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Organometallic Reagents: Grignard Reagents
Chapter 13-1
Grignard Reagents
Grignard reagents are organometallic reagents derived from an alkyl halide
and magnesium
R—X + Mg
diethyl ether
Rδ−—Mgδ+X Grignard reagent
Since the carbon carries a partial negative charge, the carbon is a strong
base and a good nucleophile.
CH3CH2—Br + Mg
diethyl ether
CH3CH2δ−—Mgδ+Br
HOH
CH3CH2—H
Reaction with epoxides
O
H
C
H
BrMgO
CH3MgBr
H
C
H
H
H
C
C H
H
CH3
HO
H+
H
C
H
H
C H
CH3
CH3–MgBr+
Because carbonyl pi bonds are polarized, they can undergo a reaction called
nucleophilic addition: the addition of a nucleophile to an electron deficient pi
bond.
O–
Oδ−
O–
C+
Cδ+
C
R1
R1
R1
R1
R1
Nu
Nucleophilic
Addition
R1
Nu:
R1
O–
Oδ−
O– MgX+
C+
Cδ+
C
R1
R1
R2
Mg-X
R1
R1
R2
R1
Nucleophilic
Addition
Chem 66H
Preparation of Alcohols from Grignard Reagents
Chapter 13-2
Chem 66H
A Grignard reaction with
1. formaldehyde produces a primary alcohol
O
1. CH3
C
H
Mg-X
+
2. H2O, H
H
Oδ−
OH
H
C
Cδ+
H
H
CH3
2. an aldehyde produces a secondary alcohol
C
CH3CH2
H
2. H2O, H+
H
C
Oδ−
CH2CH3
H
δ−
C
CH3CH2
1.(CH3)CHMgBr
+
CH3 2. H2O, H
HC
H 3C
OH
CH3
C
H
H
H
C
H
H
CH2CH3
C
H
CH(CH3)2
C
CH(CH3)2
CH2CH3
δ+
Mg-X
CH3
CH(CH3)2
O
CH3MgBr
C
OMgBr
C
– CH3O–MgBr+
OCH3
CH3
O
OH
1. 2 CH3MgBr
OCH3
2. H2O, H+
CH3
C
CH3
O
CH3MgBr
C
CH3
5. ethylene oxide produces a primary alcohol
OMgBr
C
CH3
CH3
1. C6H5MgBr
2. H2O, H+
C H
H
HO
H+
CH2CH3
OCH3
C
H
CH2CH3
4. an ester produces a tertiary alcohol (addition of two molecules of
Grignard reagent)
O
C H
BrMgO
Cδ+
CH(CH3)2
3. a ketone produces a tertiary alcohol
O
H
C
HO
H+
Mg-X
OH
1. (CH3)CHMgBr
H
H
δ+
δ−
CH3
O
BrMgO
OH
OH
CH3
C
CH3
H+
Organolithium Reagents
Chapter 13-3
Organolithium Reagents
Organolithium reagents are organometallic reagents derived from an alkyl
halide and lithium metal
R—X + 2 Li
diethyl ether
Rδ−—Liδ+ = LiX
organolithium reagent
Since the carbon carries a partial negative charge, the carbon is a strong
base and a good nucleophile.
CH3CH2—Br + 2 Lidiethyl ether CH3CH2δ−—Liδ+
HOH
CH3CH2—H
n-Butyl Lithium is a common commercially available organolithium reagent
which is used primarily as a strong organic base. It also acts as a nucleophile to
add to carbonyl compunds, much like a Grignard reagent.
O
C
CH3CH2
1. CH3CH2CH2CH2Li
+
CH3 2. H2O, H
OH
CH3
C
CH2CH3
CH2CH2CH2CH3
n-Butyl Lithium can also be used to generate aryl and vinyl lthium reagents
by lithium-halogen exchange
CH3CH2CH2CH2Li
Br
Li
+
CH3CH2CH2CH2Br
(n-BuLi)
CH3CH2CH2CH2Li
Br
Li
+ BuBr
OCH3
Br
2 CH3CH2CH2CH2Li
OCH3
Li
+ BuBr + Butane
OH
OLi
Chem 66H
Alkynes: Acetylides
Chapter 13-5
H
C
+ base
C
C
H
pKa = 45
C
C
–
..
C
+ base—H
H
+ base
H
C
.
C.
–
+ base—H
pKa = 25
The relatively high acidity of the alkyne —C≡C—H bond is associated with
the large degree of s character in the sp C—H bond (50% compared with
33% in sp2 bonds). The carbon atom is more electronegative in the sp
state; thus the C—H bond is more acidic.
The acetylide ion may be formed by such strong bases as —:NH2 (pKa
33), RMgX or RLi (pKa 45-50).
R
CH
CH2
NaNH2
No reaction
NH3
R
C
C
H
NaNH2
NH3
R
C
C
H
R
C
C:– Na+ + NH3
acetylide ion
RMgX
R
C
C:– MgX+
R
C
C:– Li+
+ RH
n-BuLi
R
C
C
H
+ RH
Chem 66H
Alkynes: Acetylides
Chapter 13-6
SN2 reaction with acetylide ion
NH3
R'—C
≡C:– Na+ +
R'
R—CH2—X
O
–
C:
MgBr+ +
C
R—CH2—C≡C—R'
R'
C
–
+
C—CH
2—CH2—O MgBr
H+
R'
C
C—CH
2—CH2—OH
Nucleophilic addition reaction with acetylide ion.
O – Li+
O
R'
– +
C:
Li
C
R
C
H
R'
C C
C
R
H
n-BuLi
R'
C
H+
OH
C—H
R'
C
C C
H
O
R'
C
–
C: Li
+
+
O – Li+
C C R'
H+
OH
C
C R'
R
Chem 66H
Chem 66H
Organocuprates
Chapter 13-7
Preparation:
Organocopper reagents can be prepared from organo lithium reagents and
Grignard reagents
Cuprates: Me2 CuLi, Bu2 CuLi for common readily available organolithium
reagents.
R—X + 2Li
R—Li + LiX
CuI
R—Cu
R—Li
CO2Et
O
LiCu
HO
H
CO2Et 90%
H
R2 CuLi
O
Higher Order Cuprates: somewhat more stable than dialkylcuprates
2 R—Li + CuCN
Epoxides
PhCH2O
OH
Me
Me2CuLi
OH
PhCH2O
R2 CuCNLi2
OH
6
Substitution
6:1
OH
Organocopper reagents react with alkyl halides, epoxides, allylic halides,
propargylic halides, vinyl halides to give substitution products
R1 —X
+
1
Me
R—R1
R2 CuLi
O
X = I, Br, Cl, OTs
Me
Me2CuCNLi2
oxidative addition
HO
90%
R
R Cu X
R1
O
O
I
t-Bu
MeO2C
reductive
R—R1 +
RCuX
elimination
OTs
OTs
CO2Me
(C6 H5)2 CuLi
O
(4 equ)
O
(t-Bu)
2 CuLi
t-Bu
Ph
Ph
47%
CO2Me
76%
NHCOC6H5
(4 equ)
NHCOC6H5
X
Bu2 CuLi
n-Bu
X = OTf
100%
X = OP(O)(OPh)
60%
2
OH
PhCH2O
t-Bu
OCH2Ph
MeO2C
OCH2Ph
Organometallic Reactions: An Overview
Chapter 13-8
Organometallic complexes contain a metal and coordinated ligands.
The type and number of ligands will depend
on the metal and its oxidation state.
Typical ligands of organometallic complexes
Ligand
Charge
H–
Cl–, Br–, I–
R–, Ar–
No. of electrons donated
-1
2
-1
2
-1
2
0
2
2
0
R3P:
2
0
:O≡C:
6
-1
Oxidation state of metal is the difference between the overall charge on
the complex and the sum of the charges for each ligand.
Cl
CH3
Pd
Ph3P
PPh3
H
Ph3P
Rh
Cl
Cl
Cl: -1
CH3: -1
Ph3P: 0
Pd: +2
CO
PPh3
Ph3P: 0
H: -1
CO: 0
Cl: -1 (X2)
Rh: +3
Chem 66H
Organometallic Reactions: An Overview
Chapter 13-9
Organometallic complexes undergo three basic reactions
Oxidative addition
Migratory insertion
reductive elimination
Oxidative addition
The oxidation state and the coordnation number of the metal ion
both increase (usually by two)
Br
R
L
L
L
M
+ R–Br
M
L
L
L
L
L
CN = 4
CN = 6
L = generic ligand
reductive elimination
The oxidation state and the coordnation number of the metal ion
both decrease (usually by two)
H
L
M
L
R
L
L
L
M
L
+ R–H
L
CN = 4
L
CN = 6
Migratory insertion
L
L
L
H 2C
L
L
L
M
H
CH2
L
Rh
L
solvent
L
L
CH2–CH3
L
Migratory
insertion
no change in the metal ion oxidation state
L
Rh
S
CH2–CH3
Chem 66H
Hydrogenation with Wilkinison's Catalyst
Chapter 13-10
Wilkinson's Catalyst, (Ph3P)3RhCl functions as a catalyst in the presence of
hydrogen to convert alkenes into alkanes, ie. hydrogenation.
