Organic Chemistry Chapter 9 Aromaticity

Chapter 9 Aromaticity

Freeway at Night (Shanghai, China) David Richardson

“We come to love not by finding a perfect person, but by learning to see an imperfect person perfectly” Dr. S Dhillon

“No amount of experimentation can ever prove me right; a single experiment can prove me wrong.”

Albert Einstein

“Science, my lad, is made up of mistakes, but they are mistakes which it is useful to make, because they lead little by little to the truth.”

Jules Verne, A Journey to the Center of the Earth

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The constant movement of electrons within the π molecular orbital systems characterizes aromatic compounds as stable. This is true for conjugated nonaromatic compounds as well. The movement of electrons in the π molecular orbital systems is resonance. Resonance in aromatic and conjugated nonaromatic compounds stabilizes the compounds, i.e., the more resonance the more stable the compound.

Carbon atoms in a ring or rings of an aromatic system, as well as some contributing attached atoms and the atoms attached to the carbon, must be coplanar to capture the essence of aromaticity, i.e., to satisfy the criteria for aromaticity. Coplanarity means in the same plane.

The ring system is a heterocyclic ring if an atom other than carbon (e.g., oxygen or nitrogen) replaces one or more carbon atoms in the ring system.

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The following are examples of cyclic compounds with coplanar atoms.

So that one can observe the coplanarity of the carbon skeleton and the heteroatoms, the following illustrations omit the "p" orbitals.

The carbon frame of benzene The carbon frame of benzene is coplanar or flat, i.e., the six carbon atoms and the six hydrogen atoms are in the same plane. The molecule has a flat arrangement where the bond angles are approximately 120o.

The five-membered heterocyclic system illustrated below must also be coplanar if it is to exhibit aromaticity.

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A five-member heterocyclic ring system

Aromaticity implies a special kind of reactivity because aromatic compounds do not add to the double bond in the traditional manner that double bond systems experience, such as the addition of HX to double bonds, or vicinal diol formation, or acid hydrolysis, etc. Aromatic compounds undergo substitution reactions rather than addition reactions. The hydrogen atoms of aromatic compounds exhibit a unique property discussed in Chapter 10 (diamagnetic anisotropy).

Benzene, C6H6, is an example of an aromatic compound. The structure of benzene is

Benzene is the parent structure for many aromatic compounds. Benzene’s structural formula is

Benzene exhibits a high degree of resonance. The compound has a hexagon shape, 5

and it is flat, i.e., there is no puckering in the ring system. Analogous to unsaturated systems, all the carbon atoms in benzene are 2sp2 hybridized; therefore, there is a 2p atomic orbital containing one electron on each carbon atom perpendicular to the plane containing the six carbon atoms. The six 2p atomic orbitals linearly combine to form three bonding molecular orbitals and three antibonding molecular orbitals. The six electrons reside in three bonding molecular orbitals. The ring containing the six carbon atoms and the six hydrogen atoms must be perfectly flat for the six 2p atomic orbitals to be perfectly parallel so that they can linearly combine. The linear combination of the six 2p orbitals defines the system as aromatic. The following is a pictorial representation of the six 2p orbitals linearly combining.

Consequently, for aromatic compounds, the π bonds are not fixed but are delocalized throughout the flat ring system. The actual structure of benzene is a hybrid somewhere between the following resonance structures.

(The double headed arrow represents resonance.)

The actual structure is more appropriately represented by the following hybrid structure.

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Experimental evidence shows that all the C-C bond lengths in benzene are 1.39 Ǻ, a value intermediate between a single bond, 1.54 Ǻ, and an isolated double bond, 1.24 Ǻ (the average of 1.54 Ǻ and 1.24 Ǻ is 1.39 Ǻ).

All the carbon atoms in benzene are equivalent. One could verify this by observing that there is only one 1,2-disubstituted benzene. If one could fix the π bonds, then there would be two different structures for 1,2-disubstituted benzene.

This is not the case, because there is only one isolable structure for disubstituted benzene. The following structure would be a more appropriate representation for 1,2disubstituted benzene.

Additional evidence for resonance in benzene can be observed in the reactivity of benzene. If the bonds in benzene were fixed, as in an imaginary cyclohexatriene structure, addition reactions would be expected. However, benzene does not undergo addition reactions, because benzene prefers to undergo substitution reactions. The reactions of benzene relative to cyclohexene are illustrated in Table 9.1.

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Table 9.1 addition reactions of cyclohexene relative to nonreactivity of benzene to addition reactions

Heats of Hydrogenation of Unsaturated Systems The resonance stability of benzene and other Ď€ systems can be determined by comparing the heat liberated upon hydrogenating the unsaturated compound compared to the heat liberated upon hydrogenating cyclohexene. The larger the ∆H value, the less stable the compound. Table 9.2 compares the heats of hydrogenation of two systems with the heat of hydrogenation of cyclohexene.

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Table 9.2 heats of hydrogenation of cyclohexene, 1,3-cyclohexadiene and benzene

If the heats of hydrogenation are additive, the theoretical heat of hydrogenation of 1,3cyclohexadiene would be 240 kJ mol-1 if one considers the double bonds as isolated systems. The experimental heat of hydrogenation for 1,3-cyclohexadiene is 232 kJ mol-1. Consequently, the heat of hydrogenation for this system is 8 kJ mol-1 less than the theoretical value if one treated the two double bonds as isolated systems. This suggests that 1,3-cyclohexadiene has 8 kJ mol-1 of stability due to the resonance energy of the conjugated π bonds.

If one assumes that the heats of hydrogenation are additive, then the theoretical heat of hydrogenation of benzene would be 360 kJ mol-1 if the double bonds are isolated systems, i.e., the hypothetical cyclohexatriene system. The experimental heat of hydrogenation for benzene is 208 kJ mol-1. Consequently, the heat of hydrogenation for this system is 152 kJ mol-1 less than the theoretical value if the three double bonds are isolated systems. This suggests that benzene has 152 kJ mol-1 of stabilization due to resonance of the aromatic system.

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Table 9.3 lists the experimental values for the heats of hydrogenation, ΔHhydrogenation, of several compounds. One could use these values to obtain the approximate resonance energies of stabilization for the listed compounds.

Table 9.3 experimental heats, ΔH, of hydrogenation for selected compounds

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Continuation of Table 9.3

Table 9.3 suggests that (a) Isolated terminal π bonds have ΔHhydrogenation of approximately 126 kJ mol-1. (b) Compounds containing two isolated terminal π bonds have ΔHhydrogenation of approximately 254 kJ mol-1. (c) Internal π bonds have ΔHhydrogenation of approximately 117 kJ mol-1; therefore, one can conclude that internal π bonds are more stable than nonconjugated terminal π bonds. (d) Conjugated π bonds have ΔHhydrogenation values that are less than twice that of two nonconjugated π bonds.

Resonance stabilization energy resulting from the 2p atomic orbitals linearly combining to form bonding molecular orbitals contributes to the differences in the ΔHhydrogenation values. For example, the ΔHhydrogenation for 1,3-cyclohexadiene is 232 kJ mol-1, but the ∆Hhydrogenation for cyclohexene is 120 kJ mol-1. If the π bonds for 1,3-cyclohexadiene were isolated systems, then the theoretical value for ΔHhydrogenation for 1,3-

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cyclohexadiene would be 240 kJ mol-1. The experimental ΔHhydrogenation for 1,3cyclohexadiene is 232 kJ mol-1. Therefore, the resonance stabilization energy is 8 kJ mol-1. One can explain the stabilization energy by considering that the system is conjugated, not isolated. The four 2p atomic orbitals, perfectly parallel to each other, can linearly combine to form four molecular orbitals that result in a stabilization energy of 8 kJ mol-1.

