Transcript Chapter 14 MS
Mass Spectrometry
质 谱
4.1 Introduction
Mass spectrometry (MS) is one of the most versatile and powerful tools of chemistry analysis. It provides quantitative information of atoms or molecules and qualitative information about molecules (such as the
molecular weight
and valuable information about the
molecular formula
), using few nanomoles of the sample (as little as 10 -12 g, 10 -15 moles for a compound of mass 1000 Daltons).
It also provides the most structural information that can
confirm a structure
derived from NMR and IR spectroscopy. Mass spectrometry fundamentally differs from other forms of spectroscopy (such as UV-vis, infrared, NMR, etc.) because electromagnetic radiation is not involved. The mass spectrometer works by using magnetic and electric fields to exert forces on charged particles (ions) in a vacuum . Therefore, a compound must be charged or ionized to be analyzed by a mass spectrometer.
Furthermore, the ions must be introduced in the gas phase into the vacuum system of the mass spectrometer. The gas phase ions are sorted in the mass analyzer according to their mass-to charge (
m
/
z
) ratios and then collected by a detector. In the detector the ion flux is converted to a proportional electrical current. The data system records the magnitude of these electrical signals as a function of
m
/
z
and converts this information into a mass spectrum.
质谱图 横坐标：质荷比
(m/z)
纵坐标：相对丰度
(
最强峰强度定为
100%)
100 80 60 40 20 77 O C 105 CH 3 43 51 77 CO 105 120(M + · )
m
/
z
20 40 60 80 苯乙酮的质谱图 100 120
4.2
质谱仪与质谱分析原理
进样系统 离子源 质量分析器 检测器 1.
气体扩散 2.
直接进样 3.
气相色谱 1.
电子轰击 2.
化学电离 3.
场致电离 4.
激光 1.
单聚焦 2.
双聚焦 3.
飞行时间 4.
四极杆 1.
电子倍增管 2.
渠道式电子 倍增器阵列 质谱仪需要在高真空下工作：离子源（ 10 -3 10 -5 Pa ） 质量分析器（ 10 -6 Pa ）
(1)
大量氧会烧坏离子源的灯丝；
(2)
用作加速离子的几千 伏高压会引起放电；
(3)
引起额外的离子－分子反应，改 变裂解模型，谱图复杂化。
Figure 4.1 The scheme of electron-impact ion source and ion-accelerating system
离子源
Electron Ionization (EI) + + + + +
(
M
-
R
3 ) + (
M
-
R
2 ) + (
M
-
R
1 ) +
M
+ Mass Spectrum : R1 : R2 : R3 : R4 : e
M
e
M 2
e
Molecule Electron Cation radical Two electrons M A B Cation radical Cation Radical M C D Cation radical Cation radical Molecule
m/z 43 m/z 56
优点：
1
）稳定
,
质谱图再现性好，便于计算机检索及比较；
2
）离子碎片多，可提供较多的分子结构信息。 缺点：
1
）样品必须易于气化；
2
）当样品分子稳定性不高时，分子离子峰的强度低， 甚至不存在分子离子峰。
化学电离源（
Chemical Ionization
，
CI
）
:
离子室内的反应气（甲烷等；
10~100Pa
，样品的
10 3 ~10 5
倍），电子（
100~240eV
）轰击，产生离子， 再与试样分离碰撞，产生准分子离子（最强峰）。 谱图简单；不适用难 挥发试样；
CH 4
CH 4
CH 4 +
、
CH 3 +
；
CH 5 +
、
C 2 H 5 + XH
XH 2 +
、
XHCH 5 + XHC 2 H 5 +
、
X +
； 电子
+
准分子离子 + 气体分子
CH 4
试样分子
XH
（
M
）
(M+1) + ; (M+17) + ; (M+29) + ;
EI
源
CI
源
(
甲烷
)
(b)
CI
源
(
异丁烷
)
(a) (c) 100 149 CO 2 C 8 H 17 43 57 71 0
m
/
z
113 167 CO 2 C 8 H 17 EI 100 200 300 400 100 0
m
/
z
100 113 149 CI(CH 4 ) 261 279 100 200 300 400 CI(CHMe 3 ) 391 391 邻 苯 二 甲 酸 二 辛 酯 的 质 谱 图 0
m
/
z
100 200 300 400
Magnetic sectors
m
z H
2
r
2 2
V
This equation shows that the
m
/
z
of the ions that reach the detector can be varied by changing either
H
or
V
.
