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Solutions for Aromatic Compounds
21(b) continued
H
H
Br Br
H
Br
H Br H
Br
C
H
C
H
H Br
Br
H
BrH
H
Br
phenanthrene
plus many other
resonance forms
Bromide attack at C-9 
leaves two aromatic rings.
(d)
cis and trans
E1
(c) A typical addition of bromine occurs with a bromonium ion intermediate, which can give only 
anti addition. Addition of bromine to phenanthrene, however, generates a free carbocation because 
the carbocation is benzylic, stabilized by resonance over two rings. In the second step of the 
mechanism, bromide nucleophile can attack either side of the carbocation, giving a mixture of cis and 
trans products.
∆
resonance-stabilized
+ HBr
Br
Br
NN
HN
F
O
COOH
22
(a)
NN
HN
F
O
COOH
The second resonance form shows that the two rings in the dashed box form a 10 electron pi system, 
isoelectronic with naphthalene. Cipro is aromatic.
(b) For a nitrogen (or any atom) to be basic, it must have an 
electron pair to share by forming a sigma bond with a proton. 
The nitrogen atoms labeled 1 and 2 are basic because they have 
an unshared electron pair, but nitrogen 3 does not have an 
unshared pair; its electron pair is delocalized in the aromatic pi 
system. NN
HN
F
O
COOH
1
2 3
H
H
H
(c) The three H atoms in bold are on the aromatic ring and 
will appear in the HNMR between δ 6–8.
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