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56 CHAPTER 2 
 
negative charge on a more electronegative atom, and it is better to place a positive charge on a less electronegative 
atom. 
 
HO C NH2
O
HO C NH2
O
NH2CHO
O
NH2CHO
O
Largest contributor (#1) Major (#3) Major (#2) Minor (#4)
 
 
2.63. The only resonance pattern evident in the enamine is an allylic lone pair. After that pattern is applied, however, 
another allylic lone pair results so the resonance can ultimately involve both  bonds. There are a total of three major 
resonance forms that all have filled octets. Consideration of the hybrid of these resonance forms predicts two electron-
rich sites. 
 
N N N N
-
-
Electron-
rich sites 
 
Did you draw the following additional structure (or something similar, with C+ and C-) and wonder why it was not 
shown in this solution? 
 
 
This resonance form suffers from two major deficiencies: 1) it does not have filled octets, while the other resonance 
forms shown above all have filled octets, and 2) it has a negative charge on a carbon atom (which is not an 
electronegative atom). Either of these deficiencies alone would render the resonance form a minor contributor. But 
with both deficiencies together (C+ and C-), this resonance form is insignificant. The same is true for any resonance 
form that has both C+ and C-. Such a resonance form will generally be insignificant (there are very few exceptions, 
one of which will be seen in Chapter 17). 
 Also, note that the  bonds cannot be moved to other parts of the six-membered ring since the CH2 groups are 
sp3 hybridized. These carbon atoms cannot accommodate an additional bond without violating the octet rule. 
 
 
 
 
 
2.64. 
(a) The molecular formula is C3H6N2O2. 
(b) Each of the highlighted carbon atoms (below) has 
four sigma bonds (the bonds to hydrogen are not shown). 
As such, these two carbon atoms are sp3 hybridized. 
 
 
 
(c) There is one carbon atom that is using a p orbital to 
form a  bond. As such, this carbon atom (highlighted) 
is sp2 hybridized. 
 
 
 
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