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Mass Spectrometry:

IntroductionChem 8361/4361:

Interpretation of Organic Spectra

© 2009 Andrew Harned & Regents of the University of Minnesota

Mass Spectrometry

• NOT part of electromagnetic spectrum

• What is measured are (typically) positively charged ions

• Under certain circumstances negative ions can be detected

• Once ions are in detector, separated on the basis of mass/charge ratio (m/z)

• Many different ionization methods and detectors

• Useful for different kinds of molecules

• Vary in amount of energy• ↳Varying amounts/kinds of ions• ↳More or less fragments

Ionization Methods

• Probably the most widely used

• Vapor phase samples → from GC

• Sample bombarded with high energy electrons

• 25-80 eV (2.4-7.6 MJ/mol) → 70 eV typical

• Ejects 1 electron from molecules → positively charged ion is left

• Typical ionization energies for organic molecules ∼15 eV

• ∼50 eV excess energy

• excess energy dissipated by breaking covalent bonds

• lots of fragment ions

• fragment pattern unique to a given molecule

• many times molecular ion is not observed

Electron impact (EI)

Ionization MethodsChemical Ionization (CI)

• A method for “Soft Ionization”• Vapor phase samples

• Sample introduced to ionized reagent gas• Reagent gases: Methane, Isobutane, Ammonia, others

• Collisions between sample & gas ions cause proton transfers → produces [M+H]+ ions, not M+ ions.

• These are even electron ions

• Much less energy transferred < 5 eV• less fragmentation

• abundance of molecular ions

• good for molecular weight determination

• less information about structure

CH5+ C4H9

+ NH4+

Mass SpectrometryComparison of EI and CI Spectra

Notice lack of fragments and M+ has become M+H+

Ionization MethodsFast Atom Bombardment (FAB)

• Soft Ionization method

• Sample is mixed with a condensed phase matrix

• Mixture is ionized with high energy (6-10 keV) Xe or Ar

• Matrix protects sample from excess energy

• Ionization from protonation ([M+H]+), cation attachment

([M+23(Na)]+), or deprotonation ([M-H]+).

• High resolution masses are possible• exact mass determination

• Can be complicated by ions from matrix

Ionization MethodsMatrix-Assisted Laser Desorption Ionization (MALDI)

• Soft Ionization method

• Sample is mixed with a condensed phase matrix

• Mixture is ionized with a laser

• Charged molecules are ejected from matrix

• Little excess energy → little fragmentation

• Good for large molecules• proteins

• carbohydrates

• polymers

Ionization MethodsEvaporative Methods

• Two methods are common: Electrospray (ESI) and Atmospheric Pressure Chemical Ionization (APCI)

• Do not require vacuum pressures

• Both can be coupled to an LC system

• Multiply charged ions are possible and can be useful for large molecules (e.g. proteins).

• m/z = 100,000 Da/40 charges ⇒ 2,500 m/z

• Magnetic field is used to deflect ions around curved path

• Magnetic field is scanned to bring ions into focus

• Ions with correct m/z pass through the detector• If too heavy or too light do not make it through

DetectorsMagnetic Sector

• Ions fly through a tunnel of four charged rods

• Voltage of the rods is changed in order to focus ions

• Ions with correct m/z are able to fly through to detector• If too heavy or too light deflected into rods

DetectorsQuadrupole

DetectorsTime-of-Flight

• Ions accelerated through potential and allowed to drift down a tube to the detector

• all ions have the same energy

• Unlimited mass range• Often combined with MALDI

zeV = mv2/2

v = (2zeV/m)1/2

t = (L2m/2zeV)1/2

V = potential v = velocityL = length of tube t = time

m = mass

different masses have

different velocities

time of flight

varies with mass

As mass ↑

time ↑

Mass SpectrometryWhat does the spectrum look like?

Base Peak(Most Abundant)

100%

Most work in this class will be based on EI

Molecular Ion

Mass SpectrometryBasics of ionization and fragmentation

M + e M + 2e

Molecular ion – odd electron ion – gives exact mass – high energy, releases energy by bond cleavage

MOLECULE

one-bondclevage

two-bondclevage

MOL ECULE+

MOLEC ULE+

Not Seen

Even electron ion – "low" energy – few further fragments

Odd electron ion – still high energy – further fragments likely

Mass SpectrometryWhat does the spectrum look like?

