chapter 13 spectroscopy infrared spectroscopy
DESCRIPTION
13.1 Principles of Molecular Spectroscopy: Electromagnetic RadiationTRANSCRIPT
Chapter 13Chapter 13SpectroscopySpectroscopy
Infrared spectroscopyInfrared spectroscopy
Ultraviolet-visible spectroscopyUltraviolet-visible spectroscopy
Nuclear magnetic resonance spectroscopyNuclear magnetic resonance spectroscopy
Mass spectrometryMass spectrometry
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13.113.1Principles of Molecular Principles of Molecular
Spectroscopy:Spectroscopy:Electromagnetic RadiationElectromagnetic Radiation
Cosmic rays
Rays
X-rays
Ultraviolet light
Visible light
Infrared radiation
Microwaves
Radio waves
Figure 13.1: The Electromagnetic Spectrum
Energy
Hi , Short
Lo , Long
Is propagated at the speed of light,
has properties of particles and waves,
and the energy of a photon is proportional to its frequency.
Electromagnetic Radiation
E = h
Figure 13.1: The Electromagnetic Spectrum
400 nm 750 nm
Visible Light
Longer Wavelength ()Shorter Wavelength ()
Higher Frequency () Lower Frequency ()
Higher Energy (E) Lower Energy (E)
Ultraviolet Infrared
The UV, VIS, IR region.
13.213.2Principles of Molecular Spectroscopy: Principles of Molecular Spectroscopy:
Quantized Energy StatesQuantized Energy States
Electromagnetic radiation is absorbed when theenergy of photon corresponds to difference in energy between two states.
E = h
Effect
Electron excitation
Vibration of bonds
Rotation of molecules
Nuclear spin states
Energy
UV-Vis
infrared
microwave
radiofrequency
Energy and its Effect
13.313.3Introduction to Introduction to
11H NMR SpectroscopyH NMR Spectroscopy(Proton NMR)(Proton NMR)
1H and 13C
Both have nuclear spin = ±1/2.1H is 99% at natural abundance.13C is 1.1% at natural abundance.12C is 98.9% at natural abundance )
but is not NMR active.
The Nuclei that are Most Useful toOrganic Chemists are:
Nuclear Spin
A spinning charge, such as the nucleus of 1H or 13C, generates a magnetic field. The magnetic field generated by a nucleus of spin +1/2 is opposite in direction from that generated by a nucleus of spin –1/2.
+ +
++
+
+
+
The distribution of nuclear spins is random in the absence of an external magnetic field B0.
+
+
+
+
+
B0 = field strength
An external magnetic field (B0) causes nuclear magnetic moments to align parallel or antiparallel to applied field.
+
+
+
+
+
B0
There is a slight excess of nuclear magnetic moments aligned parallel to the applied field.
There is no difference in energy in the two spin states absence of a magnetic field. ΔE (the energy required to flip states) is proportional to strength of external magnetic field.
Energy Differences Between Nuclear Spin States
+
+
E E '
increasing field strength
Energy Required to Flip Spin States
H1 C13
Radiofreq. Ho Ho
60MHz 14091.6 56025.0 90MHz 21137.4 84037.4100MHz 23486.0 93374.9200MHz 46971.9 186749.9300MHz 70457.9 280124.8600MHz 140915.8 560249.7
Ho = B0 (Terms for the applied field, in gauss below).
Some Important Relationships in NMR
The frequency of absorbedelectromagnetic radiationis proportional to
the energy difference betweentwo nuclear spin stateswhich is proportional to
the applied magnetic field.
Units
Hz
kJ/mol(kcal/mol)
tesla (T)
Some Important Relationships in NMR
The frequency of absorbed electromagneticradiation is different for different elements, and for different isotopes of the same element.
For a field strength of 4.7 T:1H absorbs radiation having a frequencyof 200 MHz (200 x 106 s-1)13C absorbs radiation having a frequencyof 50.4 MHz (50.4 x 106 s-1)
Some Important Relationships in NMR
The frequency of the electromagnetic radiation absorbed to flip spin states for a particular nucleus (such as 1H) depends on its molecular environment.
So, hydrogens in different environments in the same compound will absorb energy of different frequencies.
This makes NMR a very useful tool for structure determination.
