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Page 1
CHAPTER 21 HW: ALDEHYDES + KETONES
NOMENCLATURE
1. Give the name for each compound (IUPAC or common name).
Structure
Name 3,3-dimethyl-2-pentanone (or 1,1-dimethylpropyl methyl
ketone) 5-hydroxy-4-methylhexanal m-nitrobenzaldehyde
Structure
Name 1-penten-3-one
or ethyl vinyl ketone 2,4-dioxohexanal
2-methyl-5-heptyn-3-one
2. Draw each compound.
Structure
Name dicyclohexyl ketone 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”.
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.
OOH
H
O
14
56
NO2
CHO
O
135H
O
O O
12 4
O7
65 4 13
O
O
F
FF
H
O CH3zame zide
HC
CC
CH
O
H H
H H
HH
Signal at 2.40 ppm (next to carbonyl)
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4. Of compounds A-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.
O
OCH3
Br
O
Br
O
OCH3
Br
O
BrA B C D
012345678910PPM
1H
1H 1H 1H
2H2H
020406080100120140160180200220PPM
H
O
a e
f
bd
c e/f assignment could be switchedc is a wider ~doublet so
must
involve a trans coupling (to b).
ab c d e
f
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Page 3
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).
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.
C=O Stretch 1666 cm-1 C=O: 1692 cm-1
WITTIG REACTION
8. Give the curved arrow mechanism for each reaction.
O O O O O
vs.
O O
a.H3C
CPPh3
H
CH3CH2Bra. PPh3
b. nBuLi H3CC
PPh3
H
CH3CH2BrPPh3SN2 H3C C
H
H
PPh
PhPh
(nBuLi)H3C C
H
PPh
PhPh
b. O H3C CH PPh3
O
H3C CH PPh3
O
CPPh3
H3CH
O
CPPh3
H3CH
CH
CH3Ph
PPh
O
Ph
Both compounds have very conjugated C=O bonds, and conjugation
lowers the IR wavenumber (more single bond character). The first
compound has twice as much conjugation / resonance, so it’s C=O
signal is the lowest.
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Page 4
8 continued
9. Draw both stereoisomers that can be formed in this Wittig
reaction.
10. Give the major product of each reaction (one stereoisomer is
sufficient).
11. Use a Wittig reaction to produce each alkene, starting from
an ylide.
c. H
OPPh3
PPh3
Ph H
O
PhH
PPh3O
PhH
PPh3O
Ph
H
O
H
Ph3P
H Htranscis
O CH3CH2CH=PPh3a.C
CH2CH3H
c. CHO OCH3
OPh3P
O
OCH3
b.O PPh3
d.H2C PPh3
O H2C
a. C CCH2
CH2 CH2
CH3
CH3 CH3
H
b.
C O
CH2
CH2
CH3
CH3Ph3P C
CH2 CH3
H
+product
C PPh3CH2
CH2
CH3
CH3O C
CH2 CH3
H
+product
Route 1
Route 2
product
product
Route 1
Route 2
O
PPh3
HH
PPh3
O
HH
PPh3
HH
or
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Page 5
12. Use a Wittig reaction to produce this alkene, starting from
an alkyl halide.
HYDRATES
13. Give the curved arrow mechanism that shows the formation of
hydrate under each set of conditions.
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.
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.
Br
BrPPh3
PPh3nBuLi
PPh3
O
PPh3 Ph3P nBuLi Ph3P
H
ORoute 1
Route 2
a.H
O trace OH-
H2O H
HO OH
H
OO H
H
HO O HO
HH
HO OH
b.H
O
H
HO OHtrace H+
H2O
H
Otrace H+
H
OH
HO
H
H
O OHH
H
HO
H
H
HO OH
O
H
O
E F
O
HO
E-hydrateF-hydrateE
F H
OHOH
OHOH
O
H
O
HOne EDG
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Page 6
15. Compound G has a smaller equilibrium constant (Keq) for
hydrate formation than compound H. Explain this trend, including
energy diagrams with your explanation.
