We have seen that benzene, C6H6, undergoes electrophilic substitution reactions when treated with a variety of Lewis acids. This reaction is typical of the aromatic ring, and derivatives of benzene, C6H5G, behave similarly. In this topic we will consider the effect that a substituent group, G, has on the rate of electrophilic aromatic substitution. In other words, we will consider the question "If you replace one of the Hs in benzene with another atom or group, will the resulting compound, C6H5G, react faster or slower than C6H6 ?" Related questions are "Is the group G electron donating or electron withdrawing?" and "Is the group G an activating group or a deactivating group?" The information required to answer these questions comes from two sources, comparison of the results of competition reactions in which two reactants compete for the same electrophile, and 1H-NMR chemical shifts.
A simple way to assess the effect of a substituent on the rate of electrophilic aromatic substitution is to react an equimolar mixture of C6H6 and C6H5G with an electrophile and to measure the relative rates at which the two reactants disappear. This might be done, for example, by GC or NMR or IR, depending upon the reaction conditions. Table 1 summarizes the results of such competition experiments for 15 different substituents. The reaction is described in general terms as shown in Figure 1. In this type of experiment it is not necessary to know the position of E relative to G in the product since you measure the rate of disappearance of starting materials. We will consider the position of E when we discuss Substituent Effects: Orientation.
Where specific values of krel are given, they refer to relative rates of nitration, i.e. E+ = NO2+. Otherwise, the identity of E+ is not specified. The table also summarizes the chemical shifts of the protons ortho, meta, and para to the substituent group, G.
1 Adapted from Organic Structural Analysis, J.L. Lambert, H.F. Shurvell, L. Verbit, R.G. Cooks, and G.H.Stout, McMillan, 1976.
Exercise 2 Identify those substituents where the atom bonded directly to the ring has a lone pair of electrons:
Exercise 3 The chemical shift of the protons in C6H6 is 7.27 ppm. The chemical shifts of the o, m, and p hydrogens in aniline, C6H5NH2, are 6.52, 7.03, and 6.64 ppm, respectively. Is the electron density around the o, m, and p hydrogens in aniline higher or lower than it is around the hydrogens in C6H6? If you attribute the change in electron density to the replacement of an H by an NH2 group, would you classify the NH2 group as electron donating electron withdrawing
Exercise 4 Compare the chemical shifts of the o, m, and p hydrogens for the groups you identified in Exercise 1 with the value of 7.27 for benzene. Identify those where the electron density is higher at all three positions.
Exercise 5 Identify any other groups where the electron density is higher at all three positions.
Exercise 6 Identify those groups where the electron density is higher at some positions but lower at others.
Exercise 7 Identify those groups that reduce the electron density around the o, m, and p hydrogens. These groups are all electron withdrawing groups. Are they activating or deactivating in terms of their effect on electrophilic aromatic substitution reactions?
The chemical shift data in Table 1 provides a direct measure of the electron density in each aromatic ring: the higher the electron density, the lower the value of the chemical shift; the lower the electron density, the higher the chemical shift. The change in electron density is attributed to the net electron donating or electron withdrawing ability of the substituent G. The word net is emphasized because two factors contribute to the electronic effect of a group. One factor is the group's inductive effect. The other factor is its resonance effect.
The inductive effect of an atom or group is related to its electronegativity. Inductive effects are transmitted through s bonds. Consider the series of s bonds shown in Figure 2.
The d+ d- symbolism indicates the electron distribution between the s-bonded atoms. In the case of a C-H bond, the electron density is higher in the vicinity of the carbon than the hydrogen because carbon is more electronegative than hydrogen. Relative to H, the inductive effect of C is electron withdrawing. To a first approximation a C-C bond is non-polar because there is no difference in electronegativity between the two carbons. The polarities of the C-N, C-O, and C-F bonds are all such that the electron density is lower around carbon. Relative to carbon, the inductive effects of N, O, and F are all electron withdrawing. Finally, in the C-Si bond, the electron density is higher around carbon than it is around silicon. Relative to Si, the inductive effect of C is electron withdrawing. Conversely, the inductive effect of Si is electron donating relative to C.
(CH3)4Si CH3F State a correlation between the inductive effects of F and Si and the chemical shifts of the methyl protons in (CH3)4Si and CH3F.
Exercise 9 What is the only atom whose inductive effect will always be electron withdrawing?
Exercise 10 One way to think about electronegativity and inductive effects is to consider the ratio of the number of protons to the number of electrons for each atom of a bonded pair. The greater the ratio, the more stable the electrons around that atom will be. Consider this example:
Both atoms have a filled valence shell. However, the carbon has only 6 protons to attract the 8 electrons in its valence shell, a ratio of 6/8 or 0.750 protons per electron. For the nitrogen atom, the corresponding ratio is 7/8 or 0.875 protons per electron. The higher value of the proton/electron ratio means that the inductive effect of the nitrogen is electron withdrawing relative to carbon. Calculate the proton/electron ratio for an oxygen atom in a C-O bond and a fluorine atom in a C-F bond .