Ph3P
Cl
Rh
Ph3P
CH3CH2OH
ligand
exchange
Cl
H
Ph3P
H
CH2
Cl
Rh
H
Ph3P
S
Rh = +3
Rh = +3
H
H
Cl
Ph3P
H
Ph3P
Rh
Ph3P
ethanol
Cl
Rh
C–C
H
Migratory
insertion
H
Ph3P
oxidative
addition
Rh = +1
– ethanol
Rh
Ph3P
S
CH2
H
Ph3P
H2
Cl
Rh
Ph3P
PPh3
Rh = +1
Ph3P
Ph3P
Ph3P
Cl
Rh
Ph3P
S
C–C
H
reductive
elimination
Ph3P
Cl
Rh
S
+
H
C–C
H
Chem 66H
Chapter 13-11
Palladium Catalyzed Carbon-Carbon Bond Formation
Suzuki Reaction
OR
Pd(Ph3P)4
R–X +
R–R' +
R' B
NaOH
Ph3P
Pd
Ph3P
Pd
Ph3P
Ph3P
R–X
PPh3
oxidative addition
Ph3P
Pd
PPh3
Ph3P
PPh3
Ph3P
OR
– PPh3
PPh3
+ NaX
HO B
R'–B(OR)2
R
Pd
Ph3P
X
"transmetallation"
step
R
Pd
R–R' + Ph P
3
R'
Ph3P
Pd
PPh3
reductve
elimination
Example
Pd(PPh3)4
Br
+ (RO) 2B
Pd(PPh3)4
+
N
B(OH) 2
Br
N
CH3
N
N
70%
CH3
Chem 66H
Spectroscopy
Chapter 14-1
Molecular Spectroscopy
electromagnetic radiation: energy that is transmitted through space in
the form of waves
wavelength: (λ): the distance from the crest of one wave to the crest of
the next wave
frequency: (ν): the number of complete cycles per second
ν=
c
λ
where c = speed of light
Electromagnetic radiation is transmitted in particle-like packets called
photons or quanta. The energy is inversely proportional to the
wavelength and directly proportional to frequency.
Ε=
hc
λ
where c = speed of light; h = Planck's constant
Ε = hν
ultraviolet
visible
h = Planck's constant
infrared
radio
decreasing energy
Absorbtion of ultraviolet light results in the promotion of an electron to a
higher energy orbital.
Absorbtion of infrared results in increased amplitudes of vibration of
bonded atoms.
Intensity of radiation is proportional to the number of photons.
Chem 66H
Mass Spectrometry
Chapter 14-2
MassSpectrometry
useful for determining molecular weight, presence of specific atoms and
also certain molecular fragments
an organic molecule can be ionized by a number of methods such as
bombardment by electron s or other high energy species.
usually the ionization results from loss of a single electron and the
production of a cation radical.
The princple of mass spectrometry is based on the fact that depending
on the mass to charge ratio of a particular cation radical, it will travel along a
different curved path when exposed to a magentic field.
m =
z
H2r2
2V
m = mass of cation radcal
z = charge (usually +1)
H + strength of the magnetic field
r = radius of the path
V = accelerating potential
placing a detector at some point along the flight path of the ion allows its
mass to charge ratio to be calculated. Since almost all the ions will have
a charge of +1, the mass to charge ratio is also the mass.
A mass spectrum produces a series of peaks which correspond to different
mass of different molecular frgaments and their relative abundance
100
M+
M+1
0
0
10
20
30
40
50
60
70
80
M+2
90 100
Chem 66H
Chapter 14-3
Mass Spectrometry
The molecular ion is the result of loss of one electron from the
parent molecule. Sometimes the molecular ion is too unstable to be
detected, but it usually is present in the mass spectrum.
Chem 66H
M+
M+
M+
M-1+
The molecular ion also fragments into various other fragments by
bond breaking processes in the gas phase.
M+
Each of the fragments which reach the detector will produce a peak in the
spectrum corresponding to its mass.
M+1
+
A peak in the region of highest m/z in a mass spectrum often corresponds
to the moecular ion.
In addition to the peak for the molecular ion, there will also be peaks of
M+1 and M+2 mass which correspond to similar ions which contain other
isotopes of specific elements,
For example, the mass spectrum of 2-butanone contains
a peak at 72 for the molecular ion 12C41H816O
and a peak at 71 for other ions such as 13C12C31H816O
or 12C42H1H716O or 12C41H817O
The base peak is the largest peak in the spectrum corresponding to
the ion which is present in the greatest abundance.
The base peak is often the molecular ion, but not always.
The base peak can be the result of a fragmentation of the molecular
ion into two other species.
To determine the molecular weight of a compound from the mass
spectrum, first look at the region of hghest m/z ratio.
It is usually a reasonable assumption that one of these peaks will be
the molecular ion.
If the molecular ion is present and no Cl, BR or S are present in the
molecule, one of four patterns are most common.
no M+2+ present
M+1+
M+2+
M-1+
M+1+
M+2+
M-2+
M+1+
M+2+
Mass Spectrometry
Chapter 14-4
To determine the molecular weight of a compound from the mass
spectrum, first look at the region of hghest m/z ratio.
Using the known relative abundance of isoptopes of different elements,
the molecular formula can be deduced.
It is usually a reasonable assumption that one of these peaks will be
the molecular ion.
For example:
if the molecular ion is 68, there are three reasonable possibilities
If the molecular ion is present and no Cl, BR or S are present in the
molecule, one of four patterns are most common.
M
+
M
+
M-1
M+
formula
+
M
M-1
M+1+
M+2+
M+1+
+
1
2
H
H
%
99.98
0.01
Isotope
31
P
32
12
M+1+
M+2+
13
14
C
C
98.89
1.11
N
N
99.63
0.37
15
35
75.53
24.47
Cl
Cl
17
O
O
18
O
99.76
0.04
0.20
79
19
100.0
127
F
100.0
95.00
0.76
4.22
0.01
37
16
%
S
S
34
S
36
S
33
Br
81
Br
I
M+1
M+2
C 3H 4N 2
4.07
0.06
C 4H 4O
4.43
0.28
C5H8
5.53
0.12
+
Assumes M+ is 100% otherwise it would be the specified
pecentage of the M+ intensity.
no M+2+ present
Isotope
Chem 66H
50.54
49.46
100.0
M-2+
M+1+
M+2+
Mass Spectrometry
Chapter 14-5
Presence of Nitrogen
Determining the presence of nitrogen is very simple if there are an odd number
of nitrogens present, because the molecular ion will be an odd mass.
For even numebrs of nitrogens, the M+, M+1, and M+2 intensties can be
examined as illustrated before.
Presence of Sulfur
Determining the presence of sulfur can usually be determined the presence
of a slightly large M+2 peak since 34S is 4.22% abundant.
Presence of Bromin and Chlorine
Determining the presence of bromine and chlorine can also be determined
from the M+2 peak since 37Cl is 24.47% abundant and 81Br is 49.46% abundant.
Thus the M+ and M+2 peaks in a compound containing chlorine will be about
a 3:1 ratio and for one containing bromine M+ to M+2 will be about 1:1.
Fragmentation Patterns
cleavage at branches
R
+.
+
+
.
R
because of cation stability, cleavage to produce stable cations is common
Chem 66H
Mass Spectrometry
Chapter 14-6
Fragmentation Patterns
α,β-cleavage
+.
O+
O
+
.
cleavage of a bond alpha and beta to a heteroatom such as oxygen is
common in carbonyl compounds
Loss of a neutral molecule
Loss of H2O, CO, HCN, HCl, NO, etc is common due to the stability
of the neutral species
CO
+
+
+ CO
McLafferty Rearrangement
O
H
H
+.
R
O
H
+.
H
R
+
R
R
McClafferty rearrangment is very common in carbonyl compounds with a
gamma hydrogen atom.
Chem 66H
Infrared Spectroscopy
Chapter 14-7
Infrared Spectroscopy
Infrared is recorded as %T versus wavelength or frequency
When a sample absorbs at a particular wavelength or frequency,
%T is reduced and a peak or band is displayed in the spectrum.
Infrared is recorded as %T versus wavelength or frequency
When a sample absorbs at a particular wavelength or frequency, %T
is reduced and a peak or band is displayed in the spectrum.
100
%T
0
frequency
Nuclei of bonded atoms undergo vibrations similar to two balls
connected by a spring. Depending on the particular atoms bonded to
each other (and their masses) the frequency of this vibration will vary.
Infrared energy is absorbed by molecules resulting in an excited
vibrational state. Vibrations occur in quantized energy levels and thus a
particular type of bond will absorb only at certain frequencies.
Both stretching and bending vibrations can be observed by infrared.
O
CH3
CH3
stretching
O
CH3
CH3
bending
Chem 66H
Infrared Spectroscopy
Chapter 14-8
Frequency of vibration will be inversly proportional to the masses of the atoms.
C
S
C
1350 cm-1
O
C
1700 cm-1
H
C
3000 cm-1
D
2200 cm-1
Frequency of vibration will be directly proportional to the strength of the bonds
C
C
C
2150 cm-1
C
C
C
1200 cm-1
1650 cm-1
Some vibrations are coupled when atoms of similar masses are involved such as
two or mopre C–H bonds such as in a methyl group where there are symmetric
and antisymmetric stretches
H
H
C
C
H
H
H
C–Hsym = 2872 cm-1
C–Hasym = 2962 cm-1
O–
H
C
N
H
H
+
O
H
N:
H
coupled vibrations are common as in functional groups above which each
have a symmetric and antisymmetric vibration. These can help dentify certain
functional groups
Chem 66H
Infrared Spectroscopy
Chapter 14-9
4000 cm-1 to 1300 cm-1 is known as the functional group region.
