6.4 O-Nucleophiles

6.4 O-Nucleophiles

Oxygen nucleophiles will be our focus in this section.

Let's look at what happens when alcohol attacks a ketone or aldehyde.

One of the longer mechanisms that you will encounter in this course is the one we are about to see.

It's important because it lays the foundation for so many other mechanisms.
You will be in good shape if you can master this mechanism.
There is no other option but to master this mechanism.
You should be prepared to read through the next several pages slowly, and then practice the material on these pages as many times as necessary until you know this mechanism extremely well.

There is a major difference here.

We argued that the carbonyl group could not re-form after the attack because there was no leaving group.
There is a leaving group in this section.

The leaving group can be aided by the attacking nucleophile.

That gets us back to where we started.
There is no net reaction when a molecule of alcohol attacks the carbonyl group.

Let's see if there is a reaction that can happen.

The attack of an alcohol is slower than the attack of a hydrogen nucleophile because alcohols do not have a negative charge.

Under these conditions, we would have the same problem that we just had.

The carbonyl group can re-forming.

We will take a different approach.

We will focus on making the electrophilic instead of making the nucleophilic.

The carbonyl group is more electron-poor because the entity bears a full positive charge.

This is important because we will see this many times throughout the chapter.
H SO is one of the acids that can be used for this purpose.

As we continue to discuss this mechanism, we will just show H--A+ as the source of the protons, and it is expected that you will understand that the identity of H--A+ is likely a protonated alcohol.

There is an equilibrium between the forward and reverse processes.

There is something else that can happen every now and then.

We are exploring whether HO- can be expelled as a leaving group, which should theoretically work because we said before that anything can be expelled except for H- and C-.

Make sure that this rule becomes part of the way you think--NEVER expel HO- into acidic conditions-- always protons it first.

We first deprotonated to form an intermediate with no charge.

We chose this order to avoid having an intermediate with two positive charges.
This is an important rule that you should incorporate into your thinking.
Intermediates have two similar charges.

It doesn't happen that way because the oxygen atom and the protons are too far apart in space to interact.

A 4-membered ring is high in energy and unlikely to occur in a transfer of protons.
If you want to draw a mechanism, you have to remove a protons, and then you have to decide if you want to use the same protons that were removed.

There is a way to remove the charge.

Let's take a close look at the whole thing to make sure we understand some of the key features.

Let's focus on the transfers in the entire mechanism first.

There are four steps that transfer protons.
Four of them involve deprotonation.
The acid isn't consumed by the reaction.

To facilitate these three steps, all of the proton transfers are used, and they make the carbonyl group more electrophilic, to produce water as a leaving group, and to avoid multiple charges.

You need to see the reaction in this way.
It will make the mechanism simpler in your mind.

The only intermediate that does not have a charge is the one that forms an acetal from a ketone.

It's a special name because it's possible to isolated it and store it in a bottle, and because it's important if you want to learn biochemistry.

An acetal does not have a carbonyl group.

If we attempt to perform this reaction in a lab, we will get very little product.

The trick is to remove water from the reaction as it proceeds.
The reverse path will be stopped if we remove water as it is formed.

We force the reaction to a point by removing water as it is forming.

Three structures are in equilibrium in the highlighted area.

Two of them are charged, and one of them is un charged.
The un charged product is now favored by this equilibrium.

The acetal can be formed by removing the water.

There is no water present because the carbonyl group has been removed.
Even though the products are less stable than the reactants, we can force the equilibrium to favor them.

In your textbook and in your lectures, you will explore the way that chemists remove water from the reaction as it proceeds.

By writing the words "Dean-Stark" you are indicating that you understand that it is necessary to remove water in order to form the acetal.

How you would reverse this reaction is something we can appreciate.

You want to convert the acetal into a ketone.
Adding water with acid would convert the acetal into a ketone.

The equilibrium will favor the formation of the ketone.

We need to be able to control the conditions to push the reaction in either direction.

There is a mechanism for the reaction.

We are going to end up with an acetal when we start with a ketone.

It is difficult to see because of the way it is happening.

The same order of steps should be followed by the mechanism.

There are three critical steps, including a loss of water and another nucleophilic attack.
There are three steps to the transfer of protons.
The carbonyl group is rendered more electrophilic with the use of a protons transfer.

We use protons to form water.

A protons transfer is used to remove the charge and generate the product.

Maybe you could draw the mechanism for this reaction on a piece of paper.

This type of mechanism is so important because there will be many more reactions that build upon the concepts that we developed in this mechanism.

You should work through the problems slowly.

Each of the following transformations has a mechanism.

Each mechanism requires a separate piece of paper.

If you want to know if you have mastered a mechanism forward and backwards, you should try to draw it backwards.

That's correct.
Draw a mechanism for the reaction.
Before attempting to draw a mechanism, read the advice below.

You should only draw the intermediates until you reach the acetal.

Working backwards from the ketone to the acetal, draw only the intermediates.

Once you have all of the intermediates drawn, you can try to fill in the arrows with the acetal.
To draw your mechanism, use a separate sheet of paper.
You can compare your answer to the one in the back of the book.

There is a mechanism in which the ketone is attacked.

The same reaction can happen when both OH groups are in the same molecule.

It would be wise to be familiar with the process of this type of reaction because it might appear several times in your lectures and textbook.

The transformation of the diol in the reaction can be very useful.

We have been able to manipulate the conditions of this reaction to control whether the acetal is favored or not.

Please be patient as we develop this concrete example, it will take us a couple of pages to complete.

Both carbonyl groups will be reduced by LiAlH.

It would seem impossible because they are less reactive than ketones.

A reagent that reduces an ester should also reduce a ketone.

There is a way to reach the goal.

The ketone is converted into an acetal.

We are able to protect the ketone.

We use water above in the second step every time we have used LiAlH.

In the next chapter, we will talk more about this strategy.

Let's focus on knowing the reactions so that we can predict products.