13.4Nuclear Shielding
and1H Chemical Shifts
What do we mean by "shielding"?What do we mean by "shielding"?What do we mean by "chemical shift"?What do we mean by "chemical shift"?
Shielding
An external magnetic field affects the motion of the electrons in a molecule, inducing a magnetic field within the molecule.
The direction of the induced magnetic field, Bi, is opposite to that of the applied field.
C H
B0
Bi
Shielding
The induced field shields the nuclei (in this case, C and H) from the applied field.
So, a stronger external field is needed in order for the energy difference between spin states to match energy of radiofrequency (rf) radiation.
C H
B0
Bi
Chemical Shift
Chemical shift is a measure of the degree to which a nucleus in a molecule is shielded.
Protons in different environments are shielded to greater or lesser degrees which results in different chemical shifts.
C H
B0
Bi
Chemical Shift
Chemical shifts () are measured relative to the protons in tetramethylsilane (TMS) as a standard.
TMS has highly shielded hydrogens and the peak for these Hs in a scan is set at 0 .
Si CH3
CH3
CH3
H3C
=position of signal - position of TMS peak
spectrometer frequencyx 106
01.02.03.04.05.06.07.08.09.010.0
Chemical shift (, ppm)measured relative to TMS
UpfieldIncreased shielding
DownfieldDecreased shielding
(CH3)4Si (TMS)
Chemical Shift
Example: The signal for the proton in chloroform (HCCl3) appears 1456 Hz downfield from TMS at a spectrometer frequency of 200 MHz.
=position of signal - position of TMS peak
spectrometer frequencyx 106
=1456 Hz - 0 Hz
200 x 106 Hxx 106
= 7.28
Chemical shift (, ppm)
7.28 ppm
H C
Cl
Cl
Cl
01.02.03.04.05.06.07.08.09.010.0
The more electron electron density that is withdrawn from a hydrogen results in a smaller field induced from the remaining electrons.
13.5Effects of Molecular Structure
on1H Chemical Shifts
Protons in different environments experience different degrees of shielding and have
different chemical shifts.
Electronegative Substituents Decreasethe Shielding of Methyl Groups
least shielded H most shielded H CH3F CH3OCH3 (CH3)3N CH3CH3 (CH3)4Si
4.3 3.2 2.2 0.9 0.0
A more electronegative atom will remove more electron density from hydrdogen.
Electronegative Substituents Decrease Shielding
H3C—CH2—CH3
O2N—CH2—CH2—CH3
0.9 0.9 1.3
1.0 4.3 2.0
CHCl3 7.3
CH2Cl2 5.3
CH3Cl 3.1
Methyl, Methylene, and Methine
CH3 more shielded than CH2 ; CH2 more shielded than CH
H3C C
CH3
CH3
H
0.9
1.6 0.8
H3C C
CH3
CH3
CH2
0.9
CH3
1.2
Protons Attached to sp2 Hybridized Carbonare Less Shielded than those Attached
to sp3 Hybridized Carbon H H
HH
H
H
C C
HH
H H
CH3CH3
7.3 5.3 0.9
But Protons Attached to sp Hybridized Carbonare More Shielded than those Attached
to sp2 Hybridized Carbon
2.4CH2OCH3C CHC C
HH
H H
5.3
Protons Attached to Benzylic and AllylicCarbons are Somewhat Less Shielded than Usual
1.5 0.8
H3C CH3
1.2
H3C CH2
2.6
H3C—CH2—CH3
0.9 0.9 1.3
Allylic
Benzylic
Alkyl
Proton Attached to C=O of Aldehydeis Most Deshielded C—H
2.4
9.7
1.1
C C
O
H
H
CH3
H3C
Table 13.1
Type of proton Chemical shift (),ppm
Type of proton Chemical shift (),ppm
CH R 0.9-1.8
1.5-2.6CH CC
2.0-2.5CH C
O
2.1-2.3CH NC
CH Ar 2.3-2.8
Table 13.1
Type of proton Chemical shift (),ppm
Type of proton Chemical shift (),ppm
CH Br 2.7-4.1
9-10C
O
H
2.2-2.9CH NR
3.1-4.1CH Cl
6.5-8.5H Ar
C C
H
4.5-6.5
3.3-3.7CH O
Table 13.1
Type of proton Chemical shift (),ppm
1-3H NR
0.5-5H OR
6-8H OAr
10-13C
O
HO
13.6Interpreting 1H NMR Spectra
1. Number of signals.
2. Position of signals.
3. Signal intensity (measured by area under peak).
4. Splitting pattern (multiplicity).
Information Contained in an NMRSpectrum Includes:
Number of Signals
Protons that have different chemical shifts are chemically nonequivalent.