400 cm-1 to 1300 cm-1 is known as the fingerprint region since it is unique for
every compound.
Interpretation of Infrared Spectra
Correlation tables
Infrared spectra of thousands of compounds have been tabulated and
general trends are known. Some common functional groups are
shown below.
C=O str
OH and NH str
CH str
C≡N str
C—O str
C=N str
C=C str
NH bend
Chem 66H
C—C and C—H Bonds
sp3 C—C
sp2 C=C
sp2 C—C (aryl)
sp C≡C
sp3 C—H
sp2 C—H
sp C—H
C(CH3)2
2500
2000
1500
1640-1820 cm–1
Aldehydes C=O;
C—H(O)
1640-1820 cm–1
Carboxylic acids C=O;
C(O)—OH
1640-1820 cm–1
Esters C=O ;
C(O)—OR
1640-1820 cm–1
1100-1300 cm–1
2800-3000 cm–1
3000-3300 cm–1
3300 cm–1
1360-1385 cm–1 (two peaks)
Alcohols and Amines
3000-3700 cm–1
900-1300 cm–1
1000 800
–1
2820-2900 and 2700-2780 cm
(weak but characteristic)
3330-2900 cm–1
2100-2250 cm–1
CH bend
Carbonyls
One of the most useful absorbtions in infrared 1640-1820 cm-1
Ketones (saturated) C=O
1450-1600 cm–1
C—C str
O—H or N—H
C—O or C—N
3000
1600-1700 cm–1
C—N str
OH bend
3500
weak, not useful
Ethers
C—O
1050-1260 strong
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-1
Nuclear Magnetic Resonance (NMR) Spectroscopy
Some atomic nuclei (1H,
have a nuclear spin.
13
C, others) behave as if they are spinning...they
Spinning of a charged particle creates a magnetic moment.
If an external magnetic field is applied, these small magnetic moments (of the
nuclei) either align with the field α
() or against the field β
(), about 50% with
and 50% against the field at any one time.
β
Ho
∆E
hν
α
β
Ho
∆E
α
Ho = the external magnetic field
Resonance: the flip of the magnetic moment from parallel to antiparallel to
the external magnetic field.
Irradiation at the frequency equal to the energy difference,∆E, causes
resonance.
∆E depends on the external magnetic field.
Protons (or other nuclei) in different magnetic environments resonate at
different field strengths.
A proton which resonates at a higher field is in a stronger magnetic
environment or shielded.
A proton which resonates at a lower magnetic field is said to be deshielded.
Different magnetic environments are created by different electron densities in
the vicinity of a proton.
Chem 66H
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-2
Chem 66H
Adjacent electron withdrawing groups, highly electronegative atoms, or the
hybridization of the carbon to which the proton is bonded can alter the
magnetic environment.
The pi system of benzene creates a magnetic field or ring current which
deshields the protons attached to the ring.
The local electrons create a small electric and magnetic field around a proton
and shield it.
Similarly, pi electrons in a C=O bond create a field which deshields the
proton bonded to the C=O of an aldehyde. This is also affected by the
inductive effect of the C=O.
The more electron density present around the proton, the greater the field and
the greater the shielding.
Resonances are reported in chemical shifts (δ) downfield from
tetramethylsilane (TMS) (CH
3)
4Si.
δ=
distance from TMS in Hz
MHz of spectrum
ppm
In methyl halides, the more electronegative the halogen, the more deshielded
the protons on the methyl. This is because F is inductively more electron
withdrawing, causing the carbon to be more positive and thus pulling more
electrons away from the hydrogen and causing it to be less shielded.
In methyl halides, the more electronegative the halogen, the more deshielded the pr
H3C—F
δ
H3C—Cl
3.0
4.3
H3C—Br
H3C—I
2.7
2.1
Pi electron effects
Magnetic fields created by pi electrons are directional and said to have an
anisotropic effect.
R
Ho
C
H
O
H
H deshielded
H deshielded
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-3
Equivalent and Nonequivalent Protons
Protons that are in the same magnetic environment are equivalent and have
the same chemical shifts.
Protons in different magnetic fields are nonequivalent and have different
chemical shifts.
Magnetic equivalence is usually the same as chemical equivalence.
Equivalence can be established by symmetry operations such as rotation,
mirror planes and centers of symmetry
Chemically equivalent protons have the same chemical shifts.
To determine if protons are chemically equivalent, replace one by a different
group, e.g. D or Br.
Then replace a different one by the same group and compare the two
compounds. If they are identical, the protons are equivalent.
H
H
H
C
C
H
H
OH
H
H
H
H
C
C
C
H
Cl H
H
equivalent, but not to CH3 protons
all six are equivalent
equivalent
Equivalent protons can be on different carbons.
Protons which are homotopic or enantiotopic resonate at the same chemical
shift in the NMR.
If protons are interconverted by rotation about a single bond, they will average
out on the NMR time scale and a single resonance will be observed.
ClH2CCH2Cl anti and gauche forms rapidly interconvert and a single
resonance is observed.
Axial and equatorial hydrogens in cyclohexane average to a single peak
because of rapid ring inversion.
Diastereotopic hydrogens are chemically nonequivalent and thus give different
chemical shifts in the NMR
Chem 66H
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-4
Intergration
The spectrometer can integrate and determine the relative number of
hydrogens associated with each resonance in the NMR spectrum by
determining the area under the peaks.
Spin-Spin Coupling
for example...
3
3
CH3CH2OCH3
2
TMS
If a proton (H
a) is bonded to a carbon which is bonded to a carbon that has
one proton (Hb), Ha will appear as a doublet
Since in half the molecules, Hb will be in the α state and in half will be in the β
state, Ha will experience two different magnetic fields and two peaks (a
doublet) will appear for Ha.
Ha without an adjacent hydrogen
For one adjacent hydrogen
α or β
Ha with Hb adjacent in theβ state
Ha with Hb adjacent in theα state
Chem 66H
Chapter 15-5
Nuclear Magnetic Resonance Spectroscopy
For two adjacent hydrogens: Hb, Hc
At any one time Hb or Hc could be in the α or β state (50:50) thus 4
combinations for Hb, Hc exist:
αbαc αbβc
βbαc
βbβc gives 1:2:1 triplet
When both Hb and Hc are α, a different field is observed than if both are β
or one is α and one is β.
When one is α and one is β, the field is the same. That is, βbαc and αbβc
produce the same field and a single signal for Ha is observed with twice
the intensity.
Thus three signals are observed in a 1:2:1 ratio: a so-called triplet
For three adjacent protons:
ααα ααβ αββ βββ 1:3:3:1 quartet
αβα βαβ
βαα ββα
Thus the splitting pattern of a particular proton or equivalent protons will be a
pattern with n+1 lines where n is the number of adjacent equivalent protons.
singlet 0 neighboring protons
doublet 1 neighboring protons
triplet 2 neighboring protons
quartet 3 neighboring protons
quintet 4 neighboring protons
sextet 5 neighboring protons
septet 6 neighboring protons
The separation of the peaks in a splitting pattern is called the coupling
constant, J.
Chem 66H
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-6
Splitting Diagrams
Splitting patterns for protons can be constructed in diagram form by starting
with one line to represent the unsplit proton resonance.
If an adjacent proton Hb affects Ha it is split into a doublet; if another
equivalent proton to Hb is present, each line of the double will be split into a
doublet, since the coupling constant J is the same, the two center lines
overlap and a only three lines are observed with the center line twice the
height.
This can be repeated for additional adjacent protons.
Ha without an adjacent hydrogen
1
1
1
splitting diagram
Ha split by one adjacent hydrogen
1
2
Ha split by a second adjacent hydrogen
Ha split by a third adjacent hydrogen
1
3
3
1
Chemical Exchange and Hydrogen Bonding
CH3OH, methanol would be expected to give an NMR spectrum of a
doublet for the CH3 and a quartet for the OH. For a dilute sample at -40° in
CCl4 this is the case.
If the NMR spectrum is run at 25° as a more concentrated sample only two
singlets are observed. This is because the intermolecular hydrogen
bonding in methanol allows the rapid exchange of the OH proton from one
CH3OH molecule to another, effectively averaging the spin states of the OH
proton and resulting in no change in the magnetic field due to the OH.
Amines and other compounds which can undergo hydrogen bonding can
also show this effect. Thus the NMR spectra of alcohols, amines and
carboxylic acids are temperature, concentration and solvent dependent.
Chem 66H
Nuclear Magnetic Resonance Spectroscopy
Chapter 15-7
CHEMICAL SHIFTS
Functional Group Shift,δ
Primary alkyl, RCH3
Secondary alkyl, RCH2R
Tertiary alkyl, R3CH
0.8-1.0
1.2-1.4
1.4-1.7
Allylic, R2C=C—CH2R
1.6-1.9
Benzylic, ArCH2R
Iodoalkane, RCH2I
Bromoalkane, RCH2Br
Chloroalkane, RCH2Cl
Ether, RCH2OR
Alcohol, RCH2OH
Ketone, RCH2C(=O)R
2.2-2.5
3.1-3.3
3.4-3.6
3.6-3.8
3.3-3.9
3.3-4.0
2.1-2.6
Aldehyde, RCH(O)
Terminal alkene, R2C=CH2
Internal alkene, R2C=CHR
Aromatic, Ar—H
Alkyne, RC≡C—H
Alcoholic hydroxy, ROH
Amine, RNH2
9.5-9.6
4.6-5.0
5.2-5.7
6.0-9.5
1.7-3.1
0.5-5.0 (variable)
0.5-5.0 (variable)
Chem 66H
The Carbonyl group
Chapter 17-1
Carbonyl Group
sp3 hybridized, trigonal planar carbonyl carbon; partial positive C and
partial negative (Lewis basic) oxygen.
pi bond
:
C
O
lone pairs
δ+ δ−
:
C
O
..
