# Organic Chemistry I & II

Lead Author(s): Steven Forsey

Top Hat Intro Course - Organic Chemistry I & II is designed for instructors who want an active, dynamic, and understandable approach to support their own efforts in the classroom. This ever-evolving textbook includes auto-graded questions, videos and approachable language in order to make difficult concepts easier to understand and implement.

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Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

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Solomons et al., “Organic Chemistry”, 12th Edition

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David R. Klein, “Organic Chemistry”, 3rd Edition

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#### $301 Hardcover print text only ### Always up-to-date content, constantly revised by community of professors Content meets standard for Introduction to Organic Chemistry course, and is updated with the latest content ### In-Book Interactivity Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost Only available with supplementary resources at additional cost ### Customizable Ability to revise, adjust and adapt content to meet needs of course and instructor ### All-in-one Platform Access to additional questions, test banks, and slides available within one platform ## Pricing Average price of textbook across most common format ### Top Hat Steven Forsey, “Organic Chemistry”, Only one edition needed #### Up to40-60%more affordable Lifetime access on any device ### McGraw-Hill Carey & Giuliano, “Organic Chemistry”, 10th Edition ####$219

Hardcover print text only

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

#### $301 Hardcover print text only ### Wiley David R. Klein, “Organic Chemistry”, 3rd Edition ####$301

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## Always up-to-date content, constantly revised by community of professors

Constantly revised and updated by a community of professors with the latest content

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## In-book Interactivity

Includes embedded multi-media files and integrated software to enhance visual presentation of concepts directly in textbook

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## Customizable

Ability to revise, adjust and adapt content to meet needs of course and instructor

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

## All-in-one Platform

Access to additional questions, test banks, and slides available within one platform

### Top Hat

Steven Forsey, “Organic Chemistry”, Only one edition needed

### McGraw-Hill

Carey & Giuliano, “Organic Chemistry”, 10th Edition

### Wiley

Solomons et al., “Organic Chemistry”, 12th Edition

### Wiley

David R. Klein, “Organic Chemistry”, 3rd Edition

#### Dr. Steven Forsey, Ph.D.University of Waterloo

Steven Forsey is currently a Professor at University of Waterloo, teaching a variety of organic chemistry courses to Chemistry, Science, Chemical Engineering, Nanotechnology and distance education students. He received his Ph.D. (2004) for Synthetic Organic Chemistry from University of Waterloo, Ontario. He is a recipient of the Excellence of Science Teaching Award and has acted as the Teaching Fellow for the Department of Chemistry since 2016.

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# Chapter 21: Reactions of Aromatic Compounds

## Learning Objectives

• Identify electrophiles and generate electrophiles from given reagents.
• Write the two-step mechanism for electrophilic aromatic substitution.
• Recognize the products of electrophilic aromatic substitution.
• Understand the effect of substituents on the rate of electrophilic aromatic substitution.
• Identify directing effects and steric effects for disubstituted and polysubstituted benzene rings.
• Evaluate directing effect and predict the product of electrophilic aromatic substitution when disubstituted benzene is used as starting material.
• Propose synthesis of disubstituted and polysubstituted benzene rings.
• Recognize activating and deactivating groups.
• Understand the limitations of electrophilic aromatic substitution.
• Recognize nucleophilic aromatic substitution by elimination-addition and write the mechanism.
• Recognize nucleophilic aromatic substitution by addition-elimination and write the mechanism.
• Identify the reagents for reactions of side chains.
• Predict the products for reactions of side chains.

## 21.1 Electrophilic Aromatic Substitution

In Chapter 20, you were introduced to the remarkable stability of benzene due to aromaticity. In effect, while alkenes and alkynes undergo electrophilic addition when treated with bromine, benzene is inert under similar reaction conditions. Electrophilic substitution is the most common reaction of aromatic compounds. In the fundamental reaction, an electrophile E+ generated from given reagents will accept a pair of electrons from the aromatic ring to form a cyclohexadienyl cation (step 1 in Figure 21.1). The cyclohexadienyl cation then loses a proton to reform the aromatic system (step 2 in Figure 21.1). The net result is the substitution of a hydrogen atom by E+.

