Benzene -
Three double bonds; six $\pi$ electrons

- Cyclooctatetraene has eight $\pi$ electrons, but when it was first prepared in 1911 by the German chemist Richard Willstätter, it was found not to be particularly stable. In fact, its $\pi$ electrons are localized into four double bonds rather than delocalized around the ring, and the molecule is tub-shaped rather than planar. It has no cyclic conjugation because neighboring $p$ orbitals don't have the necessary parallel alignment for overlap, and it resembles an open-chain polyene in its reactivity.

X-ray studies show that the $\mathrm{C}-\mathrm{C}$ single bonds in cyclooctatetraene are 147 pm long and the double bonds are 134 pm long. In addition, the ${ }^{1} \mathrm{H}$ NMR spectrum shows a single sharp resonance line at $5.78 \mathrm{\delta}$, a value characteristic of an alkene rather than an aromatic molecule.

What's so special about $4 n+2 \pi$ electrons? Why do $2,6,10,14 \pi$ electrons lead to aromatic stability, while other numbers of electrons do not? The answer comes from molecular orbital theory. When the energy levels of molecular orbitals for cyclic conjugated molecules are calculated, it turns out that there is always a single lowest-lying MO, above which the MOs come in degenerate pairs. Thus, when electrons fill the various molecular orbitals, it takes two electrons, or one pair, to fill the lowest-lying orbital and four electrons, or two pairs, to fill each of $n$ succeeding energy levels-a total of $4 n+2$. Any other number would leave an energy level partially filled.

The six $\pi$ molecular orbitals of benzene were shown previously in FIGURE 15.5, and their relative energies are shown again in FIGURE 15.6. The lowest-energy MO, $\psi_{1}$, occurs singly and contains two electrons. The next two lowest-energy orbitals, $\psi_{2}$ and $\psi_{3}$, are degenerate, and it therefore takes four electrons to fill both. The result is a stable six- $\pi$-electron aromatic molecule with filled bonding orbitals.


Benzene -
FIGURE 15.6 Energy levels of the six benzene $\boldsymbol{\pi}$ molecular orbitals. There is a single, lowest-energy orbital, above which the orbitals come in degenerate pairs.

PROBLEM To be aromatic, a molecule must have $4 n+2 \pi$ electrons and must have a planar, monocyclic system 15-5 of conjugation. Cyclodecapentaene fulfills one of these criteria but not the other and has resisted all attempts at synthesis. Explain.


Benzene - 15.4 Aromatic lons
According to the Hückel criteria for aromaticity, a molecule must be cyclic, conjugated (nearly planar with a $p$ orbital on each atom), and have $4 n+2 \pi$ electrons. Nothing in this definition says that the number of $\pi$ electrons must be the same as the number of atoms in the ring or that the substance must be neutral. In fact, the numbers can differ and the substance can be an ion. Thus, both the cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic even though both are ions and neither contains a six-membered ring.

Cyclopentadienyl anion

Cycloheptatrienyl cation

Six $\pi$ electrons; aromatic ions
To understand why the cyclopentadienyl anion and the cycloheptatrienyl cation are aromatic, imagine starting from the related neutral hydrocarbons, 1,3-cyclopentadiene and 1,3,5-cycloheptatriene, and removing one hydrogen from the saturated $\mathrm{CH}_{2}$ carbon in each. If that carbon then rehybridizes from $s p^{3}$ to $s p^{2}$, the resultant products would be fully conjugated, with a $p$ orbital on every carbon. There are three ways in which the hydrogen might be removed.

- The hydrogen can be removed with both electrons ( $\mathrm{H}:^{-}$) from the $\mathrm{C}-\mathrm{H}$ bond, leaving a carbocation as product.
- The hydrogen can be removed with one electron (H•) from the $\mathrm{C}-\mathrm{H}$ bond, leaving a carbon radical as product.
- The hydrogen can be removed with no electrons $\left(\mathrm{H}^{+}\right)$from the $\mathrm{C}-\mathrm{H}$ bond, leaving a carbanion as product.


Benzene - 15.4 Aromatic lons
All the potential products formed by removing a hydrogen from 1,3-cyclopentadiene and from 1,3,5-cycloheptatriene can be drawn with numerous resonance structures, but Hückel's rule predicts that only the six- $\pi$-electron cyclopentadienyl anion and cycloheptatrienyl cation can be aromatic. The other products are predicted by the $4 n+2$ rule to be unstable and antiaromatic (FIGURE 15.7).

1,3-Cyclopentadiene

Cyclopentadienyl cation
(four $\pi$ electrons)
or

Cyclopentadienyl radical
(five $\pi$ electrons)
or

Cyclopentadienyl anion (six $\pi$ electrons)

1,3,5-Cycloheptatriene

Cycloheptatrienyl cation
(six $\pi$ electrons)
or

Cycloheptatrienyl radical (seven $\pi$ electrons)
or

Cycloheptatrienyl anion
(eight $\pi$ electrons)

FIGURE 15.7 The aromatic six- $\pi$-electron cyclopentadienyl anion and the six- $\pi$-electron cycloheptatrienyl cation. The anion can be formed by removing a hydrogen ion $\left(\mathrm{H}^{+}\right)$from the $\mathrm{CH}_{2}$ group of 1,3-cyclopentadiene. The cation can be generated by removing a hydride ion $\left(\mathrm{H}:^{-}\right)$from the $\mathrm{CH}_{2}$ group of 1,3,5-cycloheptatriene.


Benzene - 15.4 Aromatic lons
In practice, both the four- $\pi$-electron cyclopentadienyl cation and the five- $\pi$-electron cyclopentadienyl radical are highly reactive and difficult to prepare. Neither shows any sign of the stability expected for an aromatic system. The six- $\pi$-electron cyclopentadienyl anion, by contrast, is easily prepared and remarkably stable (FIGURE 15.8a). In fact, the anion is so stable and easily formed that 1,3 -cyclopentadiene is one of the most acidic hydrocarbons known, with $\mathrm{p} K_{\mathrm{a}}=16$, a value comparable to that of water!

In the same way, the seven- $\pi$-electron cycloheptatrienyl radical and eight- $\pi$-electron anion are reactive and difficult to prepare, while the six- $\pi$-electron cycloheptatrienyl cation is extraordinarily stable (FIGURE 15.8b).

In fact, the cycloheptatrienyl cation was first prepared more than a century ago by reaction of 1,3,5-cycloheptatriene with $\mathrm{Br}_{2}$, although its structure was not recognized at the time.
(a)

$+\mathrm{H}_{2} \mathrm{O}$

1,3-Cyclo-
Cyclopentadienyl anion

Aromatic cyclopentadienyl anion with six $\pi$ electrons
(b)

Cycloheptatrienyl cation six $\pi$ electrons

FIGURE 15.8 (a) The aromatic cyclopentadienyl anion, showing cyclic conjugation and six $\boldsymbol{\pi}$ electrons in five $\boldsymbol{p}$ orbitals, and (b) the aromatic cycloheptatrienyl cation, showing cyclic conjugation and six $\boldsymbol{\pi}$ electrons in seven $\boldsymbol{p}$ orbitals. Electrostatic potential maps indicate that both ions are symmetrical, with the charge equally shared among all atoms in each ring.


Benzene - 15.4 Aromatic lons
PROBLEM Draw the five resonance structures of the cyclopentadienyl anion. Are all carbon-carbon bonds
15-6 equivalent? How many absorption lines would you expect to see in the ${ }^{1} \mathrm{H}$ NMR and ${ }^{13} \mathrm{C}$ NMR spectra of the anion?

PROBLEM Cyclooctatetraene readily reacts with potassium metal to form the stable cyclooctatetraene dianion,
15-7 $\mathrm{C}_{8} \mathrm{H}_{8}{ }^{2-}$. Why do you suppose this reaction occurs so easily? What geometry do you expect for the cyclooctatetraene dianion?

PROBLEM The relative energy levels of the five $\pi$ molecular orbitals of the cyclopentadienyl system are similar
15-8 to those in benzene. That is, there is a single lowest-energy MO, above which the orbitals come in degenerate pairs. Draw a diagram like that in Figure 15.6, and tell which of the five orbitals are occupied in the cation, radical, and anion.


Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole
Look back once again at the definition of aromaticity in Section 15.3: a cyclic, conjugated molecule containing $4 n+2 \pi$ electrons. Nothing in this definition says that the atoms in the ring must be carbon. In fact, heterocyclic compounds can also be aromatic. A heterocycle is a cyclic compound that contains atoms of more than one element in its ring, usually carbon along with nitrogen, oxygen, or sulfur. Pyridine and pyrimidine, for example, are six-membered heterocycles with carbon and nitrogen in their rings (FIGURE 15.9).

Pyridine

Pyrimidine

FIGURE 15.9 Pyridine and pyrimidine are nitrogen-containing aromatic heterocycles with $\pi$ electron arrangements like that of benzene. Both have a lone pair of electrons on nitrogen in an $s p^{2}$ orbital in the plane of the ring.
Pyridine is much like benzene in its $\pi$ electron structure. Each of the five $s p^{2}$-hybridized carbons has a $p$ orbital perpendicular to the plane of the ring, and each $p$ orbital contains one $\pi$ electron. The nitrogen atom is also $s p^{2}$-hybridized and has one electron in a $p$ orbital, bringing the total to six $\pi$ electrons. The nitrogen lone-pair electrons (red in an electrostatic potential map) are in a $s p^{2}$ orbital in the plane of the ring and are not part of the aromatic $\pi$ system. Pyrimidine, also shown in FIGURE 15.9), is a benzene analog that has two nitrogen atoms in a six-membered, unsaturated ring. Both nitrogens are $s p^{2}$-hybridized, and each contributes one electron to the aromatic $\pi$ system.


Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole
Pyrrole (spelled with two r's and one $I$ ) and imidazole are five-membered heterocycles, yet both have six $\pi$ electrons and are aromatic. In pyrrole, each of the four $s p^{2}$-hybridized carbons contributes one $\pi$ electron and the $s p^{2}$-hybridized nitrogen atom contributes the two from its lone pair, which occupies a $p$ orbital (FIGURE 15.10). Imidazole, also shown in (FIGURE 15.10, is an analog of pyrrole that has two nitrogen atoms in a fivemembered, unsaturated ring. Both nitrogens are $s p^{2}$-hybridized, but one is in a double bond and contributes only one electron to the aromatic $\pi$ system whereas the other is not in a double bond and contributes two from its lone pair.

Pyrrole

Imidazole

(Six $\pi$ electrons)

FIGURE 15.10 Pyrrole and imidazole are five-membered, nitrogen-containing heterocycles but have six- $\pi$-electron arrangements like that of the cyclopentadienyl anion. Both have a lone pair of electrons on nitrogen in a $p$ orbital perpendicular to the ring.

Note that nitrogen atoms have different roles depending on the structure of the molecule. The nitrogen atoms in pyridine and pyrimidine are both in double bonds and contribute only one $\pi$ electron to the aromatic sextet, just as a carbon atom in benzene does. The nitrogen atom in pyrrole, however, is not in a double bond and contributes two $\pi$ electrons (its lone pair) to the aromatic sextet. In imidazole, both kinds of nitrogen are present in the same molecule-a double-bonded "pyridine-like" nitrogen that contributes one $\pi$ electron and a "pyrrolelike" nitrogen that contributes two.

Pyrimidine and imidazole rings are particularly important in biological chemistry. Pyrimidine, for instance, is the parent ring system in cytosine, thymine, and uracil, three of the five heterocyclic amine bases found in nucleic acids. An aromatic imidazole ring is present in histidine, one of the 20 amino acids found in proteins.


Benzene - 15.5 Aromatic Heterocycles: Pyridine and Pyrrole
Cytosine (in DNA and RNA)

Thymine (in DNA)

Uracil (in RNA)

Histidine (an amino acid)


Accounting for the Aromaticity of a Heterocycle -
Thiophene, a sulfur-containing heterocycle, undergoes typical aromatic substitution reactions rather than addition reactions. Why is thiophene aromatic?

Thiophene


Strategy -
Recall the requirements for aromaticity-a planar, cyclic, conjugated molecule with $4 n+2 \pi$ electrons-and see how these requirements apply to thiophene.


Solution -
Thiophene is the sulfur analog of pyrrole. The sulfur atom is $s p^{2}$-hybridized and has a lone pair of electrons in a $p$ orbital perpendicular to the plane of the ring. Sulfur also has a second lone pair of electrons in the ring plane.