They exist in different molecular environments.
Protons that have the same chemical shift are chemically equivalent.
Chemical shift (, ppm)
CCH2OCH3N
OCH3
NCCH2O
Figure 13.12
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
chemically nonequivalent
Are in identical environments
Have same chemical shift
Replacement test: replacement by some arbitrary "test group" generates same compound
H3CCH2CH3
chemically equivalent
Chemically Equivalent Protons
H3CCH2CH3
chemically equivalent
CH3CH2CH2ClClCH2CH2CH3
Chemically Equivalent Protons
Replacing protons at C-1 and C-3 gives same compound (1-chloropropane).C-1 and C-3 protons are chemically equivalent and have the same chemical shift.
Diastereotopic protons are those whoswe replacement by some arbitrary test group generates diastereomers.
Diastereotopic protons can have differentchemical shifts.
Diastereotopic Protons
C C
Br
H3C
H
H
5.3 ppm
5.5 ppm
Enantiotopic protons are those whose replacement by some arbitrary test group generates enantiomers.
They are in mirror-image environments.
Enantiotopic protons have the samechemical shift.
Enantiotopic Protons
C CH2OH
H3C
HH
EnantiotopicProtons
C CH2OH
H3C
ClH
C CH2OH
H3C
HCl
R S
13.7Spin-Spin Splitting in 1H NMR
Spectroscopy
Not all peaks are singlets.Not all peaks are singlets.
Signals can be split by coupling of Signals can be split by coupling of nuclear spins.nuclear spins.
Chemical shift (, ppm)
Cl2CHCH3Figure 13.13
4 lines;quartet
2 lines;doublet
CH3CH
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
Two-bond and Three-bond Coupling
C C
H
H
C C HH
protons separated bytwo bonds
(geminal relationship)
protons separated bythree bonds
(vicinal relationship)
In order to observe splitting, protons cannot have same chemical shift.
Coupling constant (2J or 3J) is independent of field strength.
Two-bond and Three-bond Coupling
C C
H
H
C C HH
01.02.03.04.05.06.07.08.09.010.0
Chemical shift (, ppm)
Cl2CHCH3Figure 13.13
4 lines;quartet
2 lines;doublet
CH3CH
coupled protons are vicinal (three-bond coupling)CH splits CH3 into a doublet
CH3 splits CH into a quartet
Why Do the Methyl Protons of1,1-Dichloroethane Appear as a Doublet?
C C HH
Cl
Cl
H
Hsignal for methyl protons is split into a doublet
To explain the splitting of the protons at C-2, we first focus on the two possible spin orientations of the proton at C-1.
Why Do the Methyl Protons of1,1-Dichloroethane Appear as a Doublet?
A result of spin-spin spliting.
There are two orientations of the nuclear spin forthe proton at C-1. One orientation shields theprotons at C-2; the other deshields the C-2 protons.The protons at C-2 "feel" the effect of both theapplied magnetic field and the local field resultingfrom the spin of the C-1 proton.
C C HH
Cl
Cl
H
H
Why Do the Methyl Protons of1,1-Dichloroethane Appear as a Doublet?
"true" chemicalshift of methylprotons (no coupling)
This line correspondsto molecules in which
the nuclear spin of the proton at C-1 reinforcesthe applied field.
This line correspondsto molecules in which
the nuclear spin of the proton at C-1 opposesthe applied field.
C C HH
Cl
Cl
H
H
Why Does the Methine Proton of1,1-Dichloroethane Appear as a Quartet?
signal for methine proton is split into a quartet
The proton at C-1 "feels" the effect of the applied magnetic field and the local fields resulting from the spin states of the three methyl protons. The possible combinations are shown on the next slide.
