chapter 13-1 chem 66h organometallic reagents: grignard
TRANSCRIPT
Chapter 13-1 Chem 66HOrganometallic Reagents: Grignard Reagents
O
C CH
H
H
H
BrMgO
C CH
H
H
H
CH3
HO
C CH
H
H
HCH3
Grignard Reagents
CH3MgBr
CH3–MgBr+
H+
Grignard reagents are organometallic reagents derived from an alkyl halide and magnesium
Since the carbon carries a partial negative charge, the carbon is a strong base and a good nucleophile.
diethyl etherRδ−—Mgδ+X Grignard reagent R—X + Mg
HOHdiethyl etherCH3CH2
δ−—Mgδ+Br CH3CH2—Br + Mg CH3CH2—H
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.
Reaction with epoxides
R1
Cδ+
Oδ−
R1 R1
C
O –
R1NuR1
C+
O –
R1
R2 Mg-X
R1
C
O– MgX+
R1R2R1
C+
O –
R1 R1
Cδ+
Oδ−
R1
Nu:
NucleophilicAddition
NucleophilicAddition
Chapter 13-2 Chem 66HPreparation of Alcohols from Grignard Reagents
HC
O
H
CH3 Mg-X
H C
OH
H
CH3
A Grignard reaction with
1. formaldehyde produces a primary alcohol
2. H2O, H+
1.
2. an aldehyde produces a secondary alcohol
3. a ketone produces a tertiary alcohol
4. an ester produces a tertiary alcohol (addition of two molecules of Grignard reagent)
5. ethylene oxide produces a primary alcohol
CH3CH2
C
O
H
(CH3)CHMgBrH C
OH
CH2CH3
CH(CH3)22. H2O, H+
1.
CH3CH2
C
O
CH3
(CH3)CHMgBrCH3 C
OH
CH2CH3
CH(CH3)22. H2O, H+
1.
C
O
OCH3
CH3MgBrCH3 C
OH
CH3
2. H2O, H+
1. 2
O
OH
1. C6H5MgBr
2. H2O, H+
BrMgO
C CH
H
H
H
H
HO
C CH
H
H
HH
H+
HCδ+
Oδ−
H
CH3 Mg-Xδ− δ+
BrMgO
C CH(CH3)2H
CH2CH3
HO
C CH(CH3)2H
CH2CH3
H+
HCδ+
Oδ−
CH2CH3
HC Mg-Xδ− δ+
CH3H3C
C
O
OCH3
OMgBr
C OCH3
CH3
C
O
CH3
OMgBr
C CH3
CH3
CH3 C
OH
CH3
H+
CH3MgBr
CH3MgBr
– CH3O–MgBr+
Chapter 13-3 Chem 66HOrganolithium Reagents
CH3CH2
C
O
CH3
CH3CH2CH2CH2LiCH3 C
OH
CH2CH3
CH2CH2CH2CH32. H2O, H+
1.
Organolithium Reagents
Organolithium reagents are organometallic reagents derived from an alkyl halide and lithium metal
Since the carbon carries a partial negative charge, the carbon is a strong base and a good nucleophile.
diethyl etherRδ−—Liδ+ = LiX organolithium reagentR—X + 2 Li
HOHdiethyl etherCH3CH2
δ−—Liδ+CH3CH2—Br + 2 Li 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.
n-Butyl Lithium can also be used to generate aryl and vinyl lthium reagents by lithium-halogen exchange
Br CH3CH2CH2CH2Li Li
+ CH3CH2CH2CH2Br
Br CH3CH2CH2CH2Li Li
OCH3OCH3
+ BuBr
(n-BuLi)
Br 2 CH3CH2CH2CH2Li Li
OH OLi
+ BuBr + Butane
Chapter 13-5 Chem 66HAlkynes: Acetylides
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 with33% in sp2 bonds). The carbon atom is more electronegative in the spstate; thus the C—H bond is more acidic.
The acetylide ion may be formed by such strong bases as —:NH2 (pKa33), RMgX or RLi (pKa 45-50).
No reactionNaNH2
NH3
C CH
H
C C H
C CH
C C
R C C H
CH
R C C:– Na+
CH2R
pKa = 45
+ base—H
+ base—H.
–
–
..
.
+ base
+ base
+ NH3
acetylide ion
NaNH2
NH3
pKa = 25
RMgX
n-BuLi
R C C H
R C C H
R C C:– MgX+
R C C:– Li+
+ RH
+ RH
Chapter 13-6 Chem 66HAlkynes: Acetylides
R' C C:–Li+ R C
O
H R' C C C
O – Li+
H
R
R' C C C
OH
H
RR' C C—H
R' C C:– Li+
O
C C R'O – Li+
C C R'OH
H+
Nucleophilic addition reaction with acetylide ion.
H+
+
SN2 reaction with acetylide ion
R' C C:– MgBr+O
R' C C—CH2—CH2—O – MgBr+
R' C C—CH2—CH2—OH
NH3
R—CH2—C≡C—R'
+
H+
R'—C≡C:– Na+ + R—CH2—X
n-BuLi
CuR
RR1
X
O
O OTs
OTs O
O PhPh
I CO2Me
NHCOC6H5
t-Bu CO2Me
NHCOC6H5
t-Bu
X
t-Bu
n-Bu
Chapter 13-7 Chem 66H
Preparation:Organocopper reagents can be prepared from organo lithium reagents and Grignard reagentsCuprates: Me2CuLi, Bu2CuLi for common readily available organolithium reagents.
R2CuLiR—Li
R—Cu CuI
R—Li + LiXR—X + 2Li
Higher Order Cuprates: somewhat more stable than dialkylcuprates
R2CuCNLi22 R—Li + CuCN
Substitution
Organocopper reagents react with alkyl halides, epoxides, allylic halides, propargylic halides, vinyl halides to give substitution products
+ RCuXR—R1reductive
elimination
oxidative addition
X = I, Br, Cl, OTs
R—R1R1—X + R2CuLi
76%(t-Bu)2CuLi
(4 equ)
Bu2CuLi
X = OTf 100%X = OP(O)(OPh)2 60%
(4 equ)47%
(C6H5)2CuLi
Organocuprates
CO2EtHO
H
LiCuCO2Et
OH
PhCH2OOH
PhCH2OO
OH
OH
Me
PhCH2OOH
Me
OH
O
MeO2COCH2Ph
HO
MeO2COCH2Ph
Me
Epoxides
90%
Me2CuLi
6:1
90%
6
1
Me2CuCNLi2
Chapter 13-8 Chem 66HOrganometallic Reactions: An Overview
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 No. of electrons donated
H–
Cl–, Br–, I–
R–, Ar–
R3P:
:O≡C:
- 1
- 1
- 1
- 1
0
0
0
2
2
2
2
2
2
6
Oxidation state of metal is the difference between the overall charge on the complex and the sum of the charges for each ligand.
