Download OC 2/e Ch 15

16
Organic
Chemistry
William H. Brown &
Christopher S. Foote
16-1
16
Aldehydes
&
Ketones
Chapter 16
16-2
16 The Carbonyl Group
 In
this and several following chapters we study
the physical and chemical properties of classes
of compounds containing the carbonyl group,
C=O
• aldehydes and ketones (Chapter 16)
• carboxylic acids (Chapter 17)
• acid halides, acid anhydrides, esters, amides (Chapter
18)
• enolate anions (Chapter 19)
16-3
16 The Carbonyl Group
 The
carbonyl group consists of
• one sigma bond formed by the overlap of sp2 hybrid
orbitals, and
• one pi bond formed by the overlap of parallel 2p
orbitals

C  O
16-4
16 The Carbonyl Group
• pi bonding and pi antibonding MOs for formaldehyde.
16-5
16 Structure
• The functional group of an aldehyde is a carbonyl
group bonded to a H atom and a carbon atom
• The functional group of a ketone is a carbonyl group
bonded to two carbon atoms
O
O
O
HCH
CH3 CH
CH3 CCH3
Methanal
(Formaldehyde)
Ethanal
(Acetaldehyde)
Propanone
(Acetone)
16-6
16 Nomenclature
 IUPAC
names:
• the parent chain is the longest chain that contains the
functional group
• for an aldehyde, change the suffix from -e to -al
• for an unsaturated aldehyde, show the carbon-carbon
double bond by changing the infix from -an- to -en-;
the location of the suffix determines the numbering
pattern
• for a cyclic molecule in which -CHO is bonded to the
ring, name the compound by adding the suffix carbaldehyde
16-7
16 Nomenclature: Aldehydes
O
O
H
3-Methylbutanal
1
5
7
H
2-Propenal
(Acrolein)
8
6
CH3
2
CH3
2,2-Dimethylcyclo- Benzaldehyde
hexanecarbaldehyde
3
4
2
H
(2E)-3,7-Dimethyl-2,6-octadienal
(Geranial)
CHO
CHO
O
1
C6 H5
CHO
trans-3-Phenyl-2-propenal
(Cinnamaldehyde)
16-8
16 Nomenclature: Ketones
 IUPAC
names:
• select as the parent alkane the longest chain that
contains the carbonyl group
• indicate its presence by changing the suffix -e to -one
• number the chain to give C=O the smaller number
O
O
Propanone
(Acetone)
1
O
1
3
5
5
6
5-Methyl-3-hexanone
1-Phenyl-1-pentanone
16-9
16 Order of Precedence
compounds that contain more than one
functional group indicated by a suffix
Increasing precedence
 For
Functional
Group
Suffix If Higher Prefix If Lower
in Precedence in Precedence
- COOH
-oic acid
- CH O
C= O
-al
-one
oxooxo-
- OH
-ol
hydroxy-
- SH
-thiol
-amine
-sulfanyl
-amino
- NH2
16-10
16 Common Names
• for an aldehyde, the common name is derived from
the common name of the corresponding carboxylic
acid
• for a ketone, name the two alkyl or aryl groups
bonded to the carbonyl carbon and add the word
ketone
O
O
O
H
H
H
OH
Formaldehyde Formic acid
O
Ethyl isopropyl ketone
O
H
Acetaldehyde
OH
Acetic acid
O
O
Diethyl ketone
Dicyclohexyl ketone
16-11
16 Physical Properties
 Oxygen
is more electronegative than carbon (3.5
vs 2.5) and, therefore, a C=O group is polar
Polarity of a
carbonyl group
+
C
O: –
:
O
:
C
C
O:
:
+ -
More important
contributing
structure
• aldehydes and ketones are polar compounds and
interact in the pure state by dipole-dipole interaction
• they have higher boiling points and are more soluble
in water than nonpolar compounds of comparable
molecular weight
16-12
16 Reaction Themes
 One
of the most common reaction themes of a
carbonyl group is addition of a nucleophile to
form a tetrahedral carbonyl addition compound
:
: O:
R
+
C
R
O:
:
N u: -
Nu
-
C
R
R
Tetrahedral carbonyl
addition compound
16-13
16 Reaction Themes
A
second common theme is reaction with a
proton or Lewis acid to form a resonancestabilized cation