Predicting Aromaticity In 1931, Erich Hückel proposed a rule for predicting if cyclic organic compounds are aromatic. His rule states that if the number of π electrons for a monocyclic system equals 4n+2 and n equals a positive integer, then the structure is aromatic. For example, if a compound has 2 π electrons, then n=0; if it has 6 π electrons, then n=1; if it has 10 π electrons, then n=2. In these examples, n equals a positive integer. If n equals a positive integer, then the compound complies with Hückel’s rule, and the system is a candidate for aromaticity.

If a compound agrees with Hückel’s rule, it must also undergo a second scrutiny. As indicated earlier, the cyclic system must be perfectly flat so that the 2p atomic orbitals on the carbons can linearly combine. If the 2p atomic orbitals are not perfectly parallel and cannot linearly combine, then the system will not be aromatic even though the π system complies with Hückel’s rule (4n+2 = π electron where n equals a positive integer). As illustrated earlier, the six 2p atomic orbitals of benzene are parallel, and the six 2p atomic orbitals can linearly combine to form the following six molecular orbitals. Three of the molecular orbitals are bonding molecular orbitals, and three of the molecular orbitals are antibonding molecular orbitals.

These six bonding and antibonding molecular orbitals originating from the six “2p” atomic orbitals of benzene are:

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The six π electrons reside in the ψ1, ψ2 and ψ3 bonding molecular orbitals. The remaining three molecular orbitals (ψ4, ψ5 and ψ6) are antibonding molecular orbitals and are higher in energy than the bonding molecular orbitals. The antibonding

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molecular orbitals do not contain electrons unless the molecule is in an excited state.

Benzene’s 6 π electrons reside as paired electrons with opposite spins -two in the ψ1 bonding molecular orbital; two in the ψ2 bonding molecular orbital; and two in the ψ3 bonding molecular orbital; therefore, benzene is aromatic because n = 1 (4n+2= 6 where n =1).

1,3-Cyclohexadiene has 4 π electrons. 1,3-Cyclohexadiene is not aromatic because n = 0.5 ( 4n+2 = 4 where n = 0.5).

Cyclobutadiene has 4 π electrons. Cyclobutadiene does not comply with Hückel’s rule for aromaticity because n = 0.5 (4n+2 = 4 where n = 0.5); therefore, cyclobutadiene is not aromatic.

Cyclooctatetraene has 8 π electrons. Cyclooctatetraene does not comply with Hückel’s rule for aromaticity because n = 1.5 (4n+2 = 8 where n = 1.5); therefore, cyclooctatetraene is not aromatic.

Compound I, cycloheptatrienyl cation, is aromatic, because it complies with Hückel’s rule since n = 1 (4n+2 = 6 where n = 1).

Write resonance structures for this compound.

Compound II, [18]-annulene, should be aromatic because its pi system complies with Hückel’s rule since n = 4 (4n+2=18 where n=4).

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II The 18 “2p� atomic orbitals are perfectly parallel; therefore, they can linearly combine to form the molecular orbitals giving the molecule aromatic character.

The principle that the 2p atomic orbitals in the coplanar structure containing the carbon skeleton must be parallel in order to be aromatic is exhibited by [14]-annulene, compound III.

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III Even though [14]-annulene fits Hückel’s rule, 4n+2 =14 where n = 3, the internal hydrogen atoms are sterically hindered causing a twisting of the 2p orbitals so that they are not appropriately aligned for linear combination of the fourteen 2p atomic orbitals. Consequently the 14 “2p” atomic orbitals exhibit torsional strain, and are not able to form the seven bonding molecular orbitals and seven antibonding molecular orbitals that would make this system aromatic.

The 14 “2p” atomic orbitals are not perfectly parallel; therefore the compound is not aromatic.

[10] annulene, cyclodecapentaene, is not aromatic, because of the steric interaction between the two internal hydrogen atoms distorting the alignment of the 10 “2p” atomic orbitals.

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[10] annulene Cyclopentadienyl Anion, Heterocyclic Aromatic Compounds, and Biphenyl Furan, compound IV, has five molecular orbitals. Three are bonding molecular orbitals and two are antibonding molecular orbitals. These orbitals are the result of four 2p atomic orbitals of carbon and a 2p orbital of oxygen combining linearly to form five molecular orbitals. The six electrons in the carbon atoms and a lone pair of electrons on the oxygen 2p atomic orbitals can linearly combine to form five molecular orbitals (3 bonding molecular orbitals and 2 antibonding molecular orbitals). Since there are six electrons (four in four 2p atomic orbitals and two in an oxygen 2p atomic orbital) in the p orbital systems of furan, furan agrees with Hückel’s rule for aromaticity, i.e., 4n+2 = 6 where n = 1.

The cyclopentadienyl anion, compound V, has four π electrons in four 2p atomic orbitals on four carbon atoms and a lone pair of electrons in a 2p atomic orbital of carbon that can linearly combine to form five molecular orbitals (3 bonding molecular orbitals and 2 antibonding molecular orbitals). Since there are six electrons (four in four “2p” atomic orbitals and two in a “2p” atomic orbital of a fifth carbon atom) the p orbital systems of cyclopentadienyl anion, then cyclopentadienyl anion complies with Hückel’s rule for aromaticity, i.e., 4n+2 = 6 where n = 1.

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Pyrrole, compound VI, has four π electrons in four 2p atomic orbitals on four carbon atoms and a lone pair of electrons in a 2p atomic orbitals of nitrogen that can linearly combine to form five molecular orbitals (3 bonding molecular orbitals and 2 antibonding molecular orbitals). Since there are six electrons (four in four “2p” atomic orbitals and two in a “2p” atomic orbital of nitrogen) the p orbital systems of pyrrole comply with Hückel’s rule for aromaticity (i.e., 4n+2 = 6 where n = 1).

Biphenyl, compound VII, has 12 π electrons; therefore, it would appear to be nonaromatic since n would not have an integer for 4n+2=12. However, this system has two separate rings systems in which six 2p atomic orbitals linearly combine in one ring, and six 2p atomic orbitals linearly combine in the other ring. So each ring follows Hückel’s rule; consequently, compound VII is aromatic. The circles mean that the six “2p” atomic orbitals are being viewed from the top. This means that one ring in the biphenyl system is perpendicular to the other.

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Hückel’s rule should, in the strictest sense, be applied only to monocyclic double bond systems containing (4n+2) π electrons which are planar and conjugated The theoretical basis for the Hückel rule and aromaticity lies in quantum mechanics.

The concept of the linear combination of atomic orbitals to produce bonding and antibonding orbitals as described earlier for the benzene system is a partial explanation for this principle.

As previously indicated, benzene and its derivatives are not the only groups of compounds exhibiting aromatic character. Many heterocyclic compounds are aromatic in that they possess resonance stability, and they undergo reactions common to aromatic compounds. Pyridine, compound a, is an example of an aromatic heterocyclic compound:

Resonance forms c, d, and e are major contributors to resonance because of the high electronegativity of nitrogen.

Another example of a heterocyclic compound that is resonance stabilized is thiophene, compound f:

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Only one pair of electrons on sulfur, in the thiophene molecule, overlaps with the π orbitals of the ring; therefore, the four π electrons and the two electrons in the 3p atomic orbital of sulfur would comply with Hückel’s rule for aromaticity. Another way of looking at this is to envision the four electrons in the four 2p atomic orbitals of the four carbon atoms that form the flat ring of thiophene and the two electrons in the 3p atomic orbital of sulfur linearly combine to form five molecular orbitals (three bonding molecular orbitals and two antibonding molecular orbitals). The six electrons would reside in the three bonding molecular orbitals.