ClPd
CH3
Ph3P PPh3
Cl: -1CH3: -1Ph3P: 0
Pd: +2
ClRhPh3P CO
PPh3
H
Cl
Ph3P: 0H: -1CO: 0Cl: -1 (X2)Rh: +3
Chapter 13-9 Chem 66HOrganometallic Reactions: An Overview
Organometallic complexes undergo three basic reactions
Oxidative additionMigratory insertionreductive elimination
LM
L L
L
L = generic ligand
LM
L RBr
L
L
+ R–Br
CN = 4CN = 6
Oxidative additionThe oxidation state and the coordnation number of the metal ion both increase (usually by two)
reductive elimination
The oxidation state and the coordnation number of the metal ionboth decrease (usually by two)
LM
L L
LLM
L RH
L
L
+ R–H
CN = 4CN = 6
Migratoryinsertion
LM
L L
H
L
H2C CH2
LRh
L LL
CH2–CH3
Migratory insertion
solvent
LRh
L LL
CH2–CH3
S
no change in the metal ion oxidation state
Chapter 13-10 Chem 66HHydrogenation with Wilkinison's Catalyst
Wilkinson's Catalyst, (Ph3P)3RhCl functions as a catalyst in the presence of hydrogen to convert alkenes into alkanes, ie. hydrogenation.
Ph3PRh
Ph3P Cl
PPh3
Ph3PRh
Ph3P Cl
H
H
S
Migratoryinsertion
Rh = +1
Ph3PRh
Ph3P Cl
S
Rh = +1
CH3CH2OH H2
ligandexchange
Ph3PRh
Ph3P Cl
H
H– ethanol
Rh = +3 Rh = +3
CH2
CH2
oxidative addition
Ph3PRh
Ph3P Cl
H
H
Ph3PRh
Ph3P ClH
C–C H
ethanol
Ph3PRh
Ph3P ClH
C–C H
S
Ph3PRh
Ph3P Cl
Sreductiveelimination + C–C H
H
Chapter 13-11 Chem 66HPalladium Catalyzed Carbon-Carbon Bond Formation
Suzuki Reaction
Ph3PPd
Ph3P PPh3
PPh3
N NB(OH)2 Br CH3 N N CH370%
Pd(PPh3)4+
R–X + R' B R–R' HO BOR
OR+ NaX
Pd(Ph3P)4
NaOH+
Ph3PPd
Ph3PPPh3
PdPh3P PPh3
– PPh3
oxidative addition
R–X
Ph3PPd
Ph3P R
X
R'–B(OR)2
Ph3PPd
Ph3P R
R'
"transmetallation"step
reductveelimination
R–R' + PdPh3P PPh3
Example
Br+ (RO)2B
Pd(PPh3)4
Chapter 14-1 Chem 66HSpectroscopy
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
where c = speed of lightλcν =
Electromagnetic radiation is transmitted in particle-like packets called photons or quanta. The energy is inversely proportional to thewavelength and directly proportional to frequency.
where c = speed of light; h = Planck's constantλhcΕ =
h = Planck's constanthνΕ =
ultraviolet visible 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.
Chapter 14-2 Chem 66HMass Spectrometry
MassSpectrometry
useful for determining molecular weight, presence of specific atoms and also certain molecular fragments
H2r2 =m
z 2V
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 = mass of cation radcalz = charge (usually +1)H + strength of the magnetic fieldr = radius of the pathV = accelerating potential
placing a detector at some point along the flight path of the ion allows itsmass 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
0
0 10 20 30 50 70 9040 60 80 100
M+
M+1M+2
Chapter 14-3 Chem 66HMass 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.
The molecular ion also fragments into various other fragments by bond breaking processes in the gas phase.
Each of the fragments which reach the detector will produce a peak in the spectrum corresponding to its mass.
A peak in the region of highest m/z in a mass spectrum often correspondsto 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 otherisotopes of specific elements,
For example, the mass spectrum of 2-butanone contains a peak at 72 for the molecular ion 12C4
1H816O
and a peak at 71 for other ions such as 13C12C31H8
16O or 12C4
2H1H716O or 12C4
1H817O
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 massspectrum, 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 themolecule, one of four patterns are most common.
M+
M+1+
M+M+
M+
M+1+M+1+ M+1+
M+2+M+2+ M+2+
M-1+
M-1+
M-2+
no M+2+ present
Chapter 14-4 Chem 66HMass Spectrometry
To determine the molecular weight of a compound from the massspectrum, 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 themolecule, one of four patterns are most common.
M+
M+1+
M+M+
M+
M+1+M+1+ M+1+
M+2+M+2+ M+2+
M-1+
M-1+
M-2+
no M+2+ present
1H2H
12C13C
14N15N
16O17O18O
19F
99.980.01
98.891.11
99.63
0.37
99.760.040.20
100.0
Isotope %
31P
32S33S34S36S
35Cl37Cl
79Br81Br
127I
100.0
95.00
0.764.220.01
75.5324.47
50.5449.46
100.0
Isotope %
Using the known relative abundance of isoptopes of different elements, the molecular formula can be deduced.
For example:
if the molecular ion is 68, there are three reasonable possibilities
C3H4N2
C4H4O
C5H8
formula M+1 M+2
4.07
4.43
5.53
0.06
0.28
0.12
Assumes M+ is 100% otherwise it would be the specifiedpecentage of the M+ intensity.
Chapter 14-5 Chem 66HMass Spectrometry
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 determinedfrom 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
Chapter 14-6 Chem 66HMass Spectrometry
Fragmentation Patterns
α,β-cleavage
O+. O+
.+
cleavage of a bond alpha and beta to a heteroatom such as oxygen iscommon 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
OH
R
HR
OH
R
H R
+
+.+.
McClafferty rearrangment is very common in carbonyl compounds with agamma hydrogen atom.
frequency
%T
100
0
Chapter 14-7 Chem 66HInfrared 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.
O CH3
CH3
O CH3
CH3
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.
Nuclei of bonded atoms undergo vibrations similar to two ballsconnected by a spring. Depending on the particular atoms bonded toeach other (and their masses) the frequency of this vibration will vary.
Infrared energy is absorbed by molecules resulting in an excitedvibrational 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.
bendingstretching
Chapter 14-8 Chem 66HInfrared Spectroscopy
C S C O C H C D
1350 cm-1 1700 cm-1 3000 cm-1 2200 cm-1
Frequency of vibration will be inversly proportional to the masses of the atoms.
C C C C C C
2150 cm-1 1650 cm-1 1200 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
C H
H
H
C H
H
H
C–Hsym = 2872 cm-1 C–Hasym = 2962 cm-1
C
H
HN+
O–
O
N:
H
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
Frequency of vibration will be directly proportional to the strength of the bonds
800100015002000250030003500
C—C str
C—N str
C—O str
OH bend
CH bendNH bend
C=C str
C=N str
C=O str
C≡N strCH str
OH and NH str
Chapter 14-9 Chem 66HInfrared Spectroscopy
Interpretation of Infrared Spectra
Correlation tables
Infrared spectra of thousands of compounds have been tabulated and general trends are known. Some common functional groups areshown below.
sp3 C—Csp2 C=Csp2 C—C (aryl)sp C≡Csp3 C—Hsp2 C—H
sp C—HC(CH3)2
weak, not useful
1600-1700 cm–1
1450-1600 cm–1
2100-2250 cm–1
2800-3000 cm–1
3000-3300 cm–1
3300 cm–1
1360-1385 cm–1 (two peaks)
Alcohols and Amines
O—H or N—H C—O or C—N
C—C and C—H Bonds
Ethers
C—O
3000-3700 cm–1
900-1300 cm–1
1050-1260 strongCarbonyls One of the most useful absorbtions in infrared 1640-1820 cm-1
Ketones (saturated) C=O
Aldehydes C=O;
C—H(O)
Carboxylic acids C=O;
C(O)—OH
Esters C=O ;
C(O)—OR
1640-1820 cm–1
1640-1820 cm–1
2820-2900 and 2700-2780 cm–1 (weak but characteristic)
1640-1820 cm–1
3330-2900 cm–1
1640-1820 cm–1
1100-1300 cm–1
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.