R
+
C O H + :B
:
R
O: + H- B
:
C
fast
R
R
• protonation in this manner increases the electron
deficiency of the carbonyl carbon and makes it more
reactive toward nucleophiles
16-14
16 Add’n of C Nucleophiles
 Addition
of carbon nucleophiles is one of the
most important types of nucleophilic additions
to a C=O group; a new carbon-carbon bond is
formed in the process
 We study addition of these carbon nucleophiles
RMgX
A Grignard
reagent
RLi
An organolithium
reagent
RC C An anion of a
terminal alkyne
-
C N
Cyanide
ion
16-15
16 Grignard Reagents
 Given
the difference in electronegativity between
carbon and magnesium (2.5 - 1.3), the C-Mg
bond is polar covalent, with C- and Mg+
• in its reactions, a Grignard reagent behaves as a
carbanion
 Carbanion:
an anion in which carbon has an
unshared pair of electrons and bears a negative
charge
• a carbanion is a good nucleophile and adds to the
carbonyl group of aldehydes and ketones
16-16
16 Grignard Reagents
 Addition
of a Grignard reagent to formaldehyde
followed by H3O+ gives a 1° alcohol
-
-
+
O
CH3 CH2 -MgBr + H- C-H
ether
+
Formaldehyde
-
+
O [ Mg Br ]
OH
HCl
CH3 CH2 -CH2
CH3 CH2 -CH2 + Mg2+
H2 O
A magnesium
1-Propanol
alkoxide
(a primary alcohol)
16-17
16 Grignard Reagents
 Addition
to any other RCHO gives a 2° alcohol
O
-
O [ Mg Br ]
- +
ether
Mg Br + CH3 - C-H
+
+
CHCH3
A magnesium
alkoxide
Acetaldehyde
(an aldehyde)
OH
HCl
H2 O
2+
CHCH3 + Mg
1-Cyclohexylethanol
(a secondary alcohol)
16-18
16 Grignard Reagents
 Addition
to a ketone gives a 3° alcohol

O
 
C6 H5 Mg Br + CH3 -C- CH3

Acetone
O - [ MgBr ] +
C6 H5 CCH3
CH3
A magnesium
alkoxide
ether
OH
HCl
H2 O
C6 H5 CCH3 + Mg 2 +
CH3
2-Phenyl-2-propanol
(a tertiary alcohol)
16-19
16 Grignard Reagents
Problem: 2-phenyl-2-butanol can be synthesized by three
different combinations of a Grignard reagent and a
ketone. Show each combination.
OH
C-CH2 CH3
CH3
16-20
16 Organolithium Compounds
 Organolithium
compounds are generally more
reactive in C=O addition reactions than RMgX,
and typically give higher yields
-
O
Li
OH
HCl
H2 O
+
Phenyl- 3,3-Dimethyl-2lithium
butanone
+
O Li
A lithium
alkoxide
3,3-Dimethyl-2-phenyl2-butanol
16-21
16 Salts of Terminal Alkynes
of an acetylide anion followed by H3O+
gives an -acetylenic alcohol
 Addition
O
+
:
+
C
N
a
HC
Cyclohexanol
HC C O - N a+
HC C OH
HCl
H2 O
A sodium
alkoxide
1-Ethynylcyclohexanol
16-22
16 Salts of Terminal Alkynes
O
H2 O
HO

CCH3
H2 SO 4 , HgSO 4
HO
C
CH
An -hydroxyketone
O
CH2 CH

HO
1 . ( sia ) 2 BH

2 . H2 O 2 , Na OH
A -hydroxyaldehyde
16-23
16 Addition of HCN
 HCN
adds to the C=O group of an aldehyde or
ketone to give a cyanohydrin
 Cyanohydrin: a molecule containing an -OH
group and a -CN group bonded to the same
carbon
O
CH3 CH + HC N
OH
CH 3 C-C N
H
2-Hydroxypropanenitrile
(Acetaldehyde cyanohydrin)
16-24
16 Addition of HCN
 Mechanism
H3 C
-
H3 C
••
C O +
of cyanohydrin formation
C N
O:-
H3 C
C N
C N
H3 C
+
C
H3 C
C
H3 C
H3 C
O:-
H C N
O-H
+ - :C N
C
H3 C
C N
16-25
16 Cyanohydrins
 The
value of cyanohydrins
• acid-catalyzed dehydration of the 2° or 3° alcohol
OH
CH3 CHC N
acid
catalyst
CH2 = CHC N + H2 O
2-Hydroxypropanenitrile
(Acetaldehyde cyanohydrin)
Propenenitrile
(Acrylonitrile)
• catalytic reduction of the cyano group gives a 1°
amine
OH
OH
CHC N + 2 H2
Benzaldehyde cyanohydrin
Ni
CHCH2 N H2
2-Amino-1-phenylethanol
16-26
16 Wittig Reaction
 The
Wittig reaction is a very versatile synthetic
method for the synthesis of alkenes from
aldehydes and ketones.
