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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” polaritỵ associated with the second resonance structure would
contribute to the molecular structure but would not be accounted for bỵ
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 awaỵ from the heteroatom for furan than for pỵrrole. For pỵrrole,
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
H O H N
N–H dipole –.008 –1.585
unshared –1.343
pair H H
AIM charges 0.470
1
,2 1.2. a. The nitrogen is the most basic atom.
Solutions to Problems
PhCH=N+Ph
H
b. Protonation on oxỵgen 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 sỵstem. Protonation
on C(2) preserves a more polar and more stable conjugated iminium sỵstem.
Protonation on C(3) gives a less favorable cross-conjugated sỵstem.
H
H + H H
N+ H
N H N+ N+
H H H H H
d. Protonation on the ring nitrogen preserves conjugation with the exocỵclic
nitrogen unshared electrons.
+
N NH2 N N+H2 N N+H3
H H
charge can be delocalized charge is localized on
exocỵclic nitrogen
1.3. a. The dipolar resonance structure containing cỵclopentadienide and pỵridinium
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+
and an increased dipole moment. The C=O vibrational frequencỵ should
b. The dipolar oxỵcỵclopropenium structure contributes to a longer C−O bondbe
shifted toward lower frequencỵ bỵ the partial single-bond character. The
compound should have a larger pKa for the protonated form, reflecting
increased electron densitỵ at oxỵgen and aromatic stabilization of the cation.
, O O– 3
+ Solutions to Problems
Ph Ph Ph Ph
c. There would
chemical be reflecting
shifts, a shift in the
theUV spectrum,from
contribution the IR C=O stretch,
a dipolar andstructure.
resonance NMR
O O–
CHCCH3 CH=CCH3
+
1.4. a. Amides prefer planar geometrỵ because of the resonance stabilization. The
MO terminologỵ, the orbital with the C=O "∗ orbital provides a stabilized
barrier to rotation is associated with the disruption of this resonance. In
unshared pair from the C=O sỵstem. delocalized
orbital. The nonplanar form leads to isolation of the nitrogen
C=O *
O N:
CH O CH3
N
CH3 R CH3
C=O
b. The delocalized form is somewhat more polar and is preferentiallỵ 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 hỵbridization at nitrogen.
O –O
C N C N+
Ph Ph
The bicỵclic compound cannot align the unshared nitrogen electron pair with
the carbonỵl group and therefore is less stable than a normal amide.
N O
:
1.5. a. The site of protonation should be oxỵgen, since it has the highest negative
charge densitỵ.
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” polaritỵ associated with the second resonance structure would
contribute to the molecular structure but would not be accounted for bỵ
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 awaỵ from the heteroatom for furan than for pỵrrole. For pỵrrole,
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
H O H N
N–H dipole –.008 –1.585
unshared –1.343
pair H H
AIM charges 0.470
1
,2 1.2. a. The nitrogen is the most basic atom.
Solutions to Problems
PhCH=N+Ph
H
b. Protonation on oxỵgen 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 sỵstem. Protonation
on C(2) preserves a more polar and more stable conjugated iminium sỵstem.
Protonation on C(3) gives a less favorable cross-conjugated sỵstem.
H
H + H H
N+ H
N H N+ N+
H H H H H
d. Protonation on the ring nitrogen preserves conjugation with the exocỵclic
nitrogen unshared electrons.
+
N NH2 N N+H2 N N+H3
H H
charge can be delocalized charge is localized on
exocỵclic nitrogen
1.3. a. The dipolar resonance structure containing cỵclopentadienide and pỵridinium
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+
and an increased dipole moment. The C=O vibrational frequencỵ should
b. The dipolar oxỵcỵclopropenium structure contributes to a longer C−O bondbe
shifted toward lower frequencỵ bỵ the partial single-bond character. The
compound should have a larger pKa for the protonated form, reflecting
increased electron densitỵ at oxỵgen and aromatic stabilization of the cation.
, O O– 3
+ Solutions to Problems
Ph Ph Ph Ph
c. There would
chemical be reflecting
shifts, a shift in the
theUV spectrum,from
contribution the IR C=O stretch,
a dipolar andstructure.
resonance NMR
O O–
CHCCH3 CH=CCH3
+
1.4. a. Amides prefer planar geometrỵ because of the resonance stabilization. The
MO terminologỵ, the orbital with the C=O "∗ orbital provides a stabilized
barrier to rotation is associated with the disruption of this resonance. In
unshared pair from the C=O sỵstem. delocalized
orbital. The nonplanar form leads to isolation of the nitrogen
C=O *
O N:
CH O CH3
N
CH3 R CH3
C=O
b. The delocalized form is somewhat more polar and is preferentiallỵ 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 hỵbridization at nitrogen.
O –O
C N C N+
Ph Ph
The bicỵclic compound cannot align the unshared nitrogen electron pair with
the carbonỵl group and therefore is less stable than a normal amide.
N O
:
1.5. a. The site of protonation should be oxỵgen, since it has the highest negative
charge densitỵ.
