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4.7 Identifying Activators and Deactivators
We learned how to predict the directing effects in a situation where you have more than one group on the ring.
In the previous section, I had to tell you whether the group was strong or weak.
You won't have to memorize the characteristics of every possible group if we learn how to predict this in this section.
There is very little memory involved here.
Some concepts should make sense.
You should be able to identify the nature of any group even if you have never seen it before.
We will start with strong activators.
There are groups with a single pair next to the aromatic ring.
An example of this has already been seen.
Since the ring is electron-rich, it is a stronger nucleophile than benzene.
The OH group is one of the groups that have a lone pair next to the ring.
There are many examples of strong activators.
The lone pair next to the ring is already partially tied up in resonance.
The electron density is spread out because there are some in the ring and some out of the ring.
Look closely at the examples.
They all have a pair that is tied up outside of the ring.
There is only one pair that is not tied up in resonance outside of the ring.
Hyperconjugation was encountered to explain why tertiary carbocations are more stable than secondary ones.
alkyl groups exhibit an electron-donating effect and are therefore weak activators.
The different categories of activators are strong, moderate, and weak.
There is a reason for this order to be used.
The weak deactivators are the halogens.
We saw that they are deactivators.
In the case of the halogens, the competition is close and they are only weakly deactivating.
We will summarize all of this information in a single chart, but we need to look at moderate deactivators.
Moderate deactivators withdraw electron density from the ring via resonance.
Other groups can also withdraw electron density from the ring.
The substituents withdraw electron density from the ring via resonance.
They all have a pi bond to an atom.
Take a look at the last example.
A triple bond to an atom is called a cyano group.
A triple bond can be included in this category.
The nitro group is so powerful that we have already explained why.
One very powerful deactivating group is given by the effects of each chlorine atom.
There are resonance effects to consider when a halogen is connected directly to the ring.
Make sure that every category makes sense to you by taking a close look at the chart.
We gave strong arguments for each category.
You might want to read the last few pages.
We can understand why we looked at weak deactivators first.
Predict what kind of group it is, for example a strong deactivator, a weak deactivator, or a moderate deactivator.
Predict the effects if this compound undergoes an aromatic substitution reaction.
The group does not have a single pair next to the ring.
It isn't an activator.
The group has a pi bond to an oxygen atom and is a moderate deactivator.
Determine what kind of group it is for each of the following substituents.
You can place your answer on the space provided.
You won't be able to see this chart on the exam.
Try to remember the explanations we used.
Predicting the directing effects can be done with that information.
The skills we developed in this section can be used to predict the products of a reaction.
We look at the reagents to see what happens.
There are two reagents, nitric acid and sulfuric acid.
The reagents generate NO+, which is an excellent phile.
We know that the reaction will put a group on the ring.
Predicting the directing effects of the group currently on the ring is what we must do to answer this question.
We don't need a Lewis acid catalyst in this reaction.
The material in this section is what we did in the previous section.
When you have more than one group on the ring, we learned how to predict the directing effects.
We look at the reagents to see what happens.
bromine and aluminum tri bromide are the reagents.
The reagents generate an excellent phile.
We know that we are going to put a br atom on the ring.
Predicting the directing effects of the two groups currently on the ring is what we must do to answer this question.
The two groups are competing.
The last product was placed in parentheses.
This is a very small product.
In the next section, we will see why.
We expect three products for now.
In the next section, we will fine-tune this prediction.
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