Chapter 13 Spectroscopy Infrared spectroscopy

Chapter 13 Spectroscopy Infrared spectroscopy
Ultraviolet-visible spectroscopy Nuclear magnetic resonance spectroscopy Mass spectrometry Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1

13.1 Principles of Molecular Spectroscopy: Electromagnetic Radiation

Figure 13.1: The Electromagnetic Spectrum
Cosmic rays  Rays X-rays Ultraviolet light Visible light Infrared radiation Microwaves Radio waves Hi , Short  Energy Lo , Long  4

Electromagnetic Radiation
Is propagated at the speed of light, has properties of particles and waves, and the energy of a photon is proportional to its frequency. E = h 2

Figure 13.1: The Electromagnetic Spectrum
The UV, VIS, IR region. Longer Wavelength () Shorter Wavelength () 400 nm 750 nm Ultraviolet Visible Light Infrared Higher Frequency () Lower Frequency () Higher Energy (E) Lower Energy (E) 6

13.2 Principles of Molecular Spectroscopy: Quantized Energy States
5

E = h Electromagnetic radiation is absorbed when the energy of photon corresponds to difference in energy between two states. 6

Energy and its Effect Energy UV-Vis infrared microwave radiofrequency
Electron excitation Vibration of bonds Rotation of molecules Nuclear spin states 7

13.3 Introduction to 1H NMR Spectroscopy (Proton NMR)
8

The Nuclei that are Most Useful to Organic Chemists are:
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. 9

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. 10

The distribution of nuclear spins is random in the absence of an external magnetic field B0.
+ + + + + 11

An external magnetic field (B0) causes nuclear magnetic moments to align parallel or antiparallel to applied field. + + + B0 = field strength + + 11

There is a slight excess of nuclear magnetic moments aligned parallel to the applied field.
+ + + B0 + + 11

Energy Differences Between Nuclear Spin States
+ + increasing field strength 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. 12

Energy Required to Flip Spin States
Ho = B0 (Terms for the applied field, in gauss below). 6

Some Important Relationships in NMR
Units Hz kJ/mol (kcal/mol) tesla (T) The frequency of absorbed electromagnetic radiation is proportional to the energy difference between two nuclear spin states which is proportional to the applied magnetic field. 6

The frequency of absorbed electromagnetic radiation 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 frequency of 200 MHz (200 x 106 s-1) 13C absorbs radiation having a frequency of 50.4 MHz (50.4 x 106 s-1) 6

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. 6

13.4 Nuclear Shielding and 1H Chemical Shifts
What do we mean by "shielding"? What do we mean by "chemical shift"? 14

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. Bi C H B0 15

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. Bi C H B0 15

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. Bi C H B0 15

Chemical Shift Chemical shifts (d) 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 d. Si CH3 H3C d = position of signal - position of TMS peak spectrometer frequency x 106 15

Downfield Decreased shielding Upfield Increased shielding
(CH3)4Si (TMS) 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Chemical shift (, ppm) measured relative to TMS 1

Chemical Shift Example: The signal for the proton in chloroform (HCCl3) appears 1456 Hz downfield from TMS at a spectrometer frequency of 200 MHz. d = position of signal - position of TMS peak spectrometer frequency x 106 d = 1456 Hz - 0 Hz 200 x 106 Hx x 106 d = 15

The more electron electron density that is withdrawn from a hydrogen results in a smaller field induced from the remaining electrons. H C Cl  7.28 ppm 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Chemical shift (, ppm) 1

13.5 Effects of Molecular Structure on 1H Chemical Shifts
Protons in different environments experience different degrees of shielding and have different chemical shifts. 17

Electronegative Substituents Decrease the Shielding of Methyl Groups
least shielded H most shielded H CH3F CH3OCH3 (CH3)3N CH3CH3 (CH3)4Si d 4.3 d 3.2 d 2.2 d 0.9 d 0.0 A more electronegative atom will remove more electron density from hydrdogen. 21