+ O + Ph 3 P- CH2
A phosphonium
ylide
CH2
Methylenecyclohexane
+ + Ph 3 P- O
Triphenylphosphine oxide
16-27
16 Phosphonium Ylides
 Phosphonium
Step 1:
Ph 3 P :
ylides are formed in two steps:
+ CH 3 - I
SN 2
Triphenylphosphine
+
Ph 3 P- CH 3 I
An alkyltriphenylphosphonium iodide
Step 2:
:
CH 3 CH 2 CH 2 CH 2 Li
+
+
+ H- CH 2 -PPh 3 I -
:
Butyllithium
- +
CH 2 -PPh 3
+ CH 3 CH 2 CH 2 CH 3 + LiI
A phosphonium
Butane
ylide
16-28
16 Wittig Reaction
 Phosphonium
ylides react with the C=O group of
an aldehyde or ketone to give an alkene
Step 1:
O CR2
-
:
+
Ph 3 P CH2
CR2
O CR2
Ph 3 P CH2
Ph 3 P CH2
-
:O
+
An oxaphosphetane
A betaine
Step 2:
O CR2
Ph 3 P CH2
Ph 3 P= O
+
Triphenylphosphine
oxide
R2 C= CH2
An alkene
16-29
16 Wittig Reaction
• Examples:
O
+ PhCH2 CH + Ph 3 P- CHCH3
PhCH2 CH= CH CH 3 + Ph 3 P= O
1-Phenyl-2-butene
(87% Z isomer, 13% E isomer
O
+ PhCH2 CH + Ph 3 P- CHCH3
PhCH2 CH= CH CH 3 + Ph 3 P= O
1-Phenyl-2-butene
(87% Z isomer, 13% E isomer
16-30
16 Addition of H2O
 Addition
of water (hydration) to the carbonyl
group of an aldehyde or ketone gives a gem-diol,
commonly referred to as a hydrate
• when formaldehyde is dissolved in water at 20°C, the
carbonyl group is more than 99% hydrated
O
HCH + H2 O
Formaldehyde
OH
HCOH
H
Formaldehyde
hydrate
(>99%)
16-31
16 Addition of H2O
• the equilibrium concentration of a hydrated ketone is
considerably smaller
H3 C
H3 C
C
O + H2 O
H3 C
Acetone
(99.9%)
OH
C
H3 C
OH
2,2-Propanediol
(0.1%)
16-32
16 Addition of Alcohols
 Addition
of one molecule of alcohol to the C=O
group of an aldehyde or ketone gives a
hemiacetal
 Hemiacetal: a molecule containing an -OH and
an -OR or -OAr bonded to the same carbon
H
O
CH3 CCH3 + OCH2 CH3
OH
CH3 COCH 2 CH 3
CH3
A hemiacetal
16-33
16 Addition of Alcohols
 Hemiacetals
are only minor components of an
equilibrium mixture, except where a five- or sixmembered ring can form
(the model is of the trans isomer)
O
CH 3 CHCH 2 CH2 CH
OH
4-Hydroxypentanal
H3 C
O
OH
A cyclic hemiacetal
(major form present
at equilibrium)
16-34
16 Addition of Alcohols
 Formation
of a hemiacetal is base catalyzed
• Step 1: proton transfer from HOR gives an alkoxide
B H + - :OR
B: - + H OR
• Step 2: Attack of RO- on the carbonyl carbon
O:–
O
CH3 - C-CH3 +
–
:O- R
CH3 - C-CH3
OR
• Step 3: proton transfer from the alcohol to O- gives the
hemiacetal and generates a new base catalyst
O:–
CH3 - C-CH3 + H–OR
OR
OH
CH3 - C-CH3 +
OR
–
:O- R
16-35
16 Addition of Alcohols
 Formation
of a hemiacetal is also acid catalyzed
Step 1: proton transfer to the carbonyl oxygen
+ H
O
CH3 - C-CH3 + :A
O:
CH3 - C-CH3 + H- A
Step 2: attack of ROH on the carbonyl carbon
+
O
H
: OH
:
CH3 - C-CH3 + H- O-R
CH3 - C-CH3
+
O
H
R
Step 3: proton transfer from the oxonium ion to A- gives
the hemiacetal and generates a new acid catalyst
OH
CH3 - C-CH3
+
O
A :- H
R
OH
CH3 - C-CH3
: OR
+ H- A
16-36
16 Addition of Alcohols
 Hemiacetals
react with alcohols to form acetals
Acetal: a molecule containing two -OR or -OAr groups
bonded to the same carbon
OH
H
+
CH3 COCH2 CH3 + CH3 CH2 OH
CH3
OCH2 CH3
A hemiacetal
CH3 COCH2 CH3
+ H2 O
CH3
A diethyl acetal
16-37
16 Addition of Alcohols
Step 1: proton transfer from HA gives an oxonium ion
HO:
R- C-OCH3 + H A
H
H + H
O
R- C-OCH3 + A :
H
An oxonium ion
Step 2: loss of water gives a resonance-stabilized cation
H
+
+
RC-OCH
R- C OCH3
3 + H2 O
H
H
A resonance-stabilized cation
:
H + H
O
R- C-OCH3
16-38
16 Addition of Alcohols
Step 3: reaction of the cation (a Lewis acid) with
methanol (a Lewis base) gives the conjugate acid of
the acetal
H
CH3 -O :
+
+ R- C OCH3
H
H + CH3
O
R- C-OCH3
H
A protonated acetal
Step 4: (not shown) proton transfer to A- gives the acetal
and generates a new acid catalyst
16-39
16 Addition of Alcohols
 With
ethylene glycol, the product is a fivemembered cyclic acetal
O + HOCH2 CH2 OH
H
+
O CH
2
O CH2
+ H2 O
A cyclic acetal
16-40
16 Dean-Stark Trap
QuickTime™ and a
Photo - JPEG decompressor
are needed to see this picture.
16-41
16 Acetals as Protecting Grps
 Suppose
you wish to bring about a Grignard
reaction between these compounds
O
O
H
Benzaldehyde
+
Br
H
4-Bromobutanal
??
OH
O
H
5-Hydroxy-5-phenylpentanal
16-42
16 Acetals as Protecting Grps
 If
the Grignard reagent were prepared from 4bromobutanal, it would self-destruct!
