11.3 Adding H and H

We have learned the basic terminology for predicting products of addition reactions.

You should be able to predict products with ease with these three pieces of information.

All three pieces of information have been given to you.
This information won't be given to you as we progress through this chapter.
You will have to determine all three pieces of information on your own, and you will have to look at the reagents being used.
That might sound like a lot of work.
The mechanism of each reaction contains all the information you need.
You will know all three pieces of information about each reaction if you understand the mechanism for it.
We will focus on understanding.

It's possible to add H and H together.

We need to explore the chemistry of hydrogenation reactions.

Let's take a close look at how the reaction happens.

There are reagents that we use to do a hydrogenation reaction.

Pt, Pd, or Ni are some of the metal catalysts that can be used.

The hydrogen atoms are adsorbed to the metal.

Adding hydrogen atoms across the alkene is now possible.

Remember that H's don't need to be drawn in bond-line drawings, even though the hydrogen atoms that were added are not explicitly shown.

Two new stereocenters are being created.

We don't observe all four products.

We can deuterate an alkene just as we can hydrogenate one.

D and D will be added across the alkene.
We are adding two of the same group, so we don't need to worry about regiochemistry.

H and H will be added across a double bond.

We are adding two of the same group, so we don't need to worry about regiochemistry.
We must ask if we are creating two new stereocenters.
We are not creating a stereocenter.
Stereochemistry is irrelevant in this example.

Determine if you are forming two stereocenters in each example.

Stereochemistry will be irrelevant if not.

Adding a hydrogen halide across a double bond will be explored.

We must analyze the accepted mechanism in order to understand the regiochemistry and stereochemistry of HX addition.

A carbocation is generated when a protons is transferred to the alkene.

The result is the addition of H and X across the double bond.

We have used a starting alkene that avoids issues of regiochemistry and stereochemistry, and we will soon see other examples in which we must explore both of those issues.
The curved arrows are used in both steps.
It's important to master the art of drawing curved arrows.

There is one curved arrow in the second step.

All of the examples above were symmetrical alkenes.

Let's look at a case where regiochemistry is relevant.
We have to decide where to put the H and where to put the X.

The answer is contained in the mechanism.

A carbocation was formed in the first step of the mechanism.

The stability of tertiary carbocations is more stable than secondary ones.

We expect the alkene to accept the protons in such a way as to form the stable intermediate.

The carbocation is attacked in the last step of the mechanism.

The carbocation will result in the more substituted carbon.

The rule tells us to place the H on the less substituted carbon and the X on the more substituted carbon.

The rule is named after a Russian chemist, who first showed the regiochemical preference of HBr additions to alkenes.
The preference for the reaction to proceed via the stable carbocation intermediate will determine the regiochemistry.

The mechanism explains the regiochemistry of this reaction.

Don't try to make the regiochemistry of this reaction sound like it's Markovnikov.
Try to understand why it must be that way.

The mechanism should explain the regiochemistry in any reaction.

Stereochemistry is not relevant in this particular reaction.
The stereochemical outcome in a case where two new stereocenters are formed is beyond the scope of our conversation, so you will probably not see an example where two new stereocenters are formed.

The first thing we do is focus on the regiochemistry.

We have to decide where to place the H and where to place the Cl because this alkene is unsymmetrical.

We don't need to think about stereochemistry because the product doesn't have two stereocenters.

Stereochemistry is irrelevant in this example.
Stereochemistry of this reaction will not be relevant in the problems that you will encounter.

Try to draw the mechanism so that you can see why the reaction proceeds through the rule.

We focused our attention on the intermediate carbocation in order to understand the regiochemistry of HX additions to alkenes.

The reaction would proceed via stable carbocation, we argued.
This principle will help explain why some reactions will involve a rearrangement.

The product is not what we would have expected.

Again, we turn to the mechanism for an explanation.

The carbocation is about to be attacked.

Predicting when to expect a carbocation rearrangement is important.

A carbocation can be rearranged through either a hydride shift or a methyl shift.
You will have examples in your textbook.
Any reaction that involves an intermediate carbocation is possible.

Other addition reactions will follow through carbocation intermediates later in the chapter.

You will be aware that there is a chance for carbocations.

Let's practice.

We chose the way that would produce the secondary carbocation, rather than producing a primary carbocation.

We consider whether a rearrangement can take place before attacking with the halide.

We saw how to add H and X in the previous section.

Markovnikov only works with the H--X.

We will need to explore the mechanism in order to understand the answer.

This reaction uses radical intermediates instead of ionic intermediates.

You can compare the step above to a transfer of protons.

There is one important difference.

The hallmark of radical reactions are the fishhook arrows, not the one-headed curved arrows.

The more substituted carbon radical was formed by attacking the alkene at the less substituted carbon.

The stable nature of tertiary radicals is the same as the stable nature of secondary radicals.

One of the side products of this reaction is the regeneration of another br, which can go and react with another alkene.