Chapter 15-1 Chem 66HNuclear Magnetic Resonance Spectroscopy
Nuclear Magnetic Resonance (NMR) Spectroscopy
Some atomic nuclei (1H, 13C, others) behave as if they are spinning...theyhave a nuclear spin.
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% withand 50% against the field at any one time.
HoHo
α
β
∆Ehν
∆E
β
α
Resonance: the flip of the magnetic moment from parallel to antiparallel tothe external magnetic field.
Irradiation at the frequency equal to the energy difference,∆E, causesresonance.
∆E depends on the external magnetic field.
Protons (or other nuclei) in different magnetic environments resonate atdifferent field strengths.
A proton which resonates at a higher field is in a stronger magneticenvironment 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.
Ho = the external magnetic field
2.1 2.7 3.0
Pi electron effects
Magnetic fields created by pi electrons are directional and said to have an anisotropic effect.
Chapter 15-2 Chem 66H
ppm
In methyl halides, the more electronegative the halogen, the more deshielded the prIn methyl halides, the more electronegative the halogen, the more deshielded the pr
In methyl halides, the more electronegative the halogen, the more deshielded the protons on the methyl. This is because F is inductively more electronwithdrawing, causing the carbon to be more positive and thus pulling moreelectrons away from the hydrogen and causing it to be less shielded.
H3C—F H3C—Cl H3C—Br H3C—I
C OH
R
H
δ 4.3
Nuclear Magnetic Resonance Spectroscopy
distance from TMS in HzMHz of spectrumδ =
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 local electrons create a small electric and magnetic field around a proton and shield it.
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) (CH3)4Si.
H deshielded
H deshielded
Ho
The pi system of benzene creates a magnetic field or ring current which deshields the protons attached to the ring.
Similarly, pi electrons in a C=O bond create a field which deshields theproton bonded to the C=O of an aldehyde. This is also affected by theinductive effect of the C=O.
H C
H
H
C
H
H
OHH C
H
H
C
H
Cl
C
H
H
H
Equivalent and Nonequivalent Protons
Protons that are in the same magnetic environment are equivalent and havethe same chemical shifts.
Protons in different magnetic fields are nonequivalent and have differentchemical 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 twocompounds. If they are identical, the protons are equivalent.
equivalent, but not to CH3 protons
Chapter 15-3 Chem 66HNuclear Magnetic Resonance Spectroscopy
equivalent Equivalent protons can be on different carbons.
all six are equivalent
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
Chapter 15-4 Chem 66HNuclear Magnetic Resonance Spectroscopy
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...
CH3CH2OCH3
TMS
33
2
If a proton (Ha) is bonded to a carbon which is bonded to a carbon that hasone 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 (adoublet) will appear for Ha.
Ha without an adjacent hydrogen
Ha with Hb adjacent in the α state
Ha with Hb adjacent in theβ state
For one adjacent hydrogenα or β
Chapter 15-5 Chem 66HNuclear Magnetic Resonance Spectroscopy
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 protonsdoublet 1 neighboring protonstriplet 2 neighboring protonsquartet 3 neighboring protonsquintet 4 neighboring protonssextet 5 neighboring protonsseptet 6 neighboring protons
The separation of the peaks in a splitting pattern is called the coupling constant, J.
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 βbβc gives 1:2:1 triplet βbαc
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
Chapter 15-6 Chem 66HNuclear Magnetic Resonance Spectroscopy
Splitting Diagrams
Splitting patterns for protons can be constructed in diagram form by startingwith one line to represent the unsplit proton resonance.
If an adjacent proton Hb affects Ha it is split into a doublet; if anotherequivalent 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 linesoverlap and a only three lines are observed with the center line twice theheight.
This can be repeated for additional adjacent protons.
Ha without an adjacent hydrogen
Ha split by one adjacent hydrogen
Ha split by a second adjacent hydrogen
Ha split by a third adjacent hydrogen1 3 3 1
splitting diagram
1 1
1 2 1
Chemical Exchange and Hydrogen Bonding
CH3OH, methanol would be expected to give an NMR spectrum of adoublet for the CH3 and a quartet for the OH. For a dilute sample at -40° inCCl4 this is the case.
If the NMR spectrum is run at 25° as a more concentrated sample only twosinglets are observed. This is because the intermolecular hydrogenbonding in methanol allows the rapid exchange of the OH proton from oneCH3OH 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 canalso show this effect. Thus the NMR spectra of alcohols, amines andcarboxylic acids are temperature, concentration and solvent dependent.
Chapter 15-7 Chem 66HNuclear Magnetic Resonance Spectroscopy
CHEMICAL SHIFTS
Functional Group Shift,δ
Primary alkyl, RCH3 Secondary alkyl, RCH2R Tertiary alkyl, R3CH
Allylic, R2C=C—CH2R
Benzylic, ArCH2R Iodoalkane, RCH2I Bromoalkane, RCH2Br Chloroalkane, RCH2Cl Ether, RCH2OR Alcohol, RCH2OH Ketone, RCH2C(=O)R
Aldehyde, RCH(O)
Terminal alkene, R2C=CH2 Internal alkene, R2C=CHR
Aromatic, Ar—H
Alkyne, RC≡C—H
Alcoholic hydroxy, ROH Amine, RNH2
0.8-1.01.2-1.41.4-1.7
1.6-1.9
2.2-2.53.1-3.33.4-3.63.6-3.83.3-3.93.3-4.0
2.1-2.6
9.5-9.6
4.6-5.05.2-5.7 6.0-9.5
1.7-3.1
0.5-5.0 (variable)0.5-5.0 (variable)
Chapter 17-1 Chem 66HThe Carbonyl group
Carbonyl Group
sp3 hybridized, trigonal planar carbonyl carbon; partial positive C and partial negative (Lewis basic) oxygen.
C OC O C O
...... ::–+ δ − δ +
pi bond
lone pairs
:
:
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 (C—H) 2850
1H NMR R—CHO 9-10 ppm RCH2COR 2.0 - 2.6 ppm due to inductive deshielding
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.
Chapter 17-2 Chem 66HThe Nucleophilic Addition Reaction
Nucleophilic addition reactions
Addition Reactions of Aldehydes and Ketones
Carbonyl group can be attacked by nucleophiles
C OR
RC O
RR
Nu
sp3
sp2_
:....
Nu:
....
:
or undergo addition of reagents to the pi bond by electrophiles adding first
C OR
RC O
R
R HC O
RR
Nu
H....
H+ +....
. .
Nu:
C
O
R R C
O
R H C
O
H H
increasing reactivity due to steric and electronic effects
ketone aldehyde formaldehyde
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
Chapter 17-3 Chem 66HThe Nucleophilic Addition Reaction
CH3 C
O
CH3CH3 C
O
CH3
CH3 C
O
CH3
H
H
OH H
CH3 C
O
CH3
H
OH
hydrate
H2O:
H+ H2O
- 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
H H
O
H3C H
O
H3C CH3
O
F3C CF3
O
(H3C)3C H
Khydration
41 1.8 X 10 -2 4.1 X 10 -3 2.5 X 10 -522,000
CH3 C
O
CH3CH3 C
O
CH3
H
OH
hydrate
HO –
H2O
Acid catalyzed mechanism
Base catalyzed mechanism
CH3 C
O –
CH3
OH
.... H–O–H
+ HO –
Chapter 17-4 Chem 66HFormation of Acetals
with alcohols: formation of acetals
CH3 C
O
H CH3 C
O
HCH3 C
O
H
CH3 C
O
H
HH
OCH3 H
CH3 C
CH3
H
CH3
O
CH3
C
C
O—CH3
HCH3
O—CH3
C
O—H
HO—CH3
CH3 C
O
H
H
OCH3
H
O
H
H
OCH3
CH3
CH3 C
O—CH3
HO—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
+..