Electronegative Substituents Decrease Shielding
H3C—CH2—CH3 d 4.3 d 2.0 d 1.0 O2N—CH2—CH2—CH3 CHCl3  7.3 CH2Cl2  5.3 CH3Cl  3.1 21

Methyl, Methylene, and Methine
CH3 more shielded than CH2 ; CH2 more shielded than CH H3C C CH3 H d 0.9 d 1.6 d 0.8 CH2 d 1.2 21

Protons Attached to sp2 Hybridized Carbon are Less Shielded than those Attached to sp3 Hybridized Carbon H C H CH3CH3  7.3  5.3  0.9 21

But Protons Attached to sp Hybridized Carbon are More Shielded than those Attached to sp2 Hybridized Carbon  5.3 C H  2.4 CH2OCH3 C H 21

Protons Attached to Benzylic and Allylic Carbons are Somewhat Less Shielded than Usual
 1.5  0.8 H3C CH3 Allylic d 0.9 d 1.3 H3C—CH2—CH3  1.2 H3C CH2  2.6 Alkyl Benzylic 21

Proton Attached to C=O of Aldehyde is Most Deshielded C—H
H3C 21

Table 13.1 C H R 0.9-1.8 C H N 2.1-2.3 C H 1.5-2.6 C H Ar 2.3-2.8 C H
Type of proton Chemical shift (), ppm Type of proton Chemical shift (), ppm C H R C H N C H C H Ar C H O 25

Table 13.1 C H NR 2.2-2.9 C H 4.5-6.5 C H Cl 3.1-4.1 C H Br H Ar
Type of proton Chemical shift (), ppm Type of proton Chemical shift (), ppm C H NR C H C H Cl C H Br H Ar C O H C H O 9-10 25

Table 13.1 H NR 1-3 H OR 0.5-5 H OAr 6-8 C O HO 10-13 Type of proton
Chemical shift (), ppm H NR 1-3 H OR 0.5-5 H OAr 6-8 C O HO 10-13 25

13.6 Interpreting 1H NMR Spectra
17

Information Contained in an NMR Spectrum Includes:
Number of signals. Position of signals. 3. Signal intensity (measured by area under peak). 4. Splitting pattern (multiplicity). 2

They exist in different molecular environments.
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. 3

chemically nonequivalent OCH3
Figure 13.12 CCH2OCH3 N chemically nonequivalent OCH3 NCCH2O Chemical shift (, ppm) 1

Chemically Equivalent Protons
Are in identical environments Have same chemical shift Replacement test: replacement by some arbitrary "test group" generates same compound H3CCH2CH3 chemically equivalent 7

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. ClCH2CH2CH3 CH3CH2CH2Cl H3CCH2CH3 chemically equivalent 7

Diastereotopic Protons
Diastereotopic protons are those whoswe replacement by some arbitrary test group generates diastereomers. Diastereotopic protons can have different chemical shifts. C Br H3C H  5.3 ppm  5.5 ppm 9

They are in mirror-image environments.
Enantiotopic Protons Enantiotopic protons are those whose replacement by some arbitrary test group generates enantiomers. They are in mirror-image environments. Enantiotopic protons have the same chemical shift. 10

Enantiotopic Protons C CH2OH H3C H C CH2OH H3C Cl H C CH2OH H3C H Cl R
11

13.7 Spin-Spin Splitting in 1H NMR Spectroscopy
Not all peaks are singlets. Signals can be split by coupling of nuclear spins. 17

Figure 13.13 Cl2CHCH3 4 lines; quartet 2 lines; doublet CH3 CH
Chemical shift (, ppm) 1

Two-bond and Three-bond Coupling
protons separated by two bonds (geminal relationship) protons separated by three bonds (vicinal relationship) 14

Two-bond and Three-bond Coupling
In order to observe splitting, protons cannot have same chemical shift. Coupling constant (2J or 3J) is independent of field strength. 14

Figure 13.13 Cl2CHCH3 4 lines; quartet 2 lines; doublet CH3 CH
coupled protons are vicinal (three-bond coupling) CH splits CH3 into a doublet CH3 splits CH into a quartet 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 Chemical shift (, ppm) 1