Thiophene

PROBLEM Draw an orbital picture of furan to show how the molecule is aromatic.
15-9

Furan

PROBLEM Thiamin, or vitamin $B_{1}$, contains a positively charged five-membered nitrogen-sulfur heterocycle
15-10 called a thiazolium ring. Explain why the thiazolium ring is aromatic.

Thiamin

Thiazolium ring


Solution - 15.6 Polycyclic Aromatic Compounds
The Hückel rule is strictly applicable only to monocyclic compounds, but the general concept of aromaticity can be extended to include polycyclic aromatic compounds. Naphthalene, with two benzene-like rings fused together; anthracene, with three rings; benzo[a]pyrene, with five rings; and coronene, with six rings, are all well-known aromatic hydrocarbons. Benzo[a]pyrene is particularly interesting because it is one of the cancercausing substances found in tobacco smoke.

Naphthalene

Anthracene

Benzo[a]pyrene

Coronene

All polycyclic aromatic hydrocarbons can be represented by a number of different resonance forms. Naphthalene, for instance, has three.

Naphthalene
Naphthalene and other polycyclic aromatic hydrocarbons show many of the chemical properties associated with aromaticity. Thus, measurement of its heat of hydrogenation shows an aromatic stabilization energy of approximately $250 \mathrm{~kJ} / \mathrm{mol}$ ( $60 \mathrm{kcal} / \mathrm{mol}$ ). Furthermore, naphthalene reacts slowly with electrophiles such as $\mathrm{Br}_{2}$ to give substitution products rather than double-bond addition products.

The aromaticity of naphthalene is explained by the orbital picture in FIGURE 15.11. Naphthalene has a cyclic, conjugated $\pi$ electron system, with $p$ orbital overlap both along the ten-carbon periphery of the molecule and across the central bond. Because ten $\pi$ electrons is a Hückel number, there is $\pi$ electron delocalization and consequent aromaticity in naphthalene.

Naphthalene

FIGURE 15.11 An orbital picture and electrostatic potential map of naphthalene, showing that the ten $\boldsymbol{\pi}$ electrons are fully delocalized


throughout both rings. -
Just as there are heterocyclic analogs of benzene, there are also many heterocyclic analogs of naphthalene. Among the most common are quinoline, isoquinoline, indole, and purine. Quinoline, isoquinoline, and purine all contain pyridine-like nitrogens that are part of a double bond and contribute one electron to the aromatic $\pi$ system. Indole and purine both contain pyrrole-like nitrogens that contribute two $\pi$ electrons.

Quinoline

Isoquinoline

Indole

Purine

Among the many biological molecules that contain polycyclic aromatic rings, the amino acid tryptophan contains an indole ring and the antimalarial drug quinine contains a quinoline ring. Adenine and guanine, two of the five heterocyclic amine bases found in nucleic acids, have rings based on purine.

Tryptophan (an amino acid)

Adenine (in DNA and RNA)

Guanine (in DNA and RNA)

Quinine (an antimalarial agent)

PROBLEM Azulene, a beautiful blue hydrocarbon, is an isomer of naphthalene. Is azulene aromatic? Draw a
15-11 second resonance form of azulene in addition to that shown.

Azulene

PROBLEM How many electrons does each of the four nitrogen atoms in purine contribute to the aromatic $\pi$
15-12 system?

Purine


Infrared Spectroscopy -
As we saw in the brief introduction to infrared spectroscopy (Section 12.8), aromatic rings show a characteristic $\mathrm{C}-\mathrm{H}$ stretching absorption at $3030 \mathrm{~cm}^{-1}$ and a series of peaks in the 1450 to $1600 \mathrm{~cm}^{-1}$ range of the infrared spectrum. The aromatic $\mathrm{C}-\mathrm{H}$ band at $3030 \mathrm{~cm}^{-1}$ generally has low intensity and occurs just to the left of a typical saturated $\mathrm{C}-\mathrm{H}$ band.

As many as four absorptions are sometimes observed in the 1450 to $1600 \mathrm{~cm}^{-1}$ region because of the complex molecular motions of the ring itself. Two bands, one at $1500 \mathrm{~cm}^{-1}$ and one at $1600 \mathrm{~cm}^{-1}$, are usually the most intense. In addition, aromatic compounds show weak absorptions in the 1660 to $2000 \mathrm{~cm}^{-1}$ region and strong absorptions in the 690 to $900 \mathrm{~cm}^{-1}$ range due to $\mathrm{C}-\mathrm{H}$ out-of-plane bending. The exact position of both sets of absorptions is diagnostic of the substitution pattern of the aromatic ring (FIGURE 12.24 in Section 12.8).


Infrared Spectroscopy -
| Monosubstituted: | $690-710 \mathrm{~cm}^{-1}$ | $\mathbf{1 , 2 , 4 - T r i s u b s t i t u t e d :}$ | $780-830 \mathrm{~cm}^{-1}$ |
| :--- | :--- | :--- | :--- |
|  | $730-770 \mathrm{~cm}^{-1}$ |  | $870-900 \mathrm{~cm}^{-1}$ |
| o-Disubstituted: | $735-770 \mathrm{~cm}^{-1}$ | $\mathbf{1 , 2 , 3 - T r i s u b s t i t u t e d :}$ | $670-720 \mathrm{~cm}^{-1}$ |
| m-Disubstituted: | $690-710 \mathrm{~cm}^{-1}$ |  | $750-790 \mathrm{~cm}^{-1}$ |
|  | $810-850 \mathrm{~cm}^{-1}$ | $\mathbf{1 , 3 , 5 - T r i s u b s t i t u t e d :}$ | $660-700 \mathrm{~cm}^{-1}$ |
| p-Disubstituted: | $810-840 \mathrm{~cm}^{-1}$ |  | $830-900 \mathrm{~cm}^{-1}$ |

The IR spectrum of toluene in FIGURE 15.12 shows these characteristic absorptions.


Ultraviolet Spectroscopy -
Aromatic rings are detectable by ultraviolet spectroscopy because they contain a conjugated $\pi$ electron system. In general, aromatic compounds show a series of bands, with a fairly intense absorption near 205 nm and a less intense absorption in the 255 to 275 nm range. The presence of these bands in the ultraviolet spectrum of a molecule is a sure indication of an aromatic ring. FIGURE 15.13 shows the ultraviolet spectrum of benzene.

FIGURE 15.13 Ultraviolet spectrum of benzene. There are primary bands at 184 and 202 nm and secondary (fine-structure) bands at 255 nm .


Nuclear Magnetic Resonance Spectroscopy -
Aromatic hydrogens are strongly deshielded by the ring and absorb between 6.5 and $8.0 \delta$. The spins of nonequivalent aromatic protons on substituted rings often couple with each other, giving rise to spin-spin splitting patterns that can identify the substitution of the ring.

Much of the difference in chemical shift between aromatic protons (6.5-8.0 $\delta$ ) and vinylic protons (4.5-6.5 $\delta$ ) is due to a property of aromatic rings called ring-current. When an aromatic ring is oriented perpendicular to a strong magnetic field, the electrons circulate around the ring, producing a small local magnetic field. This induced field opposes the applied field in the middle of the ring but reinforces the applied field outside the ring (FIGURE 15.14). Aromatic protons therefore experience an effective magnetic field greater than the applied field and come into resonance at a lower applied field.

FIGURE 15.14 The origin of aromatic ring-current. Aromatic protons are deshielded by the induced magnetic field caused by delocalized $\pi$ electrons circulating around the aromatic ring.
Note that the aromatic ring-current produces different effects inside and outside the ring. If a ring were large enough to have both "inside" and "outside" protons, the protons on the outside would be deshielded and absorb
at a field lower than normal but those on the inside would be shielded and absorb at a field higher than normal. This prediction has been strikingly confirmed by studies on [18]annulene, an $18-\pi$-electron cyclic conjugated polyene that contains a Hückel number of electrons $(4 n+2=18$ when $n=4)$. The six inside protons of [18]annulene are strongly shielded by the aromatic ring-current and absorb at $-3.0 \delta$ (that is, 3.0 ppm upfield from TMS, off the normal chart), while the 12 outside protons are strongly deshielded and absorb in the typical aromatic region at 9.3 ppm downfield from TMS.

[18]Annulene

$$
\text { Inside } \mathrm{H}:-3.0 \delta
$$


Nuclear Magnetic Resonance Spectroscopy -
[18]Annulene

$$
\text { Inside } \mathrm{H}:-3.0 \delta
$$

Outside H: $9.3 \delta$
The presence of a ring-current is characteristic of all Hückel aromatic molecules and is a good test of aromaticity. For example, benzene, a six- $\pi$-electron aromatic molecule, absorbs at $7.37 \delta$ because of its ringcurrent, but cyclooctatetraene, an eight- $\pi$-electron nonaromatic molecule, absorbs at $5.78 \delta$.

Hydrogens on carbon next to aromatic rings-benzylic hydrogens-also show distinctive absorptions in the NMR spectrum. Benzylic protons normally absorb downfield from other alkane protons in the region from 2.3 to 3.0 $\delta$.

The ${ }^{1} \mathrm{H}$ NMR spectrum of $p$-bromotoluene, shown in FIGURE 15.15, displays many of the features just discussed. The aromatic protons appear as two doublets at 7.04 and $7.37 \mathrm{\delta}$, and the benzylic methyl protons absorb as a sharp singlet at $2.26 \delta$. Integration of the spectrum shows the expected $2: 2: 3$ ratio of peak areas.

FIGURE 15.15 The ${ }^{\mathbf{1}} \mathrm{H}$ NMR spectrum of $\boldsymbol{p}$-bromotoluene.
The carbon atoms in an aromatic ring typically absorb in the range 110 to $140 \delta$ in the ${ }^{13} \mathrm{C}$ NMR spectrum, as
indicated by the examples in FIGURE 15.16. These resonances are easily distinguished from those of alkane carbons but occur in the same range as alkene carbons. Thus, the presence of ${ }^{13} \mathrm{C}$ absorptions at 110 to $140 \delta$ does not by itself establish the presence of an aromatic ring. Supporting evidence from infrared, ultraviolet, or ${ }^{1}$ H NMR spectra is needed.


Nuclear Magnetic Resonance Spectroscopy -
Benzene
Toluene
Chlorobenzene
Naphthalene
FIGURE 15.16 Some ${ }^{13}$ C NMR absorptions of aromatic compounds ( $\delta$ units).


Aspirin, NSAIDs, and COX-2 Inhibitors -
Whatever the cause-whether tennis elbow, a sprained ankle, or a wrenched knee-pain and inflammation seem to go together. They are, however, different in their origin, and powerful drugs are available for treating each separately. Codeine, for example, is a powerful analgesic, or pain reliever, used in the management of debilitating pain, while cortisone and related steroids are potent anti-inflammatory agents, used for treating arthritis and other crippling inflammations. For minor pains and inflammation, both problems are often treated together by using a common over-the-counter medication called an NSAID, or nonsteroidal antiinflammatory drug.

One of the most common NSAIDs is aspirin, or acetylsalicylic acid, whose use goes back to the late 1800s. It had been known from before the time of Hippocrates in 400 BC that fevers could be lowered by chewing the bark of willow trees. The active agent in willow bark was found in 1827 to be an aromatic compound called salicin, which could be converted by reaction with water into salicyl alcohol and then oxidized to give salicylic acid. Salicylic acid turned out to be even more effective than salicin for reducing fevers and to have analgesic and anti-inflammatory action as well. Unfortunately, it also turned out to be too corrosive to the walls of the stomach for everyday use. Conversion of the phenol-OH group into an acetate ester, however, yielded acetylsalicylic acid, which proved just as potent as salicylic acid but less corrosive to the stomach.

FIGURE 15.17 Many athletes rely on NSAIDs to help with pain and soreness. (credit: "Dillon Gee delivers a pitch" by Arturo Pardavila III/ Wikimedia Commons, CC BY 2.0)


Aspirin, NSAIDs, and COX-2 Inhibitors -
Acetylsalicylic acid
(aspirin)
Although extraordinary in its effect, aspirin is also more dangerous than commonly believed. 15 g of aspirin would be fatal to a child of 9-10 years, and 10 g would be lethal for a 6-year-old, and aspirin can cause stomach bleeding and allergic reactions in long-term users. Even more serious is a condition called Reye's syndrome, a potentially fatal reaction to aspirin sometimes seen in children recovering from the flu. As a result of these problems, numerous other NSAIDs have been developed in the last several decades, most notably ibuprofen and naproxen.