CH3OH
+
+
..
..
CH3OH, H+
a hemiacetal an acetal
H+
+ CH3OH
..
H+ CH3OH
- H2O
+
....
+H+
......
CH3 C
O
H
H
OCH3
H+
mechanism
OHO
OH
p-toluenesulfonic acidtoluene, heat
OO
H2O, HCl or H2SO4, acetone
..
....
..
..
..
..
Chapter 17-5 Chem 66HAcetals and Cyanohydrins
Ph C
O
HPh C
O
HPh C
O
H
Ph C
O
H
CN
H
CN
Ph C
O—H
HCN
.. .. .....
....
+ HCN
- CNa cyanohydrin
....– H—CN
– CN
Formation of cyanohydrins
forms a new C–C bond and introduces a functional group which can be converted to a carboxylic acid or an amine
C OR
RC O–H
RR
NC
..
..HCN
Acetals as protecting groups
OCO2CH3
OCH2OH
CO2CH3OO
CH2OHOO
HOOH
p-toluenesulfonic acidtoluene, heat
LiAlH4
H2O, H2SO4
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
Chapter 17-6 Chem 66HPreparation of Alcohols
a) Hydrogenation
C=O bond can be hydrogentated much like a C=C bond, C=O usually requires harsher conditions
O
CH2=CHCH2CH2
H OH
C H
O
CH3CH2CH2CH2 C H
O
CH3CH2CH2CH2 C H
OH
H
Ni, H2
heat, pressure
Ni, H2, 25°C
Pt, H2
b) Metal Hydrides
LiAlH4, lithium aluminum hydrideNaBH4, sodium borohydride
CH3CH2CH2 C CH3
O
CH3CH2CH2 C CH3
OH
CH3CH2CH2 C CH3
OH
H
H
1. LiAlH4
2. H2O, H+
1. NaBH4
2. H2O, H+
LiAlH4 much more reactive also reduces esters, carboxylic acids, nitriles, amidesNaBH4 sodium borohydride reduces only aldehydes and ketones: more selective
Reduction of Carbonyl Compounds
Chapter 17-7 Chem 66HPreparation of Alcohols
O
CH2CH2CO2Et
H OHCH2CH2CO2Et
O
CH2CH2CO2Et
H OHCH2CH2CH2OH
1. LiAlH4
2. H2O, H+
1. NaBH4
2. H2O, H+
LiAlH4 and NaBH4 do not reduce isolated double bonds
For example
Mechanism for NaBH4:
C O
R
RB –
H
H
H H
C O—B – H3
R
R
H
C O)4B –R
R
H
C OH
R
R
H
+ H3BO3
δ +
δ −
H2O
Mechanism for LiAlH4:
OHH
H
OH
Al–H2(OR)Li+
H
O
H
Li + H
Al – H
HH
H2O
Chapter 17-8 Chem 66HPreparation of Alcohols fromo Grignard Reagents
HC
O
H
CH3 Mg-X
H C
OH
H
CH3
A Grignard reaction with
1. formaldehyde produces a primary alcohol
2. H2O, H+
1.
2. an aldehyde produces a secondary alcohol
3. a ketone produces a tertiary alcohol
4. an ester produces a tertiary alcohol (addition of two molecules of Grignard reagent)
5. ethylene oxide produces a primary alcohol
CH3CH2
C
O
H
(CH3)CHMgBrH C
OH
CH2CH3
CH(CH3)22. H2O, H+
1.
CH3CH2
C
O
CH3
(CH3)CHMgBrCH3 C
OH
CH2CH3
CH(CH3)22. H2O, H+
1.
C
O
OCH3
CH3MgBrCH3 C
OH
CH3
2. H2O, H+
1. 2
O
OH
1. C6H5MgBr
2. H2O, H+
Chapter 17-9 Chem 66HCarbohydrates
Carbohydrates are naturally occurring compounds with C, H, O; often with the emperical formula CH2O.
Monosaccharides...simple sugars which cannot be broken down by hydrolysis; e.g. glucose, fructose, ribose, galactose, deoxyribose, etc.
Disaccharides...dimers of monosaccharides units; e.g.sucrose is made up of glucose and fructose
Oligosaccharides...two to eight monosaccharides units
Polysaccharides...more than eight monosaccharide units; e.g. cellulose is polyglucose
Fischer Projections
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.
Fischer Projections may be rotated 180° but not 90°. A 90° rotation creates the enantiomer.
C HHO
CH2OH
CH3
HO H
CH2OH
CH3
H OH
CHO
HHO
H OH
H OH
CH2OH
H C OH
CHO
C
C
HHO
H
C
OH
H OH
CH2OH
or
Classification of Carbohydrates
The ending-ose indicates a carbohydrateAn aldose contains and aldehyde; a ketose contains a ketoneA 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.
H OH
CHO
CH2OH
D-glyceraldehyde
Now all carbohydrates with a hydroxyl to the right on the last carbon in the Fischer
Projection is designated D. If the OH is projected to the left, the sugar is an L sugar.
H OH
CHO
HHO
H OH
H OH
CH2OH
HO H
CHO
OHH
HO H
HO H
CH2OH
H OH
CHO
OHH
H
CH2OH
OH
D-riboseL-glucoseD-glucose
D and L sugars
In the 19th century (+)-glyceraldehyde was arbitrarily assigned the configuration below and designated D.
Chapter 17-10 Chem 66HAldoses
Aldohexoses
The aldohexoses have 4 asymmetric carbons and thus 24 or 16 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer
H OH
CHO
OHH
H OH
H OH
CH2OH
H OH
CHO
HHO
HO
H
H
H
OH
CHO
HHO
H OH
H OH
CH2OH
OH
CH2OH
H OH
CHO
OHH
HO H
H
HO H
CHO
HHO
H
HO H
CHO
OHH
H OH
H OH
OH
H OH
CH2OH
OH
CH2OH
CH2OH
HO H
CHO
HHO
HO H
H OH
CH2OH
HO H
CHO
OHH
HO H
H OH
CH2OH
D-idoseD-guloseD-taloseD-galactose
D-mannoseD-altrose D-glucoseD-allose
Aldohtetroses
The aldotetroses have 2 asymmetric carbons and thus 22 or 4 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer
Aldopentoses
The aldotetroses have 3 asymmetric carbons and thus 23 or 8 stereoisomers. The D- seres is shown below. Each has a corresponding L-isomer
H OH
CHO
OHH
H OH
H OH
CHO
HHO
H OH
HO H
CHO
HHO
H
HO H
CHO
OHH
H OHOH
D-lyxoseD-arabinose D-xyloseD-ribose
H OH
CHO
CH2OH
OHH
HO H
CHO
CH2OH
OHH
D-erythrose D-threose
CH2OH CH2OH CH2OH CH2OH
Chapter 17-11 Chem 66HCyclic forms of Carbohydrates
The open chain representations of the sugars shown above is for simplicity. Sugars normally exist as cyclic hemiacetals.
