elusidasi 4 ms

8/19/2019 Elusidasi 4 MS

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Mass Spectrometry


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The concept of mass spectrometry

A compound is ionised, the ions are separated on the

basis of their mass/charge ratio, and the number of ions

representing each mass/charge “unit” (m/z) is recordedas a spectrum

It is routine to couple a mass spectrometer to some form of

chromatographic instrument, such as a gas chromatograph

(GC-MS) or a liquid chromatograph (LC-MS).

The mass spectrometer analyses the masses of cations.


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• When the electron beam ionizes the molecule, thespecies that is formed is called a radical cation, andsymbolized as M+•.

• The radical cation M+• is called the molecular ion orparent ion.

• The mass of M+• represents the molecular weight of M.

• Because M is unstable, it decomposes to formfragments of radicals and cations that have a lowermolecular weight than M+•.


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• Molecular ion (M): A radical cation formed by removal of

a single electron from a parent molecule in a massspectrometer = MW.

• For our purposes, it does not matter which electron is

lost; radical cation character is delocalized throughoutthe molecule; therefore, we write the molecular formula

of the parent molecule in brackets with:

– A plus sign to show that it is a cation.

– A dot to show that it has an odd number of electrons.

• Molecular Ion

M

+•


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Partial MS of octane (C8H18, M = 114)

114

85

71

57M

+

29 (CH3CH2)14 (CH2)

The parent peak or molecular ion

C8H18e

-

70 eV+ 2 e

-[C8H18]

(M

+

)


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The largest or most abundant peak is called the base peakand is assigned an intensity of 100%, other peaks are

then fractions of that.

11485

71

57

M

+

29 (CH3CH2)

14 (CH2)

The relative abundance of all other ions is reported

as a % of abundance of the base peak.

114(M

+

,40), 85(80), 71(60), 57(100) etc.


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Masses are usually rounded off to whole numbersassuming:

H = 1, C = 12, N = 14, O = 16, F = 19 etc.

Molecular ion (parent peak)

Daughterions[C8H18]

(M+, 114)

[C6H13]

(85)

fragmentation

-CH3CH2 (29)

[C5H11]

(71)

-CH3CH2CH2 (29+14)


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M + e- !

M+

+ 2e-Molecule High Energy

Electron

Molecular

Ion

(Radical Cation)

100

9080

70

60

50

40

30

2010

0 I n

t e n

s i t

y ( %

o f

B a

s e

P e

a k

)

20 30 40 50 60 70 80 90

m / z

1-Pentanol - MW 88

CH3(CH2)3 – CH2OH

CH2OH+

M - (H2O and CH2=CH2)

M - (H2O and CH3)

M - H2O

M+ - 1

Molecular Ion Peak

Base Peak

M + e- !

M+


Electron

Molecular

Ion

(Radical Cation)

M + e- !

M+


Electron

Molecular

Ion

(Radical Cation)

100

9080

70

60

50

40

30

2010

0 I n

t e n

s i t

y ( %

o f

B a

s e

P e

a k

)

20 30 40 50 60 70 80 90

m / z

1-Pentanol - MW 88

CH3(CH2)3 – CH2OH

CH2OH+

M - (H2O and CH2=CH2)

M - (H2O and CH3)

M - H2O

M+ - 1

Molecular Ion Peak

Base Peak

What’s in a Mass Spectrum?

Mass spectrum: A plot of the relative abundance of ions versus

their mass-to-charge ratio (m/z).


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WAYS TO PRODUCE IONS

• Electron impact (EI) - vapor of sample is bombarded with

electrons: M + e 2e + M.+ fragments

• Chemical ionization (CI) - sample M collides with reagent

ions present in excess

e.g. CH4

+ e CH4

.+ CH5

+

M + CH5+ CH4 + MH

+

• Fast Atom/Ion Bombardment (FAB)

• Laser Desorption & Matrix-Assisted Laser Desorption(MALDI) - hit the sample with a laser beam

• Electrospray Ionization (ESI) - a stream of solution

passes through a strong electric field (10

6

V/m)


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1. Electron Ionization (EI)

most common ionization technique, limited to relativelylow MW compounds (


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• vaporized sample is bombarded with high energy

electrons (typically 70 eV)

• “hard” ionization method leads to significantfragmentation

• ionization is efficient but non-selective

Electron Ionization (EI)


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Electron Ionization

Advantages

• inexpensive, versatile and reproducible

• fragmentation gives structural information

• large databases if EI spectra exist and are

searchable

Disadvantages

• fragmentation at expense of molecular ion

• sample must be relatively volatile


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Chemical Ionization (CI)

Vaporized sample reacts with pre-ionized reagent gas via

proton transfer, charge exchange, electron capture,adduct formation, etc.

– Common CI reagents:

methane, ammonia, isobutane, hydrogen, methanol

• “soft” ionization gives little fragmentation

• selective ionization-only exothermic or thermoneutralion-molecule reactions will occur

• choice of reagent allows tuning of ionization


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CnHmX x Ny Oz

The Index of Hydrogen Deficiency (or IHD) is also known as

"units of unsaturation".

The index of hydrogen deficiency is a count of how many molecules

of H2 need to be added to a structure in order to obtain the

corresponding saturated, acyclic species.

Hence, the IHD is the sum of the number of rings, the number of

double bonds and twice the number of triple bonds.

Index = (n) - (m/2) - ( x /2) + (y /2) +1


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How To Determine Molecular Formulas andMolecular Weights Using Mass Spectrometry

Isotopic Peaks & the Molecular Ion


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! The presence of isotopes of carbon, hydrogen, andnitrogen in a compound gives rise to a small M + 1

peak.

! The presence of oxygen, sulfur, chlorine, or brominein a compound gives rise to an M + 2 peak

M + 1 Elements:

M + 2 Elements:

C, H, N

O, S, Br, Cl

+!

+!


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! Is M odd or even?

According to the nitrogen rule, if it is even, then thecompound must contain an even number of nitrogen

atoms (zero is an even number)

! For our unknown, M is even. The compound musthave an even number of nitrogen atoms

+!

+!

! The relative abundance of the M +2 peak indicates thepresence (or absence) of S (4.4%), Cl (33%), or Br (98%)

! For our unknown M +2 = 0.3%; thus, we can assume

that S, Cl, and Br are absent

! The molecular formula can now be established bydetermining the number of hydrogen atoms and adding the

appropriate number of oxygen atoms, if necessary

+!

+!


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! Since M is m/z 72

! molecular weight = 72

! As determined using the relative abundance of M+1 peak, number of carbons present is 4

!

Using the “nitrogen rule”, this unknown must havean even number of N. Since M.W. = 72, and thereare 4 C present, (12 x 4 = 48), adding 2 “N” will begreater than the M.W. of the unknown. Thus, this

unknown contains zero “N”

+!

+!


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! For a molecule composed of C and H only

H = 72 – (4 x 12) = 24

but C4H24 is impossible

! For a molecule composed of C, H and O

H = 72 – (4 x 12) – 16 = 8

and thus our unknown has the molecular formulaC4H8O


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C8H8O2

1.BM vs formula

2.Index ??

3.Fragment ions and fragment lost


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