C C HH
Cl
Cl
H
H
There are eight combinations of nuclear spins for the three methyl protons.These 8 combinations split the signal into a 1:3:3:1 quartet.
Why Does the Methine Proton of1,1-Dichloroethane Appear as a Quartet?
C C HH
Cl
Cl
H
H
For simple cases, the multiplicity of a signalfor a particular proton is equal to the number of equivalent vicinal protons + 1.
Called the N + 1 Rule.
The Splitting Rule for 1H NMR
13.8Splitting Patterns:The Ethyl Group
CHCH33CHCH22X is characterized by a triplet-quartet X is characterized by a triplet-quartet pattern (quartet at lower field than the triplet).pattern (quartet at lower field than the triplet).
Chemical shift (, ppm)
BrCH2CH3Figure 13.16
4 lines;quartet
3 lines;tripletCH3
CH2
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
Splitting Patterns of Common Multiplets
Number of equivalent Appearance Intensities of linesprotons to which H of multiplet in multipletis coupled
1 Doublet 1:12 Triplet 1:2:13 Quartet 1:3:3:14 Pentet 1:4:6:4:15 Sextet 1:5:10:10:5:16 Septet 1:6:15:20:15:6:1
Table 13.2
13.9Splitting Patterns:
The Isopropyl Group
(CH(CH33))22CHX is characterized by a doublet-septet CHX is characterized by a doublet-septet pattern (septet at lower field than the doublet).pattern (septet at lower field than the doublet).
Chemical shift (, ppm)
ClCH(CH3)2Figure 13.18
7 lines;septet
2 lines;doublet
CH3
CH
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
13.10Splitting Patterns:Pairs of Doublets
Splitting patterns are not always symmetrical, Splitting patterns are not always symmetrical, but lean in one direction or the other.but lean in one direction or the other.
Pairs of Doublets
Consider coupling between two vicinal protons.
If the protons have different chemical shifts, each will split the signal of the other into a doublet.
C CH H
Pairs of Doublets
Let be the difference in chemical shift in Hz between the two hydrogens.
Let J be the coupling constant between them in Hz.
C CH H
AX
When is much larger than J the signal for each proton is a doublet, the doublet is symmetrical, and the spin system is called AX.
J J
C CH H
AM
As /J decreases, the signal for each proton remains a doublet, but becomes skewed. The outer lines decrease while the inner lines increase, causing the doublets to "lean" toward each other.
J J
C CH H
AB
When and J are similar, the spin system is called AB. Skewing is quite pronounced. It is easy to mistake an AB system of two doublets for a quartet.
J J
C CH H
A2
When = 0, the two protons have the same chemical shift and don't split each other. A single line is observed. The two doublets have collapsed to a singlet.
C CH H
Chemical shift (, ppm)
Figure 13.20
OCH3
skewed doublets
H H
HH
Cl OCH3
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
13.11Complex Splitting Patterns:
Multiplets of multipletsMultiplets of multiplets
m-Nitrostyrene
Consider the proton shown in red.
It is unequally coupled to the protons shown in blue and white.
Jcis = 12 Hz; Jtrans = 16 Hz
H
HO2N
H
m-Nitrostyrene
16 Hz16 Hz
12 Hz 12 Hz
The signal for the proton shown in red appears as a doublet of doublets.
H
HO2N
H
Figure 13.21 H
HO2N
H
doublet of doublets
doublet doublet
13.121H NMR Spectra of Aldohols
What about H bonded to O ?What about H bonded to O ?
O—H
The chemical shift for O—H is variable ( 0.5-5 ppm) and depends on temperature and concentration.
Splitting of the O—H proton is sometimes observed, but often is not. It usually appears as a broad peak.
Adding D2O converts O—H to O—D. The O—H peak disappears.
C OH H
13.13NMR and Conformations
NMR is “Slow”
Most conformational changes occur faster than NMR can detect them.
An NMR spectrum shows the weighted average of the conformations.
For example: Cyclohexane gives a single peak for its H atoms in NMR. Half of the time a single proton is axial and half of the time it is equatorial. The observed chemical shift is halfway between the axial chemical shift and the equatorial chemical shift.
13.1413C NMR Spectroscopy
1H and 13C NMR Compared:
Both give us information about the number of chemically nonequivalent nuclei (nonequivalent hydrogens or nonequivalent carbons).
