forming a mercurinium ion that bridges the two former double-bond carbons, analogous to a bromonium ion. This bridged species carries positive character but is not an ordinary freely rotating carbocation. Water attacks one carbon of the bridge, opening a C–Hg connection, and subsequent deprotonation converts the initially formed oxonium into an OH group. The other former alkene carbon remains attached to a mercury-containing fragment.

For an unsymmetrical alkene, water commonly attacks the more substituted carbon because it bears greater positive character in the bridged intermediate. Thus the OH group is placed on the more substituted carbon. In the second stage, NaBH₄ reduces the organomercury intermediate and replaces its carbon–mercury bond by a carbon–hydrogen bond. The net transformation adds H and OH across C=C with Markovnikov regiochemistry . For propene, the product is propan-2-ol.

The bridge helps explain the absence of the usual hydride and alkyl shifts seen in reactions with an open carbocation. A neighbouring sigma bond cannot simply migrate into a long-lived empty p orbital in the standard mercurinium picture. Product prediction can therefore preserve the starting carbon skeleton in cases where acid hydration may rearrange. This is a practical reason to choose the sequence in synthetic planning, beyond merely memorising the reagent list.

Oxymercuration initially has an anti-like opening relationship between incoming water and mercury because water attacks opposite the bridge. However, the final demercuration is more complex and may replace mercury with H from either face. Consequently the net hydration is not generally assigned a fixed syn or anti stereochemistry . OpenStax explicitly notes that the hydrogen in the demercuration stage can attach from either side depending on circumstances. Do not carry the initial anti relationship into the final alcohol as an unconditional stereochemical claim.

The reagents have distinct roles. Hg(OAc)₂ initiates bridged-ion formation; water supplies the oxygen of OH; NaBH₄ performs demercuration. Sodium borohydride here removes the mercury-containing group; it is not the same as simply reducing a ketone to an alcohol, despite the shared reagent name. The transformation should be drawn as a sequence rather than a single arrow where mercury disappears without explanation.

Mercury compounds are hazardous and require trained handling and appropriate waste procedures in a real laboratory; the course emphasis here is on mechanism and product prediction. When comparing alternatives, hydroboration–oxidation gives an anti-Markovnikov alcohol, while acid hydration and oxymercuration–demercuration usually give Markovnikov alcohols. The difference between the latter two lies especially in intermediate structure and rearrangement potential.

Step-by-step reasoning

Mark the more and less substituted alkene carbons. Draw mercury bridging them, then let water attack mainly at the more substituted carbon from the opposite face. Show the O⁺ intermediate and its deprotonation, leaving OH at that carbon and Hg-containing group at the other. Apply NaBH₄ to replace C–Hg by C–H. Name the final alcohol and avoid assuming a fixed net syn/anti outcome.

Visual explanation

Sketch a three-membered bridge with Hg above both alkene carbons. Draw water attacking the more substituted carbon from below, opening the bridge and leaving OH there after proton loss. In a second panel labelled NaBH₄, erase Hg from the other carbon and add H. Beside it, draw the acid-hydration carbocation route to show why only that simpler route invites ordinary hydride shifts.

Real-world analogy

A temporary brace connects two adjacent posts while a worker attaches a new fitting to the more favourable post. Once the fitting is secure, the brace is removed and replaced with a plain cap on the other post. The brace represents mercury's bridged intermediate, water supplies the fitting, and demercuration supplies the final cap. The temporary brace limits rearrangement during installation.

Real-world example

Oxymercuration–demercuration of 1-methylcyclopentene gives predominantly 1-methylcyclopentan-1-ol, placing OH on the more substituted alkene carbon without a free cation shift. This makes the reaction useful when a Markovnikov alcohol is desired from a substrate whose ordinary acid hydration could rearrange. The final stereochemistry still needs separate analysis rather than a blanket syn or anti label.

Why?

Why does this route give Markovnikov orientation without a free carbocation? The mercurinium bridge distributes positive character unevenly, making the more substituted carbon more favourable for water attack. The bridge prevents the simple open-cation rearrangement path. Removing mercury afterward supplies H at the other carbon, so the OH position set during water attack remains in the final product.

Common misconception

"Because water opens a bridge anti, the final hydration must be anti." The later NaBH₄ demercuration can place hydrogen from either face in a complex reduction. Initial water-versus-mercury geometry therefore does not impose one universal OH-versus-H relationship in the final alcohol. Distinguish first-stage stereochemistry from net stereochemistry.

Worked example

Question: What alcohol results when but-1-ene is treated with (1) Hg(OAc)₂, H₂O and (2) NaBH₄? Would a typical carbocation hydride shift be expected?

Reasoning: Water attacks the more substituted C2 of the mercurinium ion, placing OH there. Mercury remains at C1 until NaBH₄ replaces it with H. The bridged intermediate avoids a freely existing cation at C2.

Answer: Butan-2-ol is the main constitutional product; an ordinary free-carbocation hydride shift is not expected in this mechanism.

Quick check

1. Which reagent replaces the carbon–mercury bond with a carbon–hydrogen bond? Answer: Sodium borohydride, NaBH₄, in the demercuration stage.

Exam focus

Write the two reagent stages in order and track where water adds. Predict OH on the more substituted former alkene carbon and H on the other. Draw a bridged mercurinium ion, not a free carbocation, and explain the lack of routine rearrangement. Avoid assigning a universal net syn or anti stereochemistry after demercuration.

Advanced insight

The demercuration stage is not a simple textbook two-electron substitution; its mechanism is complex and can involve radicals. This explains why its facial stereochemistry is less predictable than the initial opening of the mercurinium ion. Mechanistic shorthand remains useful for connectivity, but experimental stereochemical data should govern claims about a particular substrate's final configuration.

Summary

Oxymercuration–demercuration hydrates an alkene with Markovnikov OH placement. Mercury(II) forms a bridged mercurinium ion, water attacks the more substituted carbon and loses a proton, then NaBH₄ replaces the mercury-containing group with H. The route avoids the ordinary rearrangements of a free-carbocation hydration. Its net stereochemistry is not fixed as universally syn or anti because demercuration can occur from either face.

Practice questions

1. Which former alkene carbon usually receives OH in oxymercuration–demercuration? Answer: The more substituted carbon, giving Markovnikov alcohol orientation. 2. Why is carbocation rearrangement usually avoided? Answer: The initial intermediate is a bridged mercurinium ion, not a free carbocation available for ordinary hydride or alkyl shifts. 3. What does NaBH₄ do after water has added? Answer: It replaces the carbon–mercury bond with a carbon–hydrogen bond in demercuration. 4. Can the final alcohol always be labelled a net anti-addition product? Answer: No. Hydrogen replacement during demercuration can occur from either face, so the overall facial outcome varies.