K = hydrate / carbonyl; so a large Keq means a higher % of
hydrate. Carbonyl with CF3 has more hydrate.
The reactivity differences arise from different starting
carbonyl energies (the hydrate energies are also nearly
equivalent). CF3 is a strong EWG, so destabilizes the δ+ of the
carbonyl, 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.
Pair A
Brief Explanation: 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 B
Brief Explanation: 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 C
Brief Explanation: 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.
Pair D
Brief Explanation: 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.
H
O
CH3 H
O
CF3
Keq= 1.06 Keq= 2.9 x 104G H
O O
O O
H
O
CH3
H
O
OCH3
H
O
H3N
H
O
H
O
R H R
HO OH
Keq
H2O
H
O
CF3 H
O
CF3EWG
G G+H-hydrate
H
H
O
CH3
H
O
CF3
H CH3
HO OHH CF3
HO OH
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Page 7
ACETALS
17. Give the curved arrow mechanism for this acetal formation
reaction.
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 OCH2CH3OCH2CH3
H+
CH3CH2OH
OH+
OH
CH3CH2OH
OCH2CH3OH
H
CH3CH2OHOCH2CH3OH
H+
OCH2CH3OH2
OCH2CH3
OCH2CH3
CH3CH2OH CH3CH2OH
OCH2CH3OCH2CH3
H
CH3CH2OH
product
Protonate Attack Deprotonate
Leave Attack
Deprotonate
Protonate
O
H
OO
H
HO OH
H+
OO
H
O
H
H+
H
O H
HO OHO
HHO
HHO HO OH O
HHOHO
H+
O
HH2OHO
O
H
HO O
H
HO O O
H
H HO OH
Protonate Attack Deprotonate Protonate
Leave Attack Deprotonate
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Page 8
20. Give the major organic product for each reaction.
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
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Page 9
23. Give the curved arrow mechanism for each reaction.
a.OCH3
OCH3
H+
H2O
O
H+ CH3OH
OCH3
OCH3
H+ OCH3
OCH3H
OCH3 OCH3H2O
OCH3
O
H
H H2O
OCH3
OH
H+ O
OH
H CH3 O
H
H H2Oproduct
Protonate Leave Attack
Deprotonate
Protonate Leave Deprotonate
b.O
O H+
H2O
O+
OH
OH
O
O H+
O
O
H
OOHH
H2O
O
OH
O
OH
O
H
H
H2O
O
OH
OHH+
O
OH
OHH
OH
OH
H2O
acetone
c.O
OCH2CH3H+
H2O
O
HHO + CH3CH2OH
OOCH2CH3
H+
OO
CH2CH3
H O(Mech could be begun by protonating either oxygen)
H2O OO
H
H
H2O
OOH
H+
OOH
H
O
HO
HH2O O
HO H
d.O
O
OH H+
O
O OH
O
O
OH
H+
O
O OH
O
O
OHH
O
OH
HO
product
1
3
4 1
34
H HOR
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Page 10
24. Milder conditions can be used to hydrolyze acetal J than to
hydrolyze acetal K. Explain their difference in reactivity.
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.
26. The following multi-step synthesis converts benzene into a
dicarbonyl species. a. Give the reagents needed to complete each
step in the sequence.
b. Briefly explain why the synthesis below does not work
well.
OCH3
OCH3 OCH3
OCH3
J K
O
O
BrBr
O
Br
O O
O O
OHMgBr
O OO O
O
Br2
FeBr3
Cl
O
AlCl3
HOOH
H+, heat
Mg
b) H+ workup
CrO3, H+or PCCH+, H2O
heat
H
Oa)
OO
O
AlCl3
Cl
O
AlCl3
Cl
O
Br
O
HOOH O
H+
HOOH H+, heat (protect aldehyde)
BrH
O Oa) Mg
b) O
c) H+ workup
OH
H
O O
H+, H2O, heat(hydrolysis)
Step 2 should have problems. The carbonyl is a meta director,
and also Friedel Crafts reactions don’t work well on deactivated
rings.