Acidity of Phenol, A Resonance Phenomenon Phenol, compound VIII, is approximately one million times more acidic then cyclohexanol, compound IX.

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The difference in acidity between compounds VIII and IX is due to resonance.

Cyclohexanol contains sigma bonds only; therefore, resonance delocalization cannot occur, but phenol has π electrons that can participate in resonance stabilization of the resulting phenoxide ion. The phenoxide ion forms in the presence of a base, e.g., NaOH, and the resulting phenoxide ion is resonance stabilized.

phenoxide

Resonance stabilization of the resulting phenoxide anion is the driving force for the acidity of phenol.

Also, phenol undergoes resonance.

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There is charge separation in the resonance structures of phenol, and resonance delocalizes the charges about the ring. This delocalization results in the separation of unlike charges (the positive charge and the negative charge). Such separation of charges has a destabilizing effect on phenol. This destabilizing effect is absent in the phenoxide ion; therefore, resonance stabilizes the phenoxide ion more than it stabilizes phenol.

The key to the difference in the acidity of phenol and cyclohexanol is the resonance stabilization of the phenoxide anion. The ionization of phenol results in the production of the phenoxide anion which is more resonance stabilized than phenol. cyclohexanol and the cyclohexoxide anion are not resonance stabilized; therefore, resonance delocalization is not a factor for cyclohexanol or the cyclohexoxide anion.

Consequently, unlike phenol, the ionization of cyclohexanol does not produce a resonance stabilized anion. Therefore, the tendency for cyclohexanol to ionize is not as great as the tendency for phenol to ionize.

Resonance and the Basicity of Aniline Aniline is a weaker base than aliphatic amines. The electrons on the nitrogen atoms of aliphatic amines are available to function as Lewis bases. However, the electrons on nitrogen atoms attached to an aromatic ring participate in resonance with the electrons in the aromatic ring. This situation decreases the availability of the lone pair to function as Lewis bases.

For example, aniline, compound X, would have the following resonance structures:

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The possible resonance structures after protonating aniline are:

anilinium cation

The lone pair of electrons that was on the nitrogen of aniline now participates in a coordinate covalent bond in the anilinium cation. These electrons are no longer available for delocalization. Consequently, the resulting anilinium cation has two resonance structures. Aniline has five resonance structures. The number of resonance

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structures determines the relative stability of compounds; therefore, aniline is more stable than the anilinium cation since aniline has more resonance structures. This is the case even though aniline resonance structures include charge-separated species.

The free amines nor the protonated amines have resonance stabilizations; therefore, their protonated aliphatic amines and amines have the same relative stability. For this reason, aliphatic amines possess a greater tendency toward protonation than aniline.

In summary, aliphatic amines are more basic than aniline or an amine defined solely by a nitrogen atom attached to an aromatic ring.

Destabilizing and Stabilizing Groups Attached to Phenol Certain groups attached to phenol may increase or decrease the acidity of phenol through resonance interaction. Table 9.4 compares the pKa of phenol with the pKas of selected substituted phenol compounds.

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Table 9.4 comparison of pKa of phenol with the pKas of selected substituted phenols

The smaller the pKa value, the stronger the acid Resonance determines the acidities of ortho and para substituted phenols.

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Therefore, the pH of a 1.00 x 10-2 M solution of phenol would be:

1.15 x 10−10 =

x2 1.00 x 10-2 M

1.15 x 10−10 x 1.00 x 10-2 = x = [H3O+]

1.07 x 10-6 M = [H3O+]

pH = - log [H3O+]

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pH = - log (1.07 x 10-6) = 5.97 The pKa for 4-methylphenol is larger than the pKa for phenol; therefore, 4-methylphenol is less acidic than phenol.

The inductive (electron releasing) effect of the methyl

substituent explains the difference in acidity. The methyl group has a destabilizing effect on the production of the 4-methylphenoxy anion because the methyl group releases negative charges into the ring, therefore, preventing resonance dispersion of the negative charge created by the phenoxide anion. Such charge localization causes destabilization of the 4-methylphenoxide anion relative to phenol. As a result, there is a decrease in the tendency of 4-methylphenol to ionize because of the destabilization of the resulting 4-methylphenoxide ion. Since the phenoxide ion does not have a methyl group attached, there is no charge localization of the ion. For this reason, the acid strength of 4-methylphenol is less than it is for phenol.

A nitro group in the para-position of phenol has the opposite effect of a methyl group in the para position of phenol. The pH of 1.00 x 10-2 M 4-nitrophenol is:

7.24 x 10−8 =

x2 1.00 x 10−2 M

7.24 x 10−8 x 1.00 x 10−2 M = x = [H3O+] 7.24 x 10−10 = [H3O+]

2.69 x 10−5 = [H3O+]

pH =- log [H3O+] pH =- log [2.69 x 10−5] = 4.57 The Ka for 4-Nitrophenol is 630 times greater than the Ka for phenol. The reason for

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this increase in acidity is due to the nitro group in the “para” or “4” position. The nitro group in the “para” position can participate in the resonance stabilization of the 4nitrophenoxide anion produced from the ionization of 4-nitrophenol. The electronwithdrawing ability of the nitro group causes delocalization of the lone pair of electrons on the 4-nitrophenoxide anion. This results in stabilizing the phenoxide anion; therefore, promoting ionization of 4-nitrophenol to 4-nitrophenoxide and hydrogen ion.

The following illustration explains this relationship:

As indicated in Table 9.4, adding more nitro groups to the aromatic ring system

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increases the acidities of substituted phenols. For example 2,4,6-trinitrophenol, picric acid, has a Ka that is 3.62 x 109 times that of the Ka for phenol.

Similar arguments can explain the differences in the basicity of aniline, 4-methylaniline, and 4-nitroaniline. The differences would be the opposite for acidity, i.e., electronreleasing groups attached to the aromatic ring of aniline would increase the basicity of the substituted aniline, and electron-withdrawing groups would decrease the basicity of the substituted aniline.

The lone pair of electrons on the nitrogen atom of aniline functions as a Lewis base.

The effect of electron-withdrawing groups, e.g., a nitro group in the para or ortho positions of aniline would significantly decrease the availability of the lone pair of electrons on the amine group. A nitro group in the meta position also decreases the availability of the lone pair on the amine of aniline through an inductive effect.

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The nitro group in the “para� position is electron-withdrawing and through resonance removes the lone pair of electrons decreasing the availability of the electron pair to function as a Lewis base.

On the other hand, an electron releasing group such as a methyl group intensifies the availability of the electron pair on the nitrogen atom so that the electron pair will be available to function as a Lewis base.

Take time to read the interesting discussion about aromaticity at http://www.uea.ac.uk/~c286/aromaticnotes.htm

Structures of Substituted Benzenes The number of mono, di and tri substituted products possible establishes, in part, the structure of benzene. Only one mono substituted product of benzene can be isolated.

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This suggests that all carbon atoms of benzene are equivalent.

There are three disubstituted products of benzene: (a) 1,2-disubstituted or ortho (example 1,2-dichlorobenzene)

(b) 1,3-disubstituted or meta (example 1,3-dichlorobenzene)

(c) 1,4-disubstituted or para (example 1,4-dichlorbenzene)

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Electrophilic Aromatic Substitution Reactions Benzene and related aromatic compounds react with electrophiles (Lewis acids, E+) to produce substituted products. The mechanism of the reaction occurs in five steps.

