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-tocharge ( 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 CH
3
105
43
51
77

CO
105
120(M
+ ·
) m / z 20 40 60 80 100 120
苯乙酮的质谱图

4.2
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
3
+
 ；
CH
4

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
CI(CHMe
3
)
391
100 200 300 400
391
二
辛
酯
的
质
谱
图
邻
苯
二
甲
酸
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 CO C
2
H
4
N
2
1.00782 15.9949 14.0031
27.9949 28.0313
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

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

 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
C
2
H
4
O CO
2
CN
2
H
4
3C 36.0000
2C 24.0000
1C 12.0000
1C 12.0000
8H 8.0626
4H 4.0313
4H 4.0313
44.0626
1O 15.9949
2O 31.9898
2N 28.0061
44.0262
44.9898
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

 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 M + hydrogen 1 H 100.0% carbon 12 C 98.9% nitrogen 14 N 99.6% oxygen 16 O 99.8% sulfur 32 S 95.0% chlorine 35 Cl 75.5% bromine 79 Br 50.5% iodine 127 I 100.0%
13
15
33
M+1
C 1.1%
N 0.4%
S 0.8%
18
M+2
O 0.2%
34 S 4.2%
37 Cl 24.5%
81 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) 。

 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



8 (Pr
9 (
)
)

 While the M+2 = 0.006 w ( w -1) + 0.2 z , we have
0 .
9

0 .
006

8 ( 8

1 )

0 .
2
0 .
9

0 .
006

9 ( 9

1 )

0 .
2


3 (Pr

 2 (
)
)
This suggests this molecular formula of C
9
H
10
O
2

 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 E
D
B
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

Figure 4.12 The mass spectrum for 1-pentane

Figure 4.12 The mass spectrum for 2-methylbutane

Figure 4.16 The spectra of 1-heptene (a) and 2-heptene (b)

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)

3. The mass spectra of aldehydes, ketones
 3.1 The mass spectra of aldehydes

O
R C O R'
-OR'
-R
R
O
C
R' = CH m / z = M
3
-31
O
C O R' m / z = 44 + R'
R' = CH
3
, m / z = 59

C
6
H
12
O ，酮

C
6
H
12
O ，酮

C
6
H
12
O ，酮

Deduce the complete structural formula of the compound from the mass spectrum in the figure below.

 The spectrum is shown below. The molecular formula is assigned C
11
H
12
O by accurate mass
2 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 1 HNMR spectral data.
Assign the structure to the
 compound.
IR (neat)
 max
: 2950, 2850, 1730, 1480, 1460,
1400 cm -1

1 HNMR

: 1.20 (9H, s), 3.70 (3H, s)

 An organic compound A gave the following spectra data.
Deduce the structure of the compound.
 MS (m/z): 73, 91, 149, 164;


IR: 1730;
1 HNMR:

2.0 (3H, s), 2.93
t ） , 7.3(5H, s).
（ 2H, t
）
, 4.30
（ 2H,


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)

 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