+
C
–
..
O
.. :
:
C–O pi* is low-lying and therefore interacts well
with high-lying filled-nonbonding orbitals: thus nucleophilic,
not electrophilic addition reactions are charactersistic of carbonyl compounds.
Because of the polar C=O bond, boiling points are higher than nonpolar
compounds of similar molecular weights.
Aldehydes and ketones are capable of hydrogen bonding to water,
alcohols and acids
Spectral Properties
Infrared:
Ketones(C=O) 1660-1750 cm-1
shifted by 25 cm-1 if aromatic or unsaturated :
PhCHO,
CH2=CHCOCH3
Adlehydes (C=O) 1700-1740
1
(C—H) 2850
H NMR R—CHO 9-10 ppm
RCH2COR 2.0 - 2.6 ppm due to inductive deshielding
Chem 66H
The Nucleophilic Addition Reaction
Chapter 17-2
Nucleophilic addition reactions
Addition Reactions of Aldehydes and Ketones
Carbonyl group can be attacked by nucleophiles
R
sp2
C
R
:
..
O. .
R sp3
R
C
.. _
O
.. :
Nu
Nu:
or undergo addition of reagents to the pi bond by electrophiles adding first
R
C
R
..
O. .
H+
R
C
R
R
+H
O..
R
R C R
aldehyde
O
R C H
..
O
.. H
Nu
Nu:
ketone
O
C
formaldehyde
O
H C H
increasing reactivity due to steric and electronic effects
Ketones are more sterically hindered since they have two alkyl groups;
aldehydes have one H and one alkyl group.
Alkyl groups are electron releasing and make the C=O carbon less
positive.
Reaction with H2O: Formation of Hydrates
hydrates are normally transient, unstable species which are in equilibrium
with the carbonyl compound
Chem 66H
The Nucleophilic Addition Reaction
Chapter 17-3
Acid catalyzed mechanism
H+
.. ..
O
.. + H
O
H+
H 2O
CH3 C CH3
H2O:
CH3 C CH3
. H
..O.
O
- H+
H
CH3 C CH3 hydrate
O
H
CH3 C CH3
O+
H
H
the equilibrium constant for the formation of hydrates is dependent on the
carbonyl substituents: the more sterically hindered and the more electron
rich the carbonyl, the less of the hydrate that will be present, conversely,
the less sterically hindered and the more electron deficient the carbonyl
carbon, the more hydrate that will be present.
O
O
F3C
CF3
H
22,000
O
H
H 3C
O
H (H3C)3C
1.8 X 10 -2
41
O
H
H 3C
4.1 X 10 -3
CH3
2.5 X 10 -5
Khydration
Base catalyzed mechanism
O
CH3 C CH3
HO –
H 2O
.
..O. –
CH3 C CH3
O
H
H–O–H
O
H
CH3 C CH3
O
H
hydrate
+ HO –
Chem 66H
Formation of Acetals
Chapter 17-4
with alcohols: formation of acetals
.. ..
O
O—H
H+
CH3 C H
O—CH3
CH3 C H + CH3OH
CH3OH, H+
a hemiacetal
O—CH3
CH3 C H
O—CH3
an acetal
mechanism
.. ..
O
H+
CH3 C H
CH3 C H
O+
CH3
H
CH3 C H
H+
.. H
O
H . H
.
+O
CH3 C H
.
O.
CH3 . .
CH3 C H
O
CH3
CH3 .. H
O+
CH3 C H
O
CH3
. H
..O.
..
CH3OH
..
.. + H
O
H+
..
CH3OH
..
- H2 O
CH3 C H
O+
CH3
O—CH3
CH3 C H
O—CH3
if H2O is present in a large amount, the carbonyl compound will be favored.
if H2O is not present, but ROH is present, the acetal will be favored
O
HO
OH
p-toluenesulfonic acid
toluene, heat
H2O, HCl or H2SO4, acetone
O
O
Chem 66H
Acetals and Cyanohydrins
Chapter 17-5
Acetals as protecting groups
many times multifunctional compounds must be treated to convert one
functional group selectively. In the example below, direct treatment of the
keto-ester with LiAlH4 would result in reduction of both carbonyls, so the
ketone must be protected prior to reduction of the ester
O
O
CH2OH
CO2CH3
HO
OH
p-toluenesulfonic acid
toluene, heat
O
O
H2O, H2SO4
O
LiAlH4
CO2CH3
O
CH2OH
Formation of cyanohydrins
R
R
C
R
..
O..
.. ..
O
Ph C H
–
R
C
O–H
NC
.. ..
O
Ph C H
HCN
-
O—H
CN
Ph C H
CN
+ HCN
.. ... –
O.
Ph C H
CN
H—CN
a cyanohydrin
.
..O H
Ph C H
CN
CN
forms a new C–C bond and introduces a functional group which can be
converted to a carboxylic acid or an amine
Chem 66H
Preparation of Alcohols
Chapter 17-6
Reduction of Carbonyl Compounds
a) Hydrogenation
C=O bond can be hydrogentated much like a C=C bond, C=O usually
requires harsher conditions
O
H
OH
Pt, H2
O
O
CH2=CHCH2CH2
C
H
Ni, H2, 25°C
CH3CH2CH2CH2
C
H
OH
Ni, H2
heat, pressure
CH3CH2CH2CH2
C
H
H
b) Metal Hydrides
LiAlH4, lithium aluminum hydride
NaBH4, sodium borohydride
OH
O
CH3CH2CH2
C
CH3
1. LiAlH4
CH3CH2CH2
C
CH3
H
2. H2O, H+
OH
1. NaBH4
CH3CH2CH2
+
2. H2O, H
C
CH3
H
LiAlH4 much more reactive also reduces esters, carboxylic acids, nitriles,
amides
NaBH4 sodium borohydride reduces only aldehydes and ketones: more
selective
Chem 66H
Preparation of Alcohols
Chapter 17-7
For example
O
H
CH2CH2CO2Et
OH
CH2CH2CO2Et
1. NaBH4
2. H2O, H+
O
H
OH
CH2CH2CH2OH
CH2CH2CO2Et
1. LiAlH4
2. H2O, H+
LiAlH4 and NaBH4 do not reduce isolated double bonds
Mechanism for NaBH4:
R
R
δ+
C O
δ− H
R
H B– H
R
C
O—B – H3
H
H
R
R
R
C
O)4B
H 2O
–
R
C
OH
+ H3BO3
H
H
Mechanism for LiAlH4:
Li +
H
–
O
H
Al H
H H
O
H
+
Al–H2(OR)Li
H2 O
H
OH
H
H
Chem 66H
Preparation of Alcohols fromo Grignard Reagents
Chapter 17-8
A Grignard reaction with
1. formaldehyde produces a primary alcohol
O
1. CH3
C
H
2. H2O, H+
H
OH
Mg-X
H
C
H
CH3
2. an aldehyde produces a secondary alcohol
O
OH
1. (CH3)CHMgBr
C
CH3CH2
H
2. H2O, H+
H
C
CH2CH3
CH(CH3)2
3. a ketone produces a tertiary alcohol
O
C
CH3CH2
1.(CH3)CHMgBr
+
CH3 2. H2O, H
OH
CH3
C
CH2CH3
CH(CH3)2
4. an ester produces a tertiary alcohol (addition of two molecules of
Grignard reagent)
O
OH
1. 2 CH3MgBr
C
OCH3
2. H2O, H+
CH3
C
CH3
5. ethylene oxide produces a primary alcohol
O
1. C6H5MgBr
2. H2O, H+
OH
Chem 66H
Carbohydrates
Chapter 17-9
Carbohydrates are naturally occurring compounds with C, H, O; often with the
emperical formula CH2O.
D and L sugars
In the 19th century (+)-glyceraldehyde was arbitrarily assigned the
configuration below and designated D.
Monosaccharides...simple sugars which cannot be broken down by hydrolysis;
e.g. glucose, fructose, ribose, galactose, deoxyribose, etc.
CHO
Disaccharides...dimers of monosaccharides units; e.g.sucrose is made up of
glucose and fructose
H
Oligosaccharides...two to eight monosaccharides units
Projection is designated D. If the OH is projected to the left, the sugar is
an L sugar.
for simplicity carbohydrates are often represented by Fischer Projections.
in a Fischer projection the horizontal bonds are always out and the vertical bonds
are always in.
HO
C
H
CH3
HO
H
CH3
HO
CHO
OH
H
CHO
H
Fischer Projections may be rotated 180° but not 90°.
A 90° rotation creates the enantiomer.
CHO
H
D-glyceraldehyde
Now all carbohydrates with a hydroxyl to the right on the last carbon in
the Fischer
Fischer Projections
CH2OH
OH
CH2OH
Polysaccharides...more than eight monosaccharide units; e.g. cellulose is
polyglucose
CH2OH
Chem 66H
or
H
C
OH
HO
C
H
H
OH
H
C
OH
H
OH
H
C
OH
CH2OH
Classification of Carbohydrates
The ending-ose indicates a carbohydrate
An aldose contains and aldehyde; a ketose contains a ketone
A triose has three carbons, a tetrose has four carbons, a pentose has five
carbons and a hexose has six carbons.
