Mass spectroscopy

Mass Spectrometry

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?Why it is use Mass Spectroscopy ?

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Introduction to Mass Spectrometry

Sample introduction

Ionization

Minimize collisions,

interferences

Separatemasses

Count ions

Collect results

3

PrincipleIt is also called as positive ion spectra or line spectra Sample is bombarded with the high electron beam produce the positive ions.

They travel in straight path

When a magnetic field or electric field is applied then travels in curved path

The fragments of different masses are separated based on the radius of curvature. m/e α r2

4

How does a mass spectrometer work?

Sample PlateTargetHPLCGCSolids probe

MALDIESIIonSprayFABEI/CI

SFADFAQuadrupoleFTMS

Faraday cup .Electron Mult.Photomultiplier

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V i d e o

6

Mass Spectrometry Needs

Ionization-How the protein is injected in to the MS machine

Separation-Mass and Charge is determined

Activation-Protein are broken into smaller fragments (peptides/AAs)

Mass Determination- m/z ratios are determined for the ionized protein fragments/peptides

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FRAGMENTATIONThe process of Breaking Molecules /ions into

fragments is known as fragmentation.This can be seen in the form of peaks in mass spectra Methanol can be divided in to 4fragmentse.g. CH3OH CH3OH +e¯⁺

CH3OH CH3 + OH¯⁺

CH3OH CH2OH + H¯⁺

CH3OH CHO + H2¯⁺

.

5 10 15 20 25 30 35

120

100

80

60

40

20

0

CHO⁺

CH3OH⁺

CH3⁺

CH2OH⁺

m/e

inte

nsit

y

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Fragmentation rules in MS

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1. Intensity of MM.+.+ is Larger for linear chainLarger for linear chain than for branched compound

2. Intensity of MM.+.+ decreasedecrease with IncreasingIncreasing M.W.M.W. (fatty acid is an exception)

3. Cleavage is favored at branchingfavored at branching

4.4. Aromatic Rings, Double bond, Cyclic structures stabilizeAromatic Rings, Double bond, Cyclic structures stabilize MM.+.+

5. 5. Double bond favor Allylic CleavageDouble bond favor Allylic Cleavage

6. Saturated Rings lose a Alkyl Chain (case of branching)

7. 7. Aromatic Compounds Cleave in b Aromatic Compounds Cleave in b Resonance Stabilized TropyliumTropylium

8. 8. C-C C-C Next to HeteroatomNext to Heteroatom cleave leaving the charge on the charge on the HeteroatomHeteroatom

9. 9. Cleavage of small neutral molecules (COCleavage of small neutral molecules (CO22, CO, olefins, H, CO, olefins, H22O ….).O ….).

Result often from rearrangement - McLafferty rearrangement Result often from rearrangement - McLafferty rearrangement 9

General rules of Fragmentation

1.Hydrocarbons•Hydrocarbons give clusters of peaks.

•Molecular ion peaks of very low abundance are observed for linear hydrocarbons.•For branched hydrocarbons give a low intensity at M+.•Intensity of (CnH2n+1) peaks decreases with increasing mass.

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2.Cleavage at Branched carbon

C > C

H

> C

H

H

>H

C

H

H

tert. sec.primary methyl

Cleavage at branched carbon is favored due to higher stability at tertiary carbocation.


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+

cleavage at 6-1

cleavage at 6-3

cleavage at 6-2

C H

C4H9

C3H7

C H

CH3

C4H9

+

+

C H

CH3

C3H7

+

(F1)

(F2)

(F3)

H3C CH2 CH2 C

CH3

H

CH2 CH2 CH2 CH31 2 3 4 5 6 7 8

Eg.

Produces thre secondary cations, the most favored fragments at C-4 of

4- methyl octane.Note that C4 is common for fragments (F1)(F2) And (F3). 12

3.Rule of β cleavage

X C1C2 R X CH

a b

Most important rule covers 70% of mass fragmentation.

Cleavage favored at β bond leaving positive charge on C1.


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H3C CH2 O CH2 CH3

H3C CH2 O CH2 CH3

CH2 O CH2

m/e = M-15

1.

H3C

2.

H3C CH2 N CH2

CH2 CH2 CH3

NC2H5

C3H7

H2C

m-57m-29

NC2H5

H2C

H2C

NCH2

C3H7

m-15

CH2

tert.amine

B1B2B3

e.g.: A) (x) = O, N, S.

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3.

CH2 S CH2 CH2 CH3

SH2C

CH2

SH2C

C3H7M-71

M-29

B2 B1

B1B2

R

CH2CH2

+

+

m/e = ( M-R )Stablebenzylic cation

+

m/e = 91

Tropylium cationm/e = 65

cyclopentadienylcation

+

b)

b) Benzylic clevage

-(x)- =

16

Very common fragment for ester

M-31 = methyl esterM-45 = ethyl ester

C. Allylic Cleavage

H2C

R

m/e = M-R stable allyliz cation

CH3H3C

O

R CH3

O+

R C O+CO CH3

m/e = M-R m/e = M-15

Simarly for x= N & S

i)

ii)

CR OCH3

O

CR O+

m/e = M-31

+

17

4 Rule of elimination of small neutral molecule

C

H

C

OH

C C

+

+ H2O

m/e M - 18

Α) β - EliminationThe high temperature and high vacuum are quite favourable for elimination reaction

and hencei)Loss of water (H2O) for alcohols (M-18) is a prominent fragment.Tertiary alcohols lose the water so fast that in many cases M.I. Peak is absent.