The chain reaction occurs very quickly.
When peroxides are present, the reaction is much more rapid than the one we saw in the previous section.

The regiochemistry is determined by the preference for forming the most stable intermediate.

The two reactions are very similar.
Take notice of the fundamental difference.

We have focused on the chemistry of this reaction.

The stereo chemistry is beyond the scope of the course.
When two stereocenters are formed, the results are dependent on the temperature and alkene.
Problems will only be presented if no stereocenters are formed or only one stereocenter is formed.

The double bond will be adding H and Br.

We need to look at whether we are creating two new stereocenters to determine if stereochemistry is relevant.
Only one new stereocenter will be created when we place the Br on the less substituted carbon and the H on the more substituted carbon.
There are only two possible products, a pair of enantiomers, with only one stereocenter.

Both pathways are competing.

The radical reaction is much quicker.

Let's get some practice with choosing the right conditions.

When we compare the starting alkene with the desired product, we see that we need to place the Br at the more substituted carbon.

We will learn how to add H and OH across a double bond in the next two sections.

We need to choose our reagents carefully.
The reagents that give a Markovnikov addition of water will be explored in this section.

We have performed a hydration with Markovnikov regiochemistry, if we compare the starting material and product.

The reagent we used was H O+).

There are a lot of ways to show this reagent.

The brackets show that H+ isn't consumed in the reaction.

To understand why this reaction proceeds via a Markovnikov addition, we have to look at the mechanism.
The accepted mechanism of acid-catalyzed hydration is similar to the mechanism that we presented.

The first step in each mechanism is the creation of a carbocation.

The carbocation is attacked by either X- or H O.

The first reaction gives a product that is neutral.

The second reaction produced a charged species.
The positive charge must be removed at the end of the process.

Water is used as the base, not hydrox ide, when drawing the final deprotonation step.

There aren't many hydroxide ion floating around in acidic conditions.
There is plenty of water, and a mechanism must always be consistent with the conditions that are present.

We are using equilibrium arrows here.

The equilibrium arrows show that the reaction goes in both directions.
The reverse path is a reaction that we have already studied.
Follow the sequence above from the beginning to the end and convince yourself that it is the E1 process.
The truth is that most reactions represent equilibrium processes; however, organic chemists only make an effort to draw equilibrium arrows in situations where the equilibrium can be easily manipulated, allowing us to control which products are favored.
One of those situations is this one.

Le Chatelier's principle tells us that the equilibrium can be pushed toward one side or the other by removing or adding reagents.

You add water.

The system would have to adjust to the fact that the concentrations were no longer at equilibrium.
Adding water would make alkene turn into alcohol.
Diluted acid is mostly water and would be used to favor the alcohol.
Water removal would push the equilibrium to the left if we wanted to favor the alkene.
We would use concentrated acid if we wanted to form the alkene.

Carefully choosing reaction conditions can affect the outcome of a reaction.

The regiochemistry of acid-catalyzed hydration is already explained.

The stereochemistry of acid-catalyzed hydration is very similar to the stereochemistry of ionic addition of HX.

The stereochemical outcome in a case where two new stereocenters are formed is beyond the scope of our conversation, so you will probably not see an example where two new stereocenters are formed.

The reagent suggests that we have an acid-catalyzed hydration.

When two new stereocenters are formed, the stereochemistry of acid-catalyzed hydration is complex.

We are not forming new stereocenters.
We are not forming a new stereocenter.

In the previous section, we learned how to add H and OH across a double bond.

The products show that we are adding H and OH across the alkene.

An optical illusion can be seen in the example above.

The group was already there.

We added OH and H because they don't have to be drawn in a bond-line drawing.

The answers to the three questions are encapsulated in the mechanism.

To explore the accepted mechanism, we must first get to know the reagents.
The reagents are BH and THF.
Borane is the former.

There is no octet in this structure.

Borane is very active.

The solvent is called THF.

It has high electron density to fill its empty orbital.

Borane can be attacked by a pi bond because it is a site of high electron density.
This is the first step of our mechanism.

It all happens in a four-membered transition state.

Take a close look at the first step and consider the regiochemistry and stereochemistry.

The OH group will end up on less substituted carbon because of the regiochemistry.

One of the sources of this regiochemical preference can now be understood.
The double bond is being added with H and BH.

A concerted process is represented by the step above.

The formation of the trialkylborane above is called hydroboration, which occurs when you mix an alkene with BH.

We have not seen anything like this before.

Oxygen is placed between B and R when R migrates over.

The stereocenter is unaffected by the migration.

The stereocenter's configuration is preserved.

The final product is an alcohol.

We have a two-step process for converting an alkene into alcohol.

We have to decide if stereochemistry is relevant in how we draw our products.

We have to decide if stereochemistry is relevant in how we draw our products.

We aren't creating two new stereocenters.
We are not creating a stereocenter.

Whenever stereochemistry is irrelevant, the problem becomes a bit easier to solve.