A ketohexose is a six carbon sugar containing a ketone.
CH2OH
HO
OH
H
CHO
HO
H
H
OH
H
OH
H
OH
H
OH
H
OH
HO
H
H
OH
HO
H
CH2OH
D-glucose
CHO
CH2OH
CH2OH
L-glucose
D-ribose
Aldoses
Chapter 17-10
Chem 66H
Aldohtetroses
Aldohexoses
The aldotetroses have 2 asymmetric carbons and thus 22 or 4 stereoisomers.
The D- seres is shown below. Each has a corresponding L-isomer
The aldohexoses have 4 asymmetric carbons and thus 24 or 16 stereoisomers.
The D- seres is shown below. Each has a corresponding L-isomer
CHO
CHO
CHO
H
OH
HO
H
OH
H
H
OH
CH2OH
CH2OH
D-erythrose
D-threose
CHO
CHO
CHO
H
OH
HO
H
OH
H
OH
HO
H
OH
H
OH
H
D-ribose
H
CH2OH
D-arabinose
H
HO
H
OH
H
OH
HO
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
H
HO
H
H
HO
H
H
HO
H
H
CH2OH
D-xylose
OH
CHO
OH
H
OH
CH2OH
D-lyxose
H
OH
CH2OH
D-galactose
H
CH2OH
D-altrose
CHO
HO
OH
CHO
OH
CH2OH
D-allose
The aldotetroses have 3 asymmetric carbons and thus 23 or 8 stereoisomers.
The D- seres is shown below. Each has a corresponding L-isomer
CHO
H
Aldopentoses
CH2OH
CHO
H
HO
H
HO
H
H
OH
CH2OH
D-talose
HO
H
H
HO
H
CH2OH
D-glucose
CH2OH
D-mannose
CHO
CHO
HO
OH
CHO
H
OH
HO
H
OH
H
HO
H
H
OH
CH2OH
D-gulose
HO
H
H
OH
H
OH
CH2OH
D-idose
Cyclic forms of Carbohydrates
Chapter 17-11
The open chain representations of the sugars shown above is for simplicity.
Sugars normally exist as cyclic hemiacetals.
R
C
R
H
C
Pyranose Forms
O
OH
ROH, H+
O
OR
+
ROH, H
H
R
OR
C
C
H
H
OR
acetal
hemiacetal
H
CH2OH
CH2OH
O H
H
O OH
H
H
or
OH H
OH H
HO
OH
HO
H
H OH
H OH
H
OH
HO
Furanose Forms
Chem 66H
H
H
OH
H
OH
diastereomers
CH2OH
D-glucose
CHO
HO
OH
O
H
H
H
H
H
OH
H
OH
H
OH
OH
HO
HO
H
O
H
H
H
OH
β-D-ribofuranose
α-D-ribofuranose
D-ribose
CH2OH
O
OH
HO
CH2OH
Haworth projection
HO
OH
hydrogen if not labeled
OH
OH
α-D-glucopyranose
six membered ring
glucose
O
N
N
HO
O
H
O
H
H
O
H
P
O
NH
N
NH
NH2
N
HO
O
H
O-
Oa DNA nucleotide
HO
HO
H
O
H
H
O
H
OH
P
O-
Oan RNA nucleotide
CH2OH O
OH
OH
O
CH2OH up at C-5
OH at C-1 down
Chapter 17-12
Anomers and Mutarotation
Anomers: monosaccharides which differ only in their configuration at C-1
anomeric carbon
HO
HO
CH2OH O
H
OH
OH
α−D-glucopyranose
HO
HO
CH2OH OH
C O
OH
H
HO
HO
OH
H
β−D-glucopyranose
CH2OH
O
HO
OH
OH
OH
CH2OH O
OH
CH2OH
O OH
HO
OH
OH
Mutarotation
α-D-glucose has a melting point of 146° and a specific rotation of +112°.
β-D-glucose has a melting point of 150° and a specific rotation of 18.7°.
The specific rotation of a solution of either α or β-D-glucose slowly changes
until it reaches an equilibrium value of +52.6°.
This is mutarotation and is due to the conversion of α-D-glucose to β-D-glucose
or the reverse.
The two forms are in equilibrium in solution: 36%α and 64% β.
CH3OH,
CH3OH,
OH
CH
+
2
CH2OH O
O
H+
CH2OH O
H
HO
HO
HO
HO
H
HO
H
HO
OCH3
OH
OH
OH
OH
OCH3
H
methyl-β-D-glucopyranoside
α−D-glucopyranose
methyl-α-D-glucopyranoside
an acetal
Glycosides are stable under neutral and basic conditions but are interconverted in acid.
Chem 66H
Chapter 17-13
Oxidation of Monosaccharides
Oxidation
Aldoses are readily oxidized because the hemiacetal is in equilibrium with an
aldehyde, a readily oxidized functional group.
The product of the oxidation of the aldehyde of an aldose to a carboxylic acid is
called an aldonic acid.
HO
HO
CH2OH O
HO
HO
H
OH
OH
Ag(NH3)2+
CH2OH OH
H
OHC
O
HO –
or
Br2, H2O
HO
HO
CH2OH OH
O–
C
OH
O
Even ketoses can be oxidized since they are in equilibrium with the aldose through an
enediol tautomer
CH2OH
O
HO
H
HO
CHOH
CHO
C
CHOH
OH
H
HO
CO2H
CHOH
Ag(NH3)2+
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH
CH2OH
CH2OH
enediol
ketose
aldose
HO –
or
Br2, H2O
HO
H
H
OH
H
OH
CH2OH
aldonic acid
The product of oxidation of the aldehyde and the terminal primay alcohol to form a diacid is
called an aldaric acid.
CHO
H
CO2H
OH
H
H
dil. HNO3
H
OH
heat
H
OH
HO
CH2OH
D-glucose
HO
OH
H
H
OH
H
OH
CO2H
D-glucaric acid
A uronic acid results when the terminal
primary hydroxyl has been oxidized to a
carboxylic acid. This generally only
occurs enzymatically.
Chem 66H
Chapter 17-14
Reactions of the Hydroxyl Groups
Reactions of the Hydroxyl Groups
The alcohols in sugars react much like any hydroxyl functional group.
Ester formation from acetic anhydride
O
HO
HO
CH2OH O
CH3
H
OH
C
O
O
C
CH3
NaOAc, cold
OH
O
CH2OAc O
C
CH3
O
CH3
O
H
C
O
O
CH3
O
CH3 C
C
O
O
penta-O-acetyl-α-D-glucopyranose
Ether Formation
The hydroxyls of carbohydrates are more acidic than normal alcohols because of the
inductive effects of the adjacent oxygens.
These hydroxyls can be deprotonated to form alkoxides with NaOH.
The alkoxides can be alkylated (SN2 displacement) by dimethyl sulfate...sulfate is an
excellent leaving group because of the resonance stabilization of the anion.
HO
HO
CH2OH O
NaOH,
H
OH
OH
CH3OSO3CH3
CH3O
CH3O
CH2OCH3
O
H
CH3O
CH3O
all hydroxyls are converted to methyl ethers
Acetal Formation
HO
HO
CH2OH O
C6H5CHO, H+
H
OH
OH
O
C6 H5
O
HO
O
H
OH
OH
Chem 66H
Disaccharides
Chapter 17-15
Disaccharides
Maltose
α−linkage
H2O, H+
CH2OH O
HO
HO
H
CH2OH O
O
HO
α-maltose:
glycoside between C-4
OH of glucose and glucose
anomeric carbon
α-1,4-glucan
Starch
2 glucose
OH
or enzymes
OH
H
OH
α-glucosidase
maltose
enzymes
glucose
beer
maltohydrolase
Cellobiose
principal disaccharide of cellulose
HO
HO
CH2OH O
OH
H
H2O, H+
CH2OH O
O
HO
OH
H
OH
β−linkage
2 glucose
or
β-glycosidase
Lactose
only in mammals, 5% in human milk; one unit of glucose, one of galactose
HO
H
HO
CH2OH O
OH
H
H2O, H+
CH2OH O
O
HO
β−linkage
OH
H
OH
or
enzyme
glucose + galactose
Chem 66H
Biological Oxidation of Alcohols
Chapter 17-16
O
alcohol dehydrogenase
CH3
CH2OH
CH3
ethanol
acetaldehyde
–
O OO O–
P
P
O
O
OO
HO
HO
N
N
O
OH
HO
N
+
NAD+
+
N
O
NH2
H 2N
N
O
alcohol dehydrogenase
CH2OH + NAD
CH3
CH3
ethanol
O
N
H
C
H
H
R
N
H 3C
+ NAD+
C
lactic acid
CH3
+ H+
NADH
O
OH
H
O
H 2N
NAD+
C
+
H
CH3
H 2N
O
H
O
H
+
CH + NADH + H
acetaldehyde
+
R
CH
OH
lactic acid dehydrogenase
O
H 3C
C
O
OH
pyruvic acid
+ NADH + H+
Chem 66H
Oxidation of Aldehydes and Ketones
Chapter 17-17
Oxidation of Aldehydes
Aldehydes can be oxidized to carboxylic acids by KMnO4 or H2CrO4.