Like aspirin, both ibuprofen and naproxen are relatively simple aromatic compounds containing a side-chain carboxylic acid group. Ibuprofen, sold under the brand names Advil, Nuprin, Motrin, and others, has roughly the same potency as aspirin but is less prone to causing an upset stomach. Naproxen, sold under the names Aleve and Naprosyn, also has about the same potency as aspirin but remains active in the body six times longer.

Ibuprofen
(Advil, Nuprin, Motrin)

Naproxen
(Aleve, Naprosyn)

Aspirin and other NSAIDs function by blocking the cyclooxygenase (COX) enzymes that carry out the body's synthesis of prostaglandins (Section 8.11 and Section 27.4). There are two forms of the enzyme: COX-1, which carries out the normal physiological production of prostaglandins, and COX-2, which mediates the body's response to arthritis and other inflammatory conditions. Unfortunately, both COX-1 and COX-2 enzymes are blocked by aspirin, ibuprofen, and other NSAIDs, thereby shutting down not only the response to inflammation but also various protective functions, including the control mechanism for production of acid in the stomach.


Aspirin, NSAIDs, and COX-2 Inhibitors -
Medicinal chemists have therefore devised a number of drugs that act as selective inhibitors of only the COX-2 enzyme. Inflammation is thereby controlled without blocking protective functions. Originally heralded as a breakthrough in arthritis treatment, the first generation of COX-2 inhibitors, including rofecoxib, (marketed as Vioxx), celecoxib, (marketed as Celebrex), and Valdecoxib (marketed as Bextra), turned out to cause potentially serious heart problems, particularly in elderly or compromised patients. As a result, all three Cox-2 inhibitors were removed from the U.S. and other markets. Further studies, however, concluded that in proper doses the Cox-2 inhibitors were little different from other NSAIDs, and Vioxx returned to the market in 2017 for topical treatment of osteoarthritis.

Rofecoxib
(Vioxx)


Key Terms -
- antiaromatic
- Hückel $4 n+2$ rule
- arene
- meta (m)
- aromatic
- ortho (o)
- benzyl group
- para (p)
- degenerate orbitals
- phenyl group
- heterocycle
- ring-current


Summary -
Aromatic rings are a common part of many biological structures and are particularly important in nucleic acid chemistry and in the chemistry of several amino acids. In this chapter, we've seen how and why aromatic compounds are different from such apparently related compounds as cycloalkenes.

The word aromatic is used for historical reasons to refer to the class of compounds related structurally to benzene. Aromatic compounds are systematically named according to IUPAC rules, but many common names
are also used. Disubstituted benzenes are referred to as ortho ( 1,2 disubstituted), meta ( 1,3 disubstituted), or para (1,4 disubstituted) derivatives. The $\mathrm{C}_{6} \mathrm{H}_{5}$ - unit itself is referred to as a phenyl group, and the $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}-$ unit is a benzyl group.

Benzene is described by valence-bond theory as a resonance hybrid of two equivalent structures and is described by molecular orbital theory as a planar, cyclic, conjugated molecule with six $\pi$ electrons. According to the Hückel rule, a molecule must have $4 \boldsymbol{n}+2 \boldsymbol{\pi}$ electrons, where $n=0,1,2,3$, and so on, to be aromatic. Planar, cyclic, conjugated molecules with other numbers of $\pi$ electrons are antiaromatic.

Other substances besides benzene-like compounds are also aromatic. The cyclopentadienyl anion and the cycloheptatrienyl cation, for instance, are aromatic ions. Pyridine and pyrimidine are six-membered, nitrogencontaining, aromatic heterocycles. Pyrrole and imidazole are five-membered, nitrogen-containing heterocycles. Naphthalene, quinoline, indole, and many others are polycyclic aromatic compounds.

Aromatic compounds have the following characteristics:


Summary -
Aromatic compounds have the following characteristics:

- Aromatic compounds are cyclic, planar, and conjugated.
- Aromatic compounds are unusually stable. Benzene, for instance, has a heat of hydrogenation $150 \mathrm{~kJ} / \mathrm{mol}$ less than we might expect for a cyclic triene.
- Aromatic compounds react with electrophiles to give substitution products, in which cyclic conjugation is retained, rather than addition products, in which conjugation is destroyed.
- Aromatic compounds have $4 n+2 \pi$ electrons, which are delocalized over the ring.


Visualizing Chemistry -
PROBLEM Give IUPAC names for the following substances (red $=0$, blue $=\mathrm{N}$ ):


15-13 (a) -
(b)

PROBLEM All-cis cyclodecapentaene is a stable molecule that shows a single absorption in its ${ }^{1} \mathrm{H}$ NMR 15-14 spectrum at $5.67 \delta$. Tell whether it is aromatic, and explain its NMR spectrum.

PROBLEM 1,6-Methanonaphthalene has an interesting ${ }^{1} \mathrm{H}$ NMR spectrum in which the eight hydrogens
15-15 around the perimeter absorb at 6.9 to $7.3 \delta$, while the two $\mathrm{CH}_{2}$ protons absorb at $-0.5 \delta$. Tell whether it is aromatic, and explain its NMR spectrum.


1,6-Methanonaphthalene -
PROBLEM The following molecular model is that of a carbocation. Draw two resonance structures for the
15-16 carbocation, indicating the positions of the double bonds.

PROBLEM Azulene, an isomer of naphthalene, has a remarkably large dipole moment for a hydrocarbon ( $\mu=$
15-17 1.0 D). Explain, using resonance structures.


Naming Aromatic Compounds -
PROBLEM Give IUPAC names for the following compounds:


15-18 (a) -
(b)

(c)

(d)

(e)

(f)

PROBLEM Draw structures corresponding to the following names:
15-19 (a) 3-Methyl-1,2-benzenediamine (b) 1,3,5-Benzenetriol (c) 3-Methyl-2-phenylhexane
(d) $o$-Aminobenzoic acid (e) $m$-Bromophenol (f) 2,4,6-Trinitrophenol (picric acid)

PROBLEM Draw and name all possible isomers of the following:
15-20 (a) Dinitrobenzene (b) Bromodimethylbenzene
(c) Trinitrophenol

PROBLEM Draw and name all possible aromatic compounds with the formula $\mathrm{C}_{7} \mathrm{H}_{7} \mathrm{Cl}$.
15-21
PROBLEM Draw and name all possible aromatic compounds with the formula $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$. (There are 14.) 15-22


Structure of Aromatic Compounds -
PROBLEM Propose structures for aromatic hydrocarbons that meet the following descriptions:
15-23 (a) $\mathrm{C}_{9} \mathrm{H}_{12}$; gives only one $\mathrm{C}_{9} \mathrm{H}_{11} \mathrm{Br}$ product on substitution of a hydrogen on the aromatic ring with bromine
(b) $\mathrm{C}_{10} \mathrm{H}_{14}$; gives only one $\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}$ product on substitution of a hydrogen on the aromatic ring with chlorine
(c) $\mathrm{C}_{8} \mathrm{H}_{10}$; gives three $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$ products on substitution of a hydrogen on the aromatic ring with bromine
(d) $\mathrm{C}_{10} \mathrm{H}_{14}$; gives two $\mathrm{C}_{10} \mathrm{H}_{13} \mathrm{Cl}$ products on substitution of a hydrogen on the aromatic ring with chlorine

PROBLEM Look at the three resonance structures of naphthalene shown in Section 15.6, and account for the
15-24 fact that not all carbon-carbon bonds have the same length. The C1-C2 bond is 136 pm long, whereas the $\mathrm{C} 2-\mathrm{C} 3$ bond is 139 pm long.

PROBLEM Anthracene has four resonance structures, one of which is shown. Draw the other three.
15-25

Anthracene

PROBLEM Phenanthrene has five resonance structures, one of which is shown. Draw the other four.
15-26

Phenanthrene

PROBLEM Look at the five resonance structures for phenanthrene (Problem 26), and predict which of its
15-27 carbon-carbon bonds is shortest.
PROBLEM In 1932, A. A. Levine and A. G. Cole studied the ozonolysis of $o$-xylene and isolated three products:
15-28 glyoxal, 2,3-butanedione, and pyruvaldehyde:


Structure of Aromatic Compounds -
In what ratio would you expect the three products to be formed if $o$-xylene is a resonance hybrid of two structures? The actual ratio found was 3 parts glyoxal, 1 part 2,3-butanedione, and 2 parts pyruvaldehyde. What conclusions can you draw about the structure of $o$-xylene?


Aromaticity and Hückel's Rule -
PROBLEM 3-Chlorocyclopropene, on treatment with $\mathrm{AgBF}_{4}$, gives a precipitate of AgCl and a stable solution
15-29 of a product that shows a single ${ }^{1} \mathrm{H}$ NMR absorption at $11.04 \delta$. What is a likely structure for the product, and what is its relation to Hückel's rule?


3-Chlorocyclopropene -
PROBLEM Draw an energy diagram for the three molecular orbitals of the cyclopropenyl system $\left(\mathrm{C}_{3} \mathrm{H}_{3}\right)$. How
15-30 are these three molecular orbitals occupied in the cyclopropenyl anion, cation, and radical? Which of the three substances is aromatic according to Hückel's rule?

PROBLEM Cyclopropanone is highly reactive because of its large amount of angle strain.
15-31 Methylcyclopropenone, although even more strained than cyclopropanone, is nevertheless quite stable and can even be distilled. Explain, taking the polarity of the carbonyl group into account.

Cyclopropanone


Methylcyclopropenone -
PROBLEM Cycloheptatrienone is stable, but cyclopentadienone is so reactive that it can't be isolated. Explain,
15-32 taking the polarity of the carbonyl group into account.

Cycloheptatrienone

Cyclopentadienone

PROBLEM Which would you expect to be most stable, cyclononatetraenyl radical, cation, or anion? 15-33 (Cyclononatetraene has a ring of nine carbons and four double bonds.)

PROBLEM How might you convert 1,3,5,7-cyclononatetraene to an aromatic substance?
15-34
PROBLEM Calicene, like azulene (Problem 17), has an unusually large dipole moment for a hydrocarbon.
15-35 Explain, using resonance structures.


Calicene -
PROBLEM Pentalene is a most elusive molecule that has been isolated only at liquid-nitrogen temperature.
15-36 The pentalene dianion, however, is well known and quite stable. Explain.

Pentalene

Pentalene dianion

PROBLEM Indole is an aromatic heterocycle that has a benzene ring fused to a pyrrole ring. Draw an orbital 15-37 picture of indole.
(a) How many $\pi$ electrons does indole have?
(b) What is the electronic relationship of indole to naphthalene?

Indole

PROBLEM Ribavirin, an antiviral agent used against hepatitis C and viral pneumonia, contains a 1,2,4-triazole 15-38 ring. Why is the ring aromatic?


Spectroscopy -
PROBLEM Compound $\mathbf{A}, \mathrm{C}_{8} \mathrm{H}_{10}$, yields three substitution products, $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$, on reaction with $\mathrm{Br}_{2}$. Propose two
15-39 possible structures for $\mathbf{A}$. The ${ }^{1} \mathrm{H}$ NMR spectrum of $\mathbf{A}$ shows a complex four-proton multiplet at 7.0 $\delta$ and a six-proton singlet at $2.30 \delta$. What is the structure of $\mathbf{A}$ ?

PROBLEM What is the structure of a hydrocarbon that has $\mathrm{M}^{+}=120$ in its mass spectrum and has the following 15-40 ${ }^{1}$ H NMR spectrum?
$7.25 \delta(5 \mathrm{H}$, broad singlet); $2.90 \delta(1 \mathrm{H}$, septet, $J=7 \mathrm{~Hz}) ; 1.22 \delta(6 \mathrm{H}$, doublet, $J=7 \mathrm{~Hz})$


Spectroscopy -
PROBLEM Propose structures for compounds that fit the following descriptions:
15-41 (a) $\mathrm{C}_{10} \mathrm{H}_{14}$
${ }^{1} \mathrm{H}$ NMR: $7.18 \delta(4 \mathrm{H}$, broad singlet); $2.70 \delta(4 \mathrm{H}$, quartet, $J=7 \mathrm{~Hz}) ; 1.20 \delta(6 \mathrm{H}$, triplet, $J=7 \mathrm{~Hz})$ IR absorption at $745 \mathrm{~cm}^{-1}$
(b) $\mathrm{C}_{10} \mathrm{H}_{14}$
${ }^{1}$ H NMR: $7.0 \delta(4 \mathrm{H}$, broad singlet); $2.85 \delta(1 \mathrm{H}$, septet, $J=8 \mathrm{~Hz}) ; 2.28 \delta(3 \mathrm{H}$, singlet $) ; 1.20 \delta(6$ H , doublet, $J=8 \mathrm{~Hz}$ )

IR absorption at $825 \mathrm{~cm}^{-1}$


General Problems -
PROBLEM On reaction with acid, 4-pyrone is protonated on the carbonyl-group oxygen to give a stable cationic 15-42 product. Using resonance structures and the Hückel $4 n+2$ rule, explain why the protonated product is so stable.

