---
title: "Oxymercuration-Reduction | Organic Chemistry"
description: "Oxymercuration-reduction converts alkenes to alcohols in Organic Chemistry by adding water in a Markovnikov pattern without carbocation rearrangement."
canonical: "https://fiveable.me/organic-chem/key-terms/oxymercuration-reduction"
type: "key-term"
subject: "Organic Chemistry"
unit: "Unit 8"
---

# Oxymercuration-Reduction | Organic Chemistry

## Definition

Oxymercuration-reduction is a two-step Organic Chemistry reaction that turns an alkene into an alcohol. It adds the OH group in Markovnikov fashion without carbocation rearrangement.

## What It Is

Oxymercuration-reduction is a way to turn an alkene into an alcohol in Organic Chemistry, and it is known for giving the Markovnikov product without rearranging the carbon skeleton. That means the OH group ends up on the more substituted carbon of the double bond, but you do not get a hydride shift or methyl shift in the middle.

The reaction happens in two steps. First, the alkene reacts with a mercury(II) reagent such as mercury(II) acetate, which forms a bridged mercurinium ion instead of a free carbocation. That bridged intermediate is the reason the reaction is so controlled, because it blocks the kind of open, unstable intermediate that often rearranges in other hydration reactions.

In the second step, sodium borohydride reduces the carbon-mercury bond and replaces the mercury-containing group with hydrogen. After that reduction, the product is an alcohol. So the overall sequence acts like hydration of the alkene, but it does it through a mercury intermediate rather than by simply adding acid and water.

The regioselectivity comes from how the mercurinium ion opens. Water or another oxygen nucleophile attacks the more substituted carbon because that site carries more positive character in the bridged intermediate. You do not need to track a carbocation rearrangement step the way you would in acid-catalyzed hydration.

One easy way to think about it is: oxymercuration puts the OH where Markovnikov addition would predict, but it avoids the messy rearrangement problem. If you start with a terminal or internal alkene and the reaction conditions are oxymercuration followed by reduction, you should expect a predictable alcohol product instead of a mixture built from rearranged carbocations.

## Why It Matters

This reaction shows up whenever Organic Chemistry moves from alkene structure to alcohol synthesis. It gives you a clean route from a carbon-carbon double bond to an OH-containing product, which is a common step in multi-step synthesis problems.

It also teaches a bigger mechanism idea: not every addition to an alkene has to go through a free carbocation. Oxymercuration-reduction is one of the best examples of how a bridged intermediate can control regioselectivity and avoid rearrangements at the same time.

That matters when you compare alkene hydration methods. If a problem asks you to choose between acid-catalyzed hydration, hydroboration-oxidation, and oxymercuration-reduction, you need to know which one gives Markovnikov vs anti-Markovnikov placement and which one risks rearrangement.

You will also see it in synthesis planning. If the target molecule needs an alcohol at the more substituted alkene carbon and the carbon skeleton has to stay intact, oxymercuration-reduction is often the clean choice. It gives you a product you can predict from the starting alkene, which is exactly what synthesis questions reward.

## Connections

### Mercurinium Ion

This is the key intermediate formed in the first step. Instead of a free carbocation, the alkene forms a bridged mercury-containing ion, and that bridge controls where the nucleophile attacks. If you can picture that three-membered structure, the regioselectivity of oxymercuration-reduction makes a lot more sense.

### [Hydroboration-Oxidation](/organic-chem/key-terms/hydroboration-oxidation)

This is the reaction students most often compare with oxymercuration-reduction because both convert alkenes into alcohols. The big difference is regiochemistry: hydroboration-oxidation gives anti-Markovnikov alcohols, while oxymercuration-reduction gives Markovnikov alcohols. They are also useful for opposite synthetic goals.

### [Hydroxyl Group](/organic-chem/key-terms/hydroxyl-group)

The whole point of the reaction is to install a hydroxyl group on an alkene. In synthesis problems, you are often choosing a reaction based on where the OH group ends up. Oxymercuration-reduction is one of the standard ways to add that functional group in a predictable position.

### [Hydride shift](/organic-chem/key-terms/hydride-shift)

This is the rearrangement you usually do not want in alkene hydration problems. Because oxymercuration-reduction uses a bridged intermediate instead of a free carbocation, hydride shifts are avoided. That makes the reaction cleaner than acid-catalyzed hydration when rearrangement would change the product.

## On the AP Exam

A problem set question might show you an alkene and ask for the product after oxymercuration-reduction. Your job is to place the OH group on the more substituted alkene carbon, keep the carbon skeleton unchanged, and remember that no rearrangement occurs. If the question includes several hydration methods, use the product pattern to pick the right reaction.

In a mechanism quiz, you may need to identify the mercurinium ion as the intermediate and show sodium borohydride in the second step. In synthesis problems, this reaction often appears when the target molecule needs Markovnikov alcohol formation but acid-catalyzed hydration would risk a shift.

## Oxymercuration-Reduction vs Hydroboration-Oxidation

These reactions both turn alkenes into alcohols, so they are easy to mix up. Oxymercuration-reduction gives Markovnikov addition and avoids rearrangements, while hydroboration-oxidation gives anti-Markovnikov addition. The product location is the fastest way to tell them apart.

## Key Takeaways

- Oxymercuration-reduction converts an alkene into an alcohol in two steps.
- The reaction gives Markovnikov regiochemistry, so the OH ends up on the more substituted carbon.
- A bridged mercurinium ion forms first, which keeps the reaction from going through a rearranging carbocation.
- Sodium borohydride finishes the reaction by replacing the mercury-containing group with hydrogen.
- If you need a predictable alcohol product without rearrangement, this is one of the best alkene hydration methods to recognize.

## FAQs

### What is oxymercuration-reduction in Organic Chemistry?

It is a two-step reaction that converts an alkene into an alcohol. The alkene first forms a mercurinium ion with a mercury(II) reagent, then sodium borohydride reduces the intermediate to give the alcohol product. The OH ends up in the Markovnikov position.

### Why does oxymercuration-reduction not rearrange?

Because the reaction does not form a free carbocation. Instead, it goes through a bridged mercurinium ion, which changes how the nucleophile attacks and prevents hydride shifts or methyl shifts. That is why the product stays predictable.

### How is oxymercuration-reduction different from hydroboration-oxidation?

Both make alcohols from alkenes, but they place the OH group in different spots. Oxymercuration-reduction gives Markovnikov products, while hydroboration-oxidation gives anti-Markovnikov products. That difference is often the deciding clue in synthesis questions.

### How do I identify oxymercuration-reduction on a problem?

Look for an alkene starting material, a mercury(II) reagent such as mercury(II) acetate, and then sodium borohydride. If the product is an alcohol on the more substituted carbon and the skeleton has not rearranged, that is the pattern you want.

## Related Study Guides

- [8.5 Hydration of Alkenes: Addition of H2O by Hydroboration](/organic-chem/unit-8/hydration-alkenes-addition-h2o-hydroboration/study-guide/XYyEKQ4vV9x5lgUM)
- [9.9 An Introduction to Organic Synthesis](/organic-chem/unit-9/introduction-organic-synthesis/study-guide/zWSWt45h5E3taPuq)

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