R C
O
H R C
OH
H
OR
R C
OR
H
ORacetal
ROH, H+
hemiacetal
ROH, H+
H OH
C
HHO
H OH
H OH
CH2OH
O H
OH
HOOH
H
H
OH
H
OH
CH2OH
HOH
HOOH
H
H
OH
OH
H
CH2OH
H
D-glucose
or
diastereomers
OCH2OH
OHOH
OHHO
OHO
CH2OH
HOOH
OH
Haworth projection
OH at C-1 downCH2OH up at C-5
glucose
six membered ring
α-D-glucopyranose
hydrogen if not labeled
Pyranose Forms
Furanose Forms
H
OH
HHO OH
H HO
HOH OH
CHO
OHH
H OH
D-ribose
CH2OH
β-D-ribofuranose α-D-ribofuranose
OH
H
HHO OH
H HO
HO
an RNA nucleotidea DNA nucleotide
HO
NH
O
ON
O
OHO
HHHH
PO
O-
O-
HO
NH
N
N
O
NH2N
O
H
HHHH
O
PO
O-
O-
Chapter 17-12 Chem 66HAnomers and Mutarotation
Anomers: monosaccharides which differ only in their configuration at C-1
OHO
CH2OH
HOOH
OH
HC
OHHO
CH2OH
HOOH
H
O
OCH2OH
OHOH
OHHO
OCH2OH
OH
OH
OHHO
OHO
CH2OH
HOOH
H
OH
β−D-glucopyranoseα−D-glucopyranose
anomeric carbon
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.
OHO
CH2OH
HOOH
OH
OHO
CH2OH
HOOH
H
H
OCH3
OHO
CH2OH
HOOH
OCH3
H
an acetalmethyl-α-D-glucopyranosidemethyl-β-D-glucopyranoside α−D-glucopyranose
Glycosides are stable under neutral and basic conditions but are interconverted in acid.
CH3OH, H+
CH3OH, H+
Chapter 17-13 Chem 66HOxidation of Monosaccharides
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.
Oxidation
OHO
CH2OH
HOOH
OHC
OHHO
CH2OH
HOOH
H
O
O –C
OHHO
CH2OH
HOOH
O
H
or Br2, H2O
Ag(NH3)2+
HO –
Even ketoses can be oxidized since they are in equilibrium with the aldose through an enediol tautomer
CHOH
CO2H
H
C OH
CHOH
HHO
H OH
H OH
CH2OH
O
CH2OH
HHO
H OH
H OH
CH2OH
CHOH
CHO
HHO
H OH
H OH
CH2OH
HO
H OH
H OH
CH2OH
aldonic acidaldoseketose enediol
or Br2, H2O
Ag(NH3)2+
HO –
The product of oxidation of the aldehyde and the terminal primay alcohol to form a diacid is called an aldaric acid.
CHO
HHO
H OH
H OH
CH2OH
OHH
CO2H
HHO
H OH
H OH
CO2H
OHH
D-glucaric acidD-glucose
dil. HNO3
heat
A uronic acid results when the terminal primary hydroxyl has been oxidized to a carboxylic acid. This generally only occurs enzymatically.
Chapter 17-14 Chem 66HReactions 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
OHO
CH2OH
HOOH
OH
OO
CH2OAc
OO
H
O
H
CH3C
OC
CH3
O O
CCH3
O
CCH3
O CCH3O
C
O
CH3
penta-O-acetyl-α-D-glucopyranose
NaOAc, cold
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.
OHO
CH2OH
HOOH
OH
H
OCH3O
CH2OCH3
CH3OCH3O
CH3O
H
OHO
CH2OH
HOOH
OH
HO
C6H5
HOOH
OH
HO
O
all hydroxyls are converted to methyl ethersAcetal Formation
NaOH,
CH3OSO3CH3
C6H5CHO, H+
Chapter 17-15 Chem 66HDisaccharides
Disaccharides
Maltose
OHO
CH2OH
HOOH
H
OOCH2OH
HOOH
H
OH
α−linkage
enzymesα-glucosidaseα-1,4-glucan
maltohydrolase
beerglucosemaltoseStarch
H2O, H+
or enzymes
2 glucose
α-maltose:glycoside between C-4OH of glucose and glucose anomeric carbon
Cellobiose
principal disaccharide of cellulose
OHO
CH2OH
HOOH
H
OOCH2OH
HOOH
H
OH
β−linkage
H2O, H+
or β-glycosidase
2 glucose
Lactose
only in mammals, 5% in human milk; one unit of glucose, one of galactose
OH
HOCH2OH
HOOH
H
OOCH2OH
HOOH
H
OH
β−linkage
H2O, H+
or enzyme
glucose + galactose
Chapter 17-16 Chem 66HBiological Oxidation of Alcohols
CH3 CH2OH CH3 CH
O
ethanol acetaldehyde
O
N
HO
HOO
PO
PO O
OHN
HON
NN
NH2H2N
O
+
O O –O O –
NAD+
alcohol dehydrogenase
CH3 CH2OH CH3 CH
O
ethanol acetaldehyde
alcohol dehydrogenase+ NAD+ + NADH + H+
N
H2N
O
+R H C
O
H
CH3
H
N
H2N
O
R C
O
H
CH3
H
H+
+ H+
NAD+ NADH
H3C C
H
OH
O
OH
lactic acid pyruvic acid
lactic acid dehydrogenaseH3C
C
O
O
OH+ NAD+
+ NADH + H+
Chapter 17-17 Chem 66HOxidation of Aldehydes and Ketones
Oxidation of Aldehydes
Aldehydes can be oxidized to carboxylic acids by KMnO4 or H2CrO4. Ketones cannot ordinarily be oxidized further.
CH3
CHO
CH3
CO2H
H2CrO4
CH3CH2CH2CH2CO2HCH3CH2CH2CH2CHO + KMnO4, H+, H2O
O CrO
OOHCH3 CH2OH CHCH3
HCH3 CH
O
CH3 CHOH
OH
CH3 C CH3OH
HO Cr
O
OOH C—OH
O+ HCrO3
-
H2O
H2O:
Aldehydes are oxidized to carboxylic acids throught their hydrates
Baeyer-Villiger Oxidation
oxidation of a ketone to an ester (cyclic ketone to a lactone)
OO O
OO
OO
Ar
H
+ ArCO2HArCO3H
most substituted carbon (best able to stabilize a positive charge; i.e. one with highest electron density) will migrate.
OCH3
CH3
O
O
CH3CH3
retention of configurationm-CPBA
Strained systems can be oxidized with H2O2, HO – or t-BuOOH, HO – and do not require RCO3H
O OO
t-BuOOH, NaOH
Chapter 18-6
O OO -
SCH3
O
CH3
Chem 66HSulfur Ylides
O
O - O
CH2—S+(CH3)2
O -
CH2—S+(CH3)2O
O
O -
(CH3)2S+—CH2
O
Unstabilized sulfur ylides reaction with unsaturated ketones to give epoxides:
+
O
Me2S+—-CH2
Explanation:
k2
k-1
k1+ Me2S=CH2
unstabilized sulfur ylides: k-1 < k2in stabilized sulfur ylides k-1 > k2; k'2 > k'-1
k'2
k'1
Ph2S Ph
O
Ph
Ph
O
Ph
O Ph2SO
k'-1
O
k2
k1
k-1
+Me2S+—-CH2
O
Cyclopropylidine sulfur ylides
Trost J. Am. Chem. Soc. 1973, 95, 5298, 5307, 5311, 5317.