4-Pyrone
PROBLEM Bextra, a COX-2 inhibitor once used in the treatment of arthritis, contains an isoxazole ring. Why is $\mathbf{1 5 - 4 3}$ the ring aromatic?


Bextra -
PROBLEM $N$-Phenylsydnone, so-named because it was first studied at the University of Sydney, Australia, 15-44 behaves like a typical aromatic molecule. Explain, using the Hückel $4 n+2$ rule.

$N$-Phenylsydnone
PROBLEM Show the relative energy levels of the seven $\pi$ molecular orbitals of the cycloheptatrienyl system.
15-45 Tell which of the seven orbitals are filled in the cation, radical, and anion, and account for the aromaticity of the cycloheptatrienyl cation.

PROBLEM 1-Phenyl-2-butene has an ultraviolet absorption at $\lambda_{\max }=208 \mathrm{~nm}(\varepsilon=8000)$. On treatment with a
15-46 small amount of strong acid, isomerization occurs and a new substance with $\lambda_{\max }=250 \mathrm{~nm}(\varepsilon=$ 15,800 ) is formed. Propose a structure for this isomer, and suggest a mechanism for its formation.

PROBLEM Propose structures for aromatic compounds that have the following ${ }^{1} \mathrm{H}$ NMR spectra:
15-47 (a) $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Br}$
IR absorption at $820 \mathrm{~cm}^{-1}$

(b) $\mathrm{C}_{9} \mathrm{H}_{12}$

IR absorption at $750 \mathrm{~cm}^{-1}$

(c) $\mathrm{C}_{11} \mathrm{H}_{16}$

IR absorption at $820 \mathrm{~cm}^{-1}$

PROBLEM Propose a structure for a molecule $\mathrm{C}_{14} \mathrm{H}_{12}$ that has the following ${ }^{1} \mathrm{H}$ NMR spectrum and has IR 15-48 absorptions at 700, 740 , and $890 \mathrm{~cm}^{-1}$ :


Bextra -
PROBLEM The proton NMR spectrum for a compound with formula $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{O}_{2}$ is shown. The infrared spectrum
15-49 has a strong band at $1711 \mathrm{~cm}^{-1}$. The normal carbon-13 NMR spectral results are tabulated along with the DEPT-135 and DEPT-90 information. Draw the structure of this compound.

PROBLEM The proton NMR spectrum of a compound with formula $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NCl}_{2}$ is shown. The normal carbon-13 15-50 and DEPT experimental results are tabulated. The infrared spectrum shows peaks at 3432 and 3313 $\mathrm{cm}^{-1}$ and a series of medium-sized peaks between 1618 and $1466 \mathrm{~cm}^{-1}$. Draw the structure of this compound.

| Normal Carbon | DEPT-135 | DEPT-90 |
| :--- | :--- | :--- |
| 118.0 ppm | Positive | Positive |
| 119.5 | No peak | No peak |
| 128.0 | Positive | Positive |
| 140.0 | No peak | No peak |

PROBLEM Aromatic substitution reactions occur by addition of an electrophile such as $\mathrm{Br}^{+}$to an aromatic
15-51 ring to yield an allylic carbocation intermediate, followed by loss of $\mathrm{H}^{+}$. Show the structure of the intermediate formed by reaction of benzene with $\mathrm{Br}^{+}$.

PROBLEM The substitution reaction of toluene with $\mathrm{Br}_{2}$ can, in principle, lead to the formation of three
15-52 isomeric bromotoluene products. In practice, however, only $o$ - and $p$-bromotoluene are formed in substantial amounts. The meta isomer is not formed. Draw the structures of the three possible carbocation intermediates (Problem 51), and explain why ortho and para products predominate over meta products.

PROBLEM Look at the following aromatic anions and their linear counterparts, and draw all of the resonance
15-53 forms for each. What patterns emerge?
(a)

(b)


Bextra -
PROBLEM Look at the following aromatic anions and their linear counterparts, and draw all of the resonance
15-53 forms for each. What patterns emerge?
(a)

(b)

PROBLEM After the following reaction, the chemical shift of $\mathrm{H}_{\mathrm{a}}$ moves downfield from 6.98 ppm to 7.30 ppm .
15-54 Explain.

PROBLEM The following compound is the product initially formed in a Claisen rearrangement, which we'll see
15-55 in Section 18.4. This product is not isolated, but tautomerizes to its enol form. Give the structure of the enol and provide an explanation as to why the enol tautomer is favored.

PROBLEM Compounds called azo dyes are the major source of artificial color in textiles and food. Part of the
15-56 reason for their intense coloring is the conjugation from an electron-donating group through the diazo bridge ( $-\mathrm{N}=\mathrm{N}-$ ) to an electron-withdrawing group on the other side. For the following azo dyes, draw a resonance form that shows how the electron-donating group is related to the electronwithdrawing group on the other side of the diazo bridge. Used curved arrows to show how the electrons are reorganized.
(a)

Methyl Orange
(b)
C.I. Acid Red 74


CHAPTER 16 -
Chemistry of Benzene: Electrophilic Aromatic Substitution

FIGURE 16.1 In the 19th and early-20th centuries, benzene was used as an aftershave lotion because of its pleasant smell and as a solvent to decaffeinate coffee beans. Neither is a good idea. (credit: modification of work "Ye Old Way (https://loc.getarchive.net/media/ ye-old-way-geo-h-walker-and-co-lith-boston-mass-1)" by Geo. H. Walker \& Co./Library of Congress)


CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination
16.2 Other Aromatic Substitutions
16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction
16.4 Substituent Effects in Electrophilic Substitutions
16.5 Trisubstituted Benzenes: Additivity of Effects
16.6 Nucleophilic Aromatic Substitution
16.7 Benzyne
16.8 Oxidation of Aromatic Compounds
16.9 Reduction of Aromatic Compounds
16.10 Synthesis of Polysubstituted Benzenes

WHY THIS CHAPTER? This chapter continues the coverage of aromatic molecules begun in the preceding chapter, but we'll shift focus to concentrate on reactions, looking at the relationship between aromatic structure and reactivity. This relationship is critical in understanding how many biological molecules and pharmaceutical agents are synthesized and why they behave as they do.

In the preceding chapter, we looked at aromaticity-the stability associated with benzene and related compounds that contain a cyclic conjugated system of $4 n+2 \pi$ electrons. In this chapter, we'll look at some of the unique reactions that aromatic molecules undergo.

The most common reaction of aromatic compounds is electrophilic aromatic substitution, in which an electrophile ( $E^{+}$) reacts with an aromatic ring and substitutes for one of the hydrogens. The reaction is characteristic of all aromatic rings, not just benzene and substituted benzenes. In fact, the ability of a compound to undergo electrophilic substitution is a good test of aromaticity.

Many different substituents can be introduced onto an aromatic ring through electrophilic substitution. To list some possibilities, an aromatic ring can be substituted by a halogen $(-\mathrm{Cl},-\mathrm{Br},-\mathrm{I})$, a nitro group $\left(-\mathrm{NO}_{2}\right)$, a sulfonic acid group $\left(-\mathrm{SO}_{3} \mathrm{H}\right)$, a hydroxyl group $(-\mathrm{OH})$, an alkyl group ( -R ), or an acyl group ( -COR ). Starting from only a few simple materials, it's possible to prepare many thousands of substituted aromatic compounds.

Halogenation

Nitration

Acylation

Alkylation

Sulfonation

Hydroxylation


CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination
Before seeing how electrophilic aromatic substitutions occur, let's briefly recall what we said in the chapter on Alkenes: Structure and Reactivity about electrophilic alkene additions. When a reagent such as HCl adds to an alkene, the electrophilic hydrogen ion approaches the $\pi$ electrons of the double bond and forms a bond to one carbon, leaving a positive charge at the other carbon. This carbocation intermediate then reacts with the nucleophilic $\mathrm{Cl}^{-}$ion to yield the addition product.

An electrophilic aromatic substitution reaction begins in a similar way, but there are a number of differences. One difference is that aromatic rings are less reactive toward electrophiles than alkenes. For example, $\mathrm{Br}_{2}$ in $\mathrm{CH}_{2} \mathrm{Cl}_{2}$ solution reacts instantly with most alkenes but does not react with benzene at room temperature. For bromination of benzene to take place, a catalyst such as $\mathrm{FeBr}_{3}$ is needed. The catalyst makes the $\mathrm{Br}_{2}$ molecule more electrophilic by polarizing it to give a $\mathrm{FeBr}_{4}{ }^{-} \mathrm{Br}^{+}$species that reacts as if it were $\mathrm{Br}^{+}$. The polarized $\mathrm{Br}_{2}$ molecule then reacts with the nucleophilic benzene ring to yield a nonaromatic carbocation intermediate that is doubly allylic (Section 11.5) and has three resonance forms.

Although more stable than a typical alkyl carbocation because of resonance, the intermediate in electrophilic aromatic substitution is nevertheless much less stable than the starting benzene ring itself, with its $150 \mathrm{~kJ} / \mathrm{mol}$ ( $36 \mathrm{kcal} / \mathrm{mol}$ ) of aromatic stability. Thus, the reaction of an electrophile with a benzene ring is endergonic, has a substantial activation energy, and is rather slow. FIGURE 16.2 shows an energy diagram comparing the reaction of an electrophile with an alkene and with benzene. The benzene reaction is slower (higher $\Delta G^{+}$) because the starting material is more stable.


CHAPTER CONTENTS - 16.1 Electrophilic Aromatic Substitution Reactions: Bromination
Reaction progress
FIGURE 16.2 A comparison of the reactions of an electrophile ( $E^{+}$) with an alkene and with benzene: $\Delta G^{\ddagger}$ alkene $<\Delta G^{\ddagger}$ benzene. The benzene reaction is slower than the alkene reaction because of the stability of the aromatic ring.
Another difference between alkene addition and aromatic substitution occurs after the carbocation intermediate has formed. Instead of adding $\mathrm{Br}^{-}$to give an addition product, the carbocation intermediate loses $\mathrm{H}^{+}$from the bromine-bearing carbon to give a substitution product. Note that this loss of $\mathrm{H}^{+}$is similar to what occurs in the second step of an E1 reaction (Section 11.10). The net effect of reaction of $\mathrm{Br}_{2}$ with benzene is the substitution of $\mathrm{H}^{+}$by $\mathrm{Br}^{+}$by the overall mechanism shown in FIGURE 16.3.


FIGURE 16.3 MECHANISM -
The mechanism for the electrophilic bromination of benzene. The reaction occurs in two steps and involves a resonance-stabilized carbocation intermediate.

1) An electron pair from the benzene ring attacks the positively polarized bromine, forming a new $\mathrm{C}-\mathrm{Br}$ bond and leaving a nonaromatic carbocation intermediate.
(2) A base removes $\mathrm{H}^{+}$from the carbocation intermediate, and the neutral substitution product forms as two electrons from the $\mathrm{C}-\mathrm{H}$ bond move to re-form the aromatic ring.

(1) Slow

(2) Fast

Why does the reaction of $\mathrm{Br}_{2}$ with benzene take a different course than its reaction with an alkene? The answer is straightforward. If addition occurred, the $150 \mathrm{~kJ} / \mathrm{mol}$ stabilization energy of the aromatic ring would be lost and the overall reaction would be endergonic. When substitution occurs, though, the stability of the aromatic ring is retained and the reaction is exergonic. An energy diagram for the overall process is shown in FIGURE 16.4 .

FIGURE 16.4 An energy diagram for the electrophilic bromination of benzene. Because the stability of the aromatic ring is retained, the overall process is exergonic.

PROBLEM Monobromination of toluene gives a mixture of three bromotoluene products. Draw and name 16-1 them.


FIGURE 16.3 MECHANISM - 16.2 Other Aromatic Substitutions
There are many other kinds of electrophilic aromatic substitutions besides bromination, and all occur by the same general mechanism. Let's look at some of these other reactions briefly.