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C C

NH

C C + NH2

M - 46

C2H5

C2H5

ii)Loss of Ammonia (NH3)(M-17) for primary amines and primary and secondary alkyl ammonia derivatives For

C

H

C

NH2

C C +

M - 17

NH3

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iii)Elimination at Hydrogen sulphide (H2S)[M-34] confirms thiols (mercaptons)

C

H

C

SH

C C + H2S

M - 34

iv)Elimination of Hydrogen cyanide (HCN)[M-27] confirms nitriles.

C

H

C

CN

C C + HCN

M - 27

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v)Elimination of Hydrogen halide(HX),

Common for tertiary halides.

C

H

C

X

C C

m/e = M - HX X = F, Cl, Br, I

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5.Rule – retro Diel’s Alder reaction

High temperature high vacuum highly favorable for(DA) common for all these six membered cyclic mono olefins.

+

O

O

O + O

O

O

diene dienophile


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MCLAFFERTY REARRANGEMENT:-

Rearrangement ions are fragments, they are formed due to the result of intermolecular atomic rearrangement during fragmentationTo undergo this rearrangement the molecule must posses heteroatom, one double bond and hydrogen atom

McLaffertyMcLafferty

x

CH2

CH2

H

CH2

O

CY

Y Y H, R, OH, NR2 H, R, OH, NR2

Ion Stabilized Ion Stabilized by resonanceby resonance

x

CH2

CH2

H

CH2

O

CY

- CH- CH22=CH=CH22

x

CH2

O

CY

H

x

CH2+

O+

CY

H

x

CH2+

O

C+

Y

H

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It is used for determination of molecular mass of compounds and its elemental composition

Molecules having odd mass number contain odd number of nitrogen atoms.

Molecules having even mass number contain even no of nitrogen atoms.

NITROGEN RULE:-

CH3

CH3 CH3

H

MW = 59 MW = 59 (odd)(odd)

MW = 58 MW = 58 (even)(even)

Ionisation Ionisation [M+H][M+H]

[M+H][M+H]

MW = 60MW = 60

MW = 59MW = 59

CH3

N

CH3 CH3

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Nitrogen:Odd number of N = odd MW

CH3CNM+ = 41

SDBSWeb : http://riodb01.ibase.aist.go.jp/sdbs/ (National Institute of Advanced Industrial Science and Technology, 11/2/09)

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Problems and General pattern for individual Families

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Fragmentation Patterns

Alkanes:Fragmentation often splits off simple alkyl groups:

Loss of methyl M+ - 15Loss of ethyl M+ - 29Loss of propyl M+ - 43Loss of butyl M+ - 57

Branched alkanes tend to fragment forming the most stable carbocations.

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Mass spectrum of 2-methylpentane

CH2

CH3 CH3

CH2

CH2

CH2+

CH3 CH3

CH2+

CH2

CH2

CH3 CH3

CH2

CH3 CH3

CH2+

CH2

CH

CH3 CH3

Aklenes (olefins)

CH2

CH3 CH3

CH2

CH3 CH3

m/z 69 m/z 67 m/z 93



Hydroxy compounds:Hydroxy compounds:

R2 C

R3

R1

O Hx

Loss of largest groupLoss of largest group

- R3

R2

CR1

O+

HR2

C+

R1O H

If RIf R11=H m/e 45, 59, 73 …=H m/e 45, 59, 73 …

If RIf R11=alkyl m/e 59, 73, 87 …=alkyl m/e 59, 73, 87 …

x

OH

CHR

H

CHR

CHR

CHR

OH+

CHRCHR

CHR

CHR

H

CHR+

CHR

CHR

CHR CHR+

CHR

CHR

CHR

x

OH

CHR

H

CHR

CHR

CHR

CHR

CHR

M – (HM – (H22O) – (C1=C2) AlkeneO) – (C1=C2) Alkene

- H2O

- CHR=CHR

M – (HM – (H22O)O)

– – (H(H22O)O)




Aromatics may also have a peak at m/z = 77 for the benzene ring.

NO2

77M+ = 123

77


AlcoholsFragment easily resulting in very small or missing

parent ion peakMay lose hydroxyl radical or water

M+ - 17 or M+ - 18Commonly lose an alkyl group attached to the

carbinol carbon forming an oxonium ion.1o alcohol usually has prominent peak at m/z =

31 corresponding to H2C=OH+


MS for 1-propanol

M+M+-18

CH3CH2CH2OH

H2C OH



Ethers α-cleavage forming oxonium ion

Loss of alkyl group forming oxonium ion

Loss of alkyl group forming a carbocation


Aldehydes (RCHO) Fragmentation may form acylium ion

Common fragments:

M+ - 1 for M+ - 29 for

RC O

R (i.e. RCHO - CHO)

RC O


Ketones Fragmentation leads to formation of

acylium ion:

Loss of R forming

Loss of R’ forming

RC O

R'C O

RCR'O


MS for 2-pentanoneCH3CCH2CH2CH3

O

M+

CH3CH2CH2C O

CH3C O



Esters (RCO2R’) Common fragmentation patterns

include: Loss of OR’

peak at M+ - OR’

Loss of R’ peak at M+ - R’


M+ = 136

CO

O CH3

105

77 105

77


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Any Q.Thank You

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