In our discussion of resonance theory up to this point we have defined the structural requirements for resonance, we have practiced drawing resonance structures, and we have looked at how resonance theory is used to rationalize certain experimental data, most notably the chemical shift values of protons on aromatic rings. We will conclude our introductory discussion of resonance theory by delineating a set of rules that will allow us to assess the relative contribution of the individual resonance structures to a resonance hybrid. Our assessment will consider four factors; charge separation,electronegativity, the octet rule, and orbital size. As usual, our approach will be comparative.
Figure 1 depicts a p-p resonance interaction in each of two isostructural molecules, 1,3-butadiene and propenal. Because both of these molecules meet the structural requirements for p-p interactions, we assume that these interactions will occur to some degree. The question is "to what degree?"
Note that both the 1<-->1' interaction and the 2<-->2' interaction yield resonance contributors in which one atom bears a formal positive charge and another bears a formal negative charge. Structures 1 and 2 each contain two pi bonds, while in 1' and 2' there is only one pi bond; a bonding pair of electrons in 1 has become a non-bonding pair in 1'; ditto for the 2<-->2' interaction. Since electrons in bonding orbitals have lower potential energy than those in non-bonding orbitals, the structures of 1,3-butadiene and propenal should resemble 1 and 2 more closely than they do 1' and 2'. In other words, resonance interactions which result in the separation of charge produce resonance contributors that are less important than the uncharged structures from which they were generated. Now let's consider...
The only significant difference between resonance contributors 1' and 2' is the atom that bears the formal negative charge. What conclusion can we draw from this difference? Since oxygen is more electronegative than carbon, it is reasonable to conclude that the 2<-->2' interaction is more important, i.e occurs to a greater extent, than the 1<-->1' interaction. Or, in the vernacular of the organic chemist, structure 2' makes a more important contribution to the resonance hybrid of propenal than structure 1' does to the resonance hybrid of 1,3-butadiene.
Exercise 1 If structure 2' makes a more important contribution to the resonance hybrid of propenal than structure 1' does to the resonance hybrid of 1,3-butadiene, would you expect the electron density to be greater on the methylene carbon in 1,3-butadiene or the methylene carbon in propenal? 1,3-butadiene propenal
Exercise 2 The word isostructural means that two compounds have the same basic geometry, i.e. the same shape or hybridization at each comparable atom. Styrene, C6H5CH=CH2, and benzaldehyde, C6H5CH=O, are two compounds that are isostructural. What is the hybridization of the oxygen atom in benzaldehyde? sp sp2 sp3
Exercise 3 Draw the structures of these molecules:
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styrene
benzaldehyde
Exercise 4 Draw a resonance structure for styrene comparable to resonance contributor 1' in Figure 1. Draw a resonance structure for benzaldehyde comparable to resonance contributor 2' in Figure 1.
Exercise 5 Would you expect the electron density to be higher in the aromatic ring of styrene or benzaldehyde? styrene benzaldehyde
Exercise 6 Would you expect the chemical shift of the ortho protons in styrene to be higher or lower than that of the ortho protons in benzaldehyde?higher lower
Exercise 7 Consider the n-p resonance interactions shown below:
Which interaction is more important, i.e. which interaction occurs to a greater extent? N<-->N' O<-->O'
Would the electron density be higher on C-2 of 1-aminopropene or 1-propenol? 1-aminopropene 1-propenol
In the case of 1,3-butadiene, it does not matter which direction the arrows point. Resonance structure 1' in Figure 1 is identical to resonance structure 1' in Figure 2. However, the result of the 2<-->2' interaction shown in Figure 1 is different that shown in Figure 2. In the latter case, the oxygen atom has a positive charge rather than a negative. We can safely draw two conclusions: 1. The 2<-->2' interaction shown in Figure 2 is less important than the 1<-->1' interaction. 2. The 2<-->2' interaction if Figure 2 is less important than the 2<-->2' interaction in Figure 1. In other words, the most important resonance interaction is the 2<-->2' interaction shown in Figure 1, while the least important is the 2<-->2' interaction in Figure 2. Now, let's move on to ...
The beauty of resonance theory lies in its ability to help chemists make predictions about chemical processes. We have already seen how to use resonance theory to rationalize and predict chemical shifts in 1H-NMR spectra. Now let's use it to predict the outcome of a chemical reaction, specifically the protonation of our old friend methyl vinyl ether, CH3OCH=CH2.
Which should be the least reactive ? non-bonded p-bonded s-bonded
We will apply the principles of resonance theory to make our prediction, but before we do so, we should consider the details of the three alternatives shown in Figure 3. First, note that protonation of the oxygen atom generates a positive charge on the oxygen. A trivalent oxygen atom that is positively charged is called an oxonium ion. In contrast to protonation of the oxygen, protonation of C-1 results in the formation of a positive charge on C-2, i.e. the charge develops on the atom adjacent to the atom that is protonated. Likewise for protonation of C-2; the positive charge develops on C-1. The difference between these two outcomes stems from the difference in the types of electrons that are used to bond to the proton. In the case of oxygen, it is a non-bonding pair. In sharing a non-bonded pair of electrons with the proton, the oxygen atom effectively gives up its claim to one of the two electrons, thus acquiring a positive charge. When a proton forms a bond to C-1, the pi electrons that were shared between C-1 and C-2 become a sigma-bonded pair that is shared between the carbon and the hydrogen. Consequently, C-2 loses its claim to the electron that it contributed to the pi bond, and it becomes positively charged. The same idea applies to protonation of C-2, but now the charge develops on C-1. We will see these two patterns many times during our discussions of chemical reactions.