Aromatic Halogenation -
Chlorine and iodine can be introduced into aromatic rings by electrophilic substitution reactions just as bromine can, but fluorine is too reactive and only poor yields of monofluoroaromatic products are obtained by direct fluorination. Instead, other sources of " $\mathrm{F}^{+}$" are used, in which a fluorine atom is bonded to a positively charged nitrogen. One of the most common such reagents goes by the acronym F-TEDA-BF 4 in the presence of trifluoromethanesulfonic acid (TfOH). (You don't need to know the full name of F-TEDA, which is sold under the name Selectfluor.)

More than $20 \%$ of all pharmaceutical agents sold contain fluorine, including $30 \%$ of the top 100 drugs sold. Sitagliptin (Januvia), used to treat type 2 diabetes, fluoxetine (Prozac), an antidepressant, and atorvastatin (Lipitor), a statin used to lower cholesterol, are examples.

Sitagliptin
(Januvia)

Fluoxetine
(Prozac)

Aromatic rings react with $\mathrm{Cl}_{2}$ in the presence of $\mathrm{FeCl}_{3}$ catalyst to yield chlorobenzenes, just as they react with $\mathrm{Br}_{2}$ and $\mathrm{FeBr}_{3}$. This kind of reaction is used in the synthesis of numerous pharmaceutical agents, including the antiallergy medication loratadine, marketed as Claritin.

Iodine itself is unreactive toward aromatic rings, so an oxidizing agent such as hydrogen peroxide or a copper salt such as $\mathrm{CuCl}_{2}$ must be added to the reaction. These substances accelerate the iodination reaction by oxidizing $\mathrm{I}_{2}$ to a more powerful electrophilic species that reacts as if it were $\mathrm{I}^{+}$. The aromatic ring then reacts with $\mathrm{I}^{+}$in the typical way, yielding a substitution product.

Benzene

Iodobenzene (65\%)


Aromatic Halogenation -
Benzene

Iodobenzene (65\%)

Electrophilic aromatic halogenations also occur in the biosynthesis of many naturally occurring molecules, particularly those produced by marine organisms. In humans, the best-known example occurs in the thyroid gland during the biosynthesis of thyroxine, a hormone involved in regulating growth and metabolism. The amino acid tyrosine is first iodinated by thyroid peroxidase, and two of the iodinated tyrosine molecules then couple. The electrophilic iodinating agent is an $\mathrm{I}^{+}$species, perhaps hypoiodous acid (HIO), that is formed from iodide ion by oxidation with $\mathrm{H}_{2} \mathrm{O}_{2}$.

Thyroxine
(a thyroid hormone)


Aromatic Nitration -
Aromatic rings are nitrated by reaction with a mixture of concentrated nitric and sulfuric acids. The electrophile is the nitronium ion, $\mathrm{NO}_{2}{ }^{+}$, which is formed from $\mathrm{HNO}_{3}$ by protonation and loss of water. The nitronium ion reacts with benzene to yield a carbocation intermediate, and loss of $\mathrm{H}^{+}$from this intermediate gives the neutral substitution product, nitrobenzene (FIGURE 16.5).

Nitrobenzene

FIGURE 16.5 The mechanism for electrophilic nitration of an aromatic ring. An electrostatic potential map of the reactive electrophile $\mathrm{NO}_{2}{ }^{+}$shows that the nitrogen atom is most positive.

Electrophilic nitration of an aromatic ring does not occur in nature but is particularly important in the laboratory because the nitro-substituted product can be reduced by reagents such as iron, tin, or $\mathrm{SnCl}_{2}$ to yield the corresponding arylamine, $\mathrm{ArNH}_{2}$. Attachment of an amino group $\left(-\mathrm{NH}_{2}\right)$ to an aromatic ring by the two-step nitration/reduction sequence is a key part of the industrial synthesis of many dyes and pharmaceutical agents. We'll discuss this reduction and other reactions of aromatic nitrogen compounds in Chapter 24.


Aromatic Sulfonation -
Aromatic rings can be sulfonated by reaction with so-called fuming sulfuric acid, a mixture of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{SO}_{3}$. The reactive electrophile is either $\mathrm{HSO}_{3}{ }^{+}$or neutral $\mathrm{SO}_{3}$, depending on reaction conditions, and substitution occurs by the same two-step mechanism seen previously for bromination and nitration (FIGURE 16.6). Note, however, that the sulfonation reaction is readily reversible. It can occur either forward or backward, depending on the reaction conditions. Sulfonation is favored in strong acid, but desulfonation is favored in hot, dilute aqueous acid.

FIGURE 16.6 The mechanism for electrophilic sulfonation of an aromatic ring. An electrostatic potential map of the reactive electrophile $\mathrm{HOSO}_{2}{ }^{+}$shows that sulfur and hydrogen are the most positive atoms.

Aromatic sulfonation does not occur naturally but is widely used in the preparation of dyes and pharmaceutical agents. For example, the sulfa drugs, such as sulfanilamide, were among the first clinically useful antibiotics. Although largely replaced today by more effective agents, sulfa drugs are still used in the treatment of meningitis and urinary tract infections. These drugs are prepared commercially by a process that involves aromatic sulfonation as its key step.

Sulfanilamide (an antibiotic)


Aromatic Hydroxylation -
Direct hydroxylation of an aromatic ring to yield a hydroxybenzene (a phenol) is difficult and rarely done in the laboratory but occurs much more frequently in biological pathways. An example is the hydroxylation of $p$-hydroxyphenylacetate to give 3,4 -dihydroxyphenylacetate. The reaction is catalyzed by $p$-hydroxyphenylacetate-3-hydroxylase and requires molecular oxygen plus the coenzyme reduced flavin adenine dinucleotide, abbreviated $\mathrm{FADH}_{2}$.

By analogy with other electrophilic aromatic substitutions, you might expect that an electrophilic oxygen species acting as an " $\mathrm{OH}^{+}$equivalent" is needed for the hydroxylation reaction. That is just what happens, with the electrophilic oxygen arising by protonation of FAD hydroperoxide, RO-OH (FIGURE 16.7); that is, $\mathrm{RO}-\mathrm{OH}+\mathrm{H}^{+} \rightarrow \mathrm{ROH}+\mathrm{OH}^{+}$. The FAD hydroperoxide itself is formed by reaction of $\mathrm{FADH}_{2}$ with $\mathrm{O}_{2}$.


FIGURE 16.7 MECHANISM -
Mechanism for the electrophilic hydroxylation of $p$-hydroxyphenylacetate, by reaction with FAD hydroperoxide. The hydroxylating species is an " $\mathrm{OH}^{+}$equivalent" that arises by protonation of FAD hydroperoxide, $\mathrm{RO}-\mathrm{OH}+\mathrm{H}^{+} \rightarrow \mathrm{ROH}+\mathrm{OH}^{+}$. The FAD hydroperoxide itself is formed by reaction of $\mathrm{FADH}_{2}$ with $\mathrm{O}_{2}$.
(1) Reduced flavin adenine dinucleotide reacts with molecular oxygen to give a hydroperoxide intermediate.

FAD hydroperoxide

Protonation of a hydroperoxide oxygen by an acid HA makes the neighboring oxygen electrophilic and allows the aromatic ring to react, giving a carbocation intermediate.

Loss of $\mathrm{H}^{+}$from the carbocation gives the hydroxy-substituted aromatic product.
(3)

3,4-Dihydroxyphenylacetate

PROBLEM Propose a mechanism for the electrophilic fluorination of benzene with SelectFluor.
16-2
PROBLEM How many products might be formed on chlorination of $o$-xylene ( $o$-dimethylbenzene), $m$-xylene,
16-3 and $p$-xylene?
PROBLEM When benzene is treated with $\mathrm{D}_{2} \mathrm{SO}_{4}$, deuterium slowly replaces all six hydrogens in the aromatic 16-4 ring. Explain.


FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction
Among the most useful electrophilic aromatic substitution reactions in the laboratory is alkylation-the introduction of an alkyl group onto the benzene ring. Called the Friedel-Crafts reaction after its founders in 1877, Charles Friedel and James Crafts, the reaction is carried out by treating an aromatic compound with an alkyl chloride, RCl , in the presence of $\mathrm{AlCl}_{3}$ to generate a carbocation electrophile, $\mathrm{R}^{+}$. Aluminum chloride catalyzes the reaction by helping the alkyl halide to generate a carbocation in much the same way that $\mathrm{FeBr}_{3}$ catalyzes aromatic brominations by polarizing $\mathrm{Br}_{2}$ (Section 16.1). Loss of $\mathrm{H}^{+}$then completes the reaction (FIGURE 16.8).

FIGURE 16.8 MECHANISM
Mechanism for the Friedel-Crafts alkylation reaction of benzene with 2-chloropropane to yield isopropylbenzene (cumene). The electrophile is a carbocation, generated by $\mathbf{A l C l}_{\mathbf{3}}$-assisted dissociation of an alkyl halide.

(1) An electron pair from the aromatic ring attacks the carbocation, forming a $\mathrm{C}-\mathrm{C}$ bond and yielding a new carbocation intermediate.

(1)

Loss of a proton then gives the neutral alkylated substitution product.

Despite its utility, the Friedel-Crafts alkylation has several limitations. For one thing, only alkyl halides can be used. Aromatic (aryl) halides and vinylic halides don't react because aryl and vinylic carbocations are too high in energy to form under Friedel-Crafts conditions.

An aryl halide

A vinylic halide


FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction
An aryl halide

A vinylic halide

Not reactive
Another limitation is that Friedel-Crafts reactions don't succeed on aromatic rings that are substituted either by a strongly electron-withdrawing group such as carbonyl $(\mathrm{C}=0)$ or by a basic amino group that can be protonated. We'll see in the next section that the presence of a substituent group already on a ring can have a dramatic effect on that ring's reactivity to further electrophilic substitution. Rings that contain any of the substituents listed in FIGURE 16.9 do not undergo Friedel-Crafts alkylation.

FIGURE 16.9 Limitations on the aromatic substrate in Friedel-Crafts reactions. No reaction occurs if the substrate has either an electronwithdrawing substituent or a basic amino group.

A third limitation to the Friedel-Crafts alkylation is that it's often difficult to stop the reaction after a single substitution. Once the first alkyl group is on the ring, a second substitution reaction is facilitated for reasons we'll discuss in the next section. Thus, we often observe polyalkylation. Reaction of benzene with 1 mol equivalent of 2 -chloro-2-methylpropane, for example, yields $p$-di-tert-butylbenzene as the major product, along with small amounts of tert-butylbenzene and unreacted benzene. A high yield of mono-alkylation product is obtained only when a large excess of benzene is used.

A final limitation to the Friedel-Crafts reaction is that a skeletal rearrangement of the alkyl carbocation electrophile sometimes occurs during reaction, particularly when a primary alkyl halide is used. Treatment of benzene with 1 -chlorobutane at $0^{\circ} \mathrm{C}$, for instance, gives an approximately $2: 1$ ratio of rearranged (sec-butyl) to unrearranged (butyl) products.


FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction
The carbocation rearrangements that accompany Friedel-Crafts reactions are like those that accompany electrophilic additions to alkenes (Section 7.11) and occur either by hydride shift or alkyl shift. For example, the relatively unstable primary butyl carbocation produced by reaction of 1-chlorobutane with $\mathrm{AlCl}_{3}$ rearranges to the more stable secondary butyl carbocation by the shift of a hydrogen atom and its electron pair (a hydride ion, $\mathrm{H}:^{-}$) from C 2 to C 1 . Similarly, alkylation of benzene with 1-chloro-2,2-dimethylpropane yields (1,1-dimethylpropyl)benzene. The initially formed primary carbocation rearranges to a tertiary carbocation by shift of a methyl group and its electron pair from C 2 to C 1 .

Benzene
(1,1-Dimethylpropyl)benzene

Just as an aromatic ring is alkylated by reaction with an alkyl chloride, it is acylated by reaction with a carboxylic acid chloride, RCOCl , in the presence of $\mathrm{AlCl}_{3}$. That is, an acyl group ( -COR ; pronounced a-sil) is substituted onto the aromatic ring. For example, reaction of benzene with acetyl chloride yields the ketone acetophenone.

The mechanism of Friedel-Crafts acylation is similar to that of Friedel-Crafts alkylation, and the same limitations on the aromatic substrate noted previously in FIGURE 16.9 for alkylation also apply to acylation. The reactive electrophile is a resonance-stabilized acyl cation, generated by reaction between the acyl chloride and $\mathrm{AlCl}_{3}$ (FIGURE 16.10). As the resonance structures in the figure indicate, an acyl cation is stabilized by interaction of the vacant orbital on carbon with lone-pair electrons on the neighboring oxygen. Because of this stabilization, no carbocation rearrangement occurs during acylation.