Both give us information about the environment of the nuclei (hybridization state, attached atoms, etc.).
It is convenient to use FT-NMR techniques for 1H; it is standard practice for 13C NMR.
1H and 13C NMR Compared:
13C requires FT-NMR because the signal for a carbon atom is 10-4 times weaker than the signal for a hydrogen atom.
A signal for a 13C nucleus is only about 1% as intense as that for 1H because of the magnetic properties of the nuclei, and
at the "natural abundance" level only 1.1% of all the C atoms in a sample are 13C (most are 12C).
1H and 13C NMR Compared:
13C signals are spread over a much wider range than 1H signals, making it easier to identify and count individual nuclei.
Figure 13.26 (a) shows the 1H NMR spectrum of 1-chloropentane; Figure 13.26 (b) shows the 13C spectrum. It is much easier to identify the compound as 1-chloropentane by its 13C spectrum than by its 1H spectrum.
Chemical shift (, ppm)
ClCH2
Figure 13.26(a)
CH3ClCH2CH2CH2CH2CH3
1H
01.02.03.04.05.06.07.08.09.010.0 01.02.03.04.05.06.07.08.09.010.0
Chemical shift (, ppm)
Figure 13.26(b)
ClCH2CH2CH2CH2CH3
020406080100120140160180200
13C
CDCl3
A separate, distinct peak appears for each of the 5 carbons.
13.1513C Chemical Shifts
Are measured in ppm (Are measured in ppm ())from the carbons of TMSfrom the carbons of TMS
13C Chemical Shifts are Most Affected By:
• Electronegativity of groups attached to carbon • Hybridization state of carbon
Electronegativity has an even greater effect on 13C chemical shifts than it does on 1H chemical shifts.
Types of Carbons
(CH3)3CH
CH4
CH3CH3
CH3CH2CH3
(CH3)4C
primary
secondary
tertiary
quaternary
Classification Chemical shift, 1H 13C
0.2
0.9
1.3
1.7
-2
8
16
25
28
Replacing H with C (more electronegative) deshieldsC to which it is attached.
Electronegativity Effects on CH3
CH3F
CH4
CH3NH2
CH3OH
Chemical shift,
1H
0.2
2.5
3.4
4.3
13C
-2
27
50
75
Electronegativity Effects and Chain Length
Chemicalshift,
Cl CH2 CH2 CH2 CH2 CH3
45 33 29 22 14
Deshielding effect of Cl decreases as number of bonds between Cl and C increases.
Hybridization Effects
sp3 hybridized carbon is more shielded than sp2.
114
138
36
36 126-142sp hybridized carbon is more shielded than sp2, but less shielded than sp3.
CH3H C C CH2 CH2
68 84 22 20 13
Carbonyl Carbons are Especially Deshielded O
CH2 C O CH2 CH3
127-13441 1461171
Table 13.3
Type of carbon Chemical shift (),ppm
Type of carbon Chemical shift (),ppm
RCH3 0-35
CR2R2C
65-90CRRC
R2CH2 15-40
R3CH 25-50
R4C 30-40
100-150 110-175
Table 13.3
Type of carbon Chemical shift (),ppm
Type of carbon Chemical shift (),ppm
RCH2Br 20-40
RCH2Cl 25-50
35-50RCH2NH2
50-65RCH2OH
RCH2OR 50-65
RCOR
O
160-185
RCR
O
190-220
RC N 110-125
13.1613C NMR and Peak Intensities
Pulse-FT NMR distorts intensities of signals. Pulse-FT NMR distorts intensities of signals. Therefore, peak heights and areas can be Therefore, peak heights and areas can be deceptive.deceptive.
CH3
OH
Figure 13.27
Chemical shift (, ppm)
7 carbons give 7 signals, but intensities are not equal.
020406080100120140160180200 020406080100120140160180200
13.1713C-H Coupling
13C—13C splitting is not seen because theprobability of two 13C nuclei being in the samemolecule is very small.13C—1H splitting is not seen because spectrumis measured under conditions that suppress this splitting (broadband decoupling).
