Covers All Chapters
SOLUTIONS MANUAL
,Solutions to Problems
Chapter 1
1.1. a. A dipolar resonance structure has aromatic character in both rings and would
be expected to make a major contribution to the overall structure.
– +
b. The “extra” polarity associated with the second resonance structure would
contribute to the molecular structure but would not be accounted for by
standard group dipoles.
O H –O H
N+ N N+ N+
–O H –O
H
c. There are three major factors contributing to the overall dipole moments: (1)
the !-bond dipole associated with the C−O and C−N bonds; (2) the "-bond
dipole associated with delocalization of " electrons from the heteroatom to
the ring; and (3) the dipole moment associated with the unshared electron pair
(for O) or N−H bond (for N). All these factors have a greater moment toward
rather than away from the heteroatom for furan than for pyrrole. For pyrrole,
the C−N " dipole should be larger and the N−H moment in the opposite
direction from furan. These two factors account for the reversal in the direction
of the overall dipole moment. The AIM charges have been calculated.
electrons O < N electrons –0.008 –.029
H 0.085
H 0.062
0.027
bond O > N bond 0.567 0.532
O N H O H N
N–H dipole –.008 –1.585
unshared –1.343
H H
pair 0.470
AIM charges
1
,2 1.2. a. The nitrogen is the most basic atom.
Solutions to Problems
PhCH=N+Ph
H
b. Protonation on oxygen preserves the resonance interaction with the nitrogen
unshared electron pair.
O+ – H O–H
CH3C CH3C
NH2 N+H2
c. Protonation on nitrogen limits conjugation to the diene system. Protonation
on C(2) preserves a more polar and more stable conjugated iminium system.
Protonation on C(3) gives a less favorable cross-conjugated system.
H
H + H H
N+ N
H N+ N+
H
H H H H H
d. Protonation on the ring nitrogen preserves conjugation with the exocyclic
nitrogen unshared electrons.
+
N NH2 N N+H2 N N+H3
H H
charge can be delocalized charge is localized on
exocyclic nitrogen
1.3. a. The dipolar resonance structure containing cyclopentadienide and pyridinium
rings would be a major resonance contributor. The dipole moments and bond
lengths would be indicative. Also, the inter-ring “double bond” would have a
reduced rotational barrier.
–
C2H5 C2H5
N N+
b. The dipolar oxycyclopropenium structure contributes to a longer C−O bond
and an increased dipole moment. The C=O vibrational frequency should
be shifted toward lower frequency by the partial single-bond character. The
compound should have a larger pKa for the protonated form, reflecting
increased electron density at oxygen and aromatic stabilization of the cation.
, O O– 3
+ Solutions to Problems
Ph Ph Ph Ph
c. There would be a shift in the UV spectrum, the IR C=O stretch, and NMR
chemical shifts, reflecting the contribution from a dipolar resonance structure.
O O–
CHCCH3 CH=CCH3
+
1.4. a. Amides prefer planar geometry because of the resonance stabilization. The
barrier to rotation is associated with the disruption of this resonance. In
MO terminology, the orbital with the C=O " ∗ orbital provides a stabilized
delocalized orbital. The nonplanar form leads to isolation of the nitrogen
unshared pair from the C=O system.
C=O *
O CH3 N:
CH3 O
N N
R CH3 R CH3
C=O
b. The delocalized form is somewhat more polar and is preferentially stabilized
in solution, which is consistent with the higher barrier that is observed.
c. Amide resonance is reduced in the aziridine amide because of the strain
associated with sp2 hybridization at nitrogen.
O –O
C N C N+
Ph Ph
The bicyclic compound cannot align the unshared nitrogen electron pair with
the carbonyl group and therefore is less stable than a normal amide.
N O
:
1.5. a. The site of protonation should be oxygen, since it has the highest negative
charge density.
SOLUTIONS MANUAL
,Solutions to Problems
Chapter 1
1.1. a. A dipolar resonance structure has aromatic character in both rings and would
be expected to make a major contribution to the overall structure.
– +
b. The “extra” polarity associated with the second resonance structure would
contribute to the molecular structure but would not be accounted for by
standard group dipoles.
O H –O H
N+ N N+ N+
–O H –O
H
c. There are three major factors contributing to the overall dipole moments: (1)
the !-bond dipole associated with the C−O and C−N bonds; (2) the "-bond
dipole associated with delocalization of " electrons from the heteroatom to
the ring; and (3) the dipole moment associated with the unshared electron pair
(for O) or N−H bond (for N). All these factors have a greater moment toward
rather than away from the heteroatom for furan than for pyrrole. For pyrrole,
the C−N " dipole should be larger and the N−H moment in the opposite
direction from furan. These two factors account for the reversal in the direction
of the overall dipole moment. The AIM charges have been calculated.
electrons O < N electrons –0.008 –.029
H 0.085
H 0.062
0.027
bond O > N bond 0.567 0.532
O N H O H N
N–H dipole –.008 –1.585
unshared –1.343
H H
pair 0.470
AIM charges
1
,2 1.2. a. The nitrogen is the most basic atom.
Solutions to Problems
PhCH=N+Ph
H
b. Protonation on oxygen preserves the resonance interaction with the nitrogen
unshared electron pair.
O+ – H O–H
CH3C CH3C
NH2 N+H2
c. Protonation on nitrogen limits conjugation to the diene system. Protonation
on C(2) preserves a more polar and more stable conjugated iminium system.
Protonation on C(3) gives a less favorable cross-conjugated system.
H
H + H H
N+ N
H N+ N+
H
H H H H H
d. Protonation on the ring nitrogen preserves conjugation with the exocyclic
nitrogen unshared electrons.
+
N NH2 N N+H2 N N+H3
H H
charge can be delocalized charge is localized on
exocyclic nitrogen
1.3. a. The dipolar resonance structure containing cyclopentadienide and pyridinium
rings would be a major resonance contributor. The dipole moments and bond
lengths would be indicative. Also, the inter-ring “double bond” would have a
reduced rotational barrier.
–
C2H5 C2H5
N N+
b. The dipolar oxycyclopropenium structure contributes to a longer C−O bond
and an increased dipole moment. The C=O vibrational frequency should
be shifted toward lower frequency by the partial single-bond character. The
compound should have a larger pKa for the protonated form, reflecting
increased electron density at oxygen and aromatic stabilization of the cation.
, O O– 3
+ Solutions to Problems
Ph Ph Ph Ph
c. There would be a shift in the UV spectrum, the IR C=O stretch, and NMR
chemical shifts, reflecting the contribution from a dipolar resonance structure.
O O–
CHCCH3 CH=CCH3
+
1.4. a. Amides prefer planar geometry because of the resonance stabilization. The
barrier to rotation is associated with the disruption of this resonance. In
MO terminology, the orbital with the C=O " ∗ orbital provides a stabilized
delocalized orbital. The nonplanar form leads to isolation of the nitrogen
unshared pair from the C=O system.
C=O *
O CH3 N:
CH3 O
N N
R CH3 R CH3
C=O
b. The delocalized form is somewhat more polar and is preferentially stabilized
in solution, which is consistent with the higher barrier that is observed.
c. Amide resonance is reduced in the aziridine amide because of the strain
associated with sp2 hybridization at nitrogen.
O –O
C N C N+
Ph Ph
The bicyclic compound cannot align the unshared nitrogen electron pair with
the carbonyl group and therefore is less stable than a normal amide.
N O
:
1.5. a. The site of protonation should be oxygen, since it has the highest negative
charge density.