FIGURE 16.10 Mechanism of the Friedel-Crafts acylation reaction. The electrophile is a resonance-stabilized acyl cation, whose electrostatic potential map indicates that carbon is the most positive atom.


FIGURE 16.7 MECHANISM - 16.3 Alkylation and Acylation of Aromatic Rings: The Friedel-Crafts Reaction
FIGURE 16.10 Mechanism of the Friedel-Crafts acylation reaction. The electrophile is a resonance-stabilized acyl cation, whose electrostatic potential map indicates that carbon is the most positive atom.

Unlike the multiple substitutions that often occur in Friedel-Crafts alkylations, acylations never occur more than once on a ring because the product acylbenzene is less reactive than the nonacylated starting material. We'll account for this reactivity difference in the next section.

Aromatic alkylations occur in numerous biological pathways, although there is of course no $\mathrm{AlCl}_{3}$ present in living systems to catalyze the reaction. Instead, the carbocation electrophile is typically formed by dissociation of an organodiphosphate, as we saw in Section 11.6. The dissociation is usually assisted by complexation to a divalent metal cation such as $\mathrm{Mg}^{2+}$, just as dissociation of an alkyl chloride is assisted by $\mathrm{AlCl}_{3}$.

An example of a biological Friedel-Crafts reaction occurs during the biosynthesis of phylloquinone, or vitamin $\mathrm{K}_{1}$, the human blood-clotting factor. Phylloquinone is formed by reaction of 1,4-dihydroxynaphthoic acid with phytyl diphosphate. Phytyl diphosphate first dissociates to a resonance-stabilized allylic carbocation, which then substitutes onto the aromatic ring in the typical way. Several further transformations lead to phylloquinone (FIGURE 16.11).

FIGURE 16.11 Biosynthesis of phylloquinone (vitamin $\mathbf{K}_{\mathbf{1}}$ ) from 1,4-dihydroxynaphthoic acid. The key step that joins the 20-carbon phytyl side chain to the aromatic ring is a Friedel-Crafts-like electrophilic substitution reaction with a diphosphate ion as the leaving group.


Predicting the Product of a Carbocation Rearrangement -
The Friedel-Crafts reaction of benzene with 2 -chloro-3-methylbutane in the presence of $\mathrm{AlCl}_{3}$ occurs with a carbocation rearrangement. What is the structure of the product?


Strategy -
A Friedel-Crafts reaction involves initial formation of a carbocation, which can rearrange by either a hydride shift or an alkyl shift to give a more stable carbocation. Draw the initial carbocation, assess its stability, and see if the shift of a hydride ion or an alkyl group from a neighboring carbon will result in increased stability. In the present instance, the initial carbocation is a secondary one that can rearrange to a more stable tertiary one by a hydride shift.

Secondary
carbocation
Tertiary carbocation

Use this more stable tertiary carbocation to complete the Friedel-Crafts reaction.


Solution -
PROBLEM Which of the following alkyl halides would you expect to undergo Friedel-Crafts reaction with 16-5 rearrangement and which without? Explain.
(a) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}$
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}(\mathrm{Cl}) \mathrm{CH}_{3}$
(c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}$
(d) $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CCH}_{2} \mathrm{Cl}$
(e) Chlorocyclohexane

PROBLEM What is the major monosubstitution product from the Friedel-Crafts reaction of benzene with 16-6 1-chloro-2-methylpropane in the presence of $\mathrm{AlCl}_{3}$ ?

PROBLEM Identify the carboxylic acid chloride that might be used in a Friedel-Crafts acylation reaction to 16-7 prepare each of the following acylbenzenes:
(a)

(b)


Solution - 16.4 Substituent Effects in Electrophilic Substitutions
Only one product can form when an electrophilic substitution occurs on benzene, but what would happen if we were to carry out a reaction on an aromatic ring that already has a substituent? The initial presence of a substituent on the ring has two effects.

- Substituents affect the reactivity of the aromatic ring. Some substituents activate the ring, making it more reactive than benzene, and some deactivate the ring, making it less reactive than benzene. In aromatic nitration, for instance, an -OH substituent makes the ring 1000 times more reactive than benzene, while an $-\mathrm{NO}_{2}$ substituent makes the ring more than 10 million times less reactive.

Relative rate of nitration
$6 \times 10^{-8}$

0.033

1

1000


Reactivity -
- Substituents affect the orientation of the reaction. The three possible disubstituted products-ortho, meta, and para-are usually not formed in equal amounts. Instead, the nature of the substituent initially present on the benzene ring determines the position of the second substitution. An -OH group directs substitution toward the ortho and para positions, for instance, while a carbonyl group such as -CHO directs substitution primarily toward the meta position. TABLE 16.1 lists experimental results for the nitration of some substituted benzenes.

| TABLE 16.1 Orientation Substituted Benzenes |  | of Nitration |  |
| :---: | :---: | :---: | :---: |
| <smiles>[Y]c1ccccc1</smiles> | $\xrightarrow[\mathrm{O}_{4}, 25^{\circ} \mathrm{C}]{ }$ | <smiles>[Y]c1ccccc1</smiles> | $-\mathrm{NO}_{2}$ |
|  | Product (\%) |  |  |
|  | Ortho | Meta | Para |
| Meta-directing deactivators |  |  |  |
| $-\stackrel{+}{\mathrm{N}}\left(\mathrm{CH}_{3}\right)_{3}$ | 2 | 87 | 11 |
| $-\mathrm{NO}_{2}$ | 7 | 91 | 2 |
| $-\mathrm{CO}_{2} \mathrm{H}$ | 22 | 76 | 2 |
| -CN | 17 | 81 | 2 |
| $-\mathrm{CO}_{2} \mathrm{CH}_{3}$ | 28 | 66 | 6 |
| $-\mathrm{COCH}_{3}$ | 26 | 72 | 2 |
| - CHO | 19 | 72 | 9 |
| Ortho- and para-directing deactivators |  |  |  |
| -F | 13 | 1 | 86 |
| -Cl | 35 | 1 | 64 |
| -Br | 43 | 1 | 56 |
| -I | 45 | 1 | 54 |


Reactivity -
Substituents can be classified into three groups, as shown in FIGURE 16.12: ortho- and para-directing activators, ortho- and para-directing deactivators, and meta-directing deactivators. There are no metadirecting activators. Notice how the directing effect of a group correlates with its reactivity. All meta-directing groups are strongly deactivating, and most ortho- and para-directing groups are activating. The halogens are unique in being ortho- and para-directing but weakly deactivating.

FIGURE 16.12 Classification of substituent effects in electrophilic aromatic substitution. All activating groups are ortho- and paradirecting, and all deactivating groups other than halogen are meta-directing. Halogens are unique in being deactivating but ortho- and paradirecting.


Predicting the Product of an Electrophilic Aromatic Substitution Reaction -
Predict the major product of the sulfonation of toluene.


Strategy -
Identify the substituent present on the ring, and decide whether it is ortho- and para-directing or metadirecting. According to FIGURE 16.12, an alkyl substituent is ortho- and para-directing, so sulfonation of toluene will primarily give a mixture of $o$-toluenesulfonic acid and $p$-toluenesulfonic acid.


Solution -
PROBLEM Rank the compounds in each of the following groups in order of their reactivity to electrophilic 16-8 substitution:
(a) Nitrobenzene, phenol, toluene, benzene (b) Phenol, benzene, chlorobenzene, benzoic acid
(c) Benzene, bromobenzene, benzaldehyde, aniline

PROBLEM Predict the major products of the following reactions:
16-9 (a) Nitration of bromobenzene (b) Bromination of nitrobenzene (c) Chlorination of phenol
(d) Bromination of aniline


Activating and Deactivating Effects -
What makes a group either activating or deactivating? The common characteristic of all activating groups is that they donate electrons to the ring, thereby making the ring more electron-rich, stabilizing the carbocation intermediate, and lowering the activation energy for its formation. Conversely, the common characteristic of all deactivating groups is that they withdraw electrons from the ring, thereby making the ring more electron-poor, destabilizing the carbocation intermediate, and raising the activation energy for its formation.
Reactivity

| is donates electrons; |
| :--- |
| isbocation intermediate |
| is more stable, and ring |

Compare the electrostatic potential maps of benzaldehyde (deactivated), chlorobenzene (weakly deactivated), and phenol (activated) with that of benzene. As shown in FIGURE 16.13, the ring is more positive (yellow-green) when an electron-withdrawing group such as -CHO or -Cl is present and more negative (red) when an electrondonating group such as -OH is present.

Benzaldehyde

Chlorobenzene

Benzene

Phenol

FIGURE 16.13 Electrostatic potential maps of benzene and several substituted benzenes show that an electron-withdrawing group $(-\mathrm{CHO}$ or -Cl$)$ makes the ring more electron-poor, while an electron-donating group $(-\mathrm{OH})$ makes the ring more electron-rich.


Activating and Deactivating Effects -
The withdrawal or donation of electrons by a substituent group is controlled by an interplay of inductive effects and resonance effects. As we saw in Section 2.1, an inductive effect is the withdrawal or donation of electrons through a $\sigma$ bond due to electronegativity. Halogens, hydroxyl groups, carbonyl groups, cyano groups, and nitro groups inductively withdraw electrons through the $\sigma$ bond linking the substituent to a benzene ring. This effect is most pronounced in halobenzenes and phenols, in which the electronegative atom is directly attached to the ring, but is also significant in carbonyl compounds, nitriles, and nitro compounds, in which the electronegative atom is farther removed. Alkyl groups, on the other hand, inductively donate electrons. This is the same hyperconjugative donating effect that causes alkyl substituents to stabilize alkenes (Section 7.6) and carbocations (Section 7.9).

Inductive electron withdrawal

Inductive electron
donation
A resonance effect is the withdrawal or donation of electrons through a $\pi$ bond due to the overlap of a $p$ orbital on the substituent with a $p$ orbital on the aromatic ring. Carbonyl, cyano, and nitro substituents, for example, withdraw electrons from the aromatic ring by resonance. The $\pi$ electrons flow from the ring to the substituent, leaving a positive charge in the ring. Note that substituents with an electron-withdrawing resonance effect have the general structure $-\mathrm{Y}=\mathrm{Z}$, where the Z atom is more electronegative than Y .

Conversely, halogen, hydroxyl, alkoxyl (-OR), and amino substituents donate electrons to the aromatic ring by resonance. Lone-pair electrons flow from the substituents to the ring, placing a negative charge on the ring. Substituents with an electron-donating resonance effect have the general structure - $\ddot{Y}$, where the Y atom has a lone pair of electrons available for donation to the ring.


Activating and Deactivating Effects -
One further point: inductive effects and resonance effects don't necessarily act in the same direction. Halogen, hydroxyl, alkoxyl, and amino substituents, for instance, have electron-withdrawing inductive effects because of the electronegativity of the $-\mathrm{X},-\mathrm{O}$, or -N atom bonded to the aromatic ring but have electron-donating resonance effects because of the lone-pair electrons on those $-\mathrm{X},-\mathrm{O}$, or -N atoms. When the two effects act in opposite directions, the stronger effect dominates. Thus, hydroxyl, alkoxyl, and amino substituents are activators because their stronger electron-donating resonance effect outweighs their weaker electronwithdrawing inductive effect. Halogens, however, are deactivators because their stronger electron-withdrawing inductive effect outweighs their weaker electron-donating resonance effect.

PROBLEM Use Figure 16.12 to explain why Friedel-Crafts alkylations often give polysubstitution but
16-10 Friedel-Crafts acylations do not.

(Sole product)
PROBLEM An electrostatic potential map of (trifluoromethyl)benzene, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CF}_{3}$, is shown. Would you expect
16-11 (trifluoromethyl)benzene to be more reactive or less reactive than toluene toward electrophilic substitution? Explain.

(Trifluoromethyl)benzene

Toluene


Ortho- and Para-Directing Activators: Alkyl Groups -
Inductive and resonance effects account not only for reactivity but also for the orientation of electrophilic aromatic substitutions. Take alkyl groups, for instance, which have an electron-donating inductive effect and are ortho and para directors. The results of toluene nitration are shown in FIGURE 16.14.

FIGURE 16.14 Carbocation intermediates in the nitration of toluene. Ortho and para intermediates are more stable than the meta intermediate because the positive charge is on a tertiary carbon rather than a secondary carbon.