Base Peak

(Most Abundant)

100%

Molecular

Ion

Mass Spectrometry

• *Keep in Mind* m/z = mass over charge

• this means isotopes have different m/z values and show up as different peaks

• Samples are actually a mixture of isotopic species

• To get formulas/structures must use monoisotopic masses, not those from periodic table which are weighted averages of all isotopes

• For example:

Effects of Isotope Differences

CH4

12CH4 Monoisotopicmass

13CH4

MW = 16

MW = 17

Polyisotopicmass

100

16

17

Mass SpectrometryRelative Abundance for Common Elements

1H

12C

16O

14N

28Si

31P

32S

19F

35Cl

79Br

127I

100

100

100

100

100

100

100

100

100

100

100

2H

13C

17O

15N

29Si

–

–

–

–

–

–

0.016

1.11

0.04

0.38

5.1

–

–

–

–

–

–

–

–

18O

–

30Si

–

34S

–

37Cl

81Br

–

–

–

0.02

–

3.35

–

4.4

–

32.5

98.0

–

Distinct isotopepatterns

Mass SpectrometryIsotope Patterns for Multiple Halogens

Adapted from: Crews, P.; Rodriguez, J.; Jaspars, M. Organic Structure Analysis; Oxford University Press: New York, 1998, p 257.

Mass SpectrometryA More Complicated Molecule

Why donʼt we see other combinations?

-low probablility (“non-existant”)

12C →

16O →

M2+ →

12C16O →

12C17O + 13C16O →

12C18O →

M+ →

m/z

12

16

22

28

29

30

44

45

Rel. Ab.

8.7

9.6

1.9

9.8

0.13

0.02

100

1.2

CO2 MW = 44

← 13C16O2 + 12C16O17O

← 12C16O2

Determination of FormulaThe Rule of 13

Assumes CnHn and amu equivalent (13 for n=1) is

present in all molecular fragment ions

Step 1: Divide M+ mass by 13, this gives n

Step 2: Any remainder represents count of additional Hʼs

example 1: for M+ = 78

78 ÷ 13 = 6 → n = 6 → C6H6


92 ÷ 13 = 7.077 → n = 7

7 x 13 = 91 → 1 extra H is present

Formula is C7H7+1 = C7H8



161 ÷ 13 = 12.385 → n = 12

12 x 13 = 156 → 5 extra Hʼs are present


What about heteroatoms?

Step 1: First derive formula as above

Step 2: Next, modify using CnHm equivalents

1H12

16O

14N

16O14N

19F

28Si

C

CH4

CH2

C2H6

CH7

C2H4

Element CnHm Equiv

31P

32S

16O32S

35Cl

79Br

127I

C2H7

C2H8

C4

C2H11

C6H7

C10H7

Element CnHm Equiv



108 ÷ 13 = 8.308 → n = 8

8 x 13 = 104 → 4 extra Hʼs are present


Possible candidates with heteroatoms

C8H12 – CH4 + O = C7H8O

C8H12 – 2(CH4) + O2 = C6H4O2

C8H12 – CH4 – CH2 + O + N = C6H6ON

MS will only give you molecular weight data. Must use

other spectral techniques to gleen information about

the presence of heteroatoms.

Determination of FormulaThe Nitrogen Rule

• A molecule with even numbered MW must contain either no N or even number of N

• A molecule with odd numbered MW must contain an odd number of N

• Holds for all compounds with C, N, O, S, X, P, B, Si, As, & alkaline earths

• Corollary: for fragmentation at a single bond

- even # M+ will give odd # fragment ion

- odd # M+ will give even # fragment ion

- fragment ion must contain all N if any

Determination of FormulaThe Nitrogen Rule

Back to example of MW = 108...

Formula of C6H6ON must be a fragment ion, not M+

C8H12 – CH4 – CH2 + O + N = C6H6ON

However...C8H12 – 2(CH2) + N2 = C6H8N2 is possible M+

Examples: Fragment Ions:

CH3CH2OH

CH4

CH3CH2NH2

NH2CH2CH2NH2

46

16

45

60

CH3CH3

30

CH3 CH3+

15

CH3NHCH3

45

CH3NH CH3+

30

CH3NH CH3+

15

Determination of Formula

How do we know if we have the molecular ion

or a fragment ion?