Why Do the Methyl Protons of 1,1-Dichloroethane Appear as a Doublet?
Cl signal 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. 19

Cl A result of spin-spin spliting. There are two orientations of the nuclear spin for the proton at C-1. One orientation shields the protons at C-2; the other deshields the C-2 protons. The protons at C-2 "feel" the effect of both the applied magnetic field and the local field resulting from the spin of the C-1 proton. 19

Cl "true" chemical shift of methyl protons (no coupling) This line corresponds to molecules in which the nuclear spin of the proton at C-1 reinforces the applied field. This line corresponds to molecules in which the nuclear spin of the proton at C-1 opposes the applied field. 19

Why Does the Methine Proton of 1,1-Dichloroethane Appear as a Quartet?
Cl 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. 19

Why Does the Methine Proton of 1,1-Dichloroethane Appear as a Quartet?
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. C H Cl 20

The Splitting Rule for 1H NMR
For simple cases, the multiplicity of a signal for a particular proton is equal to the number of equivalent vicinal protons + 1. Called the N + 1 Rule. 22

13.8 Splitting Patterns: The Ethyl Group
CH3CH2X is characterized by a triplet-quartet pattern (quartet at lower field than the triplet). 17

Figure 13.16 BrCH2CH3 4 lines; quartet 3 lines; triplet CH3 CH2

Splitting Patterns of Common Multiplets
Table 13.2 Splitting Patterns of Common Multiplets Number of equivalent Appearance Intensities of lines protons to which H of multiplet in multiplet is coupled Doublet 1:1 2 Triplet 1:2:1 3 Quartet 1:3:3:1 4 Pentet 1:4:6:4:1 5 Sextet 1:5:10:10:5:1 6 Septet 1:6:15:20:15:6:1 23

13.9 Splitting Patterns: The Isopropyl Group
(CH3)2CHX is characterized by a doublet-septet pattern (septet at lower field than the doublet). 17

Figure 13.18 ClCH(CH3)2 2 lines; doublet 7 lines; septet CH3 CH

13.10 Splitting Patterns: Pairs of Doublets
Splitting patterns are not always symmetrical, but lean in one direction or the other. 17

Consider coupling between two vicinal protons.
Pairs of Doublets C H H Consider coupling between two vicinal protons. If the protons have different chemical shifts, each will split the signal of the other into a doublet. 6

Let J be the coupling constant between them in Hz.
Pairs of Doublets C H H Let  be the difference in chemical shift in Hz between the two hydrogens. Let J be the coupling constant between them in Hz. 6

AX C H H J J  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. 6

AM C H H J J  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. 6

C AB H H J J  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. 6

C A2 H H 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. 6

Figure 13.20 H Cl OCH3 skewed doublets OCH3 Chemical shift (, ppm) 1

13.11 Complex Splitting Patterns:
Multiplets of multiplets 17

Consider the proton shown in red.
m-Nitrostyrene H O2N Consider the proton shown in red. It is unequally coupled to the protons shown in blue and white. Jcis = 12 Hz; Jtrans = 16 Hz 6

H O2N m-Nitrostyrene 16 Hz The signal for the proton shown in red appears as a doublet of doublets. 12 Hz 12 Hz 6

Figure 13.21 H O2N doublet doublet doublet of doublets 1

13.12 1H NMR Spectra of Aldohols
What about H bonded to O ? 17

Adding D2O converts O—H to O—D. The O—H peak disappears.
The chemical shift for O—H is variable ( 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. 6

13.13 NMR and Conformations 17

Most conformational changes occur faster than NMR can detect them.
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. 6

C NMR Spectroscopy 17

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. 6

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). 6

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 (a) shows the 1H NMR spectrum of 1-chloropentane; Figure (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. 6

1H Figure 13.26(a) ClCH2 CH3 ClCH2CH2CH2CH2CH3 Chemical shift (, ppm)