Peaks in a 13C NMR Spectrum are TypicallySinglets
13.18Using DEPT to Count the Hydrogens
Attached to 13C
DDistortionless istortionless EEnhancement nhancement of of PPolarization olarization TTransferransfer
1. Equilibration of the nuclei between the lower and higher spin states under the influence ofa magnetic field
2. Application of a radiofrequency pulse to givean excess of nuclei in the higher spin state
3. Acquisition of free-induction decay dataduring the time interval in which the equilibriumdistribution of nuclear spins is restored
4. Mathematical manipulation (Fourier transform) of the data to plot a spectrum
Measuring a 13C NMR Spectrum Involves
In DEPT, a second transmitter irradiates 1H during the sequence, which affects the appearanceof the 13C spectrum.
Some 13C signals stay the same.Some 13C signals disappear.Some 13C signals are inverted.
Measuring a 13C NMR Spectrum Involves
Chemical shift (, ppm)
Figure 13.29 (a)
O
C
C
CH
CH CH
CH2
CH2
CH2
CH3
CCH2CH2CH2CH3
O
020406080100120140160180200
DEPT 45
Chemical shift (, ppm)
Figure 13.29 (b)
CH
CH CH
CH2 CH2
CH2
CH3
CCH2CH2CH2CH3
O
CH and CH3 unaffected
C and C=O nulledCH2 inverted
020406080100120140160180200 020406080100120140160180200
DEPT 135
13.20Introduction to Infrared Spectroscopy
Gives information about the functional groups in a molecule
The region of infrared that is most useful lies between 2.5-16 m (4000-625 cm-1).
IR absorption depends on transitions between vibrational energy states (bond stretching and bending).
Bond stretching vibrations require more energy than bond bending vibrations.
Infrared Spectroscopy
Stretching Vibrations of a CH2 Group
Symmetric Antisymmetric
These vibrations are analogous to the stretching motion of two springs.
In-plane Bending Vibrations of a CH2 Group
Antisymmetric,“rocking”
Symmetric,“sissoring”
These vibrations are analogous to the in-plane bending motion of two springs.
Out-of-plane Bending Vibrations of a CH2 Group
Antisymmetric,“twisting”
Symmetric,“wagging”
These vibrations are analogous to the out-of-plane bending motion of two springs.
13.21Infrared Specta
Characteristic functional groups usually found between 4000-1600 cm-1. The fingerprint region is from 1300-625 cm-1.
Infrared Spectroscopy
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.35(a): Infrared Spectrum of Hexane
Figure 13.35(b): Infrared Spectrum of 1-Hexene
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.35(c): Infrared Spectrum of Benzene
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.35(d): Infrared Spectrum of Hexylbenzene
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
13.22Characteristic Absorption Frequencies
Stretching vibrations (single bonds)
Structural unit Frequency, cm-1
sp C—H 3310-3320
sp2 C—H 3000-3100
sp3 C—H 2850-2950
sp2 C—O 1200
sp3 C—O 1025-1200
Table 13.4 Infrared Absorption Frequencies
Figure 13.36(f): Infrared Spectrum of Dihexyl Ether
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Stretching vibrations (multiple bonds)
Structural unit Frequency, cm-1
Table 13.4Infrared Absorption Frequencies
C C 1620-1680
—C N
—C C— 2100-2200
2240-2280
Figure 13.36(b): Infrared Spectrum of Hexanenitrile
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Stretching vibrations (carbonyl groups)
Structural unit Frequency, cm-1
Aldehydes and ketones 1710-1750
Carboxylic acids 1700-1725
Acid anhydrides 1800-1850 and 1740-1790
Esters 1730-1750
Amides 1680-1700
Table 13.4 Infrared Absorption Frequencies
C O
Figure 13.36(c): Infrared Spectrum of Hexanoic Acid
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.36(d): Infrared Spectrum of 2-Hexanone
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.36(e): Infrared Spectrum of Methyl Hexanoate
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Bending vibrations of alkenes
Structural unit Frequency, cm-1
Table 13.4 Infrared Absorption Frequencies
CH2RCH
CH2R2C
CHR'cis-RCH
CHR'trans-RCH
CHR'R2C
910-990
890
665-730
960-980
790-840
Bending vibrations of derivatives of benzene
Structural unit Frequency, cm-1
Monosubstituted 730-770 and 690-710
Ortho-disubstituted 735-770
Meta-disubstituted 750-810 and 680-730
Para-disubstituted 790-840
Table 13.4 Infrared Absorption Frequencies
Stretching vibrations (single bonds)
Structural unit Frequency, cm-1
O—H (alcohols) 3200-3600
O—H (carboxylic acids) 3000-3100
N—H 3350-3500
Table 13.4 Infrared Absorption Frequencies
Figure 13.36(a): Infrared Spectrum of 1-Hexanol
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.36(g): Infrared Spectrum of Hexylamine
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
Figure 13.36(h): Infrared Spectrum of Hexanamide
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved.
