Resonance hybrid minor contributors

As noted in Figure 3.16, acetic acid has only a small amount of resonance stabilization because the lower structure is only a minor contributor to the resonance hybrid. Acetate ion has a large amount of resonance stabilization because it has two equivalent contributors to the 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. 

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. 

Structures B and C have charge separation and fewer bonds than A, making them less stable resonance structures and only minor contributors to the resonance hybrid. B and C both contain a double bond between the centrsil carbon atoms, however, so the hybrid must have a partial double bond there. This makes the c entral C—C bond shorter than a C-C single bond in an alkane. 

Some data for BF3 show a shorter than expected B—F bond. Shorter bonds indicate double-bond character, so the structure with the B=F bond may be a minor contributor to a resonance hybrid.) The main way electron-deficient atoms attain an octet is by forming additional bonds in reactions. When BF3 reacts with ammonia, for instance, a compound forms in which boron attains its octet ... 

In addition to the Kekule-type contributors to the pyridine resonance hybrid shown on page 391, there are three minor dipolar contributors. Draw their structures. Do they suggest a reason why pyridine is deactivated (relative to benzene) toward reaction with electrophiles and a reason why substitution, when it does occur, takes place at the 3-position ... 

Problem 1.17 Draw a second resonance structure for nitrous acid. Label each resonance structure as a major, minor, or equal contributor to the hybrid. Then draw the resonance hybrid. 

The thiazolium ion has an unusual feature a relatively acidic proton located between the two heteroatoms (at C2). The corresponding carbanion is stabilized inductively by the adjacent positive charge that is distributed over both heteroatoms by resonance. The charge-neutral carbene form (Section 12-9) is a minor contributor to the resonance hybrid. 

A better resonance structure is one that has more bonds and fewer charges. The better structure is the major contributor and all others are minor contributors. To draw the resonance hybrid, use dashed lines for bonds that are in only one resonance structure, and use partial charges when the charge is on different atoms in the resonance structures. 

Furan is less resonance stabilized than pyrrole because its O atom is less basic, so it donates electron density less "willingly." Thus, charge-separated resonance forms are more minor contributors to the hybrid than the charge-separated resonance forms of pyrrole. 

We can show this resonance stabilization for conjugated double bonds of butadiene in terms of Lewis structures, as shown below. The dipolar form is the minor contributor to the resonance hybrid because it takes energy to separate charges. 

The hydroxyl oxygen atom can donate an electron pair to carbon to give resonance structure 3 in which every atom has a Lewis octet. This stabilizes the C=0 group, and the carbonyl carbon atom is less electrophilic than that of aldehydes or ketones. However, since this contributing structure has a positively charged oxygen atom, it is only a minor contributor to the resonance hybrid. 

The second resonance form of formamide is a minor but significant resonance contributor. It shows that the nitrogen-carbon bond has some double bond character, requiring that the nitrogen be sp hybridized with bond angles approaching 120°. 

Problem 1.16 Label the resonance structures In each pair as major, minor, or equal contributors to the hybrid... 

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