13C Figure 13.26(b) ClCH2CH2CH2CH2CH3
A separate, distinct peak appears for each of the 5 carbons. CDCl3 20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

Are measured in ppm () from the carbons of TMS
C Chemical Shifts Are measured in ppm () from the carbons of TMS 17

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. 6

1H 13C Types of Carbons Classification Chemical shift, d (CH3)3CH CH4
0.2 -2 8 16 25 28 primary secondary tertiary quaternary 0.9 1.3 1.7 Replacing H with C (more electronegative) deshields C to which it is attached. 6

Electronegativity Effects on CH3
Chemical shift, d 1H 0.2 2.5 3.4 4.3 13C -2 27 50 75 CH4 CH3NH2 CH3OH CH3F 6

Electronegativity Effects and Chain Length
Cl CH2 CH3 Chemical shift, d 45 33 29 22 14 Deshielding effect of Cl decreases as number of bonds between Cl and C increases. 6

Hybridization Effects
114 138 36 sp3 hybridized carbon is more shielded than sp2. sp hybridized carbon is more shielded than sp2, but less shielded than sp3. CH3 H C CH2 68 84 22 20 13 6

Carbonyl Carbons are Especially Deshielded
CH2 C O CH2 CH3 41 171 61 14 6

Table 13.3 RCH3 0-35 CR RC 65-90 R2CH2 15-40 CR2 R2C 100-150 R3CH
Type of carbon Chemical shift (), ppm Type of carbon Chemical shift (), ppm RCH3 0-35 CR RC 65-90 R2CH2 15-40 CR2 R2C R3CH 25-50 R4C 30-40 25

Table 13.3 RCH2Br 20-40 RC N 110-125 O RCH2Cl 25-50 RCOR 160-185
Type of carbon Chemical shift (), ppm Type of carbon Chemical shift (), ppm RCH2Br 20-40 RC N O RCH2Cl 25-50 RCOR RCH2NH2 35-50 RCH2OH 50-65 O RCH2OR 50-65 RCR 25

13.16 13C NMR and Peak Intensities
Pulse-FT NMR distorts intensities of signals. Therefore, peak heights and areas can be deceptive. 17

7 carbons give 7 signals, but intensities are not equal.
Figure 13.27 CH3 OH 7 carbons give 7 signals, but intensities are not equal. Chemical shift (, ppm) 1

C-H Coupling 17

Peaks in a 13C NMR Spectrum are Typically Singlets
13C—13C splitting is not seen because the probability of two 13C nuclei being in the same molecule is very small. 13C—1H splitting is not seen because spectrum is measured under conditions that suppress this splitting (broadband decoupling). 9

13.18 Using DEPT to Count the Hydrogens Attached to 13C
Distortionless Enhancement of Polarization Transfer 17

Measuring a 13C NMR Spectrum Involves
1. Equilibration of the nuclei between the lower and higher spin states under the influence of a magnetic field 2. Application of a radiofrequency pulse to give an excess of nuclei in the higher spin state 3. Acquisition of free-induction decay data during the time interval in which the equilibrium distribution of nuclear spins is restored 4. Mathematical manipulation (Fourier transform) of the data to plot a spectrum 9

Measuring a 13C NMR Spectrum Involves
In DEPT, a second transmitter irradiates 1H during the sequence, which affects the appearance of the 13C spectrum. Some 13C signals stay the same. Some 13C signals disappear. Some 13C signals are inverted. 9

Figure 13.29 (a) CCH2CH2CH2CH3 O DEPT 45 CH CH CH2 CH CH2 O CH2 CH3 C
20 40 60 80 100 120 140 160 180 200 Chemical shift (, ppm) 1

Figure 13.29 (b) CH and CH3 unaffected C and C=O nulled CH2 inverted
CCH2CH2CH2CH3 O DEPT 135 CH CH CH3 CH CH and CH3 unaffected C and C=O nulled CH2 inverted CH2 CH2 CH2 Chemical shift (, ppm) 1