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Page 11
IMINES + ENAMINES
27. Give the curved arrow mechanism for each reaction.
Oa.
CH3NH2trace H+ N
CH3
+ H2O
O
NCH3
+ H2ONH CH3
H+ O H
CH3NH2HO N CH3
HHCH3NH2 HO N CH3
H
H+
H2O N CH3H
NCH3H
CH3NH2
O
HNCH3
CH3trace H+ N
H3C CH3
+ H2Ob.
O H+ O H
HNCH3
CH3HO N H
H3CCH3
HNCH3
CH3
HO N
H3CCH3
H+
H2O N
H3CCH3 N CH3H3C
H
HNCH3
CH3
NCH3H3C
c. H
O
O
H
NH2
NH2trace H+ N
N+ H2O
NH2
NH2 NH2
NH H
O
OH
H
H
O
O
HH+ H
O
O
H
H
RNH2
NH2
HN
O
OH
H
H+
NH2
HN
O
OH2
HNH2
N
O
H
H
RNH2
NH2
N
O
H
NH2
N
O
H
H+
NH2
N
O
H
H
(Half-way point)
N
N
H HOH
RNH2
NH
N
OHH+
NH
N
OH2N
N
HRNH2
N
N
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Page 12
28. Draw both stereoisomers that can be formed in this
reaction.
29. Give the major organic product for each reaction (one
stereoisomer is sufficient).
30. Give the curved arrow mechanism for each reaction.
O
NH2
mild acid NN
E Z
a. CHOCH3NH2trace H+ N
CH3 d.
Otrace H+
NH2N
O(CH3)2NH
pH 5b.
N
CH3
CH3O
e.
trace H+
NH
N
NH
O
H
trace H+c. N
H
f.O NH3
trace H+
NH
+ CH3CH2NH2a.H+
H2ON
CH2CH3O
NCH2CH3
H+
O
NCH2CH3
HNCH2CH3
H
OHH
NCH2CH3
H
OH
H+
NCH2CH3
H
OHH O O
H H
H2O
H2O
H2O
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Page 13
30 continued
31. Give the major organic product for each reaction.
b. NH+
H2O
O
H+ CH3CH2NH2
N
+ CH3CH2NH2
H+
N
H
H2ON
H
OHH
H2O
N
H
OH H+
N
OH
HH
OH OH
HH
H2O O
H
NH3C CH3
c. H+
H2OO
+HNCH3
CH3
NMeMe
H+
NMeMe
H
H2O
NMeMe
OH H
H2O
NMeMe
OH
H+
NMeMe
OH
H
OH
OH H2O
O
a. NCH3
O
+ CH3NH2
H+
H2O d. H+
H2OPh H
NHN
CH3
Ph H
O+ CH3NH-NH2
b.N O NH2H+
H2Oe.
NH+
H2O
NH2
H
O
NCH3
CH3c.
H+
H2OO N
CH3
CH3
H NH3C Ph
H+
H2Of. H
NH3C Ph O
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Page 14
REACTION SUMMARY
32. Give the major organic product for each reaction.
KOHa. OH
O
HO
O
O
HO
b.a. NaNH2
b. cyclopentanonec. H+ workup
O OH
c.H+O
HCH3CH2OH H
OCH2CH3OCH2CH3
d. Bra. Li (s)b. pentanalc. H+ workup
Li H
OOH
e.
O
Ph3P=CHCH3
H3C
(The Z isomer is also OK)
NaBD4f.
CH3CH2ODO
OD
D
g.cat. H+
CH3CH2NH2
O NCH2CH3
h.H+
H2O
O O+ 2 CH3CH2OH
H
O
O
Ha. (CH3CH2)2CuLi
b. H+ workupi.
O
H
