Resonance major contributors

Examine the geometry and electrostatic potential map for acetone enolate. Are the CC and CO bonds in the enolate more similar to those in acetone or propen-2-ol precursors Is the negative charge primarily located on oxygen or on carbon Assuming this enolate is a hybrid of the two resonance contributors as shown above, which, if either, appears to be the major contributor ... 

A similar concept exists when comparing resonance structures. One compound might have three resonance structures, but all three resonance structures might not contribute equally to the overall resonance hybrid. One resonance structure might be the major contributor (like the peach), while another resonance structure might be insignificant (like the kiwi). In order to understand the true nature of the compound, we must be able to compare the resonance structures and determine which structures are major contributors and which structures are not significant. 

Recently, a variety of (3-silylated carboxonium ions have been prepared and characterized by NMR spectroscopy.541 Kira et al.631 used the Corey hydride transfer method, whereas Olah, Prakash, and co-workers applied triphenylmethyl tetrakis (pentafluorophenyl)borate to silylate esters,632 ketones, enones, and carbonates633 [Eq. (3.91)]. The ions thus produced are resonance hybrids of oxocarbenium (327b) and carboxonium (327a) ions with the latter as the major contributors. Calculated (DFT/IGLO) 29 Si NMR chemical shifts agree well with the experimental data. 

A molecule or ion for which two or more valid Lewis structures can be drawn, differing only in the placement of the valence electrons. These Lewis structures are called resonance forms or resonance structures. Individual resonance forms do not exist, but we can estimate their relative energies. The more important (lower-energy) structures are called major contributors, and the less important (higher-energy) structures are called minor contributors. When a charge is spread over two or more atoms by resonance, it is said to be delocalized and the molecule is said to be resonance stabilized, (pp. 14-18)... 

For each of these ions, draw the important resonance forms and predict which resonance form is likely to be the major contributor. 

The phosphorus ylide has two resonance forms one with a double bond between carbon and phosphorus, and another with charges on carbon and phosphorus. The double-bonded resonance form requires ten electrons in the valence shell of phosphorus, using a d orbital. The pi bond between carbon and phosphorus is weak, and the charged structure is the major contributor. The carbon atom actually bears a partial negative charge, balanced by a corresponding positive charge on phosphorus. 

These structures are not equal in estimated energy. The first structure has the positive charge on nitrogen. The second has the positive charge on carbon, and the carbon atom does not have an octet. The first structure is more stable because it has an additional bond and all the atoms have octets. Many stable ions have a positive charge on a nitrogen atom with four bonds (see the Summary Table, page 13). We call the more stable resonance form the major contributor, and the less stable form is the minor contributor. The structure of the actual compound resembles the major contributor more than it does the minor contributor. 

The first structure, with more bonds and less charge separation, is possible because sulfur is a third-row element with accessible d orbitals, giving it an expandable valence. For example, SFg is a stable compound with 12 electrons around sulfur. Theoretical calculations suggest that the last structure, with octets on all atoms, may be the major resonance contributor, however. We cannot always predict the major contributor of a resonance hybrid. 

When two resonance structures are different, the hybrid looks more like the better resonance structure. The better resonance structure is called the major contributor to the hybrid, and all others are minor contributors. The hybrid is the weighted average of the contributing resonance structures. What makes one resonance structure better than another There are many factors, but for now, we will learn just two. 

Comparing resonance structures X and Y, X is the major contributor because it has mote bonds and fewer charges. Thus, the hybrid looks more like X than Y. 

Although the resonance hybrid is some combination of all of its valid resonance structures, the hybrid more closely resembles the most stable resonance structure. Recall from Section 1.5C that the most stable resonance structure is called the major contributor to the hybrid, and the less stable resonance structures are called the minor contributors. Two identical resonance structures are equal contributors to the hybrid. 

Because A contains a positive charge and a ione pair on adjacent atoms, a second resonance structure B can be drawn. Because B has more bonds and aii second-row atoms have octets, B is more stabie than A, making it the major contributor to the hybrid C. Because the hybrid is more stable than either resonance contributor, the order of stability is ... 

The two resonance structures that contain an intact aromatic ring and place a negative charge on an O atom are major contributors to the hybrid. Resonance stabilizes phenoxide but not as much as... 

The 2 polymorphs of CaCOs, aragonite and calcite, can readily be distinguished on the basis of their " Ca MAS spectra (Figure 8.28B). Calcite shows a characteristic second-order quadrupolar lineshape from which the NMR parameters can he extracted by spectral simulation. The narrower Ca MAS resonance from aragonite shows no discernible structure, but the corresponding static aragonite spectrum is about 20 times broader than under MAS conditions. This suggests that CSA is a major contributor to the static linewidth, which is confirmed by satisfactory simulation of the spectrum... 

Structures with the lowest formal charges usually have the lowest energy (major contributors to the resonance hybrid). 

For each of the resonance pairs below, determine which is the major contributor. 

Resonance form A must be the major contributor. If B were the major contributor, the value of the charge separation would be between 0.5 and 1.0. Even though B is "minor", it is quite significant, explaining in part the high polarity of the C=0. 

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