In order to decide where a proton will react with methyl vinyl ether, we need to evaluate the relative stabilities of the cations that are formed in each of the alternatives. Our assumption is that the preferred reaction will be the one that produces the most stable cationic intermediate. The most stable intermediate should be the one that enjoys the greatest degree of resonance stabilization. Figure 4 offers some alternatives.
The left-hand portion of the figure depicts two possibilities for the intermediate formed by protonation of the oxygen atom of methyl vinyl ether. The arrows labeled 1 and 2 in the top panel represent an n-p interaction. There are several problems with this possibility however. First, note that arrow 1 shows the lone pair of electrons on the oxygen atom moving away from the positive charge. Such a movement violates Coulomb's Law. Second, this movement results in a separation of charge; the formal charge on the oxygen atom increases from +1 to +2 while that on the carbon changes from 0 to -1. On both counts this interaction is not favorable and would not result in any stabilization of the intermediate oxonium ion. Structure O-A' is not a reasonable resonance contributor.
In the lower left-hand panel the sense of "electron flow" is correct, i.e. electrons are moving toward the positively charged oxygen atom. Furthermore, this interaction does not result in the separation of charge. However, the structure O-B'violates the octet rule; there are 10 electrons around the oxygen atom and there are only 6 electrons around the the positively charged carbon. Therefore, you should not expect this interaction to afford significant stabilization of the intermediate oxonium ion. In other words, O-B should be more stable than O-B', so there should be no tendency for the interaction shown to occur. (Note-There are two ways to violate the octet rule: 1. Have fewer than 8 electrons around an atom. 2.. Have more than 8 electrons around an atom. The first violation is common. The second does not occur with first-row elements.)
The only thing close to correct about the upper-right portion of the figure is the sense of "electron flow". Everything else is dreadful. The interaction shown results 10 electrons around C-1 of structure C2'. In addition, there is greater charge separation in structure C2' than in C2. Structure C2' on the right is not a resonance contributor. Conclusion: The ion C2 would not be stabilized by any resonance interaction.
Unlike the other cations, the cation produced by protonation of methyl vinyl ether at C-2 is stabilized by resonance. As the positive charge develops on the carbon in structure C1, it draws the lone pair of electrons from the adjacent oxygen atom towards itself as indicated by the arrow. While this interaction might appear to be unfavorable because the positive charge is transferred from the less electronegative carbon atom to the more electronegative oxygen atom, it is, in fact, quite favorable. In resonance structure C1' every atom has a filled valence shell. Indeed, resonance structure C1' makes a greater contribution to the hybrid than structure C1.
So, the bottom line is that resonance theory predicts that protonation of methyl vinyl ether should occur preferentially on carbon 2. Experimental fact, protonation of methyl vinyl ether occurs preferentaily on carbon 2!
Since resonance interactions involve orbital overlap, it should not come as a surprise that the extent of stabilization afforded by resonance depends, at least in part, on the degree of overlap between orbitals on adjacent atoms. In general, maximum overlap occurs between atoms of approximately the same size. This means atoms in the same row of the periodic table. Thus the n-p orbital overlap in vinyl fluoride, CH2=CH-F, is greater than it is in vinyl chloride, CH2=CH-Cl, since carbon and fluorine are similar in size. While at first glance the interactions shown in Figure 5 might suggest that resonance is equally important in both vinyl chloride and vinyl fluoride, structure F-1' is a more important contributor to the F-1/F-1' pair than structure Cl-1' is to the Cl-1/Cl-1' pair.
What this means is that resonance theory predicts that the electron density should be higher at C-2 of vinyl fluoride than it is at C-2 of vinyl chloride. In fact, NMR data supports this prediction as the chemical shifts in Figure 6 indicate. Even though fluorine is more electronegative than chlorine, the electrons density at C-2 of vinyl fluoride is higher than it is in vinyl chloride. Clearly the resonance effect of the fluorine must more than offset its inductive effect.
Exercise 10 Draw a resonance contributor for vinyl bromide comparable to resonance contributors F-1' and Cl-1' in Figure 5.
vinyl bromide
Exercise 11 Is the n-p orbital overlap in vinyl bromide greater or less than it is in vinyl chloride? greater less
Exercise 12 The chemical shifts of the ortho hydrogens in C6H5X are 6.97, 7.27, and 7.43 ppm. When d = 6.97, X = F Cl Br
when d = 7.27, X = F Cl Br
Exercise 13 Predict which of the vinyl halides should react fastest with a proton. vinyl bromide vinyl chloride vinyl fluoride
Exercise 14 Draw the structure of the cation that you would expect to be produced by protonation of vinyl bromide.
initial cation
Draw the structure of the resonance contributor that would help stabilize the initially formed cation.
resonance contributor