Exercise 11 Consider the following Sn1 type of reaction: . In the reactant the proton/electron ratio for the carbon atom is 0.750, while that for the fluorine is 1.125. The difference in these ratios, (1.125-0.750) = 0.375, provides a measure of the electron withdrawing effect of a fluorine compared to that of a carbon. What is the proton/electron ratio for the positively charged carbon in the product of the reaction? What is the difference between this ratio and the proton/electron ratio for the fluorine? Is the inductive effect of the fluorine constant? Yes No
Data like that in Table 2 has been used to establish a list of group electronegativities comparable to the single atom electronegativities mentioned earlier. Note that there is no possibility for a resonance interaction between the methyl group and the substituent G for the compounds listed in Table 2.
Exercise 11 Relative to H, the inductive effect of all of the groups listed in Table 1 is electron withdrawing electron donating
Unlike inductive effects which are transmitted through s bonds, resonance effects involve non-bonding electrons and pi electrons. Since neither lone pairs nor pi electrons are held as tightly to the nucleus as s electrons, resonance effects are generally more important than inductive effects, provided, of course, that the structural requirements for resonance are met. Like inductive effects, resonance effects may be electron donating or electron withdrawing.
According to the data in Table 1, nitration of phenol, C6H5-OH, occurs 1000 times faster than nitration of benzene, C6H5-H. The increase in reaction rate suggests that the electron density in the aromatic ring is higher in phenol than it is in benzene. In other words, phenol is more basic, more nucleophilic than benzene. The NMR data in Table 1 also indicates that the electron density is higher in the aromatic ring of phenol than it is in benzene. Presumably the presence of the oxygen atom is responsible for the increase in electron density in the ring. Figure 3 presents the standard resonance description of the electron donating effect of the OH group in phenol.
The orbital containing one of the lone pairs of electrons on the oxygen atom aligns with the p orbital on the adjacent carbon in such a way as to maximize orbital overlap between these two atoms. When that happens, electron density can be transferred from the oxygen into the pi system of the aromatic ring as shown by the red arrows in the figure. As the oxygen shares the lone pair of electrons with its neighboring carbon, it acquires a positive charge. As the electron density between the C-1 and the O increases, that between the C-1 and the C-2 decreases. Otherwise the octet rule would be violated. The electrons that C-1 shared with C-2 become a lone pair on C-2 as shown in resonance structure 2. The formal charge on C-2 changes from 0 to -1. This represents an increase in electron density at C-2. A similar orbital interaction involving C-2, C-3, and C-4 transfers the negative charge from C-2 to C-4 as indicated in structure 3. The electron density at C-6 increases likewise. The electron donation by the oxygen atom increases the electron density of the entire pi system of the aromatic ring. As resonance structures 2,3, and 4 make clear, however, the increase is greatest at C-2, C-4, and C-6, the positions ortho and para to the oxygen.
At the outset of this topic we mentioned that the change in electron density in the pi system of an aromatic ring is attributed to the net electron donating or electron withdrawing ability of the substituent G. In the case of phenol, the inductive effect of the OH group is electron withdrawing. We conclude that the resonance effect must be electron donating, and that it outweighs the inductive effect because the experimental evidence indicates clearly that the electron density is higher in the pi system of phenol than it is in benzene.
Consider the last 6 entries in Table 1: G= -CN, -CHO, -C(=O)CH3, -CO2H, -CO2 CH3, and -NO2. They all share three important structural features- 1. The atom bonded to the aromatic ring does not have a lone pair of electrons. 2. The atom bonded to the aromatic ring is unsaturated; it is either sp or sp2 hybridized. In either case it has an unhybridized p orbital that can overlap with the pi system of the aromatic ring. 3. The atom directly connected to the aromatic ring is also multiply-bonded to at least one electronegative atom.
The lack of a lone pair combined with the presence of an electronegative bonding partner gives rise to the electron withdrawing potential of each of these groups. That potential is realized when the p orbitals of the multiple bond of the group align with the p orbitals of the aromatic ring so as to allow extended orbital overlap as shown in Figure 4 for the specific case of benzonitrile, C6H5-CN.
The implication of the arrow labeled 1 is that the nitrogen draws electron density away from the carbon, making the carbon electron deficient as indicated by the positive charge in resonance structure 2. Electron density from the pi system of the aromatic ring is drawn toward the developing positive charge as shown by the arrow labeled 2. This reduces the electron density at C-2 as can be seen in structure 3. Structures 4 and 5 show that the decrease in charge is transferred from C-2 to C-4 and on to C-6. The electron withdrawl by the cyano group decreases the electron density of the entire pi system of the aromatic ring. As resonance structures 3,4, and 5 suggest, however, the decrease is greatest at C-2, C-4, and C-6, the positions ortho and para to the cyano group.