• first protect the -CHO group as an acetal
O
Br
+ HO
H
OH
H
O
+
Br
O
A cyclic acetal
+ H2 O
• then do the Grignard reaction
-
+
O Mg Br O
O
Br
1 . Mg, e t he r
O
O 2 . C H CHO
6 5
A cyclic acetal
• hydrolysis (not shown) gives the target molecule
16-43
16 Acetals as Protecting Grps
 Tetrahydropyranyl
(THP) protecting group
THP group
RCH2 OH +
H+
O
Dihydropyran
RCH2 O
O
A tetrahydropyranyl
ether
• the THP group is an acetal and, therefore, stable to
neutral and basic solutions and to most oxidizing and
reducting agents
• it is removed by acid-catalyzed hydrolysis
16-44
16 Add’n of S Nucleophiles
 Thiols,
like alcohols, add to the C=O of
aldehydes and ketones to give tetrahedral
carbonyl addition products
 The sulfur atom of a thiol is a better nucleophile
than the oxygen atom of an alcohol
 A common sulfur nucleophile used for this
purpose is 1,3-propanedithiol
• the product is a 1,3-dithiane
O
RCH
+
HS
SH
H
+
R
C2
H
An aldehyde
1,3-Propanedithiol
S3
1
+ H2 O
S
A 1,3-dithiane
(a cyclic thioacetal)
16-45
16 Add’n of S Nucleophiles
 The
hydrogen on carbon 2 of the 1,3-dithiane
ring is weakly acidic, pKa approximately 31
S H
S
+
+ Bu:- Li +
+ BuH
C
C: Li
S R
S R
A 1,3-dithiane Butyllithium A lithio-1,3-dithiane
Butane
(stronger acid) (stronger base)
(weaker base)
(weaker acid)
pKa 31
pKa 51
16-46
16 Add’n of S Nucleophiles
• a 1,3-dithiane anion is a good nucleophile and
undergoes SN2 reactions with methyl, 1° alkyl, allylic,
and benzylic halides
• hydrolysis gives a ketone
S
SN 2
C: Li + + R' CH2 - Br
S R
Lithium salt of
S CH2 R' H O, Hg Cl
a 1,3-dithiane
2
2
C
CH3 CN
S R
O
R- C-CH2 R'
16-47
16 Add’n of S Nucleophiles
 Treatment
of the 1,3-dithiane anion with an
aldehyde or ketone gives an -hydroxyketone
O
C: Li + + H- C-R'
S R
Lithium salt of
a 1,3-dithiane
S
O:- Li
S CH- R'
C
S R
H2 O, HgCl 2
CH3 CN
+
O OH
R C CH-R'
An -hydroxyketone
16-48
16 Add’n of N Nucleophiles
 Ammonia,
1° aliphatic amines, and 1° aromatic
amines react with the C=O group of aldehydes
and ketones to give imines (Schiff bases)
O
CH3 CH + H2 N
Acetaldehyde
O
H
+
Aniline
+
N H3
Cyclohexanone Ammonia
CH3 CH =N
+ H2 O
An imine
(a Schiff base)
H
+
N H + H2 O
An imine
(a Schiff base)
16-49
16 Add’n of N Nucleophiles
 Formation
of an imine occurs in two steps
Step 1: carbonyl addition followed by proton transfer
O:- H
+
C N -R
O
:
C
+ H2 N- R
H
O
C
H
N -R
H
A tetrahedral carbonyl
addition compound
Step 2: loss of H2O and proton transfer to solvent
H
+
O H +
H
:O
H
C N -R
H
H
+
H
O
C N -R + H2 O
C N -R
H
:O
H
H
An imine
16-50
16 Add’n of N Nucleophiles
• a value of imines is that the carbon-nitrogen double
bond can be reduced to a carbon-nitrogen single bond
O +
H+
- H2 O
H 2N
Cyclohexanone Cyclohexylamine
H
N
(An imine)
H2 / N i
N
Dicyclohexylamine
16-51
16 Add’n of N Nucleophiles
 Rhodopsin
(visual purple) is the imine formed
between 11-cis-retinal (vitamin A aldehyde) and
the protein opsin
11
1
12
+ H2 N- OPSIN
5
11-cis-Retinal
H
O
Rhodopsin
(Visual purple) H
N -OPSIN
16-52
16 Add’n of N Nucleophiles
 Secondary
amines react with the C=O group of
aldehydes and ketones to form enamines
O
Cyclohexanone
+
H-N
H
+
Piperidine
(a secondary amine)
N
+ H2 O
An enamine
• the mechanism of enamine formation involves
formation of a tetrahedral carbonyl addition
compound followed by its acid-catalyzed dehydration
• we discuss the chemistry of enamines in more detail
in Chapter 19
16-53
16 Add’n of N Nucleophiles
 The
carbonyl group of aldehydes and ketones
reacts with hydrazine and its derivatives in a
manner similar to its reactions with 1° amines
O +
H2 NNH2
NNH2
Hydrazine
+ H2 O
A hydrazone
• hydrazine derivatives include
H 2 N-OH
Hydroxylamine
H2 N-NH
Phenylhydrazine
16-54
16 Acidity of -Hydrogens
 Hydrogens
alpha to a
carbonyl group are more
acidic than hydrogens of
alkanes, alkenes, and
alkynes but less acidic
than the hydroxyl hydrogen
of alcohols
Type of Bond pKa
CH3 CH2 O-H
16
O
CH3 CCH2 -H
20
CH3 C C-H
25
CH2 = CH- H
44
CH3 CH2 -H
51
16-55
