Resonance structures lone pair

Steric effects on the nucleophile, aniline, were clearly evident. Rate constants for bimolecular attack of 2,6-dimethyl- 70a, 2,6-diethyl- 70b, and 3,5-dimethylaniline 70c at 308 K indicate that the ort/zo-substituted anilines react more than an order of magnitude slower at the same temperature (Table 7). Structure 70c must be able approach the reactive nitrogen more closely.42,43 A comparison of the rate constants for reaction of aniline 72c, /V-methyl- 71a and /V-phenylaniline 71b provides further evidence of steric effects although the very small rate constant for the diphenylamine could also be accounted for by reduced nucleophilicity on account of lone pair resonance into the additional phenyl ring. 

Figure 2-51. a) The rotational barrier in amides can only be explained by VB representation using two resonance structures, b) RAMSES accounts for the (albeit partial) conjugation between the carbonyl double bond and the lone pair on the nitrogen atom. 

The corresponding resonance description shows the delocalization of the nitrogen lone pair electrons m terms of contributions from dipolar structures... 

The table gives the computed spin densities for each atom (the value in parenthese following the substituent is its electronegativity). The illustrations are Lewis doi structures showing the primary resonance form for each structure and indicatinj unpaired electrons and lone pairs. 

Resonance forms differ only in the placement of their tt or nonbonding electrons. Neither the position nor the hybridization of any atom changes from one resonance form to another. In the acetate ion, for example, the carbon atom is sp2-hybridized and the oxygen atoms remain in exactly the same place in both resonance forms. Only the positions of the r electrons in the C=0 bond and the lone-pair electrons on oxygen differ from one form to another. This movement of electrons from one resonance structure to another can be indicated by using curved arrows. A curved arrow always indicates the movement of electrons, not the movement of atoms. An arrow shows that a pair of electrons moves from the atom or bond at the tail of the arrow to the atom or bond at the head of the arrow. 

The atoms X, Y, and Z in the general structure might be C, N, O, F, or S, and the asterisk ( ) might mean that the p orbital on atom Z is vacant, that it contains a single electron, or that it contains a lone pair of electrons. The two resonance forms differ simply by an exchange in position of the multiple bond and the asterisk from one end to the other. 

The 2,4-pentanedione anion has a lone pair of electrons and a formal negative charge on the central carbon atom, next to a C=0 bond on the left. The 0=C-C grouping is a typical one for which two resonance structures can be drawn. 

Just as there is a C=0 bond to the left of the lone pair, there is a second C=0 bond to the right. Thus, we can draw a total of three resonance structures for the 2,4-pentanedione anion. 

Exchanging the position of the double bond and an electron lone pair in each grouping generates three resonance structures. 

As the following resonance structures indicate, enamines are electronically similar to enolate ions. Overlap of the nitrogen lone-pair orbital with the double-bond p orbitals leads to an increase in electron density on the a carbon atom, making that carbon nucleophilic. An electrostatic potential map of N,N-6imethyl-aminoethvlene shows this shift of electron density (red) toward the a position. 

Draw the most important Lewis structure for each of the following ring molecules (which have been drawn without showing the locations of the double bonds). Show all lone pairs and nonzero formal charges. If there are equivalent resonance... 

The curved arrows show how one resonance structure relates to another. Notice that the formal negative charge is located on the ortho and para positions, exactly where reaction takes place most quickly. Other ortho- and para-directing groups include —NH2, —Cl, and —Br. All have an atom with a lone pair of electrons next to the ring, and all accelerate reaction. 

Now that we know how to identify good arrows and bad arrows, we need to get some practice drawing arrows. We know that the tail of an arrow must come either from a bond or a lone pair, and that the head of an arrow must go to form a bond or a lone pair. If we are given two resonance structures and are asked to show the arrow(s) that get us from one resonance structure to the other, it makes sense that we need to look for any bonds or lone pairs that are appearing or disappearing when going from one structure to another. For example, consider the following resonance structures ... 

In this example, we can see that one of the lone pairs on oxygen is coming down to form a bond, and the C=C double bond is being pushed to form a lone pair on a carbon atom. When both arrows are pushed at the same time, we are not violating either of the two commandments. So, let s focus on how to draw the resonance structure. Since we know what arrows mean, it is easy to follow the arrows. We just get rid of one lone pair on oxygen, place a double bond between carbon and oxygen, get rid of the carbon-carbon double bond, and place a lone pair on carbon ... 

PROBLEMS For each of the structures below, draw the resonance structure that you get when you push the arrows shown. Be sure to include formal charges. (Hint In some cases the lone pairs are drawn and in other cases they are not drawn. Be sure to take them into account even if they are not drawn—you need to train yourself to see lone pairs when they are not drawn.)... 

Once you learn to recognize this pattern (a lone pair next to a pi bond), you will be able to save time in calculating formal charges and determining if the octet rule is being violated. You will be able to push the arrows and draw the new resonance structure without thinking about it. 

Both compounds have a pi bond between a carbon atom and an electronegative atom (C=0), and both compounds have a lone pair next to the pi bond. So we would expect their resonance structures to be similar, and we would expect these compounds to have the same number of significant resonance structures. But they do not. Let s see why. Let s start by drawing the resonance structures of the first compound ... 

Inspection of the second resonance structure reveals that this nitrogen atom is actually sp hybridized, not sp. It might look like it is sp hybridized in the first resonance structure, but it isn t. Here is the general rule a lone pair that participates in resonance must occupy ap orbital. In other words, the nitrogen atom in the compound above is sp hybridized. And as a result, this nitrogen atom is trigonal planar rather than trigonal pyramidal. 

The structure of ozone, (a) The Lewis structure shows resonance that involves the bond (blue lines) and one lone pair (red dots), (b) The a bonding framework. 

A clue to the nature of the third itt MO can be found in the placement of electrons in the two resonance structures for ozone, which are shown with color highlights in Figure 10-36a. Notice that in one resonance structure, the left outer atom has three lone pairs and a single bond, while the right outer atom has two lone pairs and a double bond. In the other resonance structure, the third lone pair is on the right outer atom, with the double bond to the left outer atom. The double bond appears in different positions in the two stmctures, and one of the lone pairs also appears in different positions. These variations signal delocalized orbitals. 

Follow the four-step procedure for the composite model of bonding. Use localized bonds and hybrid orbitals to describe the bonding framework and the inner atom lone pairs. Next, analyze the system, paying particular attention to resonance structures or conjugated double bonds. Finally, make sure the bonding inventory accounts for all the valence electrons and all the valence orbitals. 

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