Ketones cannot ordinarily be oxidized further.
CH3CH2CH2CH2CO2H
CH3
O
H2CrO4
CHO
CO2H
Aldehydes are oxidized to carboxylic acids throught their hydrates
O
O
H 2O
CH3 CH O Cr OH
CH3 CH2OH
CH3 CH
H
O
H2O:
CH3
H
O
CH3 C O Cr OH
OH
O
OH
CH
OH
CH3
O
C—OH
+ HCrO3-
Baeyer-Villiger Oxidation
oxidation of a ketone to an ester (cyclic ketone to a lactone)
Ar
O
O
ArCO3H
H
O
O
O
O
O
+ ArCO2H
most substituted carbon (best able to stabilize a positive charge; i.e. one with
highest electron density) will migrate.
O
O
CH3
m-CPBA
O
CH3
retention of configuration
CH3
CH3
Strained systems can be oxidized with H2O2, HO – or t-BuOOH, HO – and do not
require RCO3H
t-BuOOH, NaOH
CH3CH2CH2CH2CHO + KMnO4, H+, H2O
CH3
Chem 66H
O
O
Chapter 18-6
Sulfur Ylides
Unstabilized sulfur ylides reaction with unsaturated ketones to give epoxides:
O
O
Me2S+—-CH2
O
O-
CH3 S+ CH3
O
Explanation:
k1
O + Me2S=CH2
CH2—S+(CH3)2 k2
O
O-
k-1
unstabilized sulfur ylides: k-1 < k2
in stabilized sulfur ylides k-1 > k2; k'2 > k'-1
O
O
O +
k1
Me2S+—-CH2
k-1
k'
1
CH2—S+(CH3)2
k2
O-
O
k'
-1
O-
k'
2
O
(CH3)2S+—CH2
Cyclopropylidine sulfur ylides
O
Ph
+
Ph2S
Ph
O
O
+ Ph2S
Ph
Ph
O
Trost J. Am. Chem. Soc.1973, 95, 5298, 5307, 5311, 5317.
Chem 66H
Chapter 18-5
Sulfur Ylides
Sulfur Ylides
Sulfur ylides react with aldehydes and ketones to give epoxides rather than alkenes
CH3—I
CH3—S—CH3
ICH3 S+ CH3
CH3
sulfide
—
NaH
CH3 S+ CH2
DMSO
CH3
sulfonium salt
O
CH3—S—CH3
CH3—I
I- O
CH3 S+ CH3
CH3
slower
sulfide
CH3 S CH2
CH3
sulfur ylide
O
NaH
O
—
CH3 S+ CH2
DMSO
CH3 S CH2
CH3
CH3
sulfoxonium salt
oxygen anion in intermediate displaces the sulfide to form an epoxide
—
O
O–
CH3 S+ CH2 +
CH3
O
CH3 S+ CH2
CH3
+ CH3—S—CH3
Examples:
O
CH3 S CH2
CH3
O
CHO
H
OH
BF3-OEt2
O+B-F3
H
H
O+B-F3
H
OB-F3
Chem 66H
Chapter 18-4
Wittig Reaction
Stereocontrol in the Wadsworth-Emmons
Me
Ph3P
CHO
CO2Et
CO2Et
O
O
Me
Me
Me
(MeO)
2(O)P
Me
70% 95:5 E:Z
Kishi JACS1979, 101, 259.
CO
2Me
CHO
Me
KO t-Bu, THF
Me
Me
CO2Me
5:95 E:Z
Me
CHO
(MeO)
2(O)P
CO
2CHMe2
CO2CHMe2
KO t-Bu, THF
Me
Me
Me
95:5 E:Z
Kishi Tetrahedron Lett.1981, 37, 3873.
O
CF3CH2O P
CF3CH2O
CO2Et
R
THF
18-crown-6
RCHO
Me
O
CF3CH2O P
CF3CH2O
KN(SiMe3)2
CO2Et
Me
30-50:1 Z:E
CO2Me
KN(SiMe3)2
R
CO2Me
THF
18-crown-6
RCHO
4-50:1 Z:E
Still Tetrahedron Lett.1983, 24, 4405.
O
CH3CH2O P
CH3CH2O
CO2Et
MgBr2
THF
Et3N
RCHO
R
CO2Me
>97:3 E:Z
Rathke J. Org. Chem 1986, 50, 2624.
Chem 66H
Chapter 18-3
Wittig Reaction
Special Ylides
one carbon homologation:
H2O; H+
Ph3P=CHOCH3 +
O
OHC
OCH3
Corey-Fuchs
CBr4 + Ph3P
Zn°
RCHO
Ph3P=CBr2
E+
R C C Li
RCH=CBr2
2 BuLi
E+ = H, ClCO2CH3, Cl—SiMe3
R C C E
Wadsworth-Emmons Reaction
Uses phosphonate anions instead of ylides
Nucleophile is an anion, not an ylide and it is thus more reactive
Phosphonate is formed by the reaciton of a trialkyl phosphite and and alkyl halide:
the Arbuzov Reaction:
BrCH3CH2 O
(EtO)
3P:
+ Br—CH2CO2Et
O
base
(EtO)
2P CH2CO2Et
(EtO)
CH2CO2Et
2P
+
O
(EtO)
2P CH—CO
2Et
_
phosphonate anion
O
CO2Et
CO2Et
P(OEt)
2
O–
O
+
O
(EtO)
2P O
water soluble
Chem 66H
Wittig Reaction
Chapter 18-2
Chem 66H
The Wittig Reaction
The position of the alkene is unambiguous in the product.
R1
– +
R1
O
R1
R2
R2 C PR3
R2
+
R1
carbonyl
O
P R
R
R2 R
phosphorous ylide
R1
R2
R1
R2
+ Ph3P=O
H 3C
C O
H 3C
H 3C
CH2CH3
C CH
H 3C
Driving force is the formation of the very strong P—O bond
Phosphonium salts are readily formed from triphenyl phosphine and primary or
secondary alkyl halides. Tertiary alkyl halides are not useful since the reaction is
an SN2 reaction.
R1
X C R2
H
+
R2
:PPh3
R2
Ph3P C H
R2
BASE
+
pKa = 23
t-BuO- K+
BuLi
NaH
EtO- Na+
CH3SOCH2-
R2
Ph3P C
R2
- R2
Ph3P C
R2
dπ−pπ
Phosphorane (yilde)
Acidity of carbon adjacent to+PPh3 is due to a combination of inductive and
resonance effects. C lone pair P-antibonding overlap.
Mechanism:
oxaphosphetanes
R1
O
H
+
– +
R2 C PR3
H
R1
R2
O
P R
R
R
Otrans
R2
C C
H
H
R1
R1
O
P R
R
R
+
R2
Ocis
+
CH2CH3
HC
O
better since ylide (phosphorane
is derived from a primary
halide)
ylide derived from a secondary halide: substitution will be more difficult.
R1 Ph
C P+ Ph
H Ph X–
X = I, Br, OTs
+
H 3C
C PPh3
H 3C
CH2CH3
+ HC
PPh3
+
H
R1
C C
H
R2
Addition Elimination Reactions
Chapter 18-1
Addition-Elimination Reactions
secondary amines produce enamines
primary amines produce Imines
O
H
C N
H 2N
CHO
Chem 66H
CH2CH3
+
N
H
CH2CH3
CH2CH3
N– CH2CH3
H+
enamine
an imine
H+
mechanism of enamine formation
if the reaction is too acidic the amine is completely protonated and will not add
if the reaction is not acidic, OH2 is not eliminated
..O..
mechanism of imine formation
.. ..
O
O
- H2 O
R
R
R C R
R
N
..
H
H+
R
Hydrazones and Oximes
O
+ H2N—NH2
H+
R
- H+
+ H2N—OH
R
R
C
N+
H
R
N
N—NH2
hydrazone
O
H+
addition
H
O +—H
C
elimination
..
HNR2
N—OH
oxime
oximes, phenyl hydrazones, etc. are often solid and can be used to
characterize carbonyl compounds by melting points
N
..
R
H
H+
O
+
N
R
H
R
- H2 O
+
R N
H
H
addition
..