16 Acidity of -Hydrogens
 -Hydrogens
are more acidic because the
enolate anion is stabilized by
1. delocalization of its negative charge
2. the electron-withdrawing inductive effect of the
adjacent electronegative oxygen
O
CH3 - C-CH2 -H + :A-
:
O-
O
:
CH3 - C CH2
CH3 - C= CH2 + H- A
Enolate anion
16-56
16 Keto-Enol Tautomerism
• protonation of the enolate anion on oxygen gives the
enol form; protonation on carbon gives the keto form
O
CH3 - C-CH2
OCH3 - C= CH2
Enolate anion
O
A
-
H- A
+ CH3 - C-CH3
Keto form
H- A
OH
CH3 - C= CH2
+A-
Enol form
16-57
16 Keto-Enol Tautomerism
• acid-catalyzed equilibration of keto and enol
tautomers occurs in two steps
Step 1: proton transfer to the carbonyl oxygen
O:
CH3 - C-CH3 + H- A
Keto form
+
fast
O
H
CH3 - C-CH3
+
A:-
The conjugate acid
of the ketone
Step 2: proton transfer to the base A+ H
O
CH3 - C-CH2 - H
+
:A-
slow
: OH
CH3 - C= CH2
+ H- A
Enol form
16-58
16 Keto-Enol Tautomerism
 Keto-enol
equilibria for
simple
aldehydes
and ketones
lie far toward
the keto
form
Keto form
Enol form
% Enol at
Equilibrium
O
OH
CH2 = CH
6 x 10 -5
CH 3 CH
O
OH
CH 3 CCH 3
CH3 C= CH 2
O
OH
O
OH
6 x 10 -7
1 x 10 -6
4 x 10 -5
16-59
16 Keto-Enol Tautomerism
 For
certain types of molecules, however, the
enol is the major form present at equilibrium
• for -diketones, the enol is stabilized by conjugation
of the pi system of the carbon-carbon double bond
and the carbonyl group
conjugated
O
O
O
OH
1,3-Cyclohexanedione
system
H
H
H
O
H
HO
H
16-60
16 Keto-Enol Tautomerism
-diketones are further stabilized by
intramolecular hydrogen bonding
 Open-chain
hydrogen
bonding
O
O
O
20%
2,4-Pentanedione
(Acetylacetone)
+
H
O
80%
16-61
16 Racemization
at an -carbon may be catalyzed
by either acid or base
 Racemization
Ph
O
C
C
acid
or
OH
Ph
C
C
acid
or
Ph
O
C
C
H3 C
H
CH3 base
CH3
CH3 base H3 C
H
H3 C
(R)-3-Phenyl-2An achiral enol
(S)-3-Phenyl-2butanone
butanone
16-62
16 Deuterium Exchange
exchange at an -carbon may be
catalyzed by either acid or base
 Deuterium
O
CH3 CCH3
Acetone
+ 6 D2 O
D
+
or OD
-
O
CD3 CCD3 + 6 HOD
Acetone-d6
16-63
16 -Halogenation
 -Halogenation:
aldehydes and ketones with at
least one -hydrogen react at an  -carbon with
Br2 and Cl2
O
CCH3 + Br2
CH3 COOH
Acetophenone
O
CCH2 Br + HBr
• reaction is catalyzed by both acid and base
16-64
16 -Halogenation
 Acid-catalyzed
-halogenation
Step 1: acid-catalyzed enolization
OH
R' -C- C-R
R
H -O
slow
R
C
R'
C
R
Step 2: nucleophilic attack of the enol on halogen
H -O
R
C C + Br
R'
R
Br
fast
O
Br
C C R + H+ + Br: R'
R
16-65
16 -Halogenation
 Base-promoted
-halogenation
Step 1: formation of an enolate anion
OH
slow
:
-
R' -C- C-R + :OH
O
C
O:
R
C
C
R
C
+ H2 O
R'
R'
R
R
Resonance-stabilized enolate anion
R
Step 2: nucleophilic attack of the enolate anion on
halogen
O:C
R'
R
C
R
+ Br
Br
fast
O
R'
Br
C C R + :Br R
16-66
16 -Halogenation
 Acid-catalyzed
halogenation:
• introduction of a second halogen is slower than the
first
• introduction of the electronegative halogen on the carbon decreases the basicity of the carbonyl oxygen
toward protonation
 Base-promoted
-halogenation:
• each successive halogenation is more rapid than the
previous one
• the introduction of the electronegative halogen on the
-carbon increases the acidity of the remaining hydrogens and, thus, each successive -hydrogen is
removed more rapidly than the previous one
16-67
16 Haloform Reaction
 In
the presence of base, a methyl ketone reacts
with three equivalents of halogen to give a 1,1,1trihaloketone, which then reacts with an
additional mole of hydroxide ion to form a
carboxylic salt and a trihalomethane
O
RCCH3
3 Br2
3 NaOH
O
O
NaOH
+
RCCBr3
RCO Na + CHBr3
Tribromomethane
(Bromoform)
O
O
1 . Cl 2 / N aOH
2 . HCl/ H2 O
5-Methyl-3-hexen-2-one
OH
+
CHCl 3
4-Methyl-2-pentenoic Trichloromethane
acid
(Chloroform)
16-68
16 Haloform Reaction
 The
final stage is divided into two steps
Step 1: addition of OH- to the carbonyl group gives a
tetrahedral carbonyl addition intermediate and is
followed by its collapse
