Download File

C HAPTER 21 HW: A LDEHYDES + K ETONES
NOMENCLATURE
1. Give the name for each compound (IUPAC or common name).
CHO
OH
6
O
5
Structure
1 H
4
NO2
O
3,3-dimethyl-2-pentanone
Name
(or 1,1-dimethylpropyl methyl ketone)
3
5
Structure
5-hydroxy-4-methylhexanal
O
1
H 1
1-penten-3-one
or ethyl vinyl ketone
2
4
O
O
6
5
4
7
O
Name
m-nitrobenzaldehyde
2,4-dioxohexanal
3
1
O
2-methyl-5-heptyn-3-one
2. Draw each compound.
O
O
zame zide
F
Structure
F
dicyclohexyl ketone
Name
CH3
H
F
O
1,1,1-trifluoro-3-pentanone
(Z)-3,6-dimethyl-3-heptenal
SPECTROSCOPY
3. In the 1H NMR spectrum of butanal, the signal at 2.40 ppm is a triplet of doublets (approximate J’s
are 2 Hz and 7 Hz). Explain the splitting of this signal, including a sketch of a “tree diagram”.
O
H
C
H H
C
C
C
H H
H
Signal at 2.40 ppm
(next to carbonyl)
H
H
The CH2 next to the carbonyl is the signal at 2.40 ppm.
It is split in a large way by its two CH2 neighbors (7 Hz,
splitting first level into a triplet), and in a smaller way by
the aldehyde H (2 Hz, splitting second level into a
doublet). This results in a triplet of doublets.
Page 1
4. Of compounds A-D,
O
O
OCH3
O
O
OCH3
Br
Br
A
Br
Br
B
C
D
a. Which compound would have the following 13C NMR spectra (briefly explain)?
δ = 165.9, 132.1, 131.5, 129.1, 127.4, 51.5 ppm
Compound A
165.9 ppm is the C=O, and the low range means it could be part of an ester, carboxylic acid or
amide. The 51.5 ppm signal is the carbon next to oxygen (could be A or B). There are only 6
total 13C signals, so it is A (B would have 8 13C signals).
b. Which compound would have the following 13C NMR spectra (briefly explain)?
δ = 199.8, 140.3, 135.4, 131.5, 131.0, 127.7, 121.2, 28.6 ppm
Compound D
199.8 ppm is the C=O, and a high range means it could part of an aldehyde or ketone. The 28.6
ppm signal is low, and so is not next to oxygen (could be C or D). There are 8 total 13C signals,
so it is D (C would have 6 13C signals).
5. Below are 1H and 13C NMR spectra of a compound with a formula of C5H8O. Determine the structure,
then assign the peaks in the 1H NMR spectrum.
1H
a
10
220
b
1H
9
200
8
180
7
O
aH
5
PPM
140
120
c
f
b
PPM
4
100
3
80
2H
f
2
60
1
40
20
0
0
e/f assignment could be switched
d
e
c d
1H 1H
6
160
2H
e
c is a wider ~doublet so must
involve a trans coupling (to b).
Page 2
6. Cyclohexanone has a strong signal in its IR spectrum at 1718 cm-1, while 2-cyclohexenone has a
strong signal at 1691 cm-1. Both signals represent vibration of the same kind of bond. Explain why
the absorption in 2-cyclohexenone is at a lower wavenumber, including resonance structures.
Both signals represent the IR stretching of the C=O bonds.
2-cyclohexenone has a lower wavenumber absorbance, as it has a different C=O bond strength than
cyclohexanone. All carbonyls have a +/- resonance structure (shown with cyclohexanone below), but
2-cyclohexenone has an additional resonance structure, which causes its resonance hybrid to have
more “single bond character.” With greater “single bondedness,” the C=O is weaker, and thus
absorbs at a lower wavenumber (since v α f).
O
O
O
O
O
vs.
