Introduction-
In benzene, C6H6, all six hydrogen atoms are identical. Consequently replacement one hydrogen by an electrophilic reagent produces the same product as replacement of any other. In monosubstituted benzenes, C6H5G, the five aromatic hydrogens are no longer identical. Rather, they comprise three distinct sets of hydrogens; the two hydrogens ortho to G, the two hydrogens meta to G, and the single hydrogen para to G. Consequently, treatment of C6H5G with an electrophilic reagent "E+" may produce three isomeric substitution products, o-EC6H4G, m-EC6H4G and p-EC6H4G. These possibilities are diagrammed in Figure 1.
Figure 1
Consider the Possibilities
Evidence from a large number of electrophilic aromatic substitution reactions has shown that the postion occupied by the electrophile, E, is determined by the electronic character of G. This is called the substituent effect. There are two cases:
- G directs "E+" to positions ortho and/or para to itself. Groups that behave in this way are called an ortho/para directors.
- G directs "E+" to the position meta to itself. Groups that behave in this way are called an meta directors.
ortho/para Directors
Table 1 summarizes the majority of groups that are classified as ortho/para directors.
Table 1
You, Go There. You, Go There
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Except for the alkyl groups, it is apparent that there is a lone pair of electrons on the substituent atom directly attached to the aromatic ring. Resonance theory suggests that an interaction between the substituent and the pi system of the aromatic ring increases the electron density in the ring, and that it does so more at the positions ortho and para to the substituent that at the position meta to it. Consequently the electrophilic reagent, "E+", bonds to the ortho and para positions in preference to the meta position. Equations 1-4 provide typical product distributions for reactions with various electrophiles.
meta Directors
Table 2 lists most of the groups that are classified as meta directors.
Table 2
You, Go There
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Consideration of the examples in Table 2 reveals that for every substituent, the atom directly bonded to the aromatic ring either bears a formal positive charge or a partial positive charge. These groups are classified as electron withdrawing groups. They reduce the pi electron density in the ring. However, they lower it less at the position meta to G than they do at the positions ortho and para. Note the difference between the activating group -OC(=O)CH3 and the deactivating group -C(=O)OCH3. There is a similar difference between -NHC(=O)CH3 and -C(=O)NHCH3.
Equations 5-7 describe typical product distributions in nitration reactions involving three different electron withdrawing substitutents.
Exercise 1 Draw the product(s) you would expect in each of the following reactions. Indicate whether the electrophile will bond to the ortho/para positions or the meta position by entering op or m into the appropriate text field.
A. B. C. D. E. F.
Rates vs Orientation
Is there a correlation between a substituent's effect on the rate of electrophilic aromatic substitution and the orientation of the electrophile? Comparison of the data in Table 1 of Substiuent Effects: Relative Reactivities with the data in Table 1 above indicates clearly that there is. At least for the cases where G = R, NR2, and OR, those groups that activate the aromatic ring also orient the incoming electrophile to the ortho and para positions in preference to the meta position. Conversely, deactivating substituents direct the incoming electrophile to the meta position. These correlations are rationalized in terms of the effect that the substituent has on the activation energies for the formation of the cyclohexadienyl cation intermediate. The assumption is that the more stable intermediate will be formed more readily. Remember, when we talk about substituent effects, our reference substituent is a hydrogen atom.
Figures 2 presents an animated version of the resonance picture of the cyclohexadienyl cation intermediate that is formed in the chlorination of benzene. Notice the pattern here: As the electrophilic chlorine bonds to C-2, the formal positive charge develops at C-3. In response, electron density is drawn from the C-4-C-5 bond towards C-3, transferring the positive charge to C-5. A similar Coulombic attraction results in the transfer of charge from C-5 to C-1, the carbon bearing the substituent.
Figure 2
Get to Know This Pattern
The static depiction of the process shown in Figure 2 is presented in Figure 3.
Figure 3
Freeze Frame
A similar picture may be painted for reactions where the substituent is an electron donating group. In these cases, as Figure 4 shows, an additional resonance interaction is possible in which the lone pair of electrons on the heteroatom of the substituent interacts with the positive charge that develops on C-1 (P-3 <--> P-4). The donation of electron density from the oxygen into the ring stabilizes the intermediate, making it easier to form.
Figure 4
A Variation on a Theme
Exercise 2 Select the resonance structure in Figure 4 in which every atom has an octet.
Exercise 3 Create a scheme like the one shown in Figure 4 for the case where the electrophile bonds to the position para to the OH group. How many resonance structures can you draw for the intermediate formed in this case?
What happens if the electrophile bonds to the carbon meta to the OH group in phenol? Take a look at Figure 5.
Figure 5
Nothing New Here
Here, as in the case of benzene, there are 3 resonance contributors to the structure of the cyclohexadienyl cation, P-5, P-6, and P-7, but the pattern of alternating charge density is such that the positive charge never develops on C-1 as it did in structure P-3 in Figure 4. In other words, the OH group does not offer any extra stabilization when the electrophile bonds to a carbon in the meta position.
Exercise 4 Draw the structures of the three resonance contributors for the chlorination of toluene, C6H5CH3, at the carbon ortho to the methyl group. Label them T-1, T-2, and T-3. Draw an orbital diagram to indicate how the methyl group can offer extra stabilization of the positive charge in T-3.
Finally, let's consider the situation where the substituent on the aromatic ring is an electron withdrawing group, for example, benzaldehyde, C6H5CHO. The resonance structures for the cyclohexadienyl cation formed when an electrophilic chlorine bonds to the carbon ortho to the substituent are shown in Figure 6.
Figure 6
Is There an Echo in Here?
Note the similarity between resonance contributors Z-3 and B-3. In both cases the carbon bearing the substituent also bears a positive charge. Now note the differences; 1. In B-3 there is no interaction between the substituent, H, and C-1. In Z-3 carbonyl carbon of the substituent bears a partial positive charge as shown in the right hand panel of Figure 6. The blue arrow indicates charge-charge repulsion that occurs as the positive charge develops on C-1. This unfavorable Coulombic interaction means that resonance structure Z-3 does not make as important a contribution to the stabilization of the cyclohexadienyl cation intermediate as does B-3. The energy of the actual intermediate is therefore higher than the corresponding intermediate formed by chlorination of benzene. This is why chlorination of benzaldehyde proceeds slower than chlorination of benzene.
Exercise 5 Consider the following resonance interaction: 
Exercise 6 Draw the structures of the three resonance contributors required to describe the cyclohexadienyl cation intermediate that is formed when benzaldehyde is chlorinated at the position meta to the CHO group. Is there a resonance contributor in which the formal positive charge resides on the carbon atom bearing the substituent? Yes No Which intermediate should be more stable, the one formed by chlorination ortho to the aldehyde group or the one formed by chlorination meta to the carbonyl group? ortho meta
Exercise 7 Chlorination of styrene, C6H5CH=CH2, at the carbon ortho to the substituent generates a cyclohexadienyl cation intermediate. You should be able to draw three resonance structures, S-1, S-2, and S-3, in which the positive charge is distributed around the ring. Compare the resonance interaction shown in Exercise 5 with this one:
Which resonance interaction offers greater stabilization of the positive charge, Z-3<-->Z-4 or S-3<-->S-4 ? Would your expect styrene to undergo chlorination faster or slower than benzene? faster slower
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