Table 4-1
Exact masses of some common isotopes and simple molecular species, with the atomic mass of 12 C taken to be exactly 12 Species Mass 1 H 16 O 14 N 1.00782 15.9949 14.0031
CO C 2 H 4 27.9949 28.0313
N 2 28.0061
Resolving power
（分辨率） 低分辨
MS R<1000
高分辨
MS R
≧
10000
三重四级杆质谱
Detectors
检测器 After passage through a mass analyzer, resolved ion beams sequentially strike a detector. There are many types of detectors, but the electron multiplier, either single or multichannel, is most commonly used.
1
）电子倍增管
15
～
18
级； 可测出
10 -17 A
微弱电流；
2
）渠道式电子倍增器阵列
4.3 Determination of the Molecular Formula by Mass Spectrometry
Consider a molecular ion with a mass of 44. This approximate molecular weight might correspond to C 3 H 8 (propane), C 2 H 4 O (acetaldehyde), CO 2 , or CN 2 H 4 . Each of these molecular formulas corresponds to a different exact mass.
Table 4-2 Exact mass for C 3 H 8 (propane), C 2 H 4 O (acetaldehyde), CO 2 , or CN 2 H 4 Exact mass C 3 H 8 3C 36.0000
8H 8.0626
44.0626
C 2 H 4 O 2C 24.0000
4H 4.0313
1O 15.9949
44.0262
CO 2 1C 12.0000
2O 31.9898
44.9898
CN 2 H 4 1C 12.0000
4H 4.0313
2N 28.0061
44.0374
利用
HRMS
仪给出精确分子量，以推出分子式。 例
1. HRMS
仪测定精确质量为
166.0630
（
0.006
）
166.0570~ 166.0690
Use of Heavier isotope Peaks(LRMS)
Most elements do not consist of a single isotope but contain heavier isotopes in varying amounts. These heavier isotopes give rise to small peaks at higher mass numbers than the major M + molecular ion peak. A peak that is one mass unit heavier than the M + peak is called the
M+1 peak
; two units heavier, the
M+2 peak
; and so on.
Table 4-3 Isotopic Composition of Some Common Elements
Element hydrogen carbon nitrogen oxygen sulfur chlorine bromine iodine M + 1 H 100.0% 12 C 98.9% 14 N 99.6% 16 O 99.8% 32 S 95.0% 35 Cl 75.5% 79 Br 50.5% 127 I 100.0% 13 15 33 M+1 C 1.1% N 0.4% S 0.8% 18 81 M+2 O 0.2% 34 S 4.2% 37 Cl 24.5% Br 49.5%
Br
2
Br
Br
79 Br: 50.50% 81 Br: 49.50%
That means that you would get a set of lines in the
m
/
z
= 160 region. The relative heights of the 158, 160 and 162 lines are in the ratio 1:2:1.
Now substituting the percent abundance of each isotope ( 79 Br and 81 Br) into the expansion: Gives (
a
b
) 2
a
2 2
ab
b
2 ( 4 .
7 ) ( 0 .
505 ) 2 2 ( 0 .
505 )( 0 .
495 ) ( 0 .
495 ) 2 0 .
255 0 .
500 0 .
250 ( ( 4 .
8 ) 4 .
9 ) Which are the proportions of M:(M+1):(M+4), a triplet that is slightly distorted from a 1:2:1 pattern. When two elements with heavy isotopes are present, the binomial expansion (4.10) is used.
