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Prévia do material em texto

CHAPTER 2 51 
 
 
 
Placing the chlorine atom at position 3 would be the 
same as placing it at position 2; and placing the chlorine 
atom at position 4 would be the same as it as position 1: 
 
 
 
Next, we move on to the other carbon skeleton, 
containing a branch. Once again, there are two distinctly 
different locations where a chlorine atom can be placed: 
either at position 1 or position 2, shown here. 
 
 
 
Placing the chlorine atom on any of the peripheral 
carbon atoms will lead to the same compound: 
 
 
 
In summary, there are a total of four constitutional 
isomers with the molecular formula C4H9Cl: 
 
 
 
2.52. 
(a) This compound exhibits a lone pair next to a  bond, 
so we draw the two curved arrows associated with that 
pattern. The first curved arrow is drawn showing a lone 
pair becoming a  bond, while the second curved arrow 
shows a  bond becoming a lone pair: 
 
 
 
(b) This structure exhibits a lone pair that is adjacent to a 
positive charge, so we draw one curved arrow, showing a 
lone pair becoming a  bond: 
 
 
 
(c) This structure exhibits an allylic positive charge, so 
we draw one curved arrow showing the  bond being 
pushed over. 
 
 
 
(d) This compound exhibits a C=N bond (a  bond 
between two atoms of differing electronegativity), so we 
draw one curved arrow showing the  bond becoming a 
lone pair. 
 
 
 
(e) This compound exhibits a lone pair next to a  bond, 
so we draw two curved arrows. The first curved arrow is 
drawn showing a lone pair becoming a  bond, while the 
second curved arrow shows a  bond becoming a lone 
pair. We then draw the resulting resonance structure and 
assess whether it exhibits one of the five patterns. In this 
case, the lone pair is now next to another  bond, so once 
again, we draw the two curved arrows associated with 
that pattern. The resulting resonance structure again 
exhibits a lone pair next to a  bond. This pattern 
continues again, thereby spreading a negative charge 
over many locations, as shown here: 
 
N
H
N
H
N
H
N
H
N
H
H H H
HH
 
 
(f) This structure exhibits a lone pair next to a  bond, 
so we draw two curved arrows. The first curved arrow is 
drawn showing a lone pair becoming a  bond, while the 
second curved arrow shows a  bond becoming a lone 
pair. We then draw the resulting resonance structure and 
assess whether it exhibits one of the five patterns. In this 
case, the lone pair is now next to another  bond, so once 
again, we draw the two curved arrows associated with 
that pattern. The resulting resonance structure again 
exhibits a lone pair next to a  bond, so we draw one 
more resonance structure, as shown here: 
 
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