Chapter 20
Ketones and Aldehydes
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 20. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• The suffix “_____” indicates an aldehydic group, and the suffix “_____” is used
for ketones.
• The position of equilibrium is dependent on the ability of the nucleophile to
function as a _______________.
• In acidic conditions, an aldehyde or ketone will react with two molecules of
alcohol to form an __________.
be ____________ or should bear one ___________ charge.
• _________________ of acetals, imines, and enamines under acidic conditions
produces ketones or aldehydes.
• In acidic conditions, an aldehyde or ketone will react with two equivalents of a
thiol to form a ______________.
alcohols, accompanied by the formation of a new ___________ bond.
CHAPTER 20
469
• Grignard reactions are not reversible, because carbanions do not function as
__________________.
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 20. The answers appear in the section entitled
SkillBuilder Review.
20.1: Naming Aldehydes and Ketones
20.2: Drawing the Mechanism of Acetal Formation
OH
OR
OAH
R O H
A
DRAW A MECHANISM FOR THE ACID-CATALYZED CONVERSION OF A KETONE TO A HEMIACETAL.
MAKE SURE TO DRAW ALL CURVED ARROWS AND INTERMEDIATES.
20.3: Drawing the Mechanism of Imine Formation
H
R
OH
N
R NH
2
OAH
A
DRAW A MECHANISM FOR THE ACID-CATALYZED CONVERSION OF A KETONE TO A CARBINOLAMINE.
MAKE SURE TO DRAW ALL CURVED ARROWS AND INTERMEDIATES.
470
CHAPTER 20
20.4: Drawing the Mechanism of Enamine Formation
H
R
OH
N
OAH
R
2
NH A
DRAW A MECHANISM FOR THE ACID-CATALYZED CONVERSION OF A KETONE TO A CARBINOLAMINE.
MAKE SURE TO DRAW ALL CURVED ARROWS AND INTERMEDIATES.
20.5: Drawing the Mechanism of a Hydrolysis Reaction
STEP 1 – WORKING BACKWARDS,
DRAW ALL ___________________.
STEP 2 – DRAW ALL ___________ AND __________________________________,
USING THE FOLLOWING RULES:
IN ACIDIC CONDITIONS, ALL REAGENTS, ____________________, AND
____________________ SHOULD EITHER BE NEUTRAL OR SHOULD BEAR ONE
POSITIVE CHARGE.
20.6: Planning an Alkene Synthesis with a Wittig Reaction
20.7: Proposing a Synthesis
CHAPTER 20
471
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 20. The
answers appear in the section entitled Review of Reactions.
O
O
O
N
R R
NOH NNH
2
OH
CN
H H
NR
OH
R
SS OH
OO
RO OR
HO OH
Solutions
20.1. a) 5,5-dibromo-2,2-dimethylhexanal
b) (3R,4S)-3,4,5-trimethyl-2-hexanone
20.2.
a)
O
Br
Br
b)
O
c)
H
OBr
472
CHAPTER 20
20.5.
a)
OH O
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
c)
d)
H
O
1)R
2
B H
2) H
2
O
2
, NaOH
e)
O O
H
1) O
3
2) DMS
20.6.
a)
HOH
OC C
H
H
H
H
HO HO
20.7. The carbonyl group in hexafluoroacetone is flanked by two very powerful
CHAPTER 20
473
20.8.
a)
OH O
H
Me
OH
Me
OH
HO O Me
H
Me
OH
HO OMe
H
OHMe
b)
OH O
H
Et
OH
Et
OH
HO O Et
H
Et
OH
HO OEt
H
OHEt
c)
OH O
H
Et
OH
Et
OH
HO O Et
H
Et
OH
HO OEt
H
OHEt
474
CHAPTER 20
d)
OH O
H
Me
OH
Me
OH
HO O Me
H
Me
OH
HO OMe
H
20.9.
a)
OOHHO H OSO
3
H
OOHHO H
HO HO O
H
CHAPTER 20
475
b)
OO OHH
HOH
HO OH HO O
HHO
HO O
HO
HO OH
20.10.
a)
[ H
2
SO
4
]
– H
2
O
O
excess MeOH
MeO OMe
20.11.
a)
O O
HO OH
1) [ H
+
] , , – H
2
O
2) NaNH
2
3) Et
I
4) H
3
O
+
476
CHAPTER 20
20.12. a)
H
3
O
+
OO O
HO OH
+
20.13.
