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Solutions for Structure and Properties of Organic Molecules
14 continued
net
(e)
O
O
O
Each end oxygen has
one-half negative 
charge as it is the 
composite of two 
resonance forms; see 
solution to 7(g).
large dipole (2.95)
net
(f)
H C N
H
C
CH3
O
C
N
HH
H
large dipole
large dipole (2.72)
netnet
(h)(g)
net
N
H
H
H
N
H3C
CH3
CH3
C C
H
H
H
Cl
B F
F
F
Cl Be Cl
H
(i)
small dipole (0.67)
(j)
net
large dipole (1.45)
(k)
net dipole = 0
(l)
net dipole = 0
(m)
net dipole = 0
In (k) through (m), the symmetry 
of the molecule allows the 
individual bond dipoles to cancel.
C
H
Cl
C
H
Cl
C C
Cl
H
H
Cl
15 With chlorines on the same side of the double bond, the bond dipole moments reinforce each other, 
resulting in a large net dipole. With chlorines on opposite sides of the double bond, the bond dipole 
moments exactly cancel each other, resulting in a zero net dipole.
net dipole = 0large net dipole
(2.40)
N
H
H
CH2CH2CH3
N
CH3CH2CH2
H
HO
H
CH2CH3
CH3CH2 O
H
δ−
δ−
δ+
δ+
δ+
δ+
δ+
δ+
(b)
δ+
δ+
δ−
δ+
16
δ−
δ+
(a)
(hydrogen bonds shown as wavy bond)
small dipole (0.52)
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