4.7 Identifying Activators and Deactivators

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.