7. An IR is taken of a mixture of the two compounds below. Two strong signals are noted in the mixed
IR spectrum at 1666 and 1692 cm-1, representing the carbonyl stretching modes. Which signal
corresponds to which compound? Briefly explain.
Both compounds have very conjugated
C=O bonds, and conjugation lowers the
O
O
IR wavenumber (more single bond
character). The first compound has
twice as much conjugation / resonance,
so it’s C=O signal is the lowest.
C=O Stretch
1666 cm-1
C=O: 1692 cm-1
WITTIG REACTION
8. Give the curved arrow mechanism for each reaction.
H
a.
a. PPh3
CH3CH2Br
PPh3
H3C
b. nBuLi
CH3CH2Br
S N2
H
H3C C
O
H
PPh3
H3C
C
PPh3
Ph
P
H
b.
C
Ph
Ph
Ph
H3C C
(nBuLi)
H
P
Ph
Ph
H3C CH PPh3
O
C
H3C CH PPh3
O
O
H3C
PPh3
H
C
H3C
O
H
PPh3
H
C
CH3
Ph
P
Ph
Ph
Page 3
8 continued
O
c.
PPh3
H
PPh3
O
Ph
Ph
H
Ph
H
PPh3
O
PPh3
O
Ph
H
H
9. Draw both stereoisomers that can be formed in this Wittig reaction.
Ph3P
O
H
H
cis
trans
H
10. Give the major product of each reaction (one stereoisomer is sufficient).
H
O
CH3CH2CH=PPh3
a.
O
C
O
CH2CH3
Ph3P
c.
OCH3
CHO
PPh3
OCH3
H2C PPh3
d.
b.
O
H2C
O
11. Use a Wittig reaction to produce each alkene, starting from an ylide.
CH3 CH2
a.
C
CH3 CH2
CH2 CH3
C
CH3 CH2
C
Route 1
H
CH2 CH3
O
+
Ph3P
C
CH3 CH2
CH3 CH2
Route 2
C
CH2 CH3
PPh3
+
O
C
CH3 CH2
O
b.
H
H
H
or
PPh3
PPh3
product
H
Route 1
Route 2
product
H
H
H
product
PPh3
H
product
O
Page 4
12. Use a Wittig reaction to produce this alkene, starting from an alkyl halide.
H
Br
PPh3
PPh3
Route 1
PPh3 O
nBuLi
O
PPh3 Ph3P
Br
Route 2
nBuLi Ph3P
HYDRATES
13. Give the curved arrow mechanism that shows the formation of hydrate under each set of conditions.
O
trace OH-
HO
OH
a.
H2O
H
H
HO
O
O
H
O
H
HO
trace H+
HO
OH
H
H
H
H
O
O
OH
b.
H2O
H
H
O
O
trace
H
O
H+
H
H
H
H
H
H
H
O
OH
O
H
HO
OH
H
H
14. Compound E has a higher percentage of hydrate relative to carbonyl in aqueous solution than
compound F. Explain this trend, including energy diagrams with your explanation.
O
O
O
H
H
E
F
E
F-hydrate
E-hydrate
O
F
OH
OH
OH
OH
H
The difference in reactivity often
arises from the relative energies of
the carbonyl species (starting reactant energies). The carbonyl carbon is δ+, and EDG lower the
energy. The aldehyde has one alkyl group (EDG) attached to the C=O, but the ketone has 2 EDG.
Therefore, the ketone stabilizes the δ+ more and starts at a lower energy than the aldehyde. This
causes the hydrate reaction to be uphill for the ketone (so higher hydrate % for aldehyde).
Although it’s only a minor factor, we can also be complete by
noting that the ketone hydrate energies should also be somewhat
higher E because 2 alkyl groups are more crowded than just one.
This would make the ketone reaction even more uphill.
O
O
H
H
One EDG
Page 5
15. Compound G has a smaller equilibrium constant (Keq) for hydrate formation than compound H.
Explain this trend, including energy diagrams with your explanation.