H2N—R
O +—H
–
R C R
R C R
H
–
- H+
H
elimination
N
R + R
R
N
R
enamine
Nitriles
Chapter 19-15
Nitriles
Chem 66H
Base
CH3CO2H acetic acid
CH3CH2CH2CH2CO2H pentanoic acid
—C≡N: pKb = 24
CH3CN acetonitrile
CH3CH2CH2CH2CN pentanonitrile
HO –
NH3 pKb = 4.5
electrons more tightly held in sp orbital, 50% s character N more electronegative
Preparation
H
HO–
R C N—H
O
H 2O
H—O H
R C N—H
O–
R C N—H
O—H
- H+ transfer
–
R C O—H
O
NH2
R C O – + NH3
O
SN2 with cyanide ion
Ph—CH2—Br
–
R C N
H—O
R C N:
Na+ - CN
Ph—CH2—CN
Reduction
From benzenediazonium salts
NH2
N2+ X –
NaNO2
HCl
CH2—CN
CN
CuCN
or
H2, Ni
KCN
Dehydration of amides
O
CH3CH2CH C NH2
CH3
SOCl2
CH3CH2CH C N:
CH3
Reactions: Hydrolysis
Acid
R C N:
R C N+—H
R C N—H
O+H2
H2O:
H
+
R C N—H
O—H
H2O:
H
R C N—H
+O—H
- H+
H—O H
R C N—H
H—O
R C N + H3
+
O—H
O—H
1. LiAlH4 2. H2O
- H+ transfer
– H+
H2O+ H
R C N—H
O—H
H transfer
R C OH
O
+ NH4+
CH2—CH2—NH2
Amides
Chapter 19-14
proteins
O
N
H
N
N
H
O
O
N
H
O
H
N
O
N
H
Polyamides
x HO2C(CH2)4CO2H + X H2N(CH2)6NH2
O
O
C (CH2)4 C NH (CH2)6—NH—
nylon 66
Compounds related to amides
CH3NH
O
C
OCH3
carbamate
H 2N
O
C
NH2
urea
X
NH
C
H 2N
NH2
guanidine
CH3
O
C
O
C
NH CH3
imide
Spectra
α-hydrogens in 1H NMR
O
CH3 C N CH3 C OCH3
2.00
2.03
O
CH3 C NH2
2.08
O
O
CH3 C OH CH3 C Cl
2.10
2.67
O
C=O stretch about 1800 cm-1
R C Cl
O
-1
R C OCH3 C=O stretch about 1740 cm ; C—O 1200
O
R C OCOR C=O stretch doublet; C—O 1100
O
R C NH2
O
R C NHR
O
R C NR2
—C≡N
C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH doublet 3500
C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH 3300
C=O stretch about 1700 cm-1; no NH bend; no NH stretch
2200 cm-1
Chem 66H
Hoffmann Rearrangement
Chapter 19-13
Hoffmann Rearrangement
O
R
H2O, HO-
NaOBr
R—NH2
R N C O
NH2
isocyanate
H
R'OH
R
N
OR'
carbamate
O
O
R
O
Br+
R
NH2
.. Br
N
O
HO –
R
H
.. Br
N
..
HOH
R N C O
HO
R N C O–
–
OH
HO –
R NH C O
OH
R NH C O
O–
CO2
R—NH2
or via the nitrene
O
R
O
N
Br
R
N:
..
R N C O
H
CONH2
BrO –
NH2
Chem 66H
Amides
Chapter 19-12
Reactions
Acid Hydrolysis
LiAlH4
CH3
Not reversible since amine forms ammonium salt
O
C
+O H
C
NHCH3
H+
NHCH3
:OH2
OH
NHCH3
C
OH
NHCH3
C
OH
– H+
OH2
+
OH +
NH2CH3
C
– H+
OH
O
C
H+
OH + CH NH
3
2
H+
CH3NH3+
Base Hydrolysis
CH3CH2CH2
O
C
O–
NH2
HO –
CH3CH2CH2
O
C
–
O–H
NH2
CH3CH2CH2 C NH2
O
H
O
C
+ NH3
CH3CH2CH2
O–
Reduction
CH3CH2CH2
O
C
Chem 66H
O–
NHCH3
H3A–—H
CH3CH2CH2 C NHCH3
H
H3A –—H
O—AlH3
CH3CH2CH2 C NHCH3
H
CH3CH2CH2
C NHCH3
+
H
H
CH3CH2CH2 C NHCH3
H
N
O
CH3
N
H
H
Amides
Chapter 19-11
Amides
O
C
NH2
CH3
acetamide
O
C
O
C
NHCH3
CH3
N(CH3)2
CH3
N-methylacetamide N,N-dimethylacetamide
Amides are not as basic as amines due to the overlap of the lone pair on nitrogen
with the carbonyl pi bond.
Amide pKb's: 15 - 16; CH
3NH2: pKb: 3.34.
The result is a partial double bond between the nitrogen and the carbonyl carbon.
The barrier to rotation is about 18 kcal/mol
This is evident from the difference in chemical shifts of the two methyl groups in
dimethyl formamide in the 1H NMR.
O
C
H
O–
C + CH3
N
H
CH3
CH3
N
CH3
different chemical shifts
due to restricted rotation
Preparation
R
R
R
O
C
O
C
O
C
R2NH
Cl
R2NH
OCH3
O
O
C
R
O
C
N
R
R
R2NH
R
Reactions
Acid Hydrolysis
Not reversible since amine forms ammonium salt
Chem 66H
Esters
Chapter 19-10
Lactones
Reaction with Ammonia
CH3CH2
O
C
OCH3
+ H3N
CH3CH2
O
C
NH2
+ HOCH3
COOH
O
C
1. LiAlH4 2. H2O
OCH3
or
Na, CH3CH2OH
H
CH3(CH2)5 C OH + HOCH3
H
reduction always produces one primary alcohol plus the alcohol from the ether
linkage of the ester
Reaction with Grignard reagents
Preparation of tertiary alcohols with at least two identical R groups
CH3CH2
CH3CH2
O
C
O
C
1. 2 CH3CH2CH2MgBr
OCH3
2. H2O, H+
OCH3
CH3CH2CH2– + MgBr
CH3CH2
Lactones are cyclic esters and react like esters. They are formed
from acyclic hydroxy acids or hydroxy esters
O
O
Reduction
CH3(CH2)5
Chem 66H
O
C
CH2CH2CH3
CH3CH2
CH2CH2CH3
CH3CH2 C OH
+ HOCH3
CH2CH2CH3
O–
C OCH3
CH2CH2CH3
CH3CH2
O – MgBr+
C CH2CH2CH3
CH2CH2CH3
H+
H 2O
CH3CH2CH2– + MgBr
O—H
CH3CH2 C CH2CH2CH3
CH2CH2CH3
Addition to the ketone is faster than addition to
the ester and therefore two additions occur
A secondary alcohol results from addition to a
formate HCO2R since hydrogen was already
bonded to the carbonyl carbon.
OH
+
CO2CH3 H+, heat
H , heat
O
OH
O
Esters
Chapter 19-9
Acid hydrolysis: the reverse of esterification
CH3
O
C
+ H
O
C
OCH3
CH3
OCH3
: O—H
CH3
H2O18:
H
O
+
CH3 C OCH3
HO18 H
+O H
C 18
O H
CH3
C OCH3
+ O18H2
+ CH3OH
CH3
O
C
O18H
labelled water results in labelled carboxylic acid; no label in the alcohol
Base Hydrolysis
O
C
HO –
O
C
CH2CH3
O C H
CH3
O—H +
–
O–
CH2CH3
C
O C H
OH
CH3
O
C
CH2CH3
O C H
CH3
O–
+
CH2CH3
HO C H
CH3
The product of the base hydrolysis is the carboxylate salt and the reaction is
irreversible.
If the alcohol is chiral; retention of configuration is observed.
Thus the C—O bond of the ester is broken, not the C—O bond of the alcohol.
Transesterification
one alcohol is used in excess to drive the equilibrium in the desired direction
O
C
H+, heat
OCH3
+ HOCH2CH3
O
C
OCH2CH3
+ HOCH3
Chem 66H
Esters
Chapter 19-8
Chem 66H
Esters
Acid hydrolysis: the reverse of esterification
Preparation
(CH3)2CH
(CH3)2CH
(CH3)2CH
(CH3)2CH
O
C
O
C
O
C
O
C
H+, heat
+ HOCH2CH3
OH
(CH3)2CH
Cl + HOCH2CH3
O
O
C
(CH3)2CH
HOCH2CH3
(CH3)2CH
CH(CH3)2
O – Na+
O
C
BrCH2CH3
(CH3)2CH
O
C
O
C
O
C
OCH2CH3 + H2O
OCH2CH3 + HCl
OCH2CH3 + RCO2H
(CH3)2CH
OCH2CH3
(CH3)2CH
O
C
OCH2CH3 + NaBr
H
OCH2CH3
Nu:
In alkaline solution strong nucleophiles can effect addition-elimination
(CH3)2CH
Nu :
O
C
OCH2CH3
O–
(CH3)2CH C OCH2CH3
Nu
OCH3
O—H
+
CH3 C OCH3
HO 18H
CH3
O
C
H
CH3
OCH3
O—H
C OCH3
+O 18 H2
(CH3)2CH
+
–
O
C
Nu
OEt
+O
CH3
C
H
+ CH3OH
O 18 H
CH3
O
C
O 18 H
labelled water results in labelled carboxylic acid; no label in the alcohol
Base Hydrolysis
Acid can protonate the carbonyl oxygen and make the carbonyl carbon more
susceptible to attack by nucleophiles
+
+
H2O18:
Reactions of Esters
O
C
CH3
O
C
Acid Anhydrides
Chapter 19-7
Anhydrides
Preparation
O
C
CH3CH2
O
C
CH3CH2
+
Cl
OH
Na+ – O
+ (CH
3
O
C
O
C
CH2CH3
CH3CH2
heat
CH3CH2
)2O
O
C
O
C
O
C
O
O
O
C
CH2CH3
CH2CH3
+ CH3CO2H distilled off to
drive the equilibrium
Reactions
Same reactions as acid chlorides but with somewhat slower rates due to the
poorer leaving group ability of RCO2 –
R
O
C
O
O
C
O– O
R C
C
O
R
Nu
R
Nu:
R
O
C
+
Nu
–
Nucleophiles: H2O, ROH, ArOH, NH3, RNH2, R2NH
CH3CH2CH2
O
C
O
O
C
O
C
O
O
C
H 2O
CH2CH2CH3
CH3OH
CH3CH2CH2
O
C
OCH3
Pyridine
O
C
O
O
C
NH3
O
C
O
C
NH2
OH
O
O
C
R
Chem 66H
Acid Halides
Chapter 19-6
Fredel Crafts Reactions
+ CH3CH2COCl
O
C
AlCl3
CH2CH3
Reaction with Grignard Reagents
O
C
CH3CH2
H 2O
Cl
Rδ−Mgδ+X
CH3CH2
O–
Rδ−Mgδ+X
O
CH3CH2 C Cl
C
CH3CH2
R
R
tetrahedral intermediate
O MgX
H+, H2O
C R
R
CH3CH2
O—H
C R
R
Reaction with Lithium Dialkyl Cuprates
4 Li
2 R—X
CH3CH2
(CH3)2CH
O
C
O
C
2 R—Li + 2 LiX
R2CuLi
Cl
Cl
CH3CH2
O
C
CuI
R2CuLi +
LiI
R
O
(CH3CH2)2CuLi
C
(CH3)2CH
CH2CH3
Reduction
O
C
Cl
OC(CH3)3
Li+H Al- OC(CH3)3
OC(CH3)3
or
H2, Pd/BaSO4
O
C
H
LiAl[OC(CH
3)
3]
3 is less reactive than LiAlH4 due to steric hindrance and the
electron withdrawing effects of the oxygens
Chem 66H
Acid Halides
Chapter 19-5
Acid Halides
Chem 66H
Amide formation
Preparation
O
CH3CH2CH2 C OH
O
C OH
SOCl2, heat
O
CH3CH2CH2 C Cl
R
R
O
C
Reactions of Acid Halides
most reactive of the carboxylic acid derivatives since the halide ion is a good leaving
group.