+
+
CH3 S+ CH3
CH3
CH3 S+ CH2
CH3
CH3 S CH2
CH3
CH3 S+ CH3
CH3
CH3 S+ CH2
CH3
CH3 S CH2
CH3
OO
CH3 S+ CH2
CH3
O
CH3 S+ CH2
CH3
O– O
CH3 S CH2
CH3
O OCHO
O+B-F3 O+B-F3H
H
OB-F3H
OHH
Chapter 18-5 Chem 66HSulfur Ylides
Sulfur Ylides
Sulfur ylides react with aldehydes and ketones to give epoxides rather than alkenes
I -NaH
DMSO
—
sulfide sulfonium salt sulfur ylide
CH3—ICH3—S—CH3
CH3—S—CH3
I -NaH
DMSO
—
sulfidesulfoxonium salt
slower
OO
CH3—I
oxygen anion in intermediate displaces the sulfide to form an epoxide
+ CH3—S—CH3
+—
Examples:
BF3-OEt2
OCHO
MeO
Me
CO2Et
Me
CHO
Me Me
Me
CO2Me
CHO
Me Me
CO2CHMe2
Me
Ph3P CO2Et
Me
(MeO)2(O)P CO2Me
Me
(MeO)2(O)P CO2CHMe2
Me
P CO2Et
Me
O
Wittig Reaction
CF3CH2O
CF3CH2O
P CO2Et
Chapter 18-4
OCF3CH2O
CF3CH2O
Chem 66H
RMe
CO2Me
Stereocontrol in the Wadsworth-Emmons
R
CO2Me
KO t-Bu, THF
70% 95:5 E:Z
KO t-Bu, THF
95:5 E:Z
P CO2Et
5:95 E:Z
Kishi JACS 1979, 101, 259.
O
KN(SiMe3)2
THF18-crown-6
RCHO
CH3CH2O
CH3CH2O
KN(SiMe3)2
THF18-crown-6
RCHO
MgBr2
THFEt3N
RCHO
RCO2Me
4-50:1 Z:E
Kishi Tetrahedron Lett. 1981, 37, 3873.
>97:3 E:Z
Still Tetrahedron Lett. 1983, 24, 4405.
30-50:1 Z:E
Rathke J. Org. Chem 1986, 50, 2624.
OHCO
OCH3
C C LiR C C ER
(EtO)2P CH2CO2Et
OCH3CH2
(EtO)2P CH2CO2Et
O
(EtO)2P CH—CO2Et
O
O
O–
P(OEt)2
CO2Et
O(EtO)2P O-
CO2Et O
Special Ylides
one carbon homologation:
H2O; H++
E+E+ = H, ClCO2CH3, Cl—SiMe3
2 BuLiRCHO RCH=CBr2Ph3P=CBr2
Corey-Fuchs
Zn°CBr4 + Ph3P
Ph3P=CHOCH3
Wittig Reaction
(EtO)3P: + Br—CH2CO2Et+
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:
Chapter 18-3
base
Br-
_
Chem 66H
phosphonate anion
water soluble
+
Chapter 18-2 Chem 66HWittig Reaction
OR1
R1
R2 CR2
PR3 POR1
R1
R2
R2 R1
R1 R2
R2
RR
R
X CR1
R2
HR2 C
R1
P+H
PhPh
Ph
The Wittig Reaction
+–
Ph3P C HR2
R2
Ph3P CR2
Ph3PR2
CR2
R2
+
carbonyl phosphorous ylide + Ph3P=O
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.
X = I, Br, OTs
+ :PPh3
X–
pKa = 23
dπ−pπt-BuO- K+
BuLiNaHEtO- Na+
CH3SOCH2-
BASE -++
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:
Otrans Ocis
oxaphosphetanes
+PO
R1
R2
RR
RPO
R1
R2
RR
R
C CH
R1 R2
H
C CH
R1H
R2
OR1
H
R2 CH
PR3
+–+ +
The position of the alkene is unambiguous in the product.
H3CC
H3CCH
CH2CH3
H3CC
H3CHC
CH2CH3
H3CC
H3CHC
CH2CH3O
PPh3
O
PPh3 +
+
better since ylide (phosphorane is derived from a primary halide)
ylide derived from a secondary halide: substitution will be more difficult.
Chapter 18-1 Chem 66HAddition Elimination Reactions
Addition-Elimination Reactions
primary amines produce Imines
CHO H2NCH
N an imineH+
if the reaction is too acidic the amine is completely protonated and will not addif the reaction is not acidic, OH2 is not eliminated
H
RC
R
N +
R H
RC
R C
O
R
R
NR
R C
O +—H
R
R
N
C
O –
R
N +RH
H
R H
- H+- H2O
..
..H2N—R
....
Hydrazones and OximesO N—NH2
O N—OH
+ H2N—OHH+
oxime
hydrazoneH+
+ H2N—NH2
oximes, phenyl hydrazones, etc. are often solid and can be used to characterize carbonyl compounds by melting points
H+
mechanism of imine formation
elimination
addition
secondary amines produce enamines
O N– CH2CH3H+
enamineNCH2CH3
H CH2CH3+
CH2CH3
HN
R R
O
NR
O +—HN
O –
N +RR
H
R R
- H+- H2O
..
..HNR2
....
H+
addition
R
H
H
+
mechanism of enamine formation
eliminationenamine
Chapter 19-15 Chem 66HNitriles
Nitriles
CH3CO2H acetic acid CH3CN acetonitrile CH3CH2CH2CH2CO2H pentanoic acid CH3CH2CH2CH2CN pentanonitrile
—C≡N: pKb = 24 NH3 pKb = 4.5
electrons more tightly held in sp orbital, 50% s character N more electronegative
Preparation
SN2 with cyanide ion
Ph—CH2—CNNa+ - CN
Ph—CH2—Br
From benzenediazonium salts
NH2 N2+ X – CNCuCN
KCN
NaNO2
HCl
Dehydration of amides
CH3CH2CH C NH2
O
CH3
CH3CH2CH C N:CH3
SOCl2
Reactions: Hydrolysis
Acid
R C N: R C N+—H R C N—HO+H2
R C N—HO—H
H
R C N—HO—H
H
R C N—H
O—H
HH2O+
R C N—H
O—H
HH—OR C N+H3
O—H
RH—O
C OHO
– H+
H+ transfer+ NH4
+
- H+
- H+ transfer
+
+
Base
R C N:
R C N—H
R C NH—O
R C N—HO—H
RO
HC N—H
O –
R C O –
O
H—OR C O—H
O
H
+ NH3
– NH2
H2O–
HO –
- H+ transfer
Reduction
CH2—CN CH2—CH2—NH21. LiAlH4 2. H2O
or H2, Ni
HO–
H2O:
H2O:
Chem 66HAmides
Polyamides
NN
NN
NN
O
O
O
O
O
H
H H
H
H
proteins
C (CH2)4
OCO
NH (CH2)6—NH—X
x HO2C(CH2)4CO2H + X H2N(CH2)6NH2
nylon 66Compounds related to amides
CH3NH H2N NH2C
OCH3C
H2N CH3 NHC
NH2C
NH O
CH3COO O
imideguanidineureacarbamate
R CO
OCH3
R CO
OCOR
R CO
NH2
R CO
NHR
R CO
NR2
R CO
Cl
C=O stretch about 1700 cm-1; no NH bend; no NH stretch
C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH 3300
C=O stretch about 1700 cm-1; NH bend 1515 - 1670; NH doublet 3500
C=O stretch doublet; C—O 1100
C=O stretch about 1740 cm-1; C—O 1200
C=O stretch about 1800 cm-1
—C≡N 2200 cm-1
CH3 CO
NH2 CH3 CO
OHCH3 C N CH3 CO
OCH3 CH3 CO
Cl2.672.102.082.00 2.03
α-hydrogens in 1H NMR
Spectra
Chapter 19-14
Chapter 19-13 Chem 66HHoffmann Rearrangement
Hoffmann Rearrangement
R NH2
OR N C O
RN OR'
R
O
H
NH2
O
R N
O
Br
H
R N C OR N C O –
OH
R NH C OOH
R—NH2
CO2
HOH
HO –
R'OH
H2O, HO-
carbamate
R—NH2
isocyanate
NaOBr
..