O
H
O
O
CH3
G
Keq= 1.06
H
CF3
O
O
H
CF3
H
H
Keq= 2.9 x 104
CF3
H
R
H2O
HO
Keq
H
OH
R
EWG
K = hydrate / carbonyl; so a large Keq means a higher % of hydrate. Carbonyl with CF3 has more
hydrate.
The reactivity differences arise from
O
different starting carbonyl energies (the
H CF3 H
hydrate energies are also nearly
HO OH HO OH
equivalent). CF3 is a strong EWG, so
O
H CH3 H CF3
destabilizes the δ+ of the carbonyl,
H CH3 G
G+H-hydrate
making the CF3 carbonyl higher in
energy than the aldehyde. This makes
the hydrate reaction of the CF3 carbonyl downhill, resulting in a higher amount of hydrate.
16. In each pair, predict which would have a greater percentage of hydrate relative to carbonyl when the
two forms are at equilibrium in water. Briefly explain each answer.
Brief Explanation:
O
O
The aromatic group can participate in
resonance with the C=O which greatly
stabilizes the δ+ of the carbonyl. Therefore,
the aromatic carbonyl starts at a lower energy,
and reacts less (reaction is more uphill).
Pair
A
Brief Explanation:
O
O
C=O energies are probably similar. But
hydrate energies are more sensitive to steric
issues. The dicyclohexyl compound’s hydrate
will be higher energy, so reaction is more
uphill.
Pair
B
Brief Explanation:
O
Pair
C
O
H
H
CH3
OCH3
Brief Explanation:
O
Pair
D
O
H
H3N
Methoxy is a stronger EDG than methyl
(resonance, not hyperconjugation), so best
stabilizes the δ+ of the carbonyl.
The
methoxy carbonyl starts at a lower energy, so
is more uphill.
H
The aldehyde is more reactive (makes more
hydrate) because it has a higher energy
carbonyl form (reaction is downhill). The
aromatic has an EWG (-NH3+), which
destabilizes the δ+ of the carbonyl.
Page 6
ACETALS
17. Give the curved arrow mechanism for this acetal formation reaction.
O
OCH2CH3
H+
OCH2CH3
CH3CH2OH
H
O
O
H+
H
CH3CH2OH
Attack
Protonate
OCH2CH3
O
OH2
Leave
OCH2CH3
OCH2CH3
OH CH3CH2OH
OH
H+
Deprotonate
O
CH2CH3
OCH2CH3
Deprotonate
OCH2CH3
CH2CH3
product
H
Attack
CH3CH2OH
CH3CH2OH
Protonate
CH3CH2OH
Note: attack of either resonance structure is acceptable; you don’t need to show both.
18. Explain why acid is catalytic in the formation of an acetal. (Use the mechanism in the previous
problem.)
Acid is a catalyst because it satisfies both criteria:
•
•
Acid is unconsumed. For every protonate step where it is used, there is a deprotonate step where it
is regenerated.
Acid lowers the activation barrier of the reaction. It makes the carbonyl more reactive to attack, as
it puts charge on the carbonyl, making the carbon of the carbonyl more δ+ (starts at a higher
energy, thus lowering Ea).
19. Give the curved arrow mechanism for this reaction.
O
H+
O
H
H
Protonate
HO
H2O
H
HO
HO
O
H
Leave
OH
H
HO
HO
H
HO
O
H
H
Attack
O
O
O
H+
H
O
OH
HO
H
Attack
OH HO
HO
O
H
Deprotonate
H
O
HO
O
O HO
H
OH
Deprotonate
H+
Protonate
O
O
H
Page 7
20. Give the major organic product for each reaction.
O
CH3CH2O
OCH2CH3
H+
a.
H
O
O
d.
O
OCH3
CHO
H+
OH
H+
O
OH
O
b.
CH3CH2OH
HO
c.
H
H+
OCH3
CH3OH
O
21. In order to achieve good yields for most acetal formations, they need to be driven by Le Châtelier’s
Principle. Explain why good yields are easier to achieve when reacting aldehydes than when reacting
ketones.