R
O
C
O–
O
elimination
+ Cl R C Cl
C
R
Nu
Nu
overall net substitution
tetrahedral intermediate
of Nu for Cl
Nu:
R
H 2O
Cl
H2O:
O–
O
R C Cl
C +H
R
O
+OH2
H
tetrahedral intermediate
- H+
+ Cl –
R
Ester formation
Cl
CH3CH2CH2OH
N
O
C
OCH2CH2CH3
O
C
OH
+ HCl
Rate decreases with increasing size of R since water solubility decreases
O
C
CH3
O–
R C Cl
+NH3
+
O
– -H
+
+
Cl
C
R
NH3
tetrahedral intermediate
NH2
O
C
Cl
+ HCl
NH3
NH4Cl
+ CH3NH2
Cl
2 (CH3)2NH
CH3
O
C
O
C
NHCH3
+
CH3NH3
N CH3 + (CH3)2NH2Cl
CH3
Use of an added tertiary amine avoids the loss of a second equivalent of
nucleophilic amine
Hydrolysis
O
C
Cl
O
C
addition
Cl
H 2O
H3N:
O
C Cl
PCl3, heat
O
C
+
N+ Cl
H
pyridine reacts with HCl to remove it from the reaction
Carboxylic Acid Derivatives
Chapter 19-4
Chem 66H
Derivatives of Carboxylic Acids
Reacticvity is also related to the resonance donating ablity of the acyl
Any compound which yields a carboxylic acid on hydrolysis (acid or base) with water substituent
CH3
O
O
O
C OCH3 CH3 C NH2 CH3 C Cl
ester
amide
CH3
O
O
C O C CH3 CH3 C N
acid chloride
anhydride
R
CH3
R
leaving groups
X = OR, Cl, NH2, OCOR
X = H, R, Ar
not leaving groups
Aldehydes and ketones undergo nucleophilic addition
Carboxylic acid derivatives undergo nucleophilic substitution due to the presence
of a leaving group on the carbonyl carbon
Nucleophilic acyl substitution
CH3
Nu:
X
–
<
O
C X
O
–
R C O – < OR <
–
NH2
<
–
Cl
R
O
C
O
C
O
O
O
CH3
R
O– O
C +
R
O
R
O–
C + CH3
O
R
O
R
R
O
C
CH3
N:
CH3
O–
+ C
O
R
O–
C + CH3
N
R
CH3
best resonance donor since
nitrogen is least electronegative
and better Lewis base:
O–
O
CH3 C X
CH3 C X
Nu
tetrahedral (sp3) intermediate
O–
C +
Cl
R
nitrile
Reactivity
O
C X
O
C
+ X–
CH3
increasing basicity of acyl substituent (leaving group)
decreasing reactivity (leaving group ability)
Reactivity of carboxylc acid derivatives decreases with increasing
basicity of leaving group
Acid chlorides and anhydrides react readily with water while esters and amides are
fairly stable toward water and require acid or base to effect hydrolysis
Amides have significant C=N
double bond character and
hindered rotation about the
N–Cacrbonyl bond.
Esterification of Carboxylic acids
Chapter 19-3
Chem 66H
Reduction of Carboxylic Acids
Decarboxylation
carboxylic acids can be reduced to primary alcohhols with LiAlH4
b-keto acids lose CO2 (decarboxylate) on heating
CH3CH2CH2
O
C OH
CO2H
H
CH3CH2CH2 C OH
H
2. H2O, H+
1. LiAlH4
CH2OH
1. LiAlH4
2. H2O, H+
O
H
O
O
R
R
O
H
H
H
O
O
R
H O
enol
H
R = CH3, alkyl, OH,
R = CH3, alkyl, OH, OR
Polyfunctional Carboxylic Acids
Dicarboxylic acids are called dibasic or diprotic acids
O
O
CO2H
heat
the acidity of the first COOH to lose a proton is increased by the electron
withdrawing ability of the other COOH, but the acidity of the second is lower
(pKa increased) because of the adjacent negative charge created by the first
COOacid structure
oxalic acid
malonic acid
succinic acid
glutaric acid
adipic acid H
pKa
1
1.2
2.8
4.2
4.3
4.4
HO2C—CO2H
HO2C—CH2—CO2H
HO2C—(CH2)2—CO2H
HO2C—(CH2)3—CO2H
HO2C—(CH2)4—CO2H
2
4.2
5.7
5.6
5.4
5.4
difference between pKa1 and pKa 2 decreses as the length of the chain
increases since induction is directly dependent on distance
Anhydride formation
if a 5 or 6 membered ring can form, dicarboxylic acids form cyclic anhydrides
with loss of water upon heating
CO2H
CO2H
O
heat
or O
O
CH3 C O C CH3
O
O
+ H2O
O
CH3CH2O
O
heat
CO2H
CH3
ketone
CH3CH2O
CH3
Esterification of Carboxylic acids
Chapter 19-2
O
R1
+
C
OH
acid
HO R2
CH3CH2
C
R1
alcohol
CH3CH2
CO2H + HOCH2C6H5
O
O
H+
+ HO CH
OH
3
Rate of esterification:
O–R
C—O bond of the acid is broken, not the C—O bond of the alcohol
that is, the alcohol oxygen is incorporated into the ester not the
oxygen from the acid —OH.
+ H2 O
2
ester
O
C
O
H+
C
+ H2O
C
O—CH3
CO2CH2C6H5
+ H2 O
HO2C
R
C
OH
. .
O H
.. + H
O
R
1
C
R1 C
CH3O
OH
H
+
R1 C
CH3O
+
O—H
H
+O
R
1
C
O
OH O
O
O
25
macrolactone linkage
21
OH
R1
C
O
Me
O
OH
OH OH
.. ..
O
H
OCH3
H+
OH
Me
R1 C OH
CH3O +
H
.. ..
O H
OH
O18CH3
Me
CH3OH
.. ..
O H
C
Many naturally occurring macrolactones are known and many have
important biological activities such as the antibiotic cytovarycin
Me
Mechanism of the esterification reaction:
1
HO18CH3
intramolecular ester are called lactones
formation of five and six membered lactones is very fast: yielding stable
esters
O
steric hindrance controls the rate of the reaction
Estierification proceeds through a series of reversible steps
involving protonation and deprotonation
. .
H
OH
H+
CH3OH > 1° > 2° > 3°
+
O
H+
R3CCO2H < R2CHCO2H < RCH2CO2H < CH3CO2H <
HCO2H
.. ..
O
Chem 66H
Me
OCH3
OHO
H
O
MeO
Me
Me
OH
Me
OH
Carboxylic Acids
Chapter 19-1
Carboxylic acids contain both a carbonyl and a hydroxyl function
.. ..
O
C ..
R
O—H
..
H
+
.. ..
O
C .. –
R
O
.. :
+
.. . –
..O
.
C .. R
R
O
..
O
C
O
O: H O
C R carboxylic acid dimer:
R C
O H :O:
results in higher melting and boiling points
Spectral Properties
—OH stretch in the infrared is intense due to dimers...3300 - 3000
C=O stretch 1700 - 1725 shifted to 1680 - 1700 if conjugated
—COOH in 1H NMR at about 10 - 13 ppm as a broad singlet
Preparation
1. Hydrolysis of carboxylic acid derivatives
2. oxidation of alcohols, aldehydes, or alkenes
3. Grignard reactions
Hydrolysis of Carboxylic acid derivatives
CH3
CH3
O
C
O
C
H2O,
O—CH2CH3
Cl
H+ or HO-
CH3
O
C
O
C
O
CH3
CH3
O
C
+ HOCH2CH3
O—H
CH3C≡N
also yield acetic acid on hydrolysis
Oxidation
alcohols
CH3CH2CH2CH2OH
CH2OH
H2CrO4
H2CrO4
CH3CH2CH2COOH
COOH
Chem 66H