13.20 Introduction to Infrared Spectroscopy
Gives information about the functional groups in a molecule 17

Infrared Spectroscopy
The region of infrared that is most useful lies between m ( 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. 2

Stretching Vibrations of a CH2 Group
These vibrations are analogous to the stretching motion of two springs. Antisymmetric Symmetric 2

In-plane Bending Vibrations of a CH2 Group
These vibrations are analogous to the in-plane bending motion of two springs. Antisymmetric, “rocking” Symmetric, “sissoring” 2

Out-of-plane Bending Vibrations of a CH2 Group
These vibrations are analogous to the out-of-plane bending motion of two springs. Antisymmetric, “twisting” Symmetric, “wagging” 2

13.21 Infrared Specta 17

Infrared Spectroscopy
Characteristic functional groups usually found between cm The fingerprint region is from cm-1. 2

Figure 13.35(a): Infrared Spectrum of Hexane
Francis A. Carey, Organic Chemistry, Fifth Edition. Copyright © 2003 The McGraw-Hill Companies, Inc. All rights reserved. 8

Figure 13.35(b): Infrared Spectrum of 1-Hexene

Figure 13.35(c): Infrared Spectrum of Benzene

Figure 13.35(d): Infrared Spectrum of Hexylbenzene

13.22 Characteristic Absorption Frequencies
17

Table 13.4 Infrared Absorption Frequencies
Stretching vibrations (single bonds) Structural unit Frequency, cm-1 sp C—H sp2 C—H sp3 C—H sp2 C—O 1200 sp3 C—O 8

Figure 13.36(f): Infrared Spectrum of Dihexyl Ether

Stretching vibrations (multiple bonds) Structural unit Frequency, cm-1 C —C C— —C N 8

Figure 13.36(b): Infrared Spectrum of Hexanenitrile

Stretching vibrations (carbonyl groups) Structural unit Frequency, cm-1 Aldehydes and ketones Carboxylic acids Acid anhydrides and Esters Amides 8

Figure 13.36(c): Infrared Spectrum of Hexanoic Acid

Figure 13.36(d): Infrared Spectrum of 2-Hexanone

Figure 13.36(e): Infrared Spectrum of Methyl Hexanoate

Bending vibrations of alkenes Structural unit Frequency, cm-1 CH2 RCH CH2 R2C 890 CHR' cis-RCH CHR' trans-RCH CHR' R2C 8

Bending vibrations of derivatives of benzene Structural unit Frequency, cm-1 Monosubstituted and Ortho-disubstituted Meta-disubstituted and Para-disubstituted 8

Stretching vibrations (single bonds) Structural unit Frequency, cm-1 O—H (alcohols) O—H (carboxylic acids) N—H 8

Figure 13.36(a): Infrared Spectrum of 1-Hexanol

Figure 13.36(g): Infrared Spectrum of Hexylamine

Figure 13.36(h): Infrared Spectrum of Hexanamide

13.23 Ultraviolet-Visible (UV-VIS) Spectroscopy
Gives information about conjugated  electron systems. 17

Transitions Between Electronic Energy States
Gaps between electronic energy levels are greater than those between vibrational levels. Gap corresponds to wavelengths between 200 and 800 nm. E = h 2

X-axis is wavelength in nm (high energy at left, low energy at right).
Conventions in UV-VIS 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 3

UV Spectrum of cis,trans-1,3-Cyclooctadiene
2000 Molar absorptivity () max 230 nm max 2630 1000 Wavelength, nm 4

* Transition in cis,trans-1,3-Cyclooctadiene
     LUMO  E = h  HOMO  -Electron configuration of the ground State. Most stable -Electron configuration of the excited state. 5

* 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. 5

Table 13.5 Methyl groups on double bond cause max to shift to longer wavelengths. H H H CH3 C C H H H CH3 max 170 nm max 188 nm 5