The first step is the formation of the electrophile in the presence of a catalyst.

Experimental evidence suggests that the second step is the formation of a “picomplex� with the aromatic nucleus. The pi-complex forms a benzenonium ion that is resonance stabilized. The resonance-stabilized intermediate undergoes a fast step to the formation of a second pi complex. In a final step, the pi complex forms the electrophilic aromatic substituted product.

Following is a general equation for an electrophilic aromatic substitution reaction.

The following five steps represent a general mechanism for electrophilic aromatic substitution.

(1)

(2)

first pi complex

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(3). Slow step

The formation of the benzenonium intermediate is the slow step, because the aromatic integrity of the benzene ring has been disrupted; therefore, this would be an energy consuming step. However, the benzenonium intermediate is stabilized by resonance.

The third step of the mechanism is a fast step where the benzenonium carbocation is re-aromatized into a second pi complex.

(4)

second pi complex

(5) Fast step

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The following represents the energy profile diagram for the reaction.

The resonance stabilized benzenonium ion contributes to the driving force that leads to the formation of the substituted aromatic compound.

Examples of electrophilic aromatic substitution reactions include halogenation, nitration, sulfonation, alkylation, and acylation. A variety of electrophiles are involved in electrophilic aromatic substitution reactions, and their mechanisms follow the general five-step electrophilic aromatic substitution mechanism previously described.

The following are examples of electrophilic aromatic substitution reactions.

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Nitration

Nitric acid, HNO3, in the presence of sulfuric acid, H2SO4, generates the +NO2 electrophile. A colligative property experiment of a mixture of nitric acid and sulfuric acid gives a van Hoff factor “i” equal to 4. A van Hoff factor “i” of 4 suggests that the mixture of nitric acid and sulfuric acid produces four ions.

2 H 2SO 4 + HNO3 → H3O+ +

+

NO2 + 2HSO-4

Generation of a nitro group could be explained by the following sequence of steps.

(1)

(2)

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(3)

Once the electrophile has been generated, the following steps can explain the nitration of benzene.

(1)

(2)

(3)

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(4)

The formation of the benzenonium ion intermediate is a slow step; the loss of a proton in the re-aromatization step to form the second pi-complex that leads to nitrobenzene is a fast step. This process occurs in most electrophilic aromatic substitution reactions, i.e., an electrophile, E+, attacks the aromatic nucleus to form a pi-complex in a slow step. The pi-complex forms the benzenonium intermediate followed by a rapid re-aromatization step that produces a second pi-complex. The second pi-complex forms the substituted aromatic compound. An exception to these steps for electrophilic aromatic substitution reactions is sulfonation. Unlike most electrophilic aromatic substitution reactions, sulfonation is reversible.

Isotope Effect Since the mass of a hydrogen atom and the mass of a deuterium atom are different, the C-H and C-D bonds are different. It takes more energy to break a C-D bond than it takes to break a C-H bond. If the slow step, or the rate determining step, involves the breaking of a C-H bond, then substituting deuterium for hydrogen should slow the rate of the reaction. This would occur since the added strength of a C-D bond makes it more difficult to break; therefore, reducing the rate of the reaction. This observation is the isotope effect and corroborates the mechanism of a reaction involving the breaking of a C-H bond in a slow step.

The isotope effect does not affect the rate of the reaction of nitration of benzene or any other electrophilic aromatic substitution reaction, So, substituting deuterium on the

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aromatic nucleus does not affect the rate of the reaction. This implies that the hydrogen atom must be removed in a rapid step, not in a slow step. Since most electrophilic aromatic substitution reactions do not exhibit an isotope effect, then this observation supports the mechanism as outlined where an electrophile, E+, attacks the aromatic nucleus to form a benzenonium intermediate in a slow step followed by a rapid (fast) re-aromatization step.

Sulfonation is an exception to this observation because it is a reversible reaction. The following equations illustrate the sulfonation of benzene ring with a rationale for its isotope effect.

A mixture of sulfuric acid and SO3, generally known as fuming sulfuric acid, produces +

SO3H, the electrophile that reacts with the aromatic system.

(1)

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(2)

(3)

(4)

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(5)

Unlike the other electrophilic aromatic substitution reactions, sulfonation is reversible, i.e., the proton can displace the sulfonate group.

(1)

(2) slow step

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(3)

(4)

(5)

Since sulfonation is a reversible electrophilic aromatic substitution reaction, then it would exhibit an isotope effect.

The following is an energy profile diagram of this reaction that illustrates the equilibrium for this reversible reaction. The energy barrier from reactants to products is the same as the energy barrier for products to reactants: 41

Aromatic rings can undergo electrophilic aromatic reactions that result in producing halogens (bromine and chlorine) on the aromatic nucleus. For example, chlorine can replace a proton on the aromatic nucleus by reacting chlorine, in the presence of a Lewis catalyst, with benzene. Benzene will react with chlorine in the presence of iron(III) chloride to form chlorobenzene.

The following equations illustrate the chlorination and bromination of benzene:

Bromination of benzene can be illustrated by the following equation.

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The electrophile generated in halogenation of aromatic systems is X+. X+ is the halonium ion. If chlorine is the halogen used in the reaction, then the halonium ion is the chloronium ion. If bromine is the halogen used in the reaction, then the halonium ion is the bromonium ion. The reaction with X2 and FeX3 occurs with chlorine and bromine, and not with fluorine and iodine.

The generation of Cl+, the chloronium ion, can be envisioned in the following manner.

Notice the arrangements of the electrons about the chlorine and the iron(III) chloride.

Chlorine (with eight electrons about it) in Cl2 attacks the iron(III) chloride to form a chloronium ion and tetrachloroferrate(III) anion.

The chloronium ion, Cl+, is referred to as the electrophile. The aromatic nucleus reacts slowly with the electrophile to form a benzenonium cation intermediate via a pi complex. The following equation illustrates the reaction of the chloronium ion with benzene.

(1)

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(2) Slow step

The resulting chlorobenzenonium cation is stabilized by resonance.

In a fast third step of the pathway that describes the formation of a second pi complex, a proton is removed from the complex by tetrachloroferrate(III) anion to produce chlorobenzene and HCl, and the catalyst, iron(III) chloride, is regenerated.

(3)

(4)

The mechanism for the formation of chlorobenzene benzene and chlorine in the

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presence of a Lewis catalyst takes place in four steps. the first step is the formation of a pi-complex. In a slow step (a second step), the complex forms a resonance stabilized chlorobenzenonium ion. The third step is the formation of a second pi complex. The final step involves the removal of a proton from the pi-complex using tetrachloroferrate(III) anion to produce chlorobenzene. The products are chlorobenzene, hydrochloric acid, and the catalyst, iron(III) chloride.

The mechanisms for all electrophilic aromatic substitution reactions follow a similar pathway analogous to the mechanism for the chlorination of benzene. The only change is in the electrophile, i.e., the electrophile is different in each case; however, the process for generating the electrophile is similar.

Friedel-Crafts Alkylation Reaction Among the many types of electrophilic aromatic substitution reactions is the FriedelCrafts Alkylation reaction. The Friedel-Crafts Alkylation reaction is a reaction between alkyl halides and benzene. The result is an arene, an alkyl benzene. The following equations describe the Friedel-Crafts Alkylation reaction. The first equation is a general equation where the “R� group represents alkyl groups, i.e., a methyl or an ethyl group or an isopropyl group, etc. The second equation is an illustration of the first equation.

