Making Ethers - Alkoxymercuration - Online Tutor, Practice Problems & Exam Prep

Alkoxymercuration is a reaction similar to oxymercuration, where alcohol acts as the nucleophile instead of water, leading to ether formation. The mechanism involves the double bond attacking mercury, creating a bridged ion intermediate. Ethanol then attacks the Markovnikov site, displacing mercury and forming an ether. The subsequent reduction step, using NaBH₄, converts the positive charge on oxygen to a hydrogen, yielding the final ether product. Understanding the role of the nucleophile is crucial for predicting the product outcome in these reactions.

The Mechanism of Alkoxymercuation.

Another way that we can synthesize ether is through a reaction called alkoxymercuration. And you might immediately notice that this sounds a lot like another reaction that you're supposed to know by now and that's oxymercuration. So it turns out this is going to be the same exact mechanism as oxy oxymercuration or oxymerc except that in the top reagent in the oxymercuration step, we're going to use alcohol as our nucleophile instead of water. What you can see is that the biggest difference here is this right here. I have alcohol in place of the water and that's going to make an ether instead of an alcohol. Let me just walk you guys through this mechanism really quick.

So remember that what we get is Hg(OAc)₂. And in the first step, what we do is the double bond attacks the mercury, kicks out one of the OACs and the mercury attacks back. So what we wind up making is a bridge, an ion bridge. So what that's going to look like is like this, where now I have a dotted partial bond to Hg. It's only attached to one OAC now. Dotted partial and a positive charge. There should be actually a dotted partial bond here too and a positive charge. So now we've got our intermediate. Notice that we can't get any shifts because there's no carbocation. And now this is the part that typically water would come in and attack which side? Do you remember? In this case, I have a Markovnikov site here and I have an anti-Markovnikov site there. Which one would it attack? Markovnikov because that's going to be the one with the most positive character. But instead of water, what's actually going to attack is, in this case, ethanol. So my ethanol attacks, kicks out the Hg, and what I wind up getting is something that looks like this. I'm going to try to fit it in here.

Where now, let's say that alcohol attacks from the bottom, then I would draw that the H

Here’s what students ask on this topic:

What is the mechanism of alkoxymercuration in ether synthesis?

The mechanism of alkoxymercuration in ether synthesis involves several steps. First, the double bond of an alkene attacks a mercury acetate (Hg(OAc)₂) molecule, forming a bridged mercurinium ion intermediate. Unlike oxymercuration, where water is the nucleophile, in alkoxymercuration, an alcohol (e.g., ethanol) attacks the more substituted carbon (Markovnikov site) of the intermediate. This displaces the mercury and forms an ether. The final step is a reduction using sodium borohydride (NaBH₄), which replaces the mercury with a hydrogen atom, yielding the ether product.

How does alkoxymercuration differ from oxymercuration?

Alkoxymercuration differs from oxymercuration primarily in the nucleophile used. In oxymercuration, water acts as the nucleophile, leading to the formation of an alcohol. In alkoxymercuration, an alcohol (e.g., ethanol) is used instead, resulting in the formation of an ether. The mechanism is otherwise similar, involving the formation of a bridged mercurinium ion intermediate and subsequent nucleophilic attack at the Markovnikov site. The final reduction step with NaBH₄ is also common to both reactions.

What reagents are required for the alkoxymercuration reaction?

The reagents required for the alkoxymercuration reaction include mercury(II) acetate (Hg(OAc)₂) and an alcohol (e.g., ethanol) for the initial step. The reduction step requires sodium borohydride (NaBH₄). The overall reaction involves the alkene, Hg(OAc)₂, the chosen alcohol, and NaBH₄ to yield the ether product.

Why is the Markovnikov site attacked in alkoxymercuration?

In alkoxymercuration, the Markovnikov site is attacked because it is the more substituted carbon of the bridged mercurinium ion intermediate. This site has a partial positive charge due to the electron-withdrawing effect of the mercury atom. The nucleophile (alcohol) preferentially attacks this site because it is more electrophilic, leading to the formation of the more stable carbocation intermediate and ultimately the ether product.

What is the role of NaBH₄ in the alkoxymercuration reaction?

In the alkoxymercuration reaction, sodium borohydride (NaBH₄) serves as a reducing agent in the final step. After the formation of the ether intermediate, NaBH₄ reduces the mercury atom attached to the carbon, replacing it with a hydrogen atom. This step is crucial for completing the transformation from the intermediate to the final ether product.