Acetal formation reactions have nearly the same energetics as hydrate formation. For aldehydes, K=1
for acetal formation, while ketones K<1. Ketones have a more stabilized carbonyl (2 EDG stabilizing
the δ+ of the carbonyl), so their acetal reaction is uphill. This means it is “easier” to get good yields
with aldehydes (K=1) compared to ketones (uphill).
22. Show all organic products of these reactions.
H+
OCH3
H2O
a.
b.
O
OCH3
O
d.
H+
H3CO
O
H2O
O
c.
O
CH3OH
O
OH
f.
OH
+ CH3CHO
H3CO
+ CH3CH2OH
OH
H+
H3CO
OCH3
H+
H2O
e.
HO
H2O
O
+ 2 CH3OH
O
H
H+
OCH3
H2O
O + 2 CH OH
3
O
H+
OCH3 H2O
H3CO
H
+ 2 CH3OH
Page 8
23. Give the curved arrow mechanism for each reaction.
OCH3
OCH3
+ CH3OH
H2O
OCH3
H
OCH3
H+
OCH3
OCH3
Protonate
H
O
CH3
O
H+
OH
OH
Leave
O
+
H2O
OH
O
O
OH
OH
OH
O
H2O
acetone
O
+ CH3CH2OH
HO
H2O
H+
H
CH2CH3
OCH2CH3
O
H
O
O
O
H
H2O
O
H2O
O
(Mech could be begun by
protonating either oxygen)
OH
H
O
H+
H
OH
O
O
HO
H
d.
O
H
H
OH
H+
O
O
H
H
O
H
OCH2CH3
c.
O
OH
H+
H2O
H2O
O
O
OH
OH
OH
O
H+
O
H2O
H
Deprotonate
H
O
H
O
product
OH
Protonate
H+
O
H2O
Attack
H
OCH3
H HO
2
b.
OCH3
OCH3 Leave
H
Deprotonate
OCH3
O
O
H+
a.
O
H
HO
O
H+
OH
H2O
O
OH
O
HO
H+
H
O
OH
O
1
O
4
3
OH
O
O
4
OH
O
3
O
H
product
1
OH
HOR
Page 9
24. Milder conditions can be used to hydrolyze acetal J than to hydrolyze acetal K. Explain their
difference in reactivity.
OCH3
OCH3
OCH3
OCH3
J
K
Acetal hydrolysis reactions have similar energetics to hydrate reactions. Since “J” produces a ketone
instead of an aldehyde (where the carbonyl is more stabilized by 2 EDGs), it’s a more favorable
hydrolysis reaction (reaction is “easier” and can use milder conditions).
PROTECTING GROUPS
25. Design a synthesis that uses cyclopentanone and 4-bromobutanal to efficiently produce the aldehyde
shown.
+
O
Br
O
OH
H
H
O
OH
HO
H+, H2O, heat
(hydrolysis)
H+, heat (protect aldehyde)
H
Br
O
a) Mg
H
O
b)
OH
O
O
O
c) H+ workup
26. The following multi-step synthesis converts benzene into a dicarbonyl species.
a. Give the reagents needed to complete each step in the sequence.
O
Br
Br2
Br
Br
Cl
AlCl3
FeBr3
OH
HO
H+, heat
O
O
O
Mg
OH
O
O
H+, H2O
H
heat
b) H+ workup
O
O
O
MgBr
O
a)
CrO3, H+
or PCC
O
O
O
O
b. Briefly explain why the synthesis below does not work well.
O
O
Cl
AlCl3
O
O
Cl
AlCl3
O
Step 2 should have
problems. The carbonyl
is a meta director, and
also
Friedel
Crafts
reactions don’t work well
on deactivated rings.
Page 10
IMINES + ENAMINES
27. Give the curved arrow mechanism for each reaction.
CH3NH2
a.