(1)

(2)

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The Friedel-Crafts electrophile involves the generation of an aluminum complex that provides an appropriate alkyl electrophile, an alkyl carbocation, by reacting an alkyl halide with aluminum chloride, the Lewis base.

The first step of the mechanism is the formation of a pi-complex, and the complex forms an alkylbenzenonium carbocation ion in a second step. The third step is the formation of a second pi-complex, and the final step is the attack of aluminiun tetraaluminate anion on the pi-complex to produce ethylbenzene.

Step one, the slow step, generates the resonance stabilized alkylbenzenonium ion.

(1)

(2)

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(3)

(4)

Step 4 of the mechanism, a fast step, involves removal of a proton from the second picomplex by tetrachloroaluminate anion to produce ethylbenzene and HCl, and regeneration of the catalyst, aluminum chloride.

Since carbocations are generated in the reaction, it is not possible to produce long chain alkylbenzenes by way of this method, because the resulting primary carbocations would rearrange to produce more stable carbocations. For example, npropyl carbonium ion would rearrange to form the more stable isopropyl carbonium ion.

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Rearrangement occurs because a secondary carbocation is more stable than a primary carbocation.

The product of the reaction would be isopropylbenzene not n-propylbenzene.

In addition to the Friedel-Crafts Alkylation reaction, aromatic nuclei can undergo Friedel-Crafts Acylation reactions. The Friedel-Crafts Acylation reaction is a reaction between acyl halides and benzene. The result is a ketone as illustrated in the following chemical equations.

The acylation reaction is analogous to the alkylation ion except that an acyl group

48

replaces the alkyl group.

Acyl group

(1)

(2) Slow step

49

The resulting acylbenzenonium ion is resonance stabilized.

(3)

Step (3) of the mechanism, the fast step, involves the formation of the second picomplex, followed by the final step, the removal of a proton from the pi-complex by tetrachloroaluminate anion to produce acyllbenzene, a ketone, and HCl, and regeneration of the catalyst, aluminum chloride.

(4)

The acyl halide does not rearrange; therefore, this could be a way of putting long alkyl chains on the aromatic ring. After acylating the aromatic ring, reduce the carbonyl group attached to the aromatic ring.

A very good reducing agent for converting a carbonyl group to a methylene group is zinc amalgam, Zn(Hg), in concentrated hydrochloric acid. This reduction method is the

50

Clemmensen Reduction represented by the following equation:

Analogous to catalytic hydrogenation, the Clemmensen reduction occurs on the surface of zinc. The following series of steps account for the formation of the reduction of the carbonyl group to the methylene group, using amalgamated zinc in concentrated hydrochloric acid.

(1)

(2)

51

(3)

(4)

(5)

(6)

52

(7)

The sum of steps 1-7 would give a balanced chemical equation that illustrates the conversion of the carbonyl group attached to the aromatic nucleus to a methylene group.

The balanced net ionic equation can be written in molecular form to give a balanced molecular equation for the reduction of the carbonyl group to a methylene group by zinc and hydrochloric acid.

53

Mercury is important in this process because the amalgamated zinc helps to supply electrons in the reduction process. Steps 2,4,6, and 7 of the mechanism show zinc as a metal or as Zn+ supplying electrons to the system. Zinc mixed with mercury increase the tendency of Zn and Zn+ to release electrons to the system. Also, note that the carbonyl group is adjacent to the aromatic nucleus. Resonance stabilization facilitates the reduction when the carbonyl group is adjacent to the aromatic ring.

The Clemmensen reduction reduces a carbonyl group to a methylene group in molecules that are not sensitive to acids or contain groups that could be impacted by reductions in amalgamated zinc and concentrated hydrochloric acid (e.g., double bonds). Compounds containing amine groups will be sensitive to a reduction in concentrated hydrochloric acid. Consequently, the Clemmensen reaction will not be as effective with molecules containing carbon-carbon double bonds and amine groups.

A method for reducing carbonyl groups in compounds containing amino groups is the Wolff-Kishner reduction. The reagents for the Wolff-Kishner reduction are hydrazine and a strong base such as sodium hydroxide or potassium hydroxide. The following is an example of the Wolff-Kishner reduction.

The mechanism for the conversion of a carbonyl group to a methylene group by the Wolff-Kishner reduction is:

54

(1)

(2)

(3)

(4)

55

(5)

(6)

(7)

56

(8)

(9)

(10)

The sum of steps 1-10 gives the balanced chemical equation for the Wolff-Kishner reduction. Note that the hydroxide ions cancel when the steps are added. This indicates that the hydroxide acts as a catalyst in this reaction.

57

Showing the catalytic effect of the hydroxide ion, the equation could be written in the following manner.

In addition to aromatic ketones where the carbonyl group is adjacent to the aromatic nucleus, the Clemmensen and the Wolff-Kishner reductions reduce carbonyl groups of aliphatic compounds to methylene groups.

Limitations of Fridel-Crafts Alkylation and Acylation Reactions Friedel-Crafts alkylation and acylation reactions will not work for aromatic rings with attached groups that deactivate the aromatic nucleus. Also, Lewis catalysts such as aluminum chloride complexes with amines and phenols during attempted acylation reactions producing N or O acylation products rather than ring acylation products. The N and O acylation products make the Lewis catalyst ineffective.

As mentioned previously, alkylation reactions with alkyl halides will result in branching on the aromatic ring, because rearrangement will occur to produce the most stable carbocation electrophile.

Benzenediazonium Chlorides Benzenediazonium salts, an aryldiazonium salt, can be prepared by reacting aniline with nitrous acid. The nitrous acid can be generated in situ by reacting sodium nitrite with hydrochloric acid at about zero degrees centigrade.

The following represents the balanced molecular equation for forming the diazonium 58

salt from aniline hydrochloric acid and sodium nitrite:

Chapter 20 discusses the mechanism for the formation of a diazonium salt. The diazonium group of benzenediazonium salts can be replaced with a variety of atoms or groups (Chapter 20). If the diazonium group is replaced with a chlorine or bromine atom or a cyano, CN, group, then the reaction is the Sandmeyer Reaction (Chapter 20).

The mechanism of Sandmeyer Reaction (Chapter 20) involves free radicals rather than a simple electrophilic substitution process. The Sandmeyer Reaction uses copper(I) chloride or copper(I) bromide, or copper(I) cyanide to respectively prepare chloro, bromo, or cyano aromatic compounds. The precursor for the Sandmeyer reaction is an aryldiazonium halide such as benzenediazonium chloride compound that reacts with copper(I) chloride to produce chlorobenzene.

Aniline, the precursor for the aryldiazonium halide, can be made by the reduction of nitrobenzene. The reaction involves the reduction of nitrobenzene with tin and hydrochloric acid to form anilinium chloride. The anilinium chloride is converted to aniline with a base.

59

Nitrobenzene.

aniline

aniline

Diazo coupling This reaction works for aromatic compounds with ortho-para directors that are strong activating groups such as the -OH group.

60

The following mechanism provides an explanation for diazo coupling and a rationale for the necessity of an electron releasing group attached to the aromatic ring that reacts with the aryldiazonium salt.

(1)

61

(2)

The Kolbe Synthesis for Salicylic Acid The Kolbe reaction is specific for making salicylic acid from phenol. The reaction occurs when sodium phenoxide reacts with carbon dioxide at four to seven atmospheres at approximately 398 K to produce disodium salicylate. Disodium salicylate is acidified with mineral acid to produce salicylic acid (ortho-hydroxybenzoic acid).