O
OH
H OSO
3
H
O
OH
H
20.14.
O
OH
CHAPTER 20
477
20.15. Note: For each of the mechanisms shown below, the first two steps can be
reversed (first the amine attacks the carbonyl group, and then the tetrahedral
intermediate is protonated). It would be wise to check your lecture notes to determine
if you instructor has a strong preference for this alternate sequence of steps.
a)
b)
OH
NH
Et
H N
H
H
Et
Et
NHH
Et
NHH
OH
N
H
Et H
OH
O
H
478
CHAPTER 20
20.16.
a)
O[ H
+
]
NH
3
– H
2
O
NH
20.17.
a)
H
O[ H
+
]
– H
2
O
NH
2
N
b)
[ H
+
]
– H
2
O
O
NH
2
N
20.18.
a)
NH
2
[ H
2
SO
4
]
– H
2
O
O
+
N
20.19.
a)
OHO NH
2
[ H
+
]
– H
2
O
NOH
CHAPTER 20
479
20.20.
a)
NOH
HO NH
2
[ TsOH ]
– H
2
O
O
20.21. Note: For each of the mechanisms shown below, the first two steps can be
reversed (first the amine attacks the carbonyl group, and then the tetrahedral
intermediate is protonated). It would be wise to check your lecture notes to determine
if you instructor has a strong preference for this alternate sequence of steps.
a)
OH
NEt
Et
H N
H
Et
Et
Et
NEtH
Et
NEtH
OH
OOH
N
H
Et Et
b)
H N
H
Me
Me
Me
NMeH
Me
NMeH HO N
Me
Me
HO N
Me H
Me
OH
O
480
CHAPTER 20
20.22.
a)
O
[ H
+
]
– H
2
O
N H
N
20.23.
a)
[ H
+
]
– H
2
O
ONH N
20.24. a)
N
[ H
2
SO
4
]
– H
2
O
+NH
O
CHAPTER 20
481
20.25. Note: The first two steps of this mechanism can be reversed (first the amine
attacks the carbonyl group, and then the tetrahedral intermediate is protonated). It
would be wise to check your lecture notes to determine if you instructor has a strong
preference for this alternate sequence of steps.
R
1
R
2
NN
H
R
1
R
2
NN
H
R
1
R
2
H N N
H
OH
HOH
OH
OH
N
R
2
R
1
H
N
H N
H
H
NN
NHH
N
NHH
R
1
R
2
OH
R
1
R
2
OOH
NR
2
R
1
H
NH
H
HH
H
HHH
HH
482
CHAPTER 20
20.26.
a)
HOH
H
HOH
HOH
OEt
O O
EtH
HHO O
Et
OO
Et Et
OO
Et Et
H
– EtOH
b)
HO N
ON
H
H
N
H
N
HOH
HOH
H
HOH
c)
OH
NH
NH
NO
N
H H
H
HOH
HOH
H
HOH
CHAPTER 20
483
O
OH
OH
H
H
O
OH
OH
H
O
OH
O
H
HH
HOH
HOH
H
484
CHAPTER 20
20.27.
O O
H OSO
3
H
HO
O O
H
HO
CHAPTER 20
485
20.28.
NNN
N
HOH
H
NNN
N
HNNN
N
H
NNN
N
HH H
O
H
+
H
NNN
N
HH H
O
+
H
H
4 NH
3
+ 6 CH
2
O
20.30.
SS
486
CHAPTER 20
20.32.
a) Below is a mechanism for the Cannizzaro reaction. After a hydroxide ion attacks one
molecule of benzaldehyde, the resulting tetrahedral intermediate functions as a hydride
delivery agent to attack another molecule of benzaldehyde, giving a carboxylic acid and
an alkoxide ion. The alkoxide ion then deprotonates the carboxylic acid, generating a
more stable carboxylate ion. This carboxylate ion is then protonated when an acid is
added to the reaction mixture.
Ph
O
OOH
OH
O
20.33.
a)
HO
b)
OH
c)
HO
O
20.34.
OH OHMe
H
2
SO
4
, H
2
O
1) Na
2
Cr
2
O
7
CHAPTER 20
487
20.35.