13.23Ultraviolet-Visible (UV-VIS) Spectroscopy
Gives information about conjugated electron systems.
Gaps between electronic energy levels are greater than thosebetween vibrational levels.
Gap corresponds to wavelengthsbetween 200 and 800 nm.
Transitions Between Electronic Energy States
E = h
X-axis is wavelength in nm (high energy at left, low energy at right).
max is the wavelength of maximum absorption and is related to electronic makeup of molecule, especially electron system.
Y axis is a measure of absorption of electromagnetic radiation expressed as molar absorptivity () or it may be absorbance.
Beer’s Law: A = cl
Conventions in UV-VIS
200 220 240 260 280
1000
2000
Wavelength, nm
max 230 nmmax 2630
Molarabsorptivity ()
UV Spectrum of cis,trans-1,3-Cyclooctadiene
-Electron configuration of the ground State.
Most stable
-Electron configuration of the excited state.
* Transition in cis,trans-1,3-Cyclooctadiene
HOMO
LUMO
E = h
* Transition in Alkenes
HOMO-LUMO energy gap is affected by substituents on double bond.
As HOMO-LUMO energy difference decreases (smaller E), max shifts to longer wavelengths.
Table 13.5
Methyl groups on double bond cause max to shift to longer wavelengths.
C C
H
H
H
H CH3
max 170 nm
CH3
max 188 nm
C C
H
H
Table 13.5
Extending conjugation has a larger effect on max; shift is again to longer wavelengths.
max 170 nm
max 217 nm
C C
H
H
H
H
C C
H
H
H
C C
H
H
H
Table 13.5
max 217 nm(conjugated diene)
max 263 nmconjugated triene plus
two methyl groups
C C
H
H
H
C C
H
H
H
C C
H
H
C C
H
H
C C
H
H
CH3
CH3
Lycopene
max 505 nm
Orange-red pigment in tomatoes(Lycopene)
13.24Mass Spectrometry
Atom or molecule is hit by high-energy electron.
Principles of Electron-Impact Mass Spectrometry
e–
Electron is deflected but transfers much of its energy to the molecule.
e–
Principles of Electron-Impact Mass Spectrometry
This energy-rich species ejects an electron,
Principles of Electron-Impact Mass Spectrometry
forming a positively charged, odd-electron species called the molecular ion.The molecular ion represents the molecular weight of the compound.
e–++••
The molecular ion passes between poles of a magnet and is deflected by magnetic field.
Amount of deflection depends on mass-to-charge ratio.
Highest m/z isdeflected least.
Lowest m/z is deflected most.
Principles of Electron-Impact Mass Spectrometry
++••
Principles of Electron-Impact Mass Spectrometry
If the only ion that is present is the molecular ion, mass spectrometry provides a way to measure the molecular weight of a compound and is often used for this purpose.
However, the molecular ion commonly fragments to a mixture of species of lower m/z.
The molecular ion dissociates to a cationand a radical.
Fragmentation
++ •
Usually several fragmentation pathways are available and a mixture of ions is produced.
cation radical
++••
Only the cation fragments are deflected by the magnet.
A mixture of ions of different mass gives a separate peak for each m/z.
Intensity of a peak is proportional to thepercentage of each ion of different mass in mixture.
Separation of peaks depends on relative mass.
+
++
+
+
+
Fragmentation
After passing through the magnet each ion of different mass will be detected.
Also all ions of the same mass will be detected and will reflect the intensity of that mass peak.
+ + + +
+ +
Fragmentation
20 40 60 80 100 120 m/z
m/z = 78
100
80
60
40
200
Relative intensity
Some Molecules Undergo Very Little Fragmentation
Benzene is an example. The major peak corresponds to the molecular ion.