1. Must obey Nitrogen Rule

2. Must generate lower mass ions by logical neutral losses

Unlikely losses4 – 1421 – 25

33, 37, 38 256

226

-30 amu

- ok mass loss

- even # "M+", even # fragment

- NOT M+


256

177

-79 amu

- ok mass loss

- even # "M+", odd # fragment

- √ M+


or a fragment ion?




33, 37, 38


305

268

-37 amu

- odd # "M+", even # fragment

- bad mass loss

- NOT M+


or a fragment ion?




33, 37, 38


Stability plays a factor in whether or not we see molecular ion.

Decreasing ability to give prominent M+:

aromatics > conjugated alkenes > cyclic compounds >

organic sulfides > alkanes > mercaptans

Decreasing ability to give recognizable M+:

ketones > amines > esters > ethers >

carboxylic acids ~ aldehydes ~ amides ~ halides

M+ is frequently not detectable from:

aliphatic alcohols, nitrites, nitrates, nitro compounds,

nitriles, highly branched compounds


or a fragment ion?

Determination of FormulaWhat mass to use?

With high resolution masses you can differentiate

molecules with the same nominal mass

– For low resolution, round to nearest amu16O = 15.9949 → 1613C = 13.00336 → 13

– For high resolution, round to nearest 0.0001 amu1H = 1.00783 → 1.007813C = 13.00336 → 13.0034

Example: For MW = 28

12.0000

+ 15.9949

27.9949

14.0031

+ 14.0031

28.0062

12.0000

12.0000

+ 4.0312

28.0312

C O N N H2C CH2

← All can be distinguished!!

Exact mass – within 10 ppm

#.XXXXDifference of 2 = 1 ppm ↑

Chapter 1,

Appendix A

Mass Spectrometry Handout Mass spectrometry is used for determining the molecular weight of a compound and possibly identifying components of a compound. Mass spectrums identify a cation radical or cation. If the particle of interest is not positively charged, it will not be seen. An example of a mass spectrum is given below. There are a couple of terms that you should with which you should become familiar.

Mass spectrum of toluene. Courtesy of the Integrated Spectral Data Base System for Organic Compounds by the Japanese National Institute of Advanced Industrial Science and Technology. http://www.aist.go.jp/RIODB/SDBS/menu-e.html Many spectra will be used from this resource. Terms M/Z: mass to charge ratio. For our purposes, the charge will be assumed to be one so that the value of m/z corresponds to the mass. Mass ion peak: M+ is the molecular ion for the molecule. For toluene which has a molecular weight of 92, it is the peak at 92. For most molecules the M+ peak is seen on the mass spectrum but it is not seen if formation of a stable cation is possible. (i.e. tertiary alcohols) M+1 peak: This peak is one mass unit greater than the mass ion peak. For toluene, this peak is seen at 93. M+2 peak: This peak is two mass units greater than the mass ion peak. For toluene, this peak is barely seen at 94.

Base peak: This peak is the peak with the relative intensity of 100 %. The base peak is caused by the most stable cation. For toluene, the base peak is at 91. Resolution: Resolution is measured from peak heights and spacing. The two resolutions are low resolution and high resolution. Low resolution is to whole numbers (i.e. 44) while high resolution is to more numbers (i.e. 44.04356) Possible methods to make the cation for MS EI: Electron ionization with a stream of high energy electrons (70 electron volts). All of the spectra shown in class will be EI. CI: Chemical Ionization is caused by a carrier gas (usually methane) reacting with electrons to make primary ions (CH4+ and CH3+). These ions then react with sample. This may be used when the mass ion peak does not show up with EI. FAB: Fast atom bombardment is accomplished by using Xenon atoms. ESI: Electrospray ionization concerns using a voltage across a spray coming out of a liquid chromatograph machine. John Fenn won the Nobel price in 2002 for this technique which is used on proteins and peptides. Soft Laser Desorption: This technique is where a laser is used to form a cation from large molecules. Koichi Tanaka won the Nobel prize in 2002 for this technique and an example of this technique is MALDI. (Matrix Assisted Laser Desorption/Ionization) Types of MS MS: The first type is to use just a MS to generate a mass spectrum. This is an older technique and has been overtaken by other methods. GC/MS: This type stands for gas chromatograph/mass spectroscopy. The GC separates compounds while the MS determines the mass spectrum for each component. CBU HAS a functioning GC/MS instrument. GC/MS instruments are easy to break and VERY EXPENSIVE to repair. Therefore, most students will not get to use the GC/MS but might in advanced chemistry courses. LC/MS: This type stands for liquid chromatography/mass spectroscopy. It is a newer method and uses an HPLC to separate compounds before they are analyzed. MS/MS: This type stands for having two mass spectal machines in tandem. The first MS separates the molecule into parent ions. The parent ions are then collapsed to form daughter ions in the second machine. FT-MS: This type stands for Fourier Transform-Ion Cyclotron Resonance Mass Spectrum. TOF: Time of flight separates the particles. Quadrupole Mass Filter: This type uses four (hence quadrupole) voltage carrying rods to separate the particles. How to Read a Mass Spectrum 1) Look to see if the M+ peak is even or odd. If it is even like in toluene above (92), the molecule contains an even number of nitrogens (0 being even). If the M+ peak is odd, the molecule contains an odd number of nitrogens.