1) KCN, HCN
O
HO NH
2
20.36.
a)
OH OH O
OH
H
2
SO
4
, H
2
O
1) Na
2
Cr
2
O
7
3) H
3
O
+
2) KCN, HCl
20.37. a)
O
Ph
3
P
+
b)
OPh
3
P CH
2
+
488
CHAPTER 20
20.38.
Br
1) PPh
3
2) BuLi
1) NaOH
2) PCC, CH
2
Cl
2
20.39. a)
OH 1) PCC, CH
2
Cl
2
2) Ph
3
P=CH
2
20.40.
a)
O
O
b)
OH
O
c)
O
O
CHAPTER 20
489
20.41.
1) BH
3
THF
b)
4) H
2
O
2
, NaOH
2) Ph
3
P=CH
2
1) H
2
SO
4
, H
2
O, HgSO
4
OH
3) BH
3
THF
d)
OH
O
1) NBS,
hv
2) NaOEt
3) O
3
4) DMS
5) excess MeMgBr
6) H
2
O
7) Na
2
Cr
2
O
7
, H
2
SO
4
, H
2
O
490
CHAPTER 20
g)
OEt
O
EtO
O
OOO
HO OH
1) [ H
+
] , , – H
2
O
2) excess LAH
3) H
3
O
+
4) [ H
+
] , – H
2
O
20.42.
a)
Br BrMg
Mg H
O
OH O
NH
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
[ H
2
SO
4
]
NH
3
– H
2
O
1)
2) H
2
O
b)
OH
O
H
CH
2
Cl
2
PCC
OH
1) MeMgBr
2) H
2
O
1) EtMgBr
2) H
2
O
O
c)
Br
PBr
3
OH
O
BrMg
Mg
Br
1)
2) H
2
O
d)
MgBr
O
H
CH
2
Cl
2
PCC
OH
MgBr
Mg
Br
1)
2) H
O
O
1)
2) H
2
O
492
CHAPTER 20
e)
OH
O
H
CH
2
Cl
2
PCC
OH
1) MeMgBr
2) H
2
O
1) EtMgBr
2) H
2
O
O
f)
OH
O
H
CH
2
Cl
2
PCC
OH
1) MeMgBr
2) H
2
O
1) EtMgBr
2) H
2
O
O
g)
Br Mg BrMg
OH
O
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
H
O
1)
2) H
2
O
h)
Br Mg BrMg
OH
O
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
H
O
1)
20.43.
O
(because this carbonyl group is not conjugated)
20.44. a) (2S,3R)-3-methyl-2-propylcyclopentanone
20.45. a)
H
H
O O
b)
H
O
c)
H
O
O
20.46.
H
O
butanal
H
O
2-methylpropanal
494
CHAPTER 20
20.47.
20.48.
O
2-hexanone 3-hexanone 2-methyl-3-pentanone
OO
20.49. The carbonyl group of a ketone will never appear at C-1 because if it would did,
the compound would be called an aldehyde rather than a ketone.
20.51
OP
Ph
Ph
Ph
Ph
H
Ph Ph
+
CHAPTER 20
495
20.52.
X
X
N
20.53.
X Ph
3
P
O
1) PPh
3
2) BuLi
20.54.
a)
OH
O1) MeMgBr
2) H
2
O
496
CHAPTER 20
20.55.
OOH
H
3
O
+
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
20.56. a)
NH
b)
N
c)
EtO OEt
d)
N
20.57.
HO H
OH O
Et
H
HO H
OH
OOH
H
H
CHAPTER 20
497
20.58.
OH
O
OOEt
OEt
EtOH
HO OEt
OEt
[ H
+
] 1) MeMgBr
2) H
O
20.59.
OH
O
H
HH
O
[ H
+
] O
O
20.60. a)
1) LAH
O
2) H
2
O
OH
20.61. a)
O
1) MeMgBr conc. H
2
SO
4
2) H
2
O
OH
heat
498
CHAPTER 20
20.62.
H
O
H
O
H
O
H
O
H
H
HO
H
O
O
H
H
HOH
H
HOH
HOH
20.63.
a)
N
H
3
O
+
O
+ (CH
3
)
2
NH
CHAPTER 20
499
20.64.
excess H
3
O
+
O
O
N
NHO
HO
O
NH
2
NH
O
O
H
+ +
20.65.