: O-
O
RC- CBr 3 + - :OH
O
RC- CBr 3
OH
RC + - :CBr 3
OH Conjugate base
of bromoform
Step 2: proton transfer from the carbonyl group to the
haloform anion
O
O
RC- O- H +
-
:CBr 3
RC- O: - + H- CBr 3
Bromoform
16-69
16 Oxidation of Aldehydes
 Aldehydes
are oxidized to carboxylic acids by a
variety of oxidizing agents, including H2CrO4
CHO
H2 CrO4
Hexanal
 They
COOH
Hexanoic acid
are also oxidized by Ag(I)
• in one method, a solution of the aldehyde in aqueous
ethanol or THF is shaken with a slurry of silver oxide
CH3 O
O
CH
+ A g2 O
HO
Vanillin
T HF, H 2 O
N aOH
HCl
H2 O
CH3 O
O
COH
+ Ag
HO
Vanillic acid
16-70
16 Oxidation of Aldehydes
 Aldehydes
are oxidized by O2 in a radical chain
reaction
• liquid aldehydes are so sensitive to air that they must
be stored under N2
O
O
2
CH
Benzaldehyde
+ O2
2
COH
Benzoic acid
16-71
16 Oxidation of Ketones
• ketones are not normally oxidized by chromic acid
• they are oxidized by powerful oxidants at high
temperature and high concentrations of acid or base
O
OH
O
HN O 3 HO
Cyclohexanone
(keto form)
Cyclohexanone
(enol form)
OH
O
Hexanedioic acid
(Adipic acid)
16-72
16 Reduction
• aldehydes can be reduced to 1° alcohols
• ketones can be reduced to 2° alcohols
• the C=O group of an aldehyde or ketone can be
reduced to a -CH2- group
Aldehydes
Can Be
Reduced to
Ketones
Can Be
Reduced to
OH
O
RCH2 OH
RCH
O
RCHR'
RCR'
RCH3
RCH2 R'
16-73
16 Catalytic Reduction
 Catalytic
reductions are generally carried out at
from 25° to 100°C and 1 to 5 atm H2
OH
O
+
H2
Pt
25 o C, 2 atm
Cyclohexanone
Cyclohexanol
O
2 H2
H
Ni
trans-2-Butenal
(Crotonaldehyde)
OH
1-Butanol
16-74
16 Catalytic Reduction
A
carbon-carbon double bond may also be
reduced under these conditions
O
2 H2
H
Ni
trans-2-Butenal
(Crotonaldehyde)
OH
1-Butanol
• by careful choice of experimental conditions, it is
often possible to selectively reduce a carbon-carbon
double in the presence of an aldehyde or ketone
16-75
16 Metal Hydride Reduction
 The
most common laboratory reagents for the
reduction of aldehydes and ketones are NaBH4
and LiAlH4
• both reagents are sources of hydride ion, H:-, a very
powerful nucleophile
H
Na
+
H- B- H
H
Sodium
borohydride
H
Li
+
H- A l- H
H
Lithium aluminum
hydride (LAH)
H:
Hydride ion
16-76
16 NaBH4 Reduction
• reductions with NaBH4 are most commonly carried out
in aqueous methanol, in pure methanol, or in ethanol
• one mole of NaBH4 reduces four moles of aldehyde or
ketone
O
4 RCH + NaBH4 methanol
-
+
( RCH2 O) 4 B Na
A tetraalkyl borate
H2 O
4 RCH2 OH + borate
salts
16-77
16 NaBH4 Reduction
 The
key step in metal hydride reduction is
transfer of a hydride ion to the C=O group to
form a tetrahedral carbonyl addition compound
H
O
+
N a H- B- H + R- C-R'
H
O BH3 N a
R- C-R'
+
H
from the hydride
reducing agent
H2 O
OH
from
water
R- C-R'
H
16-78
16 LiAlH4 Reduction
• unlike NaBH4, LiAlH4 reacts violently with water,
methanol, and other protic solvents
• reductions using it are carried out in diethyl ether or
tetrahydrofuran (THF)
O
4RCR + LiAlH4
ether
(R2CHO)4Al- Li+
H2O
OH
4RCHR + aluminum salts
A tetraalkyl aluminate
16-79
16 Metal Hydride Reduction
• metal hydride reducing agents do not normally reduce
carbon-carbon double bonds, and selective reduction
of C=O or C=C is often possible
O
RCH= CHCR'
1 . Na BH 4
2 . H2 O
O
RCH= CHCR'
+
H2
Rh
OH
RCH= CHCH R'
O
RCH2 CH 2 CR'
16-80
16 Clemmensen Reduction
• refluxing an aldehyde or ketone with amalgamated
zinc in concentrated HCl converts the carbonyl group
to a methylene group
OH O
OH
Zn( H g) , HCl
16-81
16 Wolff-Kishner Reduction
• in the original procedure, the aldehyde or ketone and
hydrazine are refluxed with KOH in a high-boiling
solvent
• the same reaction can be brought about using
hydrazine and potassium tert-butoxide in DMSO
O
+ H2 NN H2
Hydrazine
KOH
diethylene glycol
(reflux)
+ N 2 + H2 O
16-82
16 Prob 16.19
Draw a structural formula for the product formed by
treating each compound with propylmagnesium bromide
followed by aqueous HCl.