In this chapter, we will consider the following electrophilic aromatic substitution reactions: halogenation, nitration, sulfonation, Friedel-Crafts alkylation, Friedel-Crafts acylation, and reactions involving the generation of carbocation intermediates. A summary of these reaction types, the reagents, and sample products are shown in Figure 21.2.

Q21.1 - Level 1

Explain why cyclohexene undergoes an addition reaction with chlorine, but benzene undergoes a substitution reaction with chlorine.

A

Cyclohexene has delocalized π electrons, while benzene has localized π electrons

B

Addition of chlorine to benzene would result in retention of aromatic stabilization

C

Substitution of chlorine to benzene would result in loss of aromatic stabilization

D

Addition of chlorine to benzene would result in loss of aromatic stabilization

Q21.2 - Level 1

What type of reaction is represented by the scheme below?

A

Nucleophilic substitution

B

Electrophilic substitution

C

D

### 21.1.1 Halogenation of Benzene

Previously, we saw that Br2 undergoes electrophilic addition on the carbon-carbon double bond of alkenes. In this reaction, we saw that the pi electron cloud of the alkene double bond induces a temporary dipole on Br⎯Br, rendering one of the Br atoms electrophilic (δ+) enough to react with the alkene double bond by electrophilic addition. However, the electrophilic Br atom is not electrophilic enough to react with benzene.

A Lewis acid catalyst (FeBr3) is needed to enhance the electrophilicity of Br atom in its reaction with benzene. The Lewis acid catalyst and Br2 form a complex, which serves as the electrophilic agent (Figure 21.3). Thus, when treated with halogen and a Lewis acid catalyst, benzene undergoes halogenation to form halobenzene.

In the mechanism of the bromination of benzene, the first step (slow step) involves the aromatic ring functioning as a nucleophile that attacks the electrophilic agent generating a resonance-stabilized arenium ion called a sigma complex. This step is the rate-limiting step that is endothermic because it involves a temporary loss of aromaticity. The second step, the fast step, involves the deprotonation of the sigma complex to restore aromaticity and regenerates the Lewis acid catalyst FeBr3 (Figure 21.3). Aluminum tribromide (AlBr3) is an alternative Lewis acid catalyst that can be used in place of FeBr3 in the bromination of benzene.

Chlorination can also be achieved by the use of Cl2 and a suitable Lewis acid catalyst to form chlorobenzene. The mechanism of chlorination is analogous to that of bromination as shown in Figure 21.3, except that Br will be replaced with Cl. How do fluorination and iodination compare to chlorination and bromination? Fluorination is extremely violent and is difficult to control. Iodination is a rather slow process and occurs with relatively poor yields.

As a result, both fluorination and iodination of benzene are less common reactions. However, there are alternative reactions, involving diazonium salt intermediates, for the synthesis of fluorobenzene and iodobenzene that will be encountered in the chapter on amines. The reaction profile for the bromination of benzene is shown in Figure 21.4.

The first step is endergonic and has high activation energy due to loss of aromaticity. This step is the rate-determining step and is slow. In contrast, the second step is fast and is exergonic because aromaticity is restored.

Q21.3 - Level 1

Which of the following represents the correct structure of a sigma complex when benzene reacts with an electrophile (E$^+$)?

A

I

B

II

C

III

D

IV

Q21.4 - Level 1

Which of the following is required with Br$_2$ to generate the active electrophile in the electrophilic bromination of benzene?

A

Lewis base

B

C

UV light

D

Lewis acid

E

None of these

Q21.5 - Level 1

Which of the following reagents can be used to carry out electrophilic chlorination of benzene?

A

Cl$_2$/FeCl$_3$

B

Cl$_2$/AlCl$_3$

C

Cl$_2$/CCl$_4$

D

Cl$_2$/UV light

E

A and B

### Sleepy From Tryptophan?

Tryptophan is commonly known as that ingredient in turkey that makes us sleepy. However, this is just a myth and turkey actually contains no more tryptophan than any other kind of poultry. Moreover, ‘food coma’ after eating a heavy meal is most likely due to the amount of carbohydrates ingested.

Tryptophan does play a role in sleep because it is a precursor to melatonin, a hormone that helps regulate our sleep-wake cycles. How is melatonin made?

Step 1: L-tryptophan undergoes an enzyme-promoted arene functionalization that is reminiscent of an EAS reaction to form 5-hydroxy-trytophan.