500
CHAPTER 20
b)
N
N H
N H
OH
N H
O
H
H
HOH
H
HOHHOH
c)
O
O
H
O
HO
OH
O
OH
O
OH
O
OH
O
H H
HOH
H
HOH
20.66.
a)
[H
+
]
O
N
NH
2
CH
3
N
N
c)
O
CH
3
CO
3
HO
O
f)
O[H
+
]
NH
2
(-H
2
O)
N
20.67.
502
CHAPTER 20
20.68.
O
OH
OH
HO
H
2
SO
4
KMnO
4
NaOH
cold
[ H
+
]
– H
O
20.69.
O
OO OO
O O
HO OH
[ H
+
]
– H
2
O
1) O
3
2) DMS
1) LAH
2) H
2
O
CHAPTER 20
503
O
O
O
HO OH
[ H
+
]
– H
2
O
1) LAH
2) H
2
O
O
O
O O
OH
OH
O O
20.70.
O
O
HO
H
3
O
+
O
OH HO OH
O
20.71.
Br
Br
1) excess NaNH
2
2) H
3
O
+
H
2
SO
4
, H
2
O
HgSO
4
O
504
CHAPTER 20
20.72.
OSO
3
HH
H
N
NH
2
HOH
O
N
N
H
O
H
H
OSO
3
H
N
NH
2
H
HOH
O
N
N
H
OSO
3
H
NH
2
NH
2
N
N
OSO
3
HH
OO
H
O O
+
– H
2
O
CHAPTER 20
505
20.73. Cyclopropanone exhibits significant ring strain, with bond angles of
approximately 60º. Some of this ring strain is relieved upon conversion to the
20.74. 1,2-dioxane has two adjacent oxygen atoms and is therefore a peroxide. Like
other peroxides, it is extremely unstable and potentially explosive.
20.75.
a)
O
O
[ H
+
]
NaOH, cold
OH
OH
H H
O
KMnO
4
– H
2
O
506
CHAPTER 20
NBS
Br
NaOEt
OH
1) BH
3
THF
2) H
2
O
2
, NaOH
hv
d)
OH O KCN
HCl
OH
CN
Na
2
Cr
2
O
7
H
2
SO
4
, H
2
O
H
3
O
+
f)
Br OK
H
O
OH
CH
2
Cl
2
2) H
2
O
2
, NaOH
PCC
1) BH
3
THF
CHAPTER 20
507
H
2
SO
4
, H
2
O
HgSO
4
O O OH OH
1) LAH
2) H
2
O
20.76.
O
Compound A Compound B
a) Three
20.77.
1) O
3
2) DMS
1) EtMgBr
2) H
2
O
O
HO
508
CHAPTER 20
20.78.
H
3
O
+
H
2
CrO
4
OH O
[ H
+
]
20.79.
Br
2
Br
Mg
MgBr
FeBr
3
HH
O
1)
2) H
2
O
OH
20.80. a)
H
O
or H
O
20.81.
O
20.82.
O
CHAPTER 20
509
20.84.
a)
O N
O N HO O O H
H
– (CH
3
)
2
NH
HOH
H
HOH
H
O
O
OH
H
H
O
H
H
O
H H
– CH
3
OH
HOH
510
CHAPTER 20
c) Note: The first two steps of the mechanism below can be reversed (first the amine
attacks the carbonyl group, and then the tetrahedral intermediate is protonated). The
same is true for attack of the second carbonyl group (half-way through the
mechanism). It would be wise to check your lecture notes to determine if you
instructor has a strong preference for this alternate sequence of steps.
H
H
O O OSO
3
HH
HH
O O
HN N
H
H H
H
H
H
O
N
NH
2
H
H
O
N
N
H
HH
OSO
3
H
OSO
3
HH
H
H
O
HO
NH
HNH
2
NN
H
HH
OH
H
CHAPTER 20
511
d)
OH
O
OH
O
O
OH
OSO
3
HH
O
O
OH
e)
HO
O
OH
O
O
OH
H
OSO
3
HH
O
O
OH
512
CHAPTER 20
f)
OO OHH
H
OCH
3
O O
HHO
CH
3
H
3
C
HO OH
20.85.
HH
O
H O
H
H
H H
OH
H H
O
HO O
HH
H H
HH
O