H
H H
HH
H
H
H H
HH
H
H
H H
HH
H
all H are 1H and all C are 12C
one C is 13C one H is 2H
Isotopic Clusters
78 79 79
93.4% 6.5% 0.1%
Isotopic Abundances
M+ % M + 1 % M + 2 %1H 100
12C 98.9 13C 1.114N 99.6 15N 0.416O 99.8 18O 0.232S 95 33S 0.8 34S 4.235Cl 75.5 37Cl 24.579Br 50.5 81Br 49.5127I 100
20 40 60 80 100 120
m/z
100
80
60
40
20
0
Relative intensity
112
114
Isotopic Clustersin Chlorobenzene
Visible in peaks for molecular ion
35Cl 37Cl
20 40 60 80 100 120
m/z
Relative intensity
77
Isotopic Clustersin Chlorobenzene
No m/z 77, 79 pair; therefore ion responsible form/z 77 peak does not contain Cl.
H
H
H
H
H +
100
80
60
40
20
0
Alkanes Undergo Extensive Fragmentation
m/z
Decane
142
4357
71
85
99
CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3
Relative intensity
100
80
60
40
20
0
20 40 60 80 100 120
Propylbenzene Fragments Mostlyat the Benzylic Position
20 40 60 80 100 120
m/z
Relative intensity
120
91 CH2—CH2CH3
100
80
60
40
20
0
13.25Molecular Formula as a Clue to Structure
Molecular Weights
One of the first pieces of information we try to obtain when determining a molecular structure is the molecular formula.
However, we can gain some information even from the molecular weight. Mass spectrometry makes it relatively easy to determine molecular weights.
The Nitrogen Rule
A molecule with an odd number of nitrogens has an odd molecular weight.
A molecule that contains only C, H, and O or which has an even number of nitrogens has an even molecular weight.
NH2 93
138
NH2O2N
183
NH2O2N
NO2
Exact Molecular Weights
CH3(CH2)5CH3
Heptane
CH3CO
O Cyclopropyl acetate
Molecular formula
Molecular weight
C7H16 C5H8O2
100 100
Exact mass 100.1253 100.0524
High Resolution Mass spectrometry can measure exact masses. Therefore, mass spectrometry can give molecular formulas.
Molecular Formulas
Knowing that the molecular formula of a substance is C7H16 tells us immediately that is an alkane because it corresponds to CnH2n+2.
C7H14 lacks two hydrogens of an alkane, therefore contains either a ring or a double bond.
Index of Hydrogen Deficiency
Relates molecular formulas to multiple bonds and rings and is also referred to as the number of unsaturations.
Index of hydrogen deficiency = ½ (number of hydrogens in the saturated formula – number of hydrogens in the compound).
Example 1
Index of hydrogen deficiency
C7H14
21 (molecular formula of alkane –
molecular formula of compound)=
21 (C7H16 – C7H14)=
21 (2) = 1=
Therefore, one ring or one double bond.
Example 2
C7H12
21 (C7H16 – C7H12)=
21 (4) = 2=
Therefore, two rings, one triple bond,two double bonds, or one double bond + one ringor two rings.
Oxygen Has No Effect
CH3(CH2)5CH2OH (1-heptanol, C7H16O) has same number of H atoms as heptane.
Neglect the oxygen in the formula below.
2
Index of hydrogen deficiency =
1 (C7H16 – C7H16O) = 0
No rings or double bonds.
2
Oxygen Has No Effect
Index of hydrogen deficiency =
1 (C5H12 – C5H8O2) = 2
One ring plus one double bond.
CH3CO
O Cyclopropyl acetate
If Halogen is Present
Treat a halogen as if it were hydrogen.
C C
CH3
ClH
H
C3H5Cl
Same index of hydrogendeficiency as for C3H6.
Rings versus Multiple Bonds
Index of hydrogen deficiency tells us the sum of the number of rings plus multiple (pi) bonds.
Catalytic hydrogenation tells us how many multiple (pi) bonds there are.
The difference between these indicates the number of rings.
End of Chapter 13End of Chapter 13SpectroscopySpectroscopy