Exercise 13 Following the example in Figure 4, show how -CHO, -C(=O)CH3, -CO2H, -CO2 CH3, and -NO2 groups withdraw electron density from the pi system of an aromatic ring.
The chemical shift data in Table 1 indicate clearly that a methyl group, and by extension, alkyl groups in general, are electron donating groups. How does this happen? In a word, hyperconjugation. Compare the orbital pictures of toluene and aniline shown in Figure 5. In aniline, as the red arrow indicates, the nitrogen atom can donate electron density to the ring when dihedral angle between the orbital containing the lone pair of electrons and the p orbital on the aromatic ring is 0o. A similar geometry is required for one of the C-H bond orbitals of the methyl group in toluene. Despite the fact that there are three conformations where this arrangement is possible in toluene, s-bonded electrons are less readily delocalized than their non-bonded counterparts, so that the transfer of electron density into the ring is less in toluene than it is in aniline.
Figure 6 compares the results of electron donation implied by the colored arrows in Figure 5. Note that the donation of a s-bonded pair results in the cleavage of the C-H bond. This is energetically unfavorable, which simply means that it does not occur to the extent that donation of the non-bonding pair on the nitrogen atom in aniline does. That's what the NMR data in Table 1 is saying: the chemical shift of the protons ortho to the amino group in aniline is 6.52 ppm compared to a value of 7.10 ppm in toluene.
In Exercise 11 we considered the idea that the difference in the proton/electron ratios of two bonded atoms was related to the inductive electron withdrawing effect of one bonded partner relative to the other. We saw that the proton/electron ratio could change during a chemical reaction. The same idea applies to resonance effects: the electron withdrawing or electron donating tendency of a group will change in response to changes in the electron density in the pi system of the aromatic ring during an electrophilic aromatic substitution reaction.
Let's consider the bromination of aniline. For the purposes of this argument we will focus on the reaction ortho to the amino group. During this reaction a cyclohexadienyl cation intermediate is formed in which the positive charge is initially localized on the carbon atom to which the amino group is attached (This is called the ipso carbon.). What is the impact of the development of a positive charge on the tendency of the amino group to share its lone pair with the ipso carbon? Figure 7 compares the resonance interaction between the lone pair of electrons on the nitrogen atom with the pi system of the ring in the starting material (A-1<->A-1') with that in the intermediate (I-1 <-> I-1').
Note in the resonance interaction A-1<->A-1' that the resonance contributor A-1' bears two opposite charges: resonance structure A-1' contributes less to the overall structure of aniline than does structure A-1. Remember, we're postulating this resonance interaction as a way of increasing the electron density in the aromatic ring. We offer such a postulate because 1. aniline reacts with dibromine faster than benzene 2. the chemical shifts of the aromatic protons in aniline all occur at higher fields than they do in benzene.
Now consider the I-1<->I-1' resonance interaction. Note that the positively charged carbon in resonance structure I-1 does not satisfy the octet rule. On the other hand, all of the atoms in resonance contributor I-1' have a filled valence shell. This means that the true structure of the cyclohexadienyl cation intemediate resembles resonance structure I-1' more than it does I-1. In other words, donation of electron density from the nitrogen into the ring occurs to a greater extent after the ring is brominated than before. The magnitude of the resonance effect is not constant. In a sense the nitrogen atom's willingness to share its lone pair of electrons depends upon the demands placed upon it .
Figure 8 compares the reaction profile diagram for the bromination of benzene to that for the bromination of aniline. The difference in the levels of the lower dashed blue line and the lower dashed red line is a measure of the resonance stabilization that the amino group affords the pi system of the aromatic ring relative to a hydrogen atom. The difference in the levels of the upper dashed blue line and the upper dashed red line is a measure of the stabilization that the amino group affords the pi system of the transition state leading to the cyclohexadienyl cation intermediate. Because the amino group donates electron density into the ring more after the positive charge has developed than before, it stabilizes the intermediate cation more than the starting material. Increased stabilization of the intermediate is accompanied by increased stabilization of the transition state, which lowers Ea relative to Ea.
The data in Table 1 suggest that substituent groups may be divided into two classes:
Most activating groups have at least one lone pair of electrons on the atom directly bonded to the aromatic ring. Alkyl groups are an exception.
Aromatic rings bearing an activating group undergo electrophilic aromatic substitution reactions faster than benzene.
Most deactivating groups have a formal positive charge or a partial positive charge on the atom directly bonded to the aromatic ring. Halogen atoms are an exception.
Aromatic rings bearing deactivating groups undergo electrophilic aromatic substitution reactions slower than benzene.
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