(
a
b
)
m
(
c
d
)
n
( 4 .
10 )
The Figure (b) and (c) spectra show that chlorine is also composed of two isotopes, the more abundant having a mass of 35 amu, and the minor isotope a mass 37 amu. The precise isotopic composition of chlorine and bromine is:
35 Cl: 75.77% 37 Cl: 24.23%
C 2 H 5 C l 49 28 29 51 64 66 C l 125 140 C 2 H 5 71 105 127 142 27 29 108 110 C 2 H 5 B r 79 81 93 95 B r 156 158 51 77
Molecular ions (or fragment ions) containing various numbers of chlorine and/or bromine atoms therefore give rise to the patterns shown in Figure 4.8.
Fig. 4.8 The abundances containing various numbers of chlorine and/or bromine atoms
For a compound C w H x O z N y , a simple formula allows one to calculate the percent of the heavy isotope contributions from a monoisotopic peak,
P M
, to the
P M
+1 peak: 100
P M
1
P M
1 .
11
w
0 .
0015
x
0 .
36
y
0 .
0037
z
1 .
11
w
0 .
36
y
( 4 .
11 ) 100
P M
2
P M
0 .
006
w
(
w
1 ) 0 .
2
z
( 4 .
12 )
Tables of abundance factors have been calculated by Beynon and Williams for all combinations of C, H, N, and O up to mass 500. For example, an isotopic cluster with most mass: 80 (M=100), 81 (M+1= 5.13
), 82 (M+2= 0.10
). For nominal mass 80, in the Beynon Table M+1 and M+2 values are listed below (Table 4-4). It is quite evident that molecular formula must be C 4 H 4 N 2 , because its M+1 and M+2 are close to that in Beynon Table. Once the empirical formula is established with reasonable assurance, hypothetical molecular structures are written.
Table 4-4 The tabular data of Beynon for nominal mass 80
Empirical formula (80) CH 4 O 4 C 3 H 2 N 3 C 4 H 2 NO C 4 H 4 N 2 C 5 H 4 O C 5 H 6 N C 6 H 8 M+1 1.30
4.42
4.78
5.15
5.51
5.88
6.61
M+2 0.80
0.08
0.29
0.11
0.32
0.14
0.18
Three general rules aid in writing formulas. (1) If the molecular weight of a C, H, O, and N compound is even, so is the number of hydrogen atoms it contains; (2) If the molecular weight is divisible by four, the number of hydrogen atoms is also divisible by four. (3) When nitrogen is known to be present in any compound of C, H, O, As, P, S, Si, and the halogens that have an odd molecular weight, the number of nitrogen atoms must be odd.
This rule is called
nitrogen rule
.
例
2
：某化合物
MS
数据：
M=181
，
P M %=100% P (M+1) %=14.68% P (M+2) %=0.97%
查
[
贝诺表
]
根据“氮规则”：
M=181
，化合物分子式为
(2)
。
Example 3
A molecular-ion cluster 150 (M=100), 151 (M+1=9.9), 152 (M+2=0.9). Here M+2 = 0.9 suggests absence of the other A+2 elements except for oxygen. The even numbered value for M means that either no or an even number of N atoms are present. According to M+1 = 1.1
w
+ 0.36
y
, we have
w
8 (Pr 9 (
esence Absence of of N two N atom
)
atoms
)
While the M+2 = 0.006
w
(
w
-1) + 0.2
z
, we have 0 .
9 0 .
006 8 ( 8 1 ) 0 .
2
z or
0 .
9 0 .
006 9 ( 9 1 ) 0 .
2
z z
3 (Pr 2 (
esence Absence of N two atom
)
N atoms
) This suggests this molecular formula of C 9 H 10 O 2
Example 4
The molecular-ion peak at
m
/
z
151 is accompanied by isotopic peaks at
m
/
z
152, 153 and 154. The intensity rations of M, M+1, M+2 and M+3 peaks are 100 :10.4: 32.1
:2.9. The high M+2 value suggests the presence of a Cl atom. The odd numbered value for M means the presence of an odd number N atom. While the M+1 value requires around 9 carbon atoms, but in consideration of the limit of molecular weight can only allow 8 carbonatoms. Thus the molecular formula can be assigned as C 8 H 6 ClN.
