Resonance contributors predicted stabilities

Obtain the energy of each cation that might be generated by electrophilic addition of Br to biphenyl (biphenyl+Br+). Which one is most stable Are there others of comparable stability Examine the structure of the most stable cation(s), and draw all of the resonance contributors needed to describe this ion(s). Predict the product(s) of biphenyl bromination. Will the reaction be highly selective, moderately selective or unselective ... 

All resonance contributors do not necessarily contribute equally to the resonance hybrid. The degree to which each resonance contributor contributes depends on its predicted stability. Because resonance contributors are not real, their stabilities cannot be measured. Therefore, the stabilities of resonance contributors have to be predicted based on molecular features that are found in real molecules. The greater the predicted stability of the resonance contributor, the more it contributes to the resonance hybrid and the more it contributes to the resonance hybrid, the more similar the contributor is to the real molecule. The examples that follow illustrate these points. 

The greater the predicted stability of the resonance contributor, the more it contributes to the structure of the resonance hybrid. 

We can predict that structure G will make only a small contribution to the resonance hybrid because it has separated charges as well as an atom with an incomplete octet. Structure E also has separated charges and an atom with an incomplete octet, but its predicted stability is even less than that of structure G because it has a positive charge on the electronegative oxygen. Its contribution to the resonance hybrid is so insignificant that we do not need to include it as one of the resonance contributors. The resonance hybrid, therefore, looks very much like structure F. 

Since the ability to delocalize electrons increases the stability of a molecule, we can conclude that a resonance hybrid is more stable than the predicted stability of any of its resonance contributors. The resonance energy associated with a compound that has delocalized electrons depends on the number and predicted stability of the resonance contributors The greater the number of relatively stable resonance... 

Then compare the predicted stabilities of the set of resonance contributors for each carbocation. 

The other factor responsible for the increased stability of the carboxylate ion is its greater resonance energy relative to that of its conjugate acid. The carboxylate ion has greater resonance energy than a carboxylic acid does, because the ion has two equivalent resonance contributors that are predicted to be relatively stable, whereas the carboxylic acid has only one (Section 7.6). Therefore, loss of a proton from a carboxylic acid is accompanied by an increase in resonance energy—in other words, an increase in stability (Figure 1.1). 

NOTE TO THE STUDENT The tutorial on page 392 will give you additional practice drawing resonance contributors and predicting their relative stabilities. 

Delocalized electrons stabilize a compound. The extra stability a compound gains from having delocalized electrons is called the delocalization energy. Electron delocalization is also called resonance, so delocalization energy is also called resonance energy. Because delocalized electrons increase the stability of a compound, we can conclude that a resonance hybrid is more stable than any of its resonance contributors is predicted to be. 

The delocalization energy associated with a compound that has delocalized electrons depends on the number and the predicted stability of the resonance contributors. 

A resonance hybrid is more stable than the predicted stability of any of its resonance contributors. 

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