MASS SPECTROMETRYProf.Kista Reddy

Dept.of Chemistry, OU

Unlike IR, UV and NMR spectroscopy, mass spectrometry is not an absorption spectroscopy.

Unlike IR, UV, NMR spectroscopy, the samples used to record the mass spectra cannot be recovered.

Mass spectroscopy is used to determine the molecular weight of the substances.

It is also used to determine the molecular formulae of the compound.

Presence of certain structural units can be known from the mass spectral fragmentation.

Small quantity of the substances ( 1 μ gm = 10-6 gm) is required to record the mass spectrum.

In ( certain cases) extreme cases the mass spectrum can be recorded even with 10-12

grams of the substance.

MASS SPECTROMETRY

MOLECULE ( M :)

(Substance in

gaseous state )

Bombarded with

electron beam at 70 ev.M

+.

( Moleclar ion

or

Parent ion )

+ e-

( Electron )

Odd number electron and positive charge

are associated with lot of energy.

M+.1 + M+

2Radical

Neutral

molecule+

Fragmentation

Fragments or

daughter ions

Molecular ion is a cationic radical / radical cation, it gets recorded according to its mass (m) and charge ( 2 ) ratio ( m/z ).

When charge = 1, then m/z = m = mass of parent ion = Molecular weight of molecule.

Molecule losses a electron and forms molecular ion. The mass of electron is negligible, then mass of molecular ion is same as mass of the molecule.

Ionization potential of organic molecule= 7 – 15 eV. ( 1 eV = 23.05 k. cal/ mole )

If we bombard the molecule with 70 eV electron beam, the parent ion formed has higher energy and it easily breakdown into smaller species.

Neutral species are not recorded in mass spectrum. Only charged species can be recorded.

Detection of cationic species, determination of m/z ratio is the basic principle involvedin the mass spectroscopy. The mass spectrum is a presentation of m/z values of cationicspecies vs their relative abundance in the form of a tabular list or a plot.

Rarely di and tri positive ions are also formed.

Uni positive ions are detected at their actual mass ( z = 1, m/z = m)

Di positive ions are discovered at their half of actual mass ( z = 2, m/z = m/2 )

Triply positive ions are detected at their 1/3rd of actual mass ( z = 3, m/z = m/3 )

Presentation of mass spectral data

It can be presented in two ways.

1. Line diagram method:

Base peak

M+.1

( molecular ion peak )

+ 2M+.

M+. + 1

( isotopic peaks )

2. Tabular presentation:

( %

)R

elat

ive

abu

nd

ance

m/z ratio

m/z - - - - - - - - - - - - - - -

Ratio - - - - - - - - - - - - - - -

General rules:

1. Even electron rule :- Even electron rule states that an even electron species will not normally fragment to two odd species. Since total energy of the product mixture is too high.

Molecular ion is a cation radical ( or ) radical cation.

+.OE

No. of electrons = odd

EECationic species

having even electrons

+.oe eea small molecule

No. of electrons = evenNo. of electrons = odd

+

+ .ee oe

No. of electrons = even No. of electrons = odd

+Cation free radical

ee + eeEven Even

.oe + oeodd odd

+.Least probable thermodynamicallybecause of higher energy contents of the products

Stability: EE > EE+ > OE+

2. Ionization : The basic principle involved in mass spectrum is the ionization of gaseous molecule. The energy needed to remove electron from gaseous molecule is 7 – 15 eV.

Type of electrons

Non - bonding

Bonding

π

σ

En

erg

y req

uire

d

If the molecule is bombarded with an electron beam whose energy is equal to the ionization potential of molecule ( 15 eV ), it is possible to decide which electron is lost( the bonding or non-bonding). However in view of the high energy conditions employed70 eV, any specificity of electron removal is totally lost.

3. Bond fission:

1. Homolytic cleavage:- Represented by fish hook arrows. It is one electron transfer.

2. Heterolytic fission : It is represented by conventional ( or ) normal ( or ) full arrow.It is two electron transfer.

A B

A BA B+

A B

A B

A B+

A B+

( B is more electronegative )

( A is more electronegative )

4. Nitrogen rule:-

A molecular ion ( hence molecule ) with 0 or even nitrogen's has even mass.

Example:- For CH3OH, m/z = 32 ( Number of nitrogen's = 0, therefore even mass. )

A molecular ion ( hence molecule ) with odd number of nitrogen's has odd mass.Example:- For CH3-NH2, m/z = 31 (Number of nitrogen's = 1, therefore odd mass. )

An ion ( molecular ion or daughter ion ) with even mass give rise to a fragment ion of oddmass by the fission of one bond and an ion of odd mass by fission of one bond results afragment ion of even mass provided nitrogen ( if any ) in original ion is retained in the fragment ion.

CH3-CH3

+. CH3 CH3+m/z = 30

even mass

m/z = 15odd mass

CH3CH2 + Hm/z = 29odd mass

1. Ph N HH

+.H + [Ph-N-H]+

m/z = 92even mass

2.

m/z = 93odd mass

By the fission of two bonds an ion of even mass resulting a fragment of even mass. An anion of odd mass results a fragment of odd mass.

Example:-

CHH

CH2OH

H3C+.

CH

CH2H3C+. + H2O

m/z = 66 m/z = 42

Number of nitrogen's Mass Type of ion

Zero/even Even Molecular ion

Zero/even Odd Daughter ion

Odd Odd Molecular ion

Odd Even Daughter ion

Recognition of the molecular ion :-

The molecular ion of nearly 20 % of organic compounds decompose within 10-5 seconds , with the result the molecular ion either fails to get recorded or its intensity is low. The peak observed at highest m/z values except for isotopic peak is known as molecular ion peak. The intensity of the molecular Ion peak depends on stability.

StrongAromatic hydrocarbons,

Hetero aromatic compounds,Aromatic fiuorides,Chlorides, nitrites,

Conjugated alkenes, Saturated cyclic compounds,Straight chain hydrocarbons.

MediumAromatic bromides, Iodides, Benzyl and

Benzoyl, Straight chainAldehydes, ketones, acids,

Esters, amides and Other alkyl halides.

Weak or AbsentAlcohols,

Aliphatic acetals,Amines, nitriles,

Nitrites, BranchedChain compounds,Nitro compounds.

Intensity of the molecular ion is influenced by

1. Increase in molecular weight2. Increase in branching3. Presence of groups that favours fragmentation.

In the case of compounds whose molecular ion is weak (or ) absent, the molecular ion peak Can be detected by

1. Decreasing the energy of electron beam i.