Step 2: A decarboxylation reaction forms serotonin, a neurotransmitter thought to contribute to feelings of happiness.

Step 3: Serotonin is then converted to melatonin over several additional steps.

### 21.1.2 Nitration of Benzene

When treated with a mixture of nitric acid and sulfuric acid, benzene undergoes nitration to form nitrobenzene. Nitration is another example of electrophilic aromatic substitution in which the electrophile is the nitronium ion (NO2+). The nitronium ion is generated from the reaction between HNO3 and H2SO4 (Figure 21.5). In the reaction between HNO3 and H2SO4 to generate the electrophilic nitronium ion, H2SO4 functions as the proton donor and HNO3 functions as the proton acceptor.

Why is H2SO4 the proton donor and HNO3 the proton acceptor? Think about the relative acidities of H2SO4 and HNO3 and think about the relationship between resonance structures and the stability of conjugate bases. H2SO4 is a stronger acid than HNO3 and thus will protonate HNO3 when they are mixed (Figure 21.5). Once the nitronium ion is generated, it goes through the two-step electrophilic aromatic substitution to give nitrobenzene as the final product (Figure 21.5).

Q21.6 - Level 1

Which of the following is the electrophile in the nitration of benzene?

A

NO$_2$

B

NO$_2$$^+$

C

HNO$_3$

D

HNO$_2$

E

H$_2$SO$_4$

Q21.7 - Level 1

Predict the product of the following reaction.

A

I

B

II

C

III

D

IV

### 21.1.3 Sulfonation of Benzene

When heated with sulfuric acid, benzene reacts slowly and reversibly to form benzenesulfonic acid. Although this reaction is reversible, a quantitative amount of benzenesulfonic acid can be obtained by removing water as it is formed according to Le Chatelier’s principle (Figure 21.6). In the first step of the mechanism one molecule of sulfuric acid protonates another and then water is lost to form a reactive electrophile. Similar to the halogenation and nitration mechanisms the second step of the mechanism is followed by the rapid loss of a proton to regenerate aromaticity.

When treated with a mixture of H2SO4 and SO3 (fuming sulfuric acid), the rate of sulfonation is much faster and the equilibrium is displaced completely to the product side (Figure 21.7).

Sulfur trioxide is the main electrophile in aromatic sulfonation, with the sulfur atom being the electrophilic site as illustrated in Figure 21.8.

The mechanism of the sulfonation of benzene by sulfur trioxide is shown by the sequence of steps in Figure 21.9. The reaction follows the two steps that are characteristic of all electrophilic aromatic substitutions. In the first step, the aromatic ring functions as a nucleophile, forming a resonance-stabilized intermediate sigma complex. This step is the slow step or rate-determining step. In the second step, the sigma complex is deprotonated to restore aromaticity.

Q21.8 - Level 1

Which of the following is the electrophile in the sulfonation of benzene?

A

SO$_3$$^+$

B

SO$_2$$^+$

C

H$_2$SO$_4$

D

SO$_3$

Q21.9 - Level 1

Predict the product of the following reaction.

A

I

B

II

C

III

D

IV

Q21.10 - Level 2

Predict the product of the following reaction.

A

I

B

II

C

III

D

IV

### 21.1.4 Friedel-Crafts Alkylation

Alkyl halides react with benzene in the presence of the Lewis acid catalyst, aluminum chloride, to yield alkylbenzenes (Figure 21.10). This reaction is called Friedel-Crafts alkylation.

The Friedel-Crafts alkylation was discovered by Charles Friedel and James Crafts in 1877 and represents one of the most useful synthetic methods in organic chemistry. Alkyl halides by themselves are not sufficiently electrophilic to react with benzene. The Lewis acid catalyst helps the alkyl halide ionize, just like FeBr3 helps Br2 ionize in the halogenation of benzene. The Lewis acid enhances the electrophilicity of the alkylating agent. Methyl and primary halides do not form carbocations when treated with the Lewis acid catalyst but rather form complexes that contain highly polarized carbon-halogen bonds. These complexes are the electrophilic species that react with benzene (Figure 21.11).

With secondary and tertiary alkyl halides, the addition of aluminum chloride leads to the formation of carbocations, which are excellent electrophiles that react with benzene by electrophilic aromatic substitution as illustrated in the mechanism of Friedel-Crafts alkylation (Figure 21.12). In the first step, the benzene ring functions as a nucleophile that reacts with the electrophilic carbocation forming a resonance-stabilized intermediate, the sigma complex. This step is the rate-determining step. In the second step, the sigma complex is deprotonated, thus restoring aromaticity.