Table 13.5 Extending conjugation has a larger effect on max; shift is again to longer wavelengths. H H H H C C H C H H H H H max 170 nm max 217 nm 5

max 217 nm (conjugated diene)
H H Table 13.5 C H C H max 217 nm (conjugated diene) H H CH3 H C H max 263 nm conjugated triene plus two methyl groups C H H C H H CH3 5

Orange-red pigment in tomatoes
Lycopene Orange-red pigment in tomatoes (Lycopene) max 505 nm 5

13.24 Mass Spectrometry 17

Principles of Electron-Impact Mass Spectrometry
Atom or molecule is hit by high-energy electron. e– 2

Electron is deflected but transfers much of its energy to the molecule. e– 2

This energy-rich species ejects an electron, e– + • forming a positively charged, odd-electron species called the molecular ion. The molecular ion represents the molecular weight of the compound. 2

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 is deflected least. Lowest m/z is deflected most. + • 5

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. 6

The molecular ion dissociates to a cation and a radical.
Fragmentation The molecular ion dissociates to a cation and a radical. + • + cation radical • Usually several fragmentation pathways are available and a mixture of ions is produced. 2

Only the cation fragments are deflected by the magnet.
Fragmentation 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 the percentage of each ion of different mass in mixture. Separation of peaks depends on relative mass. + + + + + + 6

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. + + + + + + 6

Some Molecules Undergo Very Little Fragmentation
Benzene is an example. The major peak corresponds to the molecular ion. Relative intensity 100 80 60 40 20 m/z = 78 m/z 9

Isotopic Clusters H H H 79 79 78 93.4% 6.5% 0.1%
all H are 1H and all C are 12C one C is 13C one H is 2H 10

Isotopic Abundances 10

Isotopic Clusters in Chlorobenzene
Visible in peaks for molecular ion Relative intensity 100 80 60 40 20 112 114 m/z 11

Isotopic Clusters in Chlorobenzene
+ No m/z 77, 79 pair; therefore ion responsible for m/z 77 peak does not contain Cl. Relative intensity 100 80 60 40 20 77 m/z 11

Alkanes Undergo Extensive Fragmentation
CH3—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH2—CH3 Relative intensity 43 57 100 80 60 40 20 Decane 71 85 99 142 m/z 13

Propylbenzene Fragments Mostly at the Benzylic Position
Relative intensity 100 80 60 40 20 91 CH2—CH2CH3 120 m/z 14

13.25 Molecular Formula as a Clue to Structure
17

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. 6

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 NH2 O2N 183 NH2 O2N NO2 6

Exact Molecular Weights
CH3CO O CH3(CH2)5CH3 Heptane Cyclopropyl acetate Molecular formula C7H16 C5H8O2 Molecular weight 100 100 Exact mass High Resolution Mass spectrometry can measure exact masses. Therefore, mass spectrometry can give molecular formulas. 6

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. 6

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). 6

Index of hydrogen deficiency
Example 1 C7H14 Index of hydrogen deficiency 2 1 (molecular formula of alkane – molecular formula of compound) = 2 1 (C7H16 – C7H14) = 2 1 (2) = 1 = Therefore, one ring or one double bond. 6

Example 2 C7H12 1 = (C7H16 – C7H12) 2 (4) = 2
(4) = 2 Therefore, two rings, one triple bond, two double bonds, or one double bond + one ring or two rings. 6

Neglect the oxygen in the formula below.
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. 6

Index of hydrogen deficiency =
Oxygen Has No Effect CH3CO O Cyclopropyl acetate Index of hydrogen deficiency = 1 (C5H12 – C5H8O2) = 2 2 One ring plus one double bond. 6

Treat a halogen as if it were hydrogen.
If Halogen is Present Treat a halogen as if it were hydrogen. H Cl C3H5Cl C Same index of hydrogen deficiency as for C3H6. H CH3 6

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. 6

End of Chapter 13 Spectroscopy
17

Chapter 13 Spectroscopy Infrared spectroscopy

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Chapter 13 Spectroscopy Infrared spectroscopy

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Presentation on theme: "Chapter 13 Spectroscopy Infrared spectroscopy"— Presentation transcript:

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