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224  Answers
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+
N
OO
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O N
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O+
12. (A) Answer A is not a cyclohexane but a cyclohexene. 
13. (D) The conjugate base generated by the loss of the hydrogen ion through resonance 
increases the negative charge of the aromatic system. An electron-withdrawing group will 
stabilize this anion; however, an electron-donating group would have the opposite effect 
and make the acid weaker. The amino group is an electron-donating group. The amino 
group does not destabilize the aromatic system, and it is too small to create any significant 
steric hindrance.
14. (B) The triflate anion has the ability to exist in resonance forms with delocalization of 
the negative charge. This resonance stabilizes the structure, making it more stable than the 
hydroxide ion and a better leaving group.
15. (B) Here you have a six-carbon backbone with alternating single and double bonds. 
The bond between C(4) and C(5) is a single bond. Since there are sp2 bonds on either side 
of it, this bond has more s character than the sp3 hybrid bond in a normal alkane, and 
therefore it is shorter.
16. (A) The lone pair on the nitrogen resonates to form a double bond to the carbon, with 
one of the pairs from the double bond moving to the oxygen atom to give a stabilizing 
resonance form.
17. (A) The base is a catalyst in the hydrolysis of an ester because changing its concentra-
tion has no effect on the identity of the products of the reaction, and the base is not con-
sumed in the reaction. The conversion to the carboxylate ion is a distracter.
18. (D) Bonds 1 and 2 should be of equal length, since resonance will occur and they will 
be somewhere between a single and a double bond in length. The resonance makes bonds 
1 and 2 equal in length. Bond 3 is a double bond (without significant resonance) and will 
be shorter than bonds 1 and 2.
19. (A) The hydrogen labeled 3 will have the weakest bond (making it the most acidic) 
because of the two carbonyl oxygen withdrawing electron density. In addition, the loss of 
an H+ from position 3 leaves a resonance-stabilized carbanion. The presence of the two 
adjacent carbonyls is responsible for the resonance stabilization. The H at position 2 is more 
acidic than the H at position 1 than the H at position 4 because of the high electronegativity 
of the fluorine. The adjacent carbonyl, because of resonance, makes the hydrogen at position 
2 more acidic than the hydrogen at position 1.