Although the Friedel-Crafts alkylation reaction is a useful synthetic method, it has many limitations:

• The complexes of primary alkyl halides with AlCl3 readily undergo rearrangement to form more stable secondary or tertiary carbocations. For example, when treated with AlCl3, 1-chloropropane undergoes rearrangement by a hydride shift to form a secondary carbocation. The reaction between benzene and 1-chloropropane in the presence of AlCl3 results in the formation of a mixture of products (Figure 21.13). Therefore, isomerization of carbocations is a limitation in Friedel-Crafts alkylations.
• Aryl halides and vinyl halides do not undergo Friedel-Crafts reactions because their carbocations are extremely unstable. Recall that carbocations on sp2 carbons are very unstable (Figure 21.14).
• It is hard to stop the reaction at monoalkylation because the initially formed alkylbenzene is more reactive than benzene. Therefore, polyalkylation is a limitation in Friedel-Crafts alkylations (Figure 21.15). However, if a large excess of benzene, relative to the reagent, is used, then a monoalkylated product can be formed.
• In the presence of strongly deactivating electron-withdrawing groups such as –NO2, -CN, -SO3H or –COR, Friedel-Crafts reactions don’t proceed (Figure 21.16). These groups render substituted benzenes much less reactive than benzene. The reason for this inactivity will be explored in Section 21.2.
• Friedel-Crafts alkylation cannot be used if the aromatic ring bears an –NH2, -NHR or –NR2 group because of aluminum complexes with the nitrogen and deactivates the ring towards electrophilic aromatic substitution (Section 21.2)

Q21.11 - Level 1

Which of the following compounds does not undergo Friedel-Crafts alkylation?

A

Benzene

B

Nitrobenzene

C

Bromobenzene

D

Toluene

E

None, they all undergo Friedel-Crafts alkylation

Q21.12 - Level 1

Predict the major product of the following Friedel-Crafts alkylation reaction.

A

I

B

II

C

III

D

IV

Q21.13 - Level 2

Predict the product of the following Friedel-Crafts alkylation reaction.

A

I

B

II

C

III

D

IV

Q21.14 - Level 1

Which of the following halides cannot be used for Friedel-Crafts alkylation?

A

2-Bromopropane

B

Chloroethane

C

Allylchloride

D

Vinylbromide

Q21.15 - Level 1

Which of the following halides cannot be used for Friedel-Crafts alkylation?

A

2-Bromopropane

B

Allylchloride

C

Chlorobenzene

D

1-Chloropropane

### 21.1.5 Friedel-Crafts Acylation

The Friedel-Crafts acylation reaction is closely related to the Friedel-Crafts alkylation reaction. When benzene is treated with a carboxylic acid chloride, RCOCl, in the presence of AlCl3, an acyl group, ⎯COR, is substituted onto the ring. For example, benzene reacts with ethanoyl chloride (CH3COCl) in the presence of AlCl3 to generate acetophenone (Figure 21.17).

The electrophile in Friedel-Crafts acylation reaction is an acyl cation called an acylium ion. Treating an acyl chloride with a Lewis acid catalyst, AlCl3, generates the electrophile. For example, when ethanoyl chloride is treated with AlCl3 followed by cleavage of the carbon-chlorine bond, the acylium ion is generated. Unlike carbocations, acylium ions (acyl cations) are resonance stabilized and can therefore not undergo rearrangement (Figure 21.18).

Rearrangement is not possible with the acylium ions because such a rearrangement is an endergonic process that would result in the loss of resonance stabilization. The acylium ion is an excellent electrophile for electrophilic aromatic substitution (Figure 21.19). The first step of the mechanism involves the benzene ring functioning as a nucleophile in its reaction with the acylium ion to form a resonance-stabilized sigma complex intermediate. In the second step, the sigma complex is deprotonated and aromaticity is restored.

Q21.16 - Level 1

Predict the product of the following Friedel-Crafts acylation reaction.