or via the nitrene
R N
O
Br
H
R N C OR N:
O
..
CONH2 NH2BrO –
Br+ HO –
R N
O
Br..
..
R NH C OO –
HO –
Chapter 19-12 Chem 66HAmides
ReactionsAcid Hydrolysis
Not reversible since amine forms ammonium salt
CO
NHCH3
H
C
OHNHCH3
CO
NHCH3
OH2
C
OHNHCH3
OH
C
OHNH2CH3
OHCO
OH + CH3NH2 H+
H+– H+
:OH2
H+
– H+
+
+
+
Base Hydrolysis
CH3CH2CH2CO
NH2
O
CH3CH2CH2CO
O–H CH3CH2CH2CO
O –
CH3CH2CH2 C
O –
NH2
+ NH3
– NH2
HO –
Reduction
CH3CH2CH2CO
NHCH3H
H HCH3CH2CH2 C
HCH3CH2CH2 C
O—AlH3
NHCH3 NHCH3 CH3CH2CH2
CH3CH2CH2 C
O –
NHCH3
C NHCH3
H
+
H3A –—H
CH3NH3+
H3A–—H
H
N
OCH3
NCH3
HH
LiAlH4
Chapter 19-11 Chem 66HAmides
Amides
CH3CO
NHCH3 CH3CO
N(CH3)2CH3CO
NH2
N,N-dimethylacetamideN-methylacetamideacetamide
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; CH3NH2: 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.
CH3
CH3
HCH3
CH3
HCO –
NCO
N+
different chemical shifts due to restricted rotation
Preparation
RCO
O RCO
R
R
RCO
Cl
RCO
N
RCO
OCH3
R2NH
R2NH
R2NH
ReactionsAcid Hydrolysis
Not reversible since amine forms ammonium salt
Chapter 19-10 Chem 66HEsters
Reaction with Ammonia
CH3CH2CO
OCH3 CH3CH2CO
NH2+ HOCH3+ H3N
Reduction
CH3(CH2)5CO
OCH3
CH3(CH2)5 CH
OH
H
+ HOCH3
1. LiAlH4 2. H2O
orNa, CH3CH2OH
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
CH3CH2CO
OCH3
CH3CH2 CCH2CH2CH3
OH
CH2CH2CH3
+ HOCH31. 2 CH3CH2CH2MgBr 2. H2O, H+
CH3CH2CO
OCH3 CH3CH2 CO –
OCH3
CH2CH2CH3
CH3CH2CO
CH2CH2CH3 CH3CH2 CO – MgBr+
CH2CH2CH3
CH2CH2CH3
CH3CH2 CO—H
CH2CH2CH3
CH2CH2CH3
CH3CH2CH2– + MgBr
CH3CH2CH2– + MgBr
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.
Lactones
Lactones are cyclic esters and react like esters. They are formed from acyclic hydroxy acids or hydroxy esters
OH
COOH
O
OCO2CH3
O
OH
OH+, heatH+, heat
H+
H2O
Chapter 19-9 Chem 66HEsters
Acid hydrolysis: the reverse of esterification
CH3CO
OCH3 CH3CO
OCH3
H
CH3 CO—H
OCH3
O18H2
CH3 CO
OCH3
HO18 H CH3CO
O18H
H
CH3CO
O18H+ CH3OH
++
+H2O18:
+
labelled water results in labelled carboxylic acid; no label in the alcohol
Base Hydrolysis
H
:
CO
O—H CCH2CH3
CH3
H– O
CO
O –C
CH2CH3
CH3
HHO
CO
OCO –
OCCH2CH3
CH3
H CCH2CH3
CH3
HOH
++
HO –
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
CO
OCH3CO
OCH2CH3
+ HOCH2CH3
H+, heat+ HOCH3
Chapter 19-8 Chem 66HEsters
(CH3)2CHCO
Cl
(CH3)2CH
(CH3)2CH
CO
OH
CO
O CH(CH3)2CO
(CH3)2CHCO
O – Na+
(CH3)2CHCO
OCH2CH3
(CH3)2CHCO
OCH2CH3
(CH3)2CHCO
OCH2CH3
(CH3)2CHCO
OCH2CH3
+ RCO2H
+ NaBr BrCH2CH3
HOCH2CH3
+ HCl
+ H2O
+ HOCH2CH3
H+, heat+ HOCH2CH3
Esters
Preparation
Reactions of Esters
Acid can protonate the carbonyl oxygen and make the carbonyl carbon more susceptible to attack by nucleophiles
(CH3)2CHCO
OCH2CH3 (CH3)2CHCO
OCH2CH3
H
Nu:
+
In alkaline solution strong nucleophiles can effect addition-elimination
(CH3)2CHCO
OCH2CH3(CH3)2CH C
O –
OCH2CH3
Nu(CH3)2CH
CO
Nu
+ – OEt
Nu :
Acid hydrolysis: the reverse of esterification
CH3CO
OCH3 CH3CO
OCH3
H
CH3 CO—H
OCH3O�18�H2
CH3 CO—H
OCH3HO�18H CH3
CO
O�18�H
H
CH3CO
O�18�H+ CH3OH
++
+
H2O18:
+
labelled water results in labelled carboxylic acid; no label in the alcohol
Base Hydrolysis
Chapter 19-7 Chem 66HAcid Anhydrides
Anhydrides
Preparation
CH2CH3CO
Na+ – OCH3CH2 CH3CH2
CO
Cl CO
O CH2CH3CO
+
CH3CH2CO
OH (CH3CO
)2O CH3CH2CO
O CH2CH3CO
+ CH3CO2H distilled off to drive the equilibrium
heat+
Reactions
Same reactions as acid chlorides but with somewhat slower rates due to the poorer leaving group ability of RCO2
–
RCO
O RCO
R CO –
O RCO
NuR
CO
Nu – O RCO
+
Nu:
Nucleophiles: H2O, ROH, ArOH, NH3, RNH2, R2NH
CH3CH2CH2CO
O CH2CH2CH3CO
CH3CH2CH2CO
OH
CO
OCO
CO
NH2
NH3
H2O
CO
O CO
CO
OCH3
CH3OH
Pyridine
Chapter 19-6 Chem 66HAcid Halides
Reaction with Grignard Reagents
CH3CH2CO
Cl CH3CH2 CR
O –
ClCH3CH2
CO
R
CH3CH2 CR
O MgXR CH3CH2 C
R
O—HRH+, H2O
Rδ−Mgδ+X
Rδ−Mgδ+X
H2O
tetrahedral intermediate
Reaction with Lithium Dialkyl Cuprates
CH3CH2CO
Cl
(CH3)2CH CO
Cl (CH3)2CH
CH3CH2CO
R
CO
CH2CH3
(CH3CH2)2CuLi
R2CuLi
R2CuLi + LiICuI
2 R—Li + 2 LiX4 Li
2 R—X
Reduction
CO
Cl CO
H
AlOC(CH3)3
OC(CH3)3
H OC(CH3)3-Li+
orH2, Pd/BaSO4
LiAl[OC(CH3)3]3 is less reactive than LiAlH4 due to steric hindrance and the electron withdrawing effects of the oxygens
CO
CH2CH3AlCl3+ CH3CH2COCl
Fredel Crafts Reactions
Chapter 19-5 Chem 66HAcid Halides
Acid Halides
Preparation
CCO
OH
CH3CH2CH2 CO
ClCH3CH2CH2 CO
OH
OCl
PCl3, heat
SOCl2, heat
Reactions of Acid Halides
most reactive of the carboxylic acid derivatives since the halide ion is a good leaving group.