2) Look at the M+2 peak in comparison to the M+ peak. This will help you identify which elements may be present. M+ (relative intensity) M+2 (relative intensity) Element Present 100 100 Br 100 35 Cl 100 4.4 S 3) Look at the fragmentation patterns. Functional groups fragment in predictable ways. Some examples of functional groups are below. a) Look for CH2 next to benzene m/z 91 (look back at toluene for an example.) b) Ethers fragment at the alpha carbon to furnish the stable carbocation. Look for loss of the group attached to the alpha carbon. In the example of diethylether below, -15 or methyl.

OO

c) Carbonyl compounds are carbon double bonded to an oxygen. Loss of one of the side groups forms the very stable acylium ion. Examples below are for an ester and a ketone.

O

O-OCH3

O

O

O

-CH2CH2CH3

d) Some other fragmentation patterns are rearrangements like a Diels Alder or the McLafferty rearrangement. Calculations 1) Divide the M +1 peak by 1.1 to determine the number of carbons present in the molecule. Some examples follow. a) Assume the M+ peak is 100 % relative intensity and the M +1 peak is at 10 %. Dividing the M + 1 peak by 1.1 tells you that there are 9 carbons in the compound. This method rarely gives you exact whole numbers.

b) Assume the M + peak is at 40 % relative intensity and the M +1 peak is at 4.8. The M+ must be converted to 100 %. Dividing 100 by 40 gives a factor of 2.5. Multiplying 4.8 by 2.5 gives a value of 12. Dividing 12 by 1.1 gives 10.9 carbons. The M+ peak was NOT 100 so it had to be converted to 100 and multiplying it by 2.5 made it 100. The M +1 peak had to be multiplied also. 2) Rule of 13 The rule of 13 will give you a molecular formula for the alkane (C and H) for a corresponding molecular weight. Some examples follow. a) Assume you have a molecular weight of 400. Divide 400 by 13 which gives you 30.7692. Thirty is the number of carbon atoms. Mutiply the remainder (.7692) times 13 to get 10. Add this number, 10, to the whole number, 30, to get the number of hydrogens, 40. So the molecular formula is C30H40 which equals 400. b) Assume you have a compound that you think is caffeine. You do a mass spectrum and it gives you a molecular weight of 194. Dividing 194 by 13 gives you 14.9231. Multiplying 0.9231 by 13 gives you 12. Therefore, the molecular formula of the alkane is C14H26. Caffeine has four nitrogen atoms and two oxygen atoms which equals 88. Subtract 6 carbons and 16 hydrogens to get 88. Therefore, the molecular formula for caffeine is C8H10N4O. 3) Exact calculation from a high resolution mass spectrum. N2O and CO2 both give a low resolution M+ peak at 44. However, they give different values at high resolution. N = 14.00307 O = 15.99491 C = 12.00000 CO2 is 43.9898 while N2O is 44.0011. These compounds will show up at those exact numbers at high resolution allowing you to decide the molecular formula. Want more work with MS? Try http://svmsl.chem.cmu.edu/vmsl/default.htm for the virtual mass spectrum lab.

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