A

I

B

II

C

III

D

IV

### 21.1.6 Reactions Involving the Generation of Carbocation Intermediates

In Friedel-Crafts alkylation, we saw that the electrophilic species that reacted with benzene was a carbocation generated from an alkyl halide as the carbocation precursor. Electrophilic attack on benzene by a carbocation is simply another reaction type that a carbocation undergoes. Recall nucleophilic substitution reactions involving the generation of carbocations from alkyl halides (SN1). We can intuitively infer that other carbocation precursors can be used instead of alkyl halides in electrophilic aromatic substitution. For example, in the presence of an acid catalyst, H2SO4, propene can be protonated to generate a secondary carbocation, which can alkylate benzene to give isopropyl benzene (Figure 21.20).

In the formation of the electrophile, the carbon-carbon double bond of propene is protonated at the terminal carbon, following Markovnikov’s rule, to generate a more stable secondary carbocation. The more stable secondary carbocation is used in the two-step mechanism of electrophilic aromatic substitution. The first step of the mechanism involves the benzene ring functioning as a nucleophile in its reaction with the isopropyl carbocation to form a resonance-stabilized sigma complex intermediate. In the second step, the sigma complex is deprotonated restoring the aromaticity (Figure 21.21).

Another reaction that generates an intermediate carbocation is the reaction of 2-methyl-2-propanol with H2SO4. In the presence of benzene, this carbocation will alkylate benzene to form tert-butylbenzene (Figure 21.22).

In the formation of the electrophile, the –OH group of the alcohol is protonated forming H2O, a good leaving group. The loss of H2O generates a very stable tertiary carbocation. The very stable tertiary carbocation is then used in the two-step mechanism of electrophilic aromatic substitution. The first step of the mechanism involves the benzene ring functioning as a nucleophile in its reaction with the tert-butyl carbocation to form a resonance-stabilized sigma complex intermediate. In the second step, the sigma complex is deprotonated restoring the aromaticity (Figure 21.23).

Q21.17 - Level 2

Which of the following reactions will give isopropylbenzene as the major product?

A

Treatment of benzene with 2-butene and HF

B

Treatment of benzene with ethanol and sulfuric acid

C

Treatment of benzene with 1-propyne and one equivalent of HF

D

Treatment of benzene with 1-propene and HF

Q21.18

What is the major organic product of the following reaction?

A

I

B

II

C

III

D

IV

Q21.19 - Level 1

What is the major organic product of the following reaction?

A

I

B

II

C

III

D

IV

## 21.2 Disubstitution and Polysubstitution

So far the focus has been on electrophilic aromatic substitution of benzene. When we turn our attention to electrophilic aromatic substitution on a benzene ring that already bears at least one substituent, we have to answer two important questions:

• What is the effect of a substituent on the rate of electrophilic aromatic substitution?
• What is the effect of a substituent on the regioselectivity of electrophilic aromatic substitution?

To illustrate the effect of a substituent on the rate of electrophilic aromatic substitution, consider the nitration of benzene and its derivatives on Table 21.1.

As can be seen in Table 21.1, the range of rates of nitration among the compounds is quite large. Compared to benzene, toluene undergoes nitration 25 times faster. Compared to benzene, phenol undergoes nitration 1000 times faster. Because toluene and phenol are more reactive than benzene, both the methyl group and a hydroxyl group activate the ring toward electrophilic aromatic substitution.

On the other hand, chlorobenzene undergoes nitration 33 times slower than benzene and nitrobenzene undergoes nitration 10,000,000 times slower than benzene. Because chlorobenzene and nitrobenzene are less reactive than benzene, both the chloro group and a nitro group deactivate the ring toward electrophilic aromatic substitution.

Just as these substituents have different effects on the rate of electrophilic substitution, they also have different effects on the regioselectivity. A group attached to the benzene ring destroys the six-fold symmetry of the ring. Therefore, incoming electrophiles have three isomeric positions to substitute at relative to the group that is already attached to the ring. These isomeric positions are the ortho, meta and para positions (Figure 21.24).

### 21.2.1 Effect of Activating Groups: Ortho-Para Directors

Activating groups are groups that donate electrons to the benzene ring by inductive effect or resonance, making the substituted benzene more reactive compared to unsubstituted benzene, for example, A and B in Table 21.1. All activating groups are ortho-para directors

In other words the major products result from substitution in the ortho and para positions when an activating group is present. For example, experiments have shown that upon nitration of toluene, three products are possible in different amounts: o-nitrotoluene, m-nitrotoluene, and p-nitrotoluene. The meta isomer makes up only 3% of the product mixture, while the ortho and para isomers make up 97% (Figure 21.25).