+ H2O
N
trace H+
O
CH3
H+
CH3NH2
O
N
H2O
O
H+
CH3
N
H
H
N
HO
H
H
N
CH3NH2
N
HO
H
CH3
H
CH3NH2
CH3
H
CH3
CH3
N
+ H2O
CH3
CH3
b.
H
N
CH3
trace
O
+ H2O
H+
N
H3C
CH3
CH3
CH3
O
H
O
H+
HO
H
N
CH3
H3C
N
H3C
O
c.
H
CH3
H
H
O
O
H
N
NH2
H
O
NH2
NH2
NH2
NH2
H
O
NH2
H
OH
O
H
N
O
NH2
H
O
(Half-way point)
H
N
N
N
H
H
H
OH
RNH2
N
H
OH
N
N
H
H+
O
N
OH2
H+
H
N
H
O
N
CH3
NH2
H
RNH2
H+
N
H
OH RNH
2
N
N
NH2
H
N
N
CH3
N
H
OH2
H3C
+ H2O
H
H
H+
N
N
O
H
N
H3C
trace H+
O
HO
CH3
CH3
NH2
O
CH3
H
H+
NH2
H
N
H3C
CH3
N
N
H
CH3
H
H
H2O
N
N
RNH2
Page 11
28. Draw both stereoisomers that can be formed in this reaction.
NH2
E
N
mild acid
O
Z
N
29. Give the major organic product for each reaction (one stereoisomer is sufficient).
CH3NH2
CHO
a.
CH3
N
trace H+
NH2
d.
N
trace H+
O
CH3
O
b.
N
(CH3)2NH
O
N
N
H
e.
CH3
trace H+
pH 5
O
c.
N
H
O
f.
N
H
NH
NH3
trace H+
trace H+
H
30. Give the curved arrow mechanism for each reaction.
H+
N
a.
CH2CH3
H+
N
CH2CH3
O
H2O
H
N
+ CH3CH2NH2
H
H2O
H
N
CH2CH3
O
H
H
N
CH2CH3
H2O
H+
CH2CH3
OH
H
N
OH
H
CH2CH3
O
H2O
O
H
O
H
Page 12
30 continued
O
H+
b.
N
+ CH3CH2NH2
H2O
H
H
H
H+
OH
OH
N
H
O
H
H
H
O
H2O
H
CH3
H2O
O
CH3
N
H
H
H
O
Me
Me
N
CH3
H
H
H
O
Me
Me
N
Me
Me
N
Me
Me
H+
O
O
H
O
H2O
H2O
N
H
H
+
H+
Me
N
H
H+
H3C
N
+ CH3CH2NH2
c.
N
H+
OH
H2O
H2O
N
N
H
O
N
H
Me
O
H
H2O
31. Give the major organic product for each reaction.
a.
H+
N
CH3
N
b.
O
H2O
d.
+ CH3NH2
H+
O
N
Ph
NH2
H
N
CH3
O
H+
+ CH3NH-NH2
H2O
H
N
Ph
H
NH2
H+
e.
H2O
H2O
O
H
CH3
c.
N
CH3
H+
H2O
H
O
N
CH3
CH3
H3C
f.
N
Ph
H+
H2O
H3C
N
H
Ph
O
Page 13
REACTION SUMMARY
32. Give the major organic product for each reaction.
O
O
a.
OH
KOH
O
HO
HO
O
a. NaNH2
b.
b. cyclopentanone
c. H+ workup
O
OCH2CH3
OCH2CH3
H+
c.
CH3CH2OH
H
H
OH
O
a. Li (s)
Br
d.
Li H
b. pentanal
c. H+ workup
H3C
O
Ph3P=CHCH3
e.
O
f.
(The Z isomer is
also OK)
OD
NaBD4
D
CH3CH2OD
N
O
CH2CH3
cat. H+
g.
CH3CH2NH2
O
O
H+
h.
H2O
O
i.
OH
O
+ 2 CH3CH2OH
H
O
a. (CH3CH2)2CuLi
H
b. H+ workup
H
Page 14