Nitration of toluene might occur either ortho, meta, or para to the methyl group, giving the three carbocation intermediates shown in in FIGURE 16.14. Although all three intermediates are resonance-stabilized, the ortho and para intermediates are more stabilized than the meta intermediate. For both the ortho and para reactions, but not for the meta reaction, a resonance form places the positive charge directly on the methyl-substituted carbon, where it is in a tertiary position and can be stabilized by the electron-donating inductive effect of the methyl group. The ortho and para intermediates are thus lower in energy than the meta intermediate and form faster.


Ortho- and Para-Directing Activators: OH and $\mathrm{NH}_{2}$ -
Hydroxyl, alkoxyl, and amino groups are also ortho-para activators, but for a different reason than for alkyl groups. As described earlier in this section, hydroxyl, alkoxyl, and amino groups have a strong, electrondonating resonance effect that outweighs a weaker electron-withdrawing inductive effect. When phenol is nitrated, for instance, reaction can occur either ortho, meta, or para to the -OH group, giving the carbocation intermediates shown in FIGURE 16.15. The ortho and para intermediates are more stable than the meta intermediate because they have more resonance forms, including one particularly favorable form that allows the positive charge to be stabilized by electron donation from the substituent oxygen atom. The intermediate from the meta reaction has no such stabilization.

FIGURE 16.15 Carbocation intermediates in the nitration of phenol. The ortho and para intermediates are more stable than the meta intermediate because they have more resonance forms, including one particularly favorable form that involves electron donation from the oxygen atom.

PROBLEM Acetanilide is less reactive than aniline toward electrophilic substitution. Explain.
16-12


Ortho- and Para-Directing Deactivators: Halogens -
Halogens are deactivating because their stronger electron-withdrawing inductive effect outweighs their weaker electron-donating resonance effect. Although weak, that electron-donating resonance effect is nevertheless felt only at the ortho and para positions and not at the meta position (FIGURE 16.16). Thus, a halogen substituent can stabilize the positive charge of the carbocation intermediates from ortho and para reaction in the same way that hydroxyl and amino substituents can. The meta intermediate, however, has no such stabilization and is therefore formed more slowly.

FIGURE 16.16 Carbocation intermediates in the nitration of chlorobenzene. The ortho and para intermediates are more stable than the meta intermediate because of electron donation of the halogen lone-pair electrons.

Note again that halogens, hydroxyl, alkoxyl, and amino groups all withdraw electrons inductively but donate electrons by resonance. Halogens have a stronger electron-withdrawing inductive effect but a weaker electrondonating resonance effect and are thus deactivators. Hydroxyl, alkoxyl, and amino groups have a weaker electron-withdrawing inductive effect but a stronger electron-donating resonance effect and are thus activators. All are ortho and para directors, however, because of the lone pair of electrons on the atom bonded to the aromatic ring.


Meta-Directing Deactivators -
The influence of meta-directing substituents can be explained using the same kinds of arguments used for ortho and para directors. Look at the nitration of benzaldehyde, for instance (FIGURE 16.17). Of the three possible carbocation intermediates, the meta intermediate has three favorable resonance forms, whereas the ortho and para intermediates have only two. In both ortho and para intermediates, the third resonance form is unfavorable because it places the positive charge directly on the carbon that bears the aldehyde group, where it is disfavored by a repulsive interaction with the positively polarized carbon atom of the $\mathrm{C}=\mathrm{O}$ group. Hence, the meta intermediate is more favored and is formed faster than the ortho and para intermediates.

FIGURE 16.17 Carbocation intermediates in the nitration of benzaldehyde. The ortho and para intermediates are less stable than the meta intermediate. The meta intermediate is more favorable than ortho and para intermediates because it has three favorable resonance forms rather than two.

In general, any substituent that has a positively polarized atom $\left(\delta^{+}\right)$directly attached to the ring will make one of the resonance forms of the ortho and para intermediates unfavorable and will thus act as a meta director.

PROBLEM Draw resonance structures for the intermediates from the reaction of an electrophile at the ortho,
16-13 meta, and para positions of nitrobenzene. Which intermediates are most stable?


A Summary of Substituent Effects in Electrophilic Aromatic Substitution -
A summary of the activating and directing effects of substituents in electrophilic aromatic substitution is shown in TABLE 16.2.

TABLE 16.2 Substituent Effects in Electrophilic Aromatic Substitution

| Substituent | Reactivity |  | Orienting effect |  |
| :--- | :--- | :--- | :--- | :--- |
| -CH 3 | Inductive effect | Resonance effect |  |  |
| $-\mathrm{OH},-\mathrm{NH}_{2}$ | Activating | Ortho, para | Weak withdrawing | Strong donating |
| $-\mathrm{F},-\mathrm{Cl}$ | Deactivating | Ortho, para | Strong withdrawing | Weak donating |
| $-\mathrm{Br},-\mathrm{I}$ |  |  |  |  |
| $-\mathrm{NO}_{2},-\mathrm{CN}$, | Deactivating | Meta | Strong withdrawing | Strong withdrawing |
| $-\mathrm{CHO},-\mathrm{CO}_{2} \mathrm{R}$ |  |  |  |  |
| $-\mathrm{COR},-\mathrm{CO}_{2} \mathrm{H}$ |  |  |  |  |


A Summary of Substituent Effects in Electrophilic Aromatic Substitution - 16.5 Trisubstituted Benzenes: Additivity of Effects
Electrophilic substitution of a disubstituted benzene ring is governed by the same resonance and inductive effects that influence monosubstituted rings. The only difference is that it's necessary to consider the additive effects of two different groups. In practice, this isn't as difficult as it sounds; three rules are usually sufficient.

RULE 1
If the directing effects of the two groups reinforce each other, the situation is straightforward. In $p$-nitrotoluene, for example, both the methyl and the nitro group direct further substitution to the same position (ortho to the methyl = meta to the nitro). A single product is thus formed on electrophilic substitution.

p-Nitrotoluene
2-Bromo-4-nitrotoluene
RULE 2
If the directing effects of the two groups oppose each other, the more powerful activating group has the dominant influence, but mixtures of products are often formed. For example, bromination of $p$-methylphenol yields primarily 2-bromo-4-methylphenol because - OH is a more powerful activator than $-\mathrm{CH}_{3}$.

p-Methylphenol
2-Bromo-4-methylphenol
RULE 3
Further substitution rarely occurs between the two groups in a meta-disubstituted compound because this site is too hindered. Aromatic rings with three adjacent substituents must therefore be prepared by some other route, such as by substitution of an ortho-disubstituted compound.

But:


Predicting the Product of Substitution on a Disubstituted Benzene -
What product would you expect from bromination of $p$-methylbenzoic acid?


Strategy -
Identify the two substituents present on the ring, decide the directing effect of each and, if necessary, decide which substituent is the stronger activator. In the present case, the carboxyl group $\left(-\mathrm{CO}_{2} \mathrm{H}\right)$ is a meta director and the methyl group is an ortho and para director. Both groups direct bromination to the position next to the methyl group, yielding 3-bromo-4-methylbenzoic acid.


Solution -
p-Methylbenzoic acid
3-Bromo-4-methylbenzoic acid

PROBLEM At what position would you expect electrophilic substitution to occur in each of the following 16-14 substances:
(a)

(b)

(c)

PROBLEM Show the major product(s) from reaction of the following substances with (1) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}, \mathrm{AlCl}_{3}$ and 16-15 (2) $\mathrm{HNO}_{3}, \mathrm{H}_{2} \mathrm{SO}_{4}$ :
(a)

(b)


Solution - 16.6 Nucleophilic Aromatic Substitution
Although aromatic substitution reactions usually occur by an electrophilic mechanism, aryl halides that have electron-withdrawing substituents can also undergo a nucleophilic substitution reaction. For example, $2,4,6$-trinitrochlorobenzene reacts with aqueous NaOH at room temperature to give 2,4,6-trinitrophenol. Here, the nucleophile $\mathrm{OH}^{-}$substitutes for $\mathrm{Cl}^{-}$.

2,4,6-Trinitrochlorobenzene

2,4,6-Trinitrophenol (100\%)

Nucleophilic aromatic substitution is much less common than electrophilic substitution but nevertheless does have certain uses. One such use is the reaction of proteins with 2,4-dinitrofluorobenzene, known as Sanger's reagent, to attach a "label" to the terminal $\mathrm{NH}_{2}$ group of the amino acid at one end of the protein chain.

2,4-Dinitrofluorobenzene
Although the reaction appears superficially similar to the $S_{N} 1$ and $S_{N} 2$ nucleophilic substitutions of alkyl halides discussed in the chapter on Reactions of Alkyl Halides: Nucleophilic Substitutions and Eliminations, it must be different because aryl halides are inert to both $\mathrm{S}_{\mathrm{N}} 1$ and $\mathrm{S}_{\mathrm{N}} 2$ conditions. $\mathrm{S}_{\mathrm{N}} 1$ reactions don't occur with aryl halides because dissociation of the halide is energetically unfavorable, due to the instability of the potential aryl cation product. $\mathrm{S}_{\mathrm{N}} 2$ reactions don't occur with aryl halides because the halo-substituted carbon of the aromatic ring is sterically shielded from a backside approach. For a nucleophile to react with an aryl halide, it would have to approach directly through the aromatic ring and invert the stereochemistry of the aromatic ring carbon-a geometric impossibility.

Dissociation reaction does not occur because the aryl cation is unstable; therefore, no $\mathrm{S}_{\mathrm{N}} 1$ reaction.


Solution - 16.6 Nucleophilic Aromatic Substitution
Dissociation reaction does not occur because the aryl cation is unstable; therefore, no $\mathrm{S}_{\mathrm{N}} 1$ reaction.

Backside displacement is sterically blocked; therefore, no $\mathrm{S}_{\mathrm{N}} 2$ reaction.
Instead, nucleophilic substitutions on an aromatic ring proceed by the mechanism shown in FIGURE 16.18. The nucleophile first adds to the electron-deficient aryl halide, forming a resonance-stabilized, negatively charged intermediate called a Meisenheimer complex after its discoverer. Halide ion is then eliminated.


FIGURE 16.18 MECHANISM -
Mechanism of nucleophilic aromatic substitution. The reaction occurs in two steps and involves a resonance-stabilized carbanion intermediate.
(1) Nucleophilic addition of hydroxide ion to the electron-poor aromatic ring takes place, yielding a stabilized carbanion intermediate.

(1)

2 The carbanion intermediate undergoes elimination of chloride ion in a second step to give the substitution product.
(2)

Nucleophilic aromatic substitution occurs only if the aromatic ring has an electron-withdrawing substituent in a position ortho or para to the leaving group to stabilize the anion intermediate through resonance (FIGURE 16.19). A meta substituent offers no such resonance stabilization. Thus, $p$-chloronitrobenzene and $o$-chloronitrobenzene react with hydroxide ion at $130{ }^{\circ} \mathrm{C}$ to yield substitution products, but $m$-chloronitrobenzene is inert to $\mathrm{OH}^{-}$.


Ortho -
Para

Meta

FIGURE 16.19 Nucleophilic aromatic substitution on nitrochlorobenzenes. Only in the ortho and para intermediates is the negative charge stabilized by a resonance interaction with the nitro group, so only the ortho and para isomers undergo reaction.

Note the differences between electrophilic and nucleophilic aromatic substitutions. Electrophilic substitutions are favored by electron-donating substituents, which stabilize a carbocation intermediate, while nucleophilic substitutions are favored by electron-withdrawing substituents, which stabilize a carbanion intermediate. Thus, the electron-withdrawing groups that deactivate rings for electrophilic substitution (nitro, carbonyl, cyano, and so forth) activate them for nucleophilic substitution. What's more, these functional groups are meta directors in electrophilic substitution but are ortho-para directors in nucleophilic substitution. And finally, electrophilic substitutions replace hydrogen on the ring, while nucleophilic substitutions replace a leaving group, usually halide ion.

PROBLEM The herbicide oxyfluorfen can be prepared by reaction between a phenol and an aryl fluoride.
16-16 Propose a mechanism.

Oxyfluorfen


Ortho - 16.7 Benzyne
Halobenzenes without electron-withdrawing substituents don't react with nucleophiles under most conditions. At high temperature and pressure, however, even chlorobenzene can be forced to react. Chemists at the Dow Chemical Company discovered in 1928 that phenol could be prepared on an industrial scale by treatment of chlorobenzene with dilute aqueous NaOH at $340^{\circ} \mathrm{C}$ under 170 atm pressure.

