MAss Presentation 01.04
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Transcript of MAss Presentation 01.04
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Structure Determination From
Molecule to Materials by MassSpectrometry
B S KITAWATResearch Scholar
School of Chemical Sciences
Date: April 01,2011
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Basic principles Accurate method: Determining molecular mass &
elemental composition
In this tech : Molecules are bombarded with a beam
of energetic e-
The molecules are ionized & broken up into
fragments, some of which are +ve
The masses of the fragments & their relative
abundance reveal information about the structure of
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How do achieve this?
Persuade the molecule to enter the vapor phase (can
be difficult)
Produce ions from the molecules that enter the gas
phase
Separate the ions according to their mass-to-charge
ratios (m/z)
Measure and record these ions
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H-C:H
H
H
+ e H-C
H
H
H + 2e CH4
Radical cation
H-C
H
H
HH-C
H
H+ + H
Bond-breaking CationRadical
Only cations are carried to detector
Electron Impact
H-C
H
H
H++
1 15 16
Molecular ion
fragments
base peak
m/z
.
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Typical LC-MS
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The Mass Spectrometer
Mass Spectrometer
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Ionization Methods Electron ionization (EI)=50-70eV beam, volatile solvent,
low MW compounds analyzed
Chemical ionization (CI)= Ionized gas like CH4& NH3
Desorption ionization tech.(SIMS,FAB & MALDI) for
high MW up to 20,000 Da
SIMS= Sec. ion mass spe. (1-10keV, Ar+ & Cs+beam)
FAB= Fast Atom bombardment (Ar, Xe beam)
MALDI=Matrix assisted laser desorption ionization
(Used nitrogen laser emits at 337nm)
Electrospray Ionization (ESI)= Sample contain sol. sprayed
into heated chamber, high MW biomolecules like lysozyme
14,306 Da can analyzed
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Mass Analyzer: Ions are separated according to m/z
Magnetic Sector Mass Analyzers
Double Focusing Mass Analyzers
Quadrupole Mass Analyzers
Time of Flight (TOF) Mass Analyzers
How we get a mass peak ?
In EI-MS, a molecule is bombarded by high-energy e- in
the ionization chamber
The collision between sample molecules & the e- initially
results in the sample molecule losing one e- to form aradical cation
The molecule also absorbs a considerable amount of extra
energy during its collision with the incident e-
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This extra energy places the molecular ion in a highly
excited vibrational state
The vibrationally excited molecular ion may be unstable,
& it may lose some of its extra energy by breaking apart
into fragments
If the lifetime of the molecular ion is >105 sec, a peak
corresponding to the molecular ion will appear in the
mass spectrum
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tructura na ys s ragmentat on atternsStevensons Rule: States that the most probable fragmentation
is the one that leaves the positive charge on the fragment with
the lowest ionization energy
Figure:Mass
spectra expectedfor various
combinations
of bromine &
chlorine
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Initial Ionization Event: The e- most likely to be ejectedduring the ionization event are the ones that are in the highest
potential energy molecular orbitals, that is, the e- held least
tightly by the molecule
Example:It is much easier to
eject an e- from a
orbital in
comparison to a orbital
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Retro Diels-Alder Cleavage Unsaturated six-membered rings can undergo a retro Diels
Alder fragmentation to produce the radical cation of a diene
& a neutral alkene
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McLafferty Rearrangement A H- atom on a carbon 3 atom away from the radical
cation of an alkene, arene, carbonyl, or imine (a so-called
-hydrogen) is transferred to the charge site via a six-membered T.S., with concurrent cleavage of the bond
between the & positions
This forms a new radical cation & an alkene with a bond
between what were the original and carbons
Oth Cl T
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Other Cleavage Types
Alkanes
Cleavage of the C1-C2
bond results in the lossof a methyl radical &
the formation of the
propyl carbocation (m/z
= 43). Cleavage of the
C2/C3 bond results in
the loss of a ethyl
radical and theformation of the ethyl
carbocation (m/z = 29)
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Alkenestrans-2-hexene, MW = 84
Methyllyl cation, m/z = 55Ethyl radical, m/z = 29
29
Alk
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Alkynes1-pentyne, MW = 68
Loss of ethyl radical(m/z =29) to form propargyliccation(m/z =53), m/z = 39 due to loss of [C2H5]
+
A i H d b
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Aromatic Hydrocarbons
Benzene
Toluene
m- Xylene Cumene
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Aldehydes
Mass Spectrum ofValeraldehyde m/z = 86.13
Loss of 1H m/z= 85
Due to CHO+ m/z= 29
Propyl cation m/z= 43 due to
Beta cleavage
m/z =57, [CH3CH
2CH
2CH
2]+
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Ketones
2- butanone (MW=72)
CH3CH2CO+ m/z = 57
CH3CO+ m/z = 43
Al h l
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Alcohols
CH2OH+ m/z = 31
Loss of OH+ m/z = 70
CH3CH2CH+ m/z = 42
Nit l (A id )
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Nitrogen rule (Amides) If a compound contains an even number ofN atoms (or no
N atoms), its molecular ion will appear at an even mass
number If, an odd number ofN atoms, then its molecular ion will
appear at an odd mass value
In the case of benzamide the molecular ion appears at m/z121, indicating an odd number ofN atoms in the structure
[CONH2]+ m/z = 44
[C6H5]+
m/z = 77Loss of NH2 gives m/z =
105
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