R CO
Cl R CNu
O –
ClR C
O
Nu
tetrahedral intermediate
eliminationaddition+ Cl -
Nu: overall net substitution of Nu for Cl
Hydrolysis
R CO
Cl R COH2
O –
ClR C
O
O R CO
OH
+ HCl
- H+
+
tetrahedral intermediate
H2O+
+ Cl –
H2O:
Rate decreases with increasing size of R since water solubility decreases
Ester formation
CO
Cl
N
CO
OCH2CH2CH3
N+
H
+
Cl-
CH3CH2CH2OH
pyridine reacts with HCl to remove it from the reaction
H
H
Amide formation
R CO
ClR C
NH3
O –
Cl R CO
NH3
R CO
NH2NH4Cl
NH3+ HCl
- H+
+
tetrahedral intermediate
H2O + + Cl –
H3N:
CH3CO
Cl CH3CO
NHCH3
CO
ClCO
N CH3
CH3
+ (CH3)2NH2Cl2 (CH3)2NH
+ CH3NH3+ CH3NH2
Use of an added tertiary amine avoids the loss of a second equivalent of nucleophilic amine
Chapter 19-4 Chem 66HCarboxylic Acid Derivatives
Derivatives of Carboxylic Acids
Any compound which yields a carboxylic acid on hydrolysis (acid or base) with water
CH3 CO
OCH3 CH3 CO
NH2 CH3 CO
Cl CH3 CO
O CH3CO
CH3 C N
nitrileanhydrideacid chlorideamideester
Reactivity
CH3 CO
X X = OR, Cl, NH2, OCOR leaving groupsX = 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 CO
X CH3 C
O –
XNu
CH3 CO
X + X –
tetrahedral (sp3) intermediateNu:
R CO
O –
Reactivity of carboxylc acid derivatives decreases with increasing basicity of leaving group
increasing basicity of acyl substituent (leaving group)
decreasing reactivity (leaving group ability)
<– OR < – NH2 < – CH3X – <
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
Reacticvity is also related to the resonance donating ablity of the acyl substituent
R RCO –
Cl+CO
Cl
CH3R
CH3R
CO –
OCO
O+
R RCO –
OCO
O+
CH3
CH3
RCH3
CH3
RCO –
NCO
N:+
R
O
R
O
best resonance donor since nitrogen is least electronegative and better Lewis base:
Amides have significant C=N double bond character and hindered rotation about the N–Cacrbonyl bond.
RCO –
O+
R
O
Chapter 19-3 Chem 66HEsterification of Carboxylic acids
Reduction of Carboxylic Acids
carboxylic acids can be reduced to primary alcohhols with LiAlH4
CH3CH2CH2 CO
OH CH3CH2CH2 CH
OHH
CO2H CH2OH1. LiAlH4
2. H2O, H+
1. LiAlH4
2. H2O, H+
Polyfunctional Carboxylic Acids
Dicarboxylic acids are called dibasic or diprotic acids
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 COO-
acid structure oxalic acidmalonic acid succinic acid glutaric acid adipic acid H
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
HO2C—CO2H HO2C—CH2—CO2H HO2C—(CH2)2—CO2H HO2C—(CH2)3—CO2H HO2C—(CH2)4—CO2H
1.2 4.22.8 5.7 4.2 5.64.3 5.44.4 5.4
pKa 1 2
CO2HCH3 C
CO2H
OO C
OCH3
O
O
O
+ H2Oheat
or
Decarboxylation
b-keto acids lose CO2 (decarboxylate) on heating
O OH
R OH H
O OH
R OH
H
O
CH3R
enolketone
R = CH3, alkyl, OH, ORR = CH3, alkyl, OH,
OCO2H
O
CH3CH2O CH3CH2O CH3
OCO2H
O
heat
heat
Chapter 19-2 Chem 66HEsterification of Carboxylic acids
HO R2
R1 CO–R2
O
HO CH3CH3CH2
CO—CH3
O
CH3CH2C
OH
O
R1
CO2H
COH
O
CO2CH2C6H5
+ H2O
H++ HOCH2C6H5
H+
+ + H2O
esteralcoholacid
+ H2OH+
+
Rate of esterification: CH3OH > 1° > 2° > 3°
R3CCO2H < R2CHCO2H < RCH2CO2H < CH3CO2H < HCO2H
steric hindrance controls the rate of the reactionEstierification proceeds through a series of reversible steps involving protonation and deprotonation
R1 COH
OH
R1 C OH
O
CH3O
H
HR1
R1
COH
O
C OH
O
CH3O
H
R1 C O—H
O
CH3O
H
H R1 C R1
O
OCH3
H
C
O
OCH3
....+
+
...... ..
+
.
.
.
.
CH3OH
+....H+
..
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.
COH
O
CO18CH3
O
HO18CH3
H+
H+
Mechanism of the esterification reaction:
HO2COH H + O
O
intramolecular ester are called lactones formation of five and six membered lactones is very fast: yielding stable esters
Many naturally occurring macrolactones are known and many have important biological activities such as the antibiotic cytovarycin
OO
Me
O
OHO
OH
Me
Me
OH OH
O Me
Me
OH
OHMe OH
O
Me
O
MeOH
MeO
H
2125 macrolactone linkage
Chapter 19-1 Chem 66HCarboxylic Acids
Carboxylic acids contain both a carbonyl and a hydroxyl function
R CO
O : –R CO
O—H R CO
OC
O
OR
–......
.... H + +
....
.... ..
..
.. ..
CO H
RHO:
:O:C
OR carboxylic acid dimer:
results in higher melting and boiling points
Spectral Properties
—OH stretch in the infrared is intense due to dimers...3300 - 3000C=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 alkenes3. Grignard reactions
Hydrolysis of Carboxylic acid derivatives
CH3CO
O—CH2CH3 CH3CO
O—H
CH3CO
Cl CH3CO
O CH3CO
also yield acetic acid on hydrolysis
CH3C≡N
H2O,
H+ or HO-
+ HOCH2CH3
Oxidation
CH2OH COOH
alcohols
H2CrO4
CH3CH2CH2COOHCH3CH2CH2CH2OH H2CrO4