A similar substitution reaction occurs with other strong bases. Treatment of bromobenzene with potassium amide $\left(\mathrm{KNH}_{2}\right)$ in liquid $\mathrm{NH}_{3}$ solvent, for instance, gives aniline. Curiously, though, when using bromobenzene labeled with radioactive ${ }^{14} \mathrm{C}$ at the C 1 position, the substitution product has equal amounts of the label at both C 1 and C2, implying the presence of a symmetrical reaction intermediate in which C 1 and C 2 are equivalent.

Further mechanistic evidence comes from trapping experiments. When bromobenzene is treated with $\mathrm{KNH}_{2}$ in the presence of a conjugated diene, such as furan, a Diels-Alder reaction (Section 14.4) occurs, implying that the symmetrical intermediate is a benzyne, formed by elimination of HBr from bromobenzene. Benzyne is too reactive to be isolated as a pure compound but, in the presence of water, addition occurs to give phenol. In the presence of a diene, Diels-Alder cycloaddition takes place.

The electronic structure of benzyne, shown in FIGURE 16.20, is that of a highly distorted alkyne. Although a typical alkyne triple bond uses $s p$-hybridized carbon atoms, the benzyne triple bond uses $s p^{2}$-hybridized carbons. Furthermore, a typical alkyne triple bond has two mutually perpendicular $\pi$ bonds formed by $p-p$ overlap, but the benzyne triple bond has one $\pi$ bond formed by $p-p$ overlap and one $\pi$ bond formed by $s p^{2}-s p^{2}$ overlap. The latter $\pi$ bond is in the plane of the ring and is very weak.

Side view


Ortho - 16.7 Benzyne
Side view

FIGURE 16.20 An orbital picture and electrostatic potential map of benzyne. The benzyne carbons are $s p^{2}$-hybridized, and the "third" bond results from weak overlap of two adjacent $s p^{2}$ orbitals.

PROBLEM Treatment of $p$-bromotoluene with NaOH at $300^{\circ} \mathrm{C}$ yields a mixture of two products, but treatment 16-17 of $m$-bromotoluene with NaOH yields a mixture of three products. Explain.


Oxidation of Alkyl Side Chains -
Despite its unsaturation, the benzene ring is inert to strong oxidizing agents such as $\mathrm{KMnO}_{4}$, which will cleave alkene carbon-carbon bonds (Section 8.8). It turns out, however, that the presence of the aromatic ring has a dramatic effect on the reactivity of alkyl side chains. These side chains react rapidly with oxidizing agents and are converted into carboxyl groups, $-\mathrm{CO}_{2} \mathrm{H}$.The net effect is conversion of an alkylbenzene into a benzoic acid, $\mathrm{Ar}-\mathrm{R} \rightarrow \mathrm{Ar}-\mathrm{CO}_{2} \mathrm{H}$. Butylbenzene is oxidized by aqueous $\mathrm{KMnO}_{4}$ to give benzoic acid, for instance.

A similar oxidation is employed industrially for the preparation of the terephthalic acid used in the production of polyester fibers. Worldwide, approximately 118 million tons per year of terephthalic acid is produced by oxidation of $p$-xylene, using air as the oxidant and $\mathrm{Co}(\mathrm{III})$ salts as catalyst.

$p$-Xylene $\quad$ Terephthalic acid
The mechanism of side-chain oxidation is complex and involves reaction of $\mathrm{C}-\mathrm{H}$ bonds at the position next to the aromatic ring to form intermediate benzylic radicals. tert-Butylbenzene has no benzylic hydrogens, however, and is therefore inert.

tert-Butylbenzene
Analogous side-chain oxidations occur in various biosynthetic pathways. The neurotransmitter norepinephrine, for instance, is biosynthesized from dopamine by a benzylic hydroxylation reaction. The process is catalyzed by the copper-containing enzyme dopamine $\beta$-monooxygenase and occurs by a radical mechanism. A copper-oxygen species in the enzyme first abstracts the pro-R benzylic hydrogen to give a radical, and a hydroxyl is then transferred from copper to carbon.

PROBLEM What aromatic products would you obtain from the $\mathrm{KMnO}_{4}$ oxidation of the following substances? 16-18
(a)

(b)


Bromination of Alkylbenzene Side Chains -
Side-chain bromination at the benzylic position occurs when an alkylbenzene is treated with $N$-bromosuccinimide (NBS). For example, propylbenzene gives (1-bromopropyl)benzene in 97\% yield on reaction with NBS in the presence of benzoyl peroxide, $\left(\mathrm{PhCO}_{2}\right)_{2}$, as a radical initiator. Bromination occurs exclusively in the benzylic position next to the aromatic ring and does not give a mixture of products.

The mechanism of benzylic bromination is similar to that discussed in Section 10.3 for allylic bromination of alkenes. Abstraction of a benzylic hydrogen atom first generates an intermediate benzylic radical, which then reacts with $\mathrm{Br}_{2}$ in step 2 to yield product and a Br • radical, which cycles back into the reaction to carry on the chain shown below as a summary. The $\mathrm{Br}_{2}$ needed for reaction with the benzylic radical is produced in step 3 by a concurrent reaction of HBr with NBS.

Reaction occurs exclusively at the benzylic position because the benzylic radical intermediate is stabilized by resonance. FIGURE 16.21 shows how the benzyl radical is stabilized by overlap of its $p$ orbital with the ringed $\pi$ electron system.

FIGURE 16.21 A resonance-stabilized benzylic radical. The spin-density surface shows that the unpaired electron is shared by the ortho
and para carbons of the ring.
PROBLEM Refer to Table 6.3 for a quantitative idea of the stability of a benzyl radical. How much more stable
16-19 (in $\mathrm{kJ} / \mathrm{mol}$ ) is the benzyl radical than a primary alkyl radical? How does a benzyl radical compare in stability to an allyl radical?

PROBLEM Styrene, the simplest alkenylbenzene, is prepared commercially for use in plastics manufacture
16-20 by catalytic dehydrogenation of ethylbenzene. How might you prepare styrene from benzene using reactions you've studied?

Styrene


Catalytic Hydrogenation of Aromatic Rings -
Just as aromatic rings are generally inert to oxidation, they're also inert to catalytic hydrogenation under conditions that reduce typical alkene double bonds. As a result, it's possible to reduce an alkene double bond selectively in the presence of an aromatic ring. For example, 4-phenyl-3-buten-2-one is reduced to 4-phenyl-2-butanone using a palladium catalyst at room temperature and atmospheric pressure. Neither the benzene ring nor the ketone carbonyl group is affected.

To hydrogenate an aromatic ring, it's necessary either to use a platinum catalyst with hydrogen gas at a pressure of several hundred atmospheres or to use a more effective catalyst such as rhodium on carbon. Under these conditions, aromatic rings are converted into cyclohexanes. For example, $o$-xylene yields 1,2-dimethylcyclohexane, and 4-tert-butylphenol gives 4-tert-butylcyclohexanol.


Reduction of Aryl Alkyl Ketones -
In the same way that an aromatic ring activates a neighboring (benzylic) $\mathrm{C}-\mathrm{H}$ toward oxidation, it also activates a benzylic carbonyl group toward reduction. Thus, an aryl alkyl ketone prepared by Friedel-Crafts acylation of an aromatic ring can be converted into an alkylbenzene by catalytic hydrogenation over a palladium catalyst.

Propiophenone, for instance, is reduced to propylbenzene by catalytic hydrogenation. Because the net effect of Friedel-Crafts acylation followed by reduction is the preparation of a primary alkylbenzene, this two-step sequence of reactions makes it possible to circumvent the carbocation rearrangement problems associated with direct Friedel-Crafts alkylation using a primary alkyl halide (Section 16.3).

The conversion of a carbonyl group into a methylene group $\left(\mathrm{C}=\mathrm{O} \rightarrow \mathrm{CH}_{2}\right)$ by catalytic hydrogenation is limited to aryl alkyl ketones; dialkyl ketones are not reduced under these conditions. Furthermore, the catalytic reduction of aryl alkyl ketones is not compatible with the presence of a nitro substituent on the aromatic ring because a nitro group is reduced to an amino group under reaction conditions. We'll see a more general method for reducing ketone carbonyl groups to yield alkanes in Section 19.9.

PROBLEM How would you prepare diphenylmethane, $(\mathrm{Ph})_{2} \mathrm{CH}_{2}$, from benzene and an acid chloride? 16-21


Reduction of Aryl Alkyl Ketones - 16.10 Synthesis of Polysubstituted Benzenes
As discussed in the Introduction to Organic Synthesis in Section 9.9, one of the surest ways to learn organic chemistry is to work synthesis problems. The ability to plan a successful multistep synthesis of a complex molecule requires a working knowledge of the uses and limitations of a great many organic reactions. Not only must you know which reactions to use, you must also know when to use them because the order in which reactions are carried out is often critical to the success of the overall scheme.

The ability to plan a sequence of reactions in the right order is particularly important in the synthesis of substituted aromatic rings, where the introduction of a new substituent is strongly affected by the directing effects of other substituents. Planning syntheses of substituted aromatic compounds is therefore a good way to gain confidence in using the many reactions discussed in the past few chapters.

As we said in Section 9.9, it's usually best to work a synthesis problem backward, or retrosynthetically. Look at the target molecule and ask yourself, "What is an immediate precursor of this compound?" Choose a likely answer and continue working backward, one step at a time, until you arrive at a simple starting material. Let's try some examples.


Synthesizing a Polysubstituted Benzene -
How would you synthesize 4-bromo-2-nitrotoluene from benzene?


Strategy -
Draw the target molecule, identify the substituents, and recall how each group can be introduced separately. Then plan retrosynthetically.

4-Bromo-2-nitrotoluene

The three substituents on the ring are a bromine, a methyl group, and a nitro group. A bromine can be introduced by bromination with $\mathrm{Br}_{2} / \mathrm{FeBr}_{3}$, a methyl group can be introduced by Friedel-Crafts alkylation with $\mathrm{CH}_{3} \mathrm{Cl} / \mathrm{AlCl}_{3}$, and a nitro group can be introduced by nitration with $\mathrm{HNO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}$.


Solution -
Ask yourself, "What is an immediate precursor of the target?" The final step will involve introduction of one of three groups-bromine, methyl, or nitro-so we have to consider three possibilities. Of the three, the bromination of o-nitrotoluene could be used because the activating methyl group would dominate the deactivating nitro group and direct bromination to the correct position. Unfortunately, a mixture of product isomers would be formed. A Friedel-Crafts reaction can't be used as the final step because this reaction doesn't work on a nitro-substituted (strongly deactivated) benzene. The best precursor of the desired product is probably $p$-bromotoluene, which can be nitrated ortho to the activating methyl group to give a single product.

$o$-Nitrotoluene
This ring will give a mixture of isomers on bromination.

m-Bromonitrobenzene
This deactivated ring will not undergo a Friedel-Crafts reaction.


p-Bromotoluene -
This ring will give only the desired isomer on nitration.


4-Bromo-2-nitrotoluene -
Next ask, "What is an immediate precursor of $p$-bromotoluene?" Perhaps toluene is an immediate precursor because the methyl group would direct bromination to the ortho and para positions. Alternatively, bromobenzene might be an immediate precursor because we could carry out a Friedel-Crafts methylation and obtain a mixture of ortho and para products. Both answers are satisfactory, although both would also lead unavoidably to a mixture of products that would have to be separated.

"What is an immediate precursor of toluene?" Benzene, which could be methylated in a Friedel-Crafts reaction. Alternatively, "What is an immediate precursor of bromobenzene?" Benzene, which could be brominated.

The retrosynthetic analysis has provided two valid routes from benzene to 4-bromo-2-nitrotoluene.


Synthesizing a Polysubstituted Benzene -
Synthesize 4-chloro-2-propylbenzenesulfonic acid from benzene.


Strategy -
Draw the target molecule, identify its substituents, and recall how each of the three can be introduced. Then plan retrosynthetically.


4-Chloro-2-propylbenzenesulfonic acid -
The three substituents on the ring are a chlorine, a propyl group, and a sulfonic acid group. A chlorine can be introduced by chlorination with $\mathrm{Cl}_{2} / \mathrm{FeCl}_{3}$, a propyl group can be introduced by Friedel-Crafts acylation with $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCl} / \mathrm{AlCl}_{3}$ followed by reduction with $\mathrm{H}_{2} / \mathrm{Pd}$, and a sulfonic acid group can be introduced by sulfonation with $\mathrm{SO}_{3} / \mathrm{H}_{2} \mathrm{SO}_{4}$.