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11.8 Synthesis Techniques
It is essential that you master all of the individual reactions that we have seen so far in order to begin practicing synthesis problems.
We will focus on one-step problems.
We can string the individual reactions together in various sequence to form synthesis problems if you feel comfortable with them.
Let's review what these reactions can do.
Adding two groups across a double bond is possible with addition reactions.
We must add H and OH if we compare the starting material and product.
We need a Markovnikov addition because we see that OH is ending up at the more substituted carbon.
The stereochemistry of the reaction won't matter.
We need to choose reagents that will add H and OH.
The result is a change in the position of the atom.
There is no one-step method for doing this.
If we waited long enough, the bromide could leave in an S 1 reaction and the carbocation could rear N range to become tertiary.
Waiting for an S 1 is not the best idea.
There are a few things to keep in mind when doing this type of sequence.
We can choose our base to control which product is favored.
If we use a strong base like methoxide or ethoxide, it will be favored.
The less substituted alkene will be favored.
We must carefully consider the regiochemistry of how we add HBr across the double bond after forming it.
The problem requires us to move the Br to the left.
We must be careful to control the regiochemistry in each step.
We must use a sterically hindered base in the elimination step to form the less substituted double bond.
When dealing with an OH group, let's see what to do.
The elimination reaction to give the Hofmann product is the first step of our technique.
The OH group needs to be converted into a good leaving group in order to do this.
If you haven't yet learned about tosylates in your lecture course, you should consult your textbook for more information.
We have seen how to combine two reactions into one synthetic technique.
Adding and then eliminating is a type of technique.
We are asked to move the position of a double bond.
We have not been able to do the transformation in one step.
In the first step, we need a Markovnikov addition, which we can easily accomplish by using HBr.
If we choose a base that is not sterically hindered, we can eliminate the second step and give the Zaitsev product.
We must consider the regiochemistry of each step when using this technique.
This technique can be used in the following problems.
The starting compound always had a functional group that we could manipulate.
Either the starting material had a leaving group or a double bond.
If you want to discuss radical bromination of alkanes, you should consult your textbook and/or lecture notes.
The following example will show how radical bromination can be used in a synthesis.
We can't do an elimination reaction because there are no leaving groups.
We can't do an addition because there are no double bonds.
We are stuck with nothing to do.
The only way out of this situation is to introduce a functional group into the compound.
There are a few things to think about when using this technique.
First of all, radical bromination will place a Br on the most substituted position.
The tertiary position is where the br will go.
Make sure to choose the right base in order to achieve the desired regiochemistry.
Let's practice with this.
We will analyze the regiochemistry and stereochemistry of the reaction.
The regiochemistry is irrelevant because we are adding two of the same group.
We will see that we have a pair of enantiomers if we carefully examine the products.
Many students think that the two products are the same, but they are not.
We have to use the accepted mechanism again.
In the first step, we have an alkene reacting.
A region in space of electron density is represented by the alkene's pi bond.
This means that the person is the electrophile.
An interesting thing happens when a br2 molecule approaches an alkene.
The effect becomes more pronounced as the molecule gets closer to the alkene.
There are three curved arrows here.
Students forget to draw the third curved arrow when drawing this mechanism.
This step must be a back-side attack because it is an S 2-type process.
There is a different mechanism in those cases.
When we use water as the solvent, the outcome is more interesting.
If we look at the products, we'll see that we're adding OH instead of BR.
We refer back to the accepted mechanism in order to understand what is happening.
There are now two nucleophiles present, bromide and water.
Rather than attacking the bromonium ion, a water molecule can attack, which will give the products shown above.
It has a positive charge on a bromine atom and is a very high-energy intermediate.
It's very eager to react with any nucleophile.
The first nucleophile it encounters will have an effect on it.
Since we are using water as the solvent here, the bromonium ion will most likely encounter a water molecule before it gets attacked by a bromide ion.
We didn't need to think about regiochemistry in the absence of water.
In the presence of water, we are adding two different groups.
The structure of the bromonium ion needs to be looked at more carefully to answer this question.
The bromine atom was drawn to form a perfect three-membered ring.
The tertiary carbon can handle this character.
The less substituted carbon will have more d+ than the more substituted carbon.
The geometry won't be trigonal or tetrahedral.
The geometry of the tertiary carbon atom will be between the two.
We can't have an S 2 at a tertiary center.
We don't use HO- because there isn't much of it around.
It's important to stay consistent with the conditions.
The reagents are not hydroxide.
The regiochemistry is irrelevant because we are adding two more people.
We are looking to see if we are creating two new stereocenters.
We are in this case.
The stereochemistry is relevant.
We are adding two groups in this problem.
The products will need to be drawn with the OH.
We are looking to see if we are creating two new stereocenters.
We are in this case.
We will see how to add two OH groups.
The two-step process will now be explored.
Peroxy acids have one additional oxygen atom.
Peroxy acids are strong oxidizing agents.
It is one of the most common examples in this course.
If you see it, you should immediately recognize it as an example of a peroxy acid.
An alkene will form an epoxide.
Depending on which textbook you are using, the mechanism may or may not be in it.
We will not spend a lot of time on this mechanism because it is so complicated.
Let's look at the product of this reaction.
We open it with water under acid-catalysis.
Let's look at how this happens.
The intermediate produced by this step is very similar to a bromonium ion.
We use water to deprotonate in order to stay consistent with the conditions.
Regiochemistry will be irrelevant because we are adding OH and OH.
We must carefully consider the stereochemistry of the reaction in order to draw the correct pair of enantiomers.
Only one new stereocenter is being created.
The stereochemistry in this case is irrelevant.
We have a process that adds both oxygen atoms at the same time.
We don't need to think about regio chemistry because we are adding two OH groups.
If we are forming two new stereocenters, stereochemistry will only be relevant.
There are many reagents that can be used to cleave the bond.
There are two C==O double bonds that are completely split apart.
We need to explore the reagents in order to understand how this reaction happens.
The initial product is called a molozonide, and it undergoes further rearrangements, before ultimately giving the product upon treatment with dimethyl sulfide.
A mild reducing agent is CH3 DMS.
There are many reducing agents that can be used in the final step of an ozonolysis.
The method for drawing the products of an ozonolysis is to split the bonds into two.
There are two bonds in this compound.
We place two C==O bonds in its place if we erase the C==C double bond.
The diagram shows the key reactions in the chapter.
If you have completed all of the problems in this chapter before this point, you should be able to fill in the necessary reagents.
You should study this diagram after you have filled in the necessary reagents.
Make sure that the whole thing makes sense to you.
You have to go over it ten times.
If you can reconstruct the whole diagram on a blank piece of paper, you have done your job.
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