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5.2 SNAr Mechanism
The three criteria that are necessary in order for an aromatic ring to undergo a substitution reaction were shown in the previous section.
There are some possible mechanisms for this process.
Our reaction is not an S 2 mechanism.
This kind of carbocation is not stable.
We don't expect the leaving group to leave if it means creating an unstable intermediate.
We don't expect the mechanism to be an S 1 mechanism.
It's a new mechanism called S Ar.
The intermediate is not stable.
The reaction doesn't happen.
There is a subtle point that deserves attention before you start drawing the complete mechanism of an S Ar process.
The ring is attacked by a nucleophile and the leaving group is lost to restore aromaticity.
Carefully inspect the product.
The proton is mildly acidic and can't survive in the strongly basic conditions being employed.
You might be surprised by the last two steps when you look at this mechanism.
The mechanism needs to show the last two steps.
The last two steps are necessary.
The phenolic proton does not survive under the basic reaction conditions.
Whether we like it or not, detonation occurs.
The mechanism indicates that we understand the subtle point.
By drawing the last two steps of the mechanism, you are showing that you understand why a H O+) must be added after the reaction is complete.
We are going to practice drawing an S Ar mechanism.
There is a mechanism for the reaction.
The leaving group is expelled in the second step of the mechanism.
This could be a complete mechanism.
In the previous section, we talked about the three criteria you need to get an S Ar mechanism.
There is no reaction if we try to apply heat.
This reaction is important because it is an efficient way of making phenol.
We don't need high temperatures to make aniline.
H N- is used in liquid ammonia.
The chemists used a technique called isotopic labeling to understand the mechanism.
Deuterium is an isotope of hydrogen because it has a neutron in the nucleus while hydrogen does not.
There are some important isotopes in carbon.
13C is important because we can easily determine the position of a 13C atom in a compound.
We can follow where the carbon atom goes during the reaction if we enrich a specific spot with 13C.
The 13C was placed at the site with the asterisk.
Most of the molecule in the flask have a 13C atom in that position, when we say that we enriched that spot with 13C.
As the reaction proceeds, let's see what happens to the isotopic label.
We can't explain this with a simple substitution.
We wouldn't be able to explain the isotopic labeling results if we didn't have a reservoir for the electron density during the reaction.
The proposal explains the experiments.
This creates a very strange looking intermediate called benzyne.
This time, another hydroxide ion is involved, and it's acting as a nucleophile to attack benzyne.
There is no reason to prefer one site over the other, so we must assume that these two pathways occur with equal probability.
As we saw in the previous section, phenol will undergo deprotonation as a result of the basic reaction conditions.
The elimination is followed by an addition.
When you think about it, it seems off the wall.
It looks like an intermediate.
The intermediate of this reaction has been shown by chemists with other experiments.
Evidence for the short-lived existence of benzyne will most likely be provided by your textbook or instructor.
You can look at the evidence in your textbook if you are curious.
Let's make sure that we can predict the products of elimination- addition reactions.
An elimination-addition mechanism is being dealt with.
The middle two of these products are the same.
I will give you some problems that go through an addition elimination mechanism, rather than an elimination-addition.
We have seen how to install an OH group on an aromatic ring.
A summary of how to install an OH group or NH2 group on an aromatic ring can be found here.
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