(a) CH2 O
(c)
O
(b)
(d)
O
O
16-83
16 Prob 16.20
Suggest a synthesis of each alcohol from an aldehyde or
ketone and a Grignard reagent. Under each is the
number of combinations of Grignard reagents and
aldehyde or ketone that might be used.
OH
OH
(a)
3 Combinations
OH
(b)
2 Combinations
(c)
OCH3
2 Combinations
16-84
16 Prob 16.21
Show how to prepare this alcohol from the three given
starting materials.
Br +
CHO +
O
several
steps
OH
16-85
16 Prob 16.22
Show how to synthesize 1-phenyl-2-butanol from these
starting materials.
several
steps
Br
+
Bromobenzene
1-Butene
OH
1-Phenyl-2-butanol
16-86
16 Prob 16.24
Draw the Wittig reagent formed from each haloalkane,
and for the alkene formed by treating the Wittig reagent
with acetone.
(a)
Cl
(c)
(e)
Br
(b)
Br
Br
O
(d) Cl
(f) Ph
O
Cl
16-87
16 Prob 16.25
Show how to bring about each conversion using a Wittig
reaction.
O
(a)
O
(b)
(c)
O
CH
OCH3
OCH3
16-88
16 Prob 16.26
Show two sets of reagents that might be combined in a
Wittig reaction to give this conjugated diene.
CH= CHCH= CHCH3
1-Phenyl-1,3-pentadiene
16-89
16 Prob 16.27
Wittig reactions with an -haloether can be used for the
synthesis of aldehydes and ketones. To see this, convert
each -haloether to a Wittig reagent, and react the Wittig
reagent with cyclopentanone followed by hydrolysis in
aqueous acid.
CH3
ClCH 2 OCH3
ClCHOCH 3
( A)
( B)
16-90
16 Prob 16.28
Suggest a mechanism for the reaction of a sulfur ylide
with a ketone to give an epoxide.
Ph
CH3
+
S CH
Br
-
strong
base
Ph
CH3
A sulfonium bromide salt
Ph
CH3
+ O +
S C:
Ph
CH3
Ph
CH3
+ S C:
Ph
CH3
A sulfur ylide
O
+
CH3
( Ph) 2 S
16-91
16 Prob 16.29
Propose a structural formula for compound D and for the
product C9H14O.
+
S
C6 H5
Br
-
BuLi
O
D
C9 H1 4 O
C6 H5
16-92
16 Prob 16.30
Draw a structural formula for the cyclic hemiacetal. How
many stereoisomers are possible for it? Draw alternative
chair conformations for each possible stereoisomer.
OH
O
H
5-Hydroxyhexanal
H
+
a cyclic hemiacetal
16-93
16 Prob 16.31
Draw structural formulas for the hemiacetal and acetal
formed from each pair of reagents in the presence of an
acid catalyst.
O
(a)
+ CH 3 CH 2 OH
OH
(b)
O
+ CH 3 CCH 3
OH
O
(c) CH3 CH2 CH2 CH + CH3 OH
16-94
16 Prob 16.32
Draw structural formulas for the products of hydrolysis
of each acetal in aqueous acid.
CH 3 O
OCH 3
(a)
O
(b)
OCH 3
O
H
CHO
(c)
O
16-95
16 Prob 16.33
Propose a mechanism for this reaction. If the carbonyl
oxygen is enriched with oxygen-18, will the oxygen label
appear in the cyclic acetal or in the water?
O
H + CH3 OH
H
+
OH
4-Hydroxypentanal
O
OCH3 +
H2 O
A cyclic acetal
16-96
16 Prob 16.34
Propose a mechanism for this acid-catalyzed reaction.
OCH3
+ H2 O
H
+
O
+ CH3 OH
16-97
16 Prob 16.35
Propose a mechanism for this acid-catalyzed
rearrangement.
OOH
O
H2 SO 4
OH + CH3 CCH3
CCH3
CH3
Cumene
hydroperoxide
Phenol
Acetone
16-98
16 Prob 16.37
Show how to bring about this conversion.
O
O
H
HO
H
OH
16-99
16 Prob 16.39
Which compound will cyclize to give the insect
pheromone frontalin?