Phenol reacts with sodium hydroxide to produce sodium phenoxide:

The formation of salicylic acid takes place in two steps. First, the formation of disodium salicylate, and the second step is the acidification of disodium salicylate to form salicylic acid.

62

(1)

disodium salicylate

(2)

Salicylic Acid

The following two steps rationalize the formation of disodium salicylate acid.

63

(1)

(2)

The sum of the series of elementary steps (equations 1 and 2) gives the following balanced equation.

64

The Reimer-Tiemann Reaction The Reimer-Tiemann Reaction is a reaction limited to phenol. The reaction involves treating phenol with chloroform in sodium hydroxide.

The product of reacting phenol with chloroform in base is salicylaldehyde. The salicyladehyde forms by an interesting mechanism.

The following schema explores this mechanism: (1)

The net ionic reaction is

65

(2)

(3)

(4)

66

(5)

(6)

(7)

67

(8)

The sum of steps 1-8 gives the following net ionic equation.

The molecular form of the net ionic equation is:

68

Nitrosation If the aromatic ring is activated, then nitrosation occurs at the ortho or para positions.

For example, phenol will react with nitrous acid to produce p-nitrosophenol and onitrophenol.

Aromatic compounds containing substituents that activate aromatic nuclei play a pivotal role in facilitating the attack on the electrophile as indicated by step 3 of the mechanism that explains the formation of p-nitrosophenol as the product of the reaction between phenol and nitrous acid.

(1)

(2)

69

(3)

(4)

The sum of steps 1-4 gives the following net ionic equation.

The molecular form of the above chemical equation is:

70

Protonolysis Aromatic nuclei can undergo protonolysis reactions if the sulfonic acid group is attached to the ring. Prontonolysis allows for the possibility of adding deuterium atoms to the aromatic nucleus. Deuterium in the place of hydrogen can be useful in a variety of investigations. The following are equations demonstrating the protonolysis of aromatic ring systems.

We previously discussed the series of elementary steps that rationalize the formation of the protonated aromatic ring from aryl sulfonic acids. A similar mechanism can be written to show the formation of deuterobenzene using D3O+.

71

(1)

(2)

(3)

72

(4)

(5)

The sum of steps (1) to (5) is

Subjecting monosubstituted aromatic compounds such as methylbenzene (toluene) to electrophilic aromatic substitution reactions, only three disubstituted products are possible- the ortho product, the meta product, and the para product:

73

where x + y + z = 1 In this case, (x +z) >> y The sulfonation of toluene gives rise to all three monosubstituted compounds, but produces only the ortho and para compounds in appreciable quantities. From the experiment, the methyl group on benzene directs the attacking electrophile primarily to the ortho and para positions.

Toluene is more reactive toward electrophilic aromatic substitution than benzene.

Toluene is sulfonated by fuming sulfuric acid twenty-thirty times faster than benzene is sulfonated by fuming sulfuric acid. These data indicate that the methyl group activates the benzene ring toward electrophilic aromatic substitution reactions.

On the other hand, sulfonation of benzenesulfonic acid gives primarily the meta disulfonic acid as a product.

74

where x + y + z = 1 and (x +z) << y Sulfonation of benzenesulfonic acid requires a higher temperature than the sulfonation of benzene. Sulfonation of benzene sulfonic acid is directed meta to the sulfonic acid group and the sulfonic acid group deactivates the benzene toward electrophilic aromatic substitution reactions.

Electrophilic aromatic substitution reactions with monosubstituted benzenes can lead to meta or ortho-para products. Also, the substituents attached to the benzene ring may activate or deactivate the benzene ring.

In addition to â&#x20AC;&#x201C;SO3H, Table 9.5 lists deactivating groups which are meta directors.

75

Table 9.5 deactivators and meta directors

The methoxy group in phenyl methyl ether, compound X, directs groups to the ortho

76

and para positions as the major products.

The sulfonation of phenyl methyl ether, compound X, requires a lower temperature than the sulfonation of benzene. Therefore, the methoxy group activates the benzene ring toward electrophilic aromatic substitution reactions, and the methoxy group is an ortho/para director.

In addition to â&#x20AC;&#x201C;OCH3, Table 9.6 lists strongly activating groups which are ortho-para directors.

Table 9.6 strong activators and ortho-para directors

Table 9.7 lists moderate activating and ortho-para directors

77

Table 9.7 moderate activating groups

Phenyl, C6H5 and alkyl groups, e.g., CH3, CH3CH2 groups, are weakly deactivating.

A limited number of ortho-para directors also deactivate the aromatic ring. These include the halogens, F, Cl, Br and I. Resonance provides a rationalization for why some groups direct meta and others ortho-para. For example, a partial mechanism for the sulfonation of benzenesulfonic acid can demonstrate that effect. Assume that the ortho product is produced. The intermediate formed in the production of the ortho product would include the following resonance structures 1,2, and 3.

Resonance structure â&#x20AC;&#x153;3â&#x20AC;? is particularly unstable because electrons are flowing away 78

from an electronegative S atom. The para substituted intermediate (4, 5, and 6) would exhibit the same behavior.

Resonance structure â&#x20AC;&#x153;5â&#x20AC;? is unstable.

The intermediate for the meta product does not have a structure where the positive charge is on a carbon atom which has an S atom attached. Therefore, there is a preference for the meta compound since its intermediate has resonance stabilized structures which are more stable than those for the ortho and para possibilities.

In structures 7, 8, and 9, no resonance structure has a positive charge on a carbon atom with an attached S atom.

Similar arguments, but with opposite conclusions can be used to explain the preference of the ortho and para products for activators.

79

Oxygen has a lone pair of electrons that could participate in resonance.

Unlike the atoms that are meta directors, except for alkyl groups, the ortho/para directors have lone pairs that can delocalize the incipient positive charge of the benzenonium cation as observed in structures 14 and 15, and 19 and 20.

80

Resonance structures 14 and 19 have oxygen atoms with lone pairs adjacent to carbon atoms with positive charges. The positive charge is delocalized on the oxygen atom.

These resonance structures contribute to the overall stabilization of the ortho and para products.

Aromatic Nomenclature Derivatives of benzene are generally named as derivatives of benzene. For example,

is nitrobenzene

is ethyl benzene

is bromobenzene

81

A few aromatic compounds like phenol, toluene, benzoic acid, and benzenesulfonic acid use their common names more frequently than their IUPAC names.

Two substituents on the benzene ring have names that use the terms ortho, -o; para, p; and meta, -m.

82

Electrophilic Aromatic Substitution of Disubstituted Aromatic Compounds Chlorination of o-nitrobenzenesulfonic acid results in the formation of compounds XI, XII, XIII, or XIV.

in the chlorination of o-nitrobenzenesulfonic acid, compounds XI and XIV, 3-chloro-2nitrobenzenesulfonic acid and 2-chloro-6-nitrobenzenesulfonic acid have close attached substituents. Therefore, these two compounds are minor products. The crowding of substituents on compounds XII and XIII, 4-chloro-2-nitrobenzenesulfonic acid and 5-chloro-2-nitrobenzenesulfonic acid are not close and, therefore, are major products.

Electrophilic aromatic substitution of compound XV, 4-cyano-N-acetylaniline, leads primarily to one trisubstituted benzene product, compound XVI.

83

Compound XVI is the primary product because the CN group is a meta director and the HNCOCH3 group is an ortho-para director. The reinforcement of the directing influences of CN and HNCOCH3 explains the exclusive formation of compound XVI. Electrophilic aromatic substitution on compound XVII, p-methylphenol, would lead to two possible products, compounds XVIII and XIX since the OH and the CH3 groups are both ortho-para directors.