4.4 The Mass Spectra of Organic Compounds
McLafferty rearrangement
-Cleavage of a bond with -hydrogen rearrangement to form a cation radical and a neutral molecule is called
McLafferty rearrangement
. This rearrangement is characteristic of compounds containing a -hydrogen with respect to multiple bond. The general scheme is shown as following: H A B E D C Molecular ion McLafferty rearrangement H E D C Fragment + A B Neutral fragment
R H O Molecular ion R' McLafferty rearrangement R OH + R' Fragment, where R' = H, R, OR, OH, NH 2 , etc.
R H C Molecular ion McLafferty rearrangement R + CH Fragment
1. The mass spectra of hydrocarbons
1.1 The mass spectrum of alkane
Figure 4.12 The mass spectrum for 1-pentane
Figure 4.12 The mass spectrum for 2-methylbutane
1.2 The mass spectrum of alkenes
Figure 4.16 The spectra of 1-heptene (a) and 2-heptene (b)
1.3 The mass spectra of alkynes
1.4 The mass spectra of aromatic compounds 91 134 105
2. The mass spectra of alcohols, phenol, ethers
2.1 The mass spectrum of alcohols Figure 4.24 The mass spectrum of 1-butanol
MS of 1-cetene (b) MS of 1-cetanol (a)
2.2 The mass spectra of phenols
2.3 The mass spectrum of ethers
3. The mass spectra of aldehydes, ketones
3.1 The mass spectra of aldehydes
3.2 The mass spectra of ketones
4. The mass spectra of carboxylic acids
5. The mass spectra of esters
R O C O -OR' R' -R R O C R' = CH 3
m
/
z
=
M
-31 O C O R'
m
/
z
= 44 + R' R' = CH 3 ,
m
/
z
= 59
4.5 Interpretation to MS
Example 5-7
C 6 H 12 O
，酮
C 6 H 12 O
，酮
C
6
H
12
O
，酮
Example 8
Deduce the complete structural formula of the compound from the mass spectrum in the figure below.
Example 9
The spectrum is shown below. The molecular formula is assigned C 11 H 12 O 2 by accurate mass measurement. Deduce this compound ’ s structure.
多谱联解
Example 10
An organic compound
A
(molecular formula C 6 H 12 O 2 ) showed the following IR and spectral data.
Assign the structure 1 HNMR to the compound.
IR (neat) max : 2950, 2850, 1730, 1480, 1460, 1400 cm -1 1 HNMR : 1.20 (9H, s), 3.70 (3H, s)
Example 11
An organic compound spectra data.
Deduce compound.
A
the MS (m/z): 73, 91, 149, 164; gave the following structure of the IR: 1730; 1 HNMR: 2.0 (3H, s), 2.93
（ 2H,
t
） , 7.3(5H, s).
t
） , 4.30
（ 2H,
Example 12
A colorless solid, melting at 103 to 105 C, gave the following MS and IR spectra and spectral data in the table. Deduce the structure of the compound.
MS spectral data (
m
/
z
) 29 41 44 IR spectral data 59 (base) 72 85 & 86 (small) 3366 & 3190 cm -1 (str & brd) 643 & 704 cm -1 (brd) 57 101 (v, small, P) 2850 to 2970 cm -1 1481 & 1427 cm -1
Example 13 A compound, containing carbon and hydrogen only, gave the following MS, NMR, UV-vis and IR spectra and spectral data. Deduce the structure of the compound.
MS: M + 120.0939; UV: max 270nm ( = 300)
Example 14
A low boiling liquid compound gave the following MS, NMR, UV-vis and IR spectra and spectral data. Deduce the structure of the compound.
MS: M + 74.0368; UV: Transparent above 210nm