O
O
Frontalin
O
O
O
OH
O
A
HO
O O
OH
B
C
16-100
16 Prob 16.41
Draw a structural formula for the product formed by
treating each compound with (1) the lithium salt of the
1,3-dithiane derived from acetaldehyde and then (2) H2O,
HgCl2.
(a)
O
CH
O
(b) CH2 CH
(c) ClCH 2 CH= CH2
16-101
16 Prob 16.42
Show how to bring about each conversion using a 1,3dithiane.
O
(a)
O
H
O
O
H
(b)
O
(c)
OH
Ph
Ph
O
H
OH
16-102
16 Prob 16.44
Show how each compound can be synthesized by
reductive amination of an aldehyde or ketone and an
amine.
H
N
N H2
(a)
(b)
Amphetamine
Methamphetamine
16-103
16 Prob 16.45
Show how to bring about this final step in the synthesis
of the antiviral drug rimantadine.
O
N H2
16-104
16 Prob 16.46
Draw a structural formula for the -hydroxyaldehyde and
-hydroxyketone with which this enediol is in
equilibrium.
CH- OH
-Hydroxyaldehyde
C-OH
-Hydroxyketone
CH3
An enediol
16-105
16 Prob 16.47
Propose a mechanism for the isomerism of (R)glyceraldehyde to (R,S)-glyceraldehyde and
dihydroxyacetone.
CHO
CHOH
NaOH
CH2 OH
(R)-Glyceraldehyde
CHO
CHOH
CH2 OH
(R,S)-Glyceraldehyde
CH2 OH
+
C=O
CH2 OH
Dihydroxyacetone
16-106
16 Prob 16.48
When cis--decalone is dissolved in ether containing a
trace of HCl, the following equilibrium is established.
Propose a mechanism for the isomerization and account
for the fact that the trans isomer predominates.
H
H
HCl
HO
cis-2-Decalone
HO
trans-2-Decalone
16-107
16 Prob 16.49
When this bicyclic ketone is treated with D2O in the
presence of an acid catalyst, only two of the three hydrogens exchange. Propose a mechanism for the
exchange and account for the fact that the bridgehead
hydrogen does not exchange.
H
This -hydrogen
does not exchange
H
H
These two
-hydrogens exchange
O
16-108
16 Prob 16.51
Propose a mechanism for the formation of the bracketed
intermediate and for the formation of the sodium salt of
cyclopentanecarboxylic acid.
O
O
Cl N aOH
N aOH
THF
THF
A proposed
intermediate
O
O
CO - Na +
HCl
H2 O
COH
16-109
16 Prob 16.52
If the Favorskii rearrangement is carried out using
sodium ethoxide in ethanol, the product is an ethyl ester.
Propose a mechanism for this reaction.
O
O
Cl
CH3 CH2 O - Na +
COCH 2 CH 3
CH3 CH2 OH
16-110
16 Prob 16.53
Propose a mechanism for each step in this
transformation, and account for the regioselectivity of
the HCl addition.
Cl O
O
HCl
1 . Na OH
2 . HCl
(R)-(+)-Pulegone
C
OH
O
(R)-3,7-Dimethyl-6-octenoic acid
(R)-Citronellic acid
16-111
16 Prob 16.57
Show how to convert cyclopentanone to each
compound.
(a)
OH
(b)
Cl
OH
(c)
CH- CH = CH2
(d)
16-112
16 Prob 16.59
Propose structural formulas for A, B, and C. Show how C
can also be prepared by a Wittig reaction.
O
1 . HC CH, Na NH 2
2 . H2 O
C7 H1 0 O
A
C7 H1 2 O
B
H2
Lindlar
catalyst
KHSO4
heat
C7 H1 0
C
16-113
16 Prob 16.60
Given this retrosynthetic analysis, show how to
synthesize cis-3-penten-2-ol from the three given starting
materials.
OH
OH
O
+ HCCH3
CH3 I + HC CH
16-114
16 Prob 16.61
Propose a synthesis for Oblivon from acetylene and a
ketone.
HO
Oblivon
16-115
16 Prob 16.62
Propose a synthesis for Surfynol from acetylene and a
ketone.
OH
OH
Surfynol
16-116
16 Prob 16.63
Propose a mechanism for this acid-catalyzed
rearrangement.
OH
C CH
H
+
CHO
C
H
16-117
16 Prob 16.64
Propose a mechanism for this acid-catalyzed
rearrangement.
O
O
A rSO3 H
16-118
16 Prob 16.66
Propose mechanisms for Steps (1) and (4) and reagents
for Steps (2), (3), and (5).
O
H2 SO 4
O
(1)
(2 )
-Ionone
Pseudoionone
Ph 3 P, HBr
OH
(3)
OH
(4)
O
-
Br
+
PPh 3
OCCH 3
(5)
Vitamin A acetate
16-119
16 Prob 16.68
Propose a mechanism for this Lewis acid catalyzed
isomerization. Account for the fact that only a single
stereoisomer of isopulegol is formed.
O
H
1 . SnCl 4 , CH2 Cl 2
2 . NH4 Cl
(S)-Citronellal
(C10H18O)
OH
Isopulegol
(C10H18O)
16-120
16
Aldehydes
&
Ketones
End Chapter 16
16-121