84

The OH group is a stronger activating group than the CH3; therefore the OH group has more influence in directing the incoming electrophile than the methyl group. The predominate product is compound XIX since the OH group is a stronger ring activator (resonance stabilizes the resulting benzenonium cation intermediate) than the CH3 group, stabilized by hyperconjugation, is a weaker activator.

Compound XX, p-nitro-N-acetylaniline, will undergo bromination to produce one predominate product, 2-bromo-4-nitro-N-acetylaniline.

XX

The nitro group of compound XX is a meta director and the N-acetyl group of compound XX is an ortho-para director. Therefore, the two groups reinforce one another in such a manner that the predominate product is 2-bromo-4-nitro-Nacetylaniline.

85

Compound XX1, o-hydroxybenzaldehyde, reacts with bromine to produce 5-bromo-2hydroxybenzaldehyde as the predominate product.

The -OH group of compound XXI is an ortho-para director and the â&#x20AC;&#x201C;CHO is a meta director; therefore, the two groups reinforce one another to give 2-hydroxy-5bromobenzaldehyde as the predominate product.

In synthesizing certain aromatic compounds, a careful selection of reagents and the sequence will minimize unwanted isomers. For example, the following sequence of reactions prepare p-bromonitrobenzene from benzene with minimum production of isomers:

This sequence would produce p-bromonitrobenzene and o-bromonitrobenzene. If nitration is done first, then the major product would be m-bromonitrobenzene.

Consequently, the schema for the sequence of reactions is important in determining the outcome of products.

86

The major product would be m-bromonitrobenzene.

The details of the general mechanism for ortho-para orientation of aromatic activators attached are the following series of elementary steps:

First, letâ&#x20AC;&#x2122;s explain the role of the activator. The activator, substituent on the aromatic nucleus, may have a lone pair of electrons on the atom adjacent to the aromatic carbon atom. As indicated earlier, this makes for resonance stabilization that is superior to the resonance structures that do not have electrons attached to atoms adjacent to the aromatic carbon atom. Generally, if the atom adjacent to the aromatic nucleus does not contain a lone pair of electrons, the group is a deactivator. Alkyl groups and halogens are exceptions to this observation. Alkyl groups are weak activators and they release electrons into the aromatic systems via hyperconjugation.

Halogens are electron releasing, but they deactivate aromatic nuclei by induction.

The following is a general mechanism for activators attached to the aromatic nucleus undergoing electrophilic aromatic substitution. E+ is the electrophilic reagent generated by a Lewis acid-base. For example, iron(III) bromide, a Lewis acid, can react with bromine, a Lewis base, to form the bromonium ion, the conjugated Lewis acid, and the electrophilic reagent, and tetrabromoferrate, tetrabromoiron(III) anion (the conjugate Lewis base).

87

FeBr3 Lewis Acid

+

Br2 Lewis Base

â&#x2020;&#x2019;

FeBr4-

+

Br+

Conjugate Lewis Base Conjugate Lewis Acid and electrophilic reagent, E+

Step 1 is the formation of the first pi-complex, and the pi-complex generates the benzenonium ion in a second step. These are generally the slow steps of the mechanism.

(1)

(2) Slow step

The benzenonium intermediate is resonance stabilized. The benzenonium intermediate is more stable when the electrophile is in the ortho or para positions because the incipient positive charge is more delocalized in these positions than in the meta position. This is illustrated by the following resonance structures.

88

Step 3 is the generation of a second pi-complex, and step 4 is attack of a conjugate Lewis base on the pi-complex generating the disubstituted aromatic nucleus.

(3)

The para-product

89

(4)

The â&#x20AC;&#x153;pi-complexesâ&#x20AC;? maintain the integrity of the aromatic systems.

An analogous argument can be made for the ortho product. The benzenonium intermediate is more stable when the electrophile is in the ortho or para positions, because the incipient positive charge is more delocalized in these positions than in the meta position. There are four resonance structures for the benzenonium carbocation when the electrophile is in the ortho or para positions compared to three when the electrophile is in the meta position. This is illustrated by the following resonance structures.

Step 1

90

Step 2: Slow step

Either carbon with hydrogen atoms in the ortho positions (indicated in green) can receive the electrophilic reagent. The benzenonium intermediate is resonance stabilized. The benzenonium intermediate is more stable when the electrophile is in the ortho or para positions, because of the greater delocalization of the incipient positive charge in the para and ortho positions as compared to the meta position. There are fewer resonance structures for the meta substituted product than for the ortho or para substituted products.

91

Step 3

(4)

ortho product

On the other hand, the meta product would generate fewer resonance structures to stabilize the benzenonium carbocation intermediate.

92

(1)

(2) Slow step

There are three resonance structures for the benzenonium intermediate when the electrophile is in the meta position. Therefore, the meta isomer is not as stable and would constitute the minor product in the electrophilic aromatic substation of aromatic systems containing only an activator on the aromatic ring.

93

(3)

(4)

meta

Deactivators on the Aromatic Nucleus Deactivators on the aromatic ring have the opposite effect as activators. Except for

94

the halogens, deactivators do not have lone pairs on atoms adjacent to the aromatic nucleus; therefore, the resonance structure with a positive charge on a carbon atom of the substituted benzenonium carbon atom attached to the deactivator is unfavorable.

Pathways exhibiting such a situation are not favored. To alleviate this problem, deactivators would prefer to form meta substituted compounds rather than para or ortho substituted compounds.

(1)

(2)

(3)

95

For deactivators, electrophiles in the ortho and para position lead to resonance structures that are unfavorable.

The para product (1)

(2)

An unfavorable resonance structure

(3)

96

The ortho product (1)

(2)

An unfavorable resonance structure

(3)

The series of elementary steps suggests that the meta substituted aromatic compound

97

predominate when deactivators are attached to the aromatic ring.

Also, the mechanism suggests that the ortho and para products predominate when activators are attached to the aromatic ring undergoing electrophilic aromatic substitution reactions.

98

Problems

1.

Use the data in Table 9.3 to calculate the resonance stabilization energy for benzene.

2.

Which of the following compounds are aromatic?

99

3.

Name the following compounds:

a.

Cl

c.

Cl

e.

C

Br

CH3

CH2CH3

CH3

O b.

O2N

NO2

d.

NH2

f.

OCH3

C

NO2 CH3

OH

O

100

4.

Arrange the following compounds in order of increasing acidity:

a

OH

b

OH

OH

c

Br

CH2CH3 OH f

e

OCH3

d

OH

g

OH

NO2

OH

h

OH

NO2

N

NO2

CH3

O- Na +

H

5.

Explain the following observations:

+

No Reaction

Br2/CCl4

+

+

No Reaction

HBr

KMnO4

273 K

No Reaction

101

6.

An unknown aromatic organic compound C6H4Cl2 upon treatment with HNO3/H2SO4 gave only one product. What is the structure of the aromatic compound?

7.

Draw all possible isomers for trimethylphenol.

8.

Suggest a mechanism for the following hypothetical reaction and construct an energy profile diagram for the reaction.

H +

GY

G

Z

+

HY

+

Z

9.

Using the information you have read concerning electrophilic aromatic substitution reactions, suggest products for the following reactions: a. +

+

Hg(O2CCH3)2

H

b.

+

DCl

AlCl3

102

c.

O +

Tl(OCCF3)3

CF3COOH

10.