How do we explain this experimental product distribution for the nitration of toluene? How do we explain the fact that the methyl substituent is an ortho-para director? We can explain the directing effect of the methyl group by considering the difference in stability of the sigma complex resulting from attack at the ortho, meta, and para positions (Figure 21.26).

As can be seen from Figure 21.26, the sigma complexes resulting from ortho and para attack of the nitro group each have a resonance structure with tertiary carbocation character. In these structures, the positive charge resides on the carbon that bears the methyl group.

As a result of the resonance structures with tertiary carbocation character, the intermediates leading to the ortho and para substitution are more stable and are formed faster than the one leading to the meta substitution. The three resonance forms of the intermediate leading to meta substitution are all secondary carbocations, which are less stable than the tertiary carbocations.

From these resonance structures, we can make the following observations:

• The methyl group is an activating substituent because it stabilizes the carbocation intermediate formed in the rate-determining step more than a hydrogen does.
• The methyl group is ortho-para-directing because it stabilizes the carbocation formed by electrophilic attack at these positions more than it stabilizes the intermediate formed by attack at the meta position.

A better visualization of the relative energy of each sigma complex can be achieved by comparing the activation energies for attack at each position (Figure 21.27).

As can be seen in Figure 21.27, the intermediate sigma complex formed from meta attack has the largest activation energy (Ea). This would explain why the meta product is only formed in relatively trace amounts. The major products result from both ortho and para attack, both of which have smaller activation energies compared to the meta attack.

How do we explain the fact that the para attack has smaller activation energy than the ortho attack yet it is the ortho attack that forms the most product? The major contributing factor of the methyl group is electronic, with the relative stability of carbocations being the key consideration.

To a small extent, the ortho positions are sterically hindered by the methyl group and this makes attack slightly more likely at the para position than at each ortho position. However, attack at the ortho position gives the major product because of statistical advantage; there are two equivalent ortho positions compared to only one para position.

We also saw in Table 21.1 that phenol reacted about 400 times faster than toluene although both contain activating groups. In other words, the hydroxyl group (-OH) is a stronger activating group compared to the methyl group. How do we explain this difference? A hydroxyl group attached to a benzene ring has two effects, inductive effect and resonance

Because oxygen is more electronegative than carbon, the –OH group exerts an electron-withdrawing inductive effect. Oxygen has two lone pairs of electrons, it also exerts electron-donating resonance and increases electron density in the ring. Since phenol resonance is stronger than the inductive effect of oxygen, the overall effect of the –OH group is activating, resulting in resonance having the greatest influence on the electron density of the benzene ring. Let us examine the resonance structures resulting from the placement of the –OH group on the benzene ring (Figure 21.28).

As can be seen in Figure 21.28, resonance structures resulting from the placement of electron-donating groups on the benzene ring result in negative charges at the ortho and para positions (greater electron density). Thus the ortho and para positions are more electron rich and more nucleophilic compared to the meta position. In the presence of electron-donating groups (activating groups), the ortho and para positions are more susceptible to electrophilic attack compared to the meta position.

Experimental evidence has shown the ortho-para directing effect when phenol undergoes nitration. The ortho and para products are the major nitration products with only a trace amount of the meta product. In the case of toluene, we can explain this experimental observation by comparing the stability of the sigma complexes formed for ortho, meta, and para attack. (Figure 21.26).

As seen in Figure 21.29, the sigma complex resulting from ortho and para attack have one additional resonance structure that stabilizes the sigma complex (structure highlighted in Figure 21.29). The sigma complex resulting from meta attack does not have this resonance-stabilized structure and has the highest activation energy (analogous to the representation in Figure 21.27).

In conclusion, we have demonstrated that both methyl and hydroxyl groups activate the benzene ring and are ortho-para directors. Note that when it is stated that a substituent activates the ring, it means that compound will react faster than benzene. Any alkyl group attached to the benzene ring will activate the ring by inductive effect, while any substituent attached to the benzene ring with a lone pair of electrons (except halogens) will activate the ring by resonance.

The rate of reaction for ortho-para directors follows the following trend.