Suggest syntheses for the following from benzene and any other inorganic or organic reagents.

a

c

CH2CH2CH2CH3

b

CH2OH

OH

d

CH3

NO2

e

O

COOH O

C

f

F

CH3 Br

103

g

h

COOH

D

NO2

11.

Give an explanation for the following observation: N(CH3)2

N(CH3)2

+

CH3Cl

AlCl3

N(CH3)2

CH3 +

CH3

12.

Give a rationale for the following observation: Nitration of t-butylbenzene produces 16% of the ortho product and 80% of the para product; however, nitration of methylbenzene, toluene, produces 58% of the ortho product and 38% of the para product.

13.

Indicate using an arrow the position(s) most likely to undergo electrophilic aromatic substitution in each of the following compounds: (a) o-xylene (b) m-nitrotoluene

(c) p-bromotoluene

(3) o-bromotoluene (f) m-chloronitrobenzene

(d) m-methylphenol

(g) m-xylene

104

14.

Draw structures for the principle mononitration products of: OCH3

a

H

O

N

C

c

b CH3

COOH

CH3

SO3H d

e

COOH

OCH2CH3

f

CH3O

Cl CH3

15.

Suggest a mechanism for the following reaction:

CH3

CH3 CH3

H3C

H3C +

HF

H

CH3

CH3 C

+

C

H H3C

C

CH3

CH3

CH3

16.

One mole of compound A, C16H16, reacts with Br2/CCl4, dilute KMnO4, and one mole of H2/Pt. Treatment with KMnO4 gives a dicarboxylic acid which has only one monobromo substitution product. Ozonolysis of compound A gives a compound that has two monobromo substituted products. What is the structure of A.?

105

17.

Supply structures for the following pathway: C7H8 + Br2 → A + Mg + dry ether → C7H7MgBr C7H7MgBr + D2O → C7H7D C7H7D + H2SO4 → C7H6DSO3H C7H6DSO3H + D2O → C7H6D2

18.

Suggest a synthesis for

CH2

C

from

C

C

O

19.

Give an explanation for the fact that the sulfonation reaction in aromatic substitution reactions is reversible.

106

Solutions to Problems Chapter 9

1.

Use the data in Table 9.3 to calculate the resonance stabilization energy for benzene.

Resonance Energy for Benzene = 3 ΔH hydrogenation of cyclohexene − ΔH benzene

REB = 3 x 120 kJ/mol − 208 kJ/mol = 152 kJ/mol 2.

Which of the following compounds are aromatic? (a) not aromatic (b) not aromatic (c) not aromatic (d) aromatic (e) not aromatic (f) aromatic (g) not aromatic (h) not aromatic

107

3.

Name the following compounds: The following names are not exhaustive for these compounds,

i.e.,

other names are also acceptable. a. (a)

3-chlorobromobenzene or m-chlorobromobenzene

b. (b)

1,3,5-trinitrobenzene or picric acid

c. (c)

3-chloroacetophenone or m-chloroacetophenone

d. (d)

4-methylaniline or p-methylaniline

e. (e)

3-ethyltoluene or m-ethyltoluene

f. (f) 3-methoxybenzoic acid or m-methoxybenzoic acid

4.

Arrange the following compounds in order of increasing acidity: f>d>a>b>c>g>e>h

108

5.

Explain the following observations:

+

No Reaction

Br2/CCl4

+

+

No Reaction

HBr

KMnO4

273 K

No Reaction

Benzene is aromatic, i.e., the electrons in the six â&#x20AC;&#x153;2pâ&#x20AC;? parallel atomic orbitals perpendicular to six coplanar carbon atoms interact in such a manner that the resulting structure does not consist of isolated double bonds, but a structure that is stabilized by forming three bonding molecular orbitals and 3 antibonding molecular orbitals. This aromatic structure does not undergo addition reactions.

6.

An unknown aromatic organic compound C6H4Cl2 upon treatment with HNO3/H2SO4 gave only one product. What is the structure of the aromatic compound?

109

7.

Draw all possible isomers for trimethylphenol.

8.

Suggest a mechanism for the following hypothetical reaction and construct an energy profile diagram for the reaction.

Where Z is a catalyst 1.

GY + Z â&#x2020;&#x2019; YZ â&#x2C6;&#x2019; + G +

110

2.

3.

4.

(5)

(6) HYZ ➝ HY + Z

111

9.

Using the information you have read concerning electrophilic aromatic substitution reactions, suggest products for the following reactions: a.

112

b.

c.

10.

Suggest syntheses for the following from benzene and any other inorganic or organic reagents. a.

113

b.

c.

114

d.

e.

115

f.

ammonium polysulfide will reduce one nitro group when two nitro groups are attached to an aromatic ring.

116

9.

117

h.

11.

Give an explanation for the following observation:

The ortho and para products predominate, because the methyl group assists in the delocalization of the positively charge benzenonium intermediate through hyperconjugation. Meta attack would not allow the methyl group to participate in the delocalization of the positive charge.

118

and

119

12.

Give a rationale for the following observation: Nitration of t-butylbenzene produces 16% of the ortho product and 80% of the para product; however, nitration of methylbenzene, toluene, produces 58% of the ortho product and 38% of the para product.

The t-butyl group sterically inhibits attack at the ortho position; therefore, the para product would be favored. This is not the case for the smaller methyl group; therefore, statistically, the ortho product would be favored since there are two sites available for the formation of the ortho product and one site available for the formation of the para product.

13.

Indicate using an arrow the position(s) most likely to undergo electrophilic aromatic substitution in each of the following compounds:

120

(a) o-xylene (b) m-nitrotoluene

(c) p-bromotoluene

(a) o-bromotoluene (f) m-chloronitrobenzene

(d) m-methylphenol

(g) m-xylene

(a)

b.

The following is theoretically possible, but the groups would be crowded; therefore, it would be a minor product due to steric crowding:

121

c.

d.

The following is theoretically possible, but the groups would be crowded; therefore, it would be a minor product due to steric crowding:

(e)

The following is theoretically possible, but the groups would be crowded; therefore, it would be a minor product due to steric crowding:

122

(f)

The following is theoretically possible, but the groups would be crowded; therefore, it would be a minor product due to steric crowding:

(g)

The following is theoretically possible, but the groups would be crowded; therefore, it would be a minor product due to steric crowding:

123

14.

Draw structures for the principle mononitration products of: (a)

(b)

(c)

124

(d)

(e)

(f)

125

15.

Suggest a mechanism for the following reaction:

1.

2.

126

3.

4.

16.

One mole of compound A, C16H16, reacts with Br2/CCl4, dil KMnO4, and one mole of H2/Pt. Treatment with hot concentrated KMnO4 gives a dicarboxylic acid which has only one monobromo substitution product. Ozonolysis of compound A gives a compound that has two monobromo substituted products. What is the structure of A.?

Following is one possible solution. Can you think of another?

1. reaction with bromine in carbon tetrachloride:

127

2. reaction with dilute potassium permanganate

(a) Reaction with one mole of hydrogen:

4. Reaction with concentrated hot potassium permanganate in acidic solution.

128

1. Ozonolysis

17.

Supply structures for the following pathway: C7H8 + Br2/light → A + Mg + dry ether → C7H7MgBr C7H7MgBr + D2O → C7H7D C7H7D + H2S2O7 → C7H6DSO3H C7H6DSO3H + D2O → C7H6D2

129

statistically anticipated yields

130

18.

Suggest a synthesis for

CH2

C

from

C

C

O

19. `Give an explanation for the fact that the sulfonation reaction in aromatic substitution reactions is reversible.

The energy barrier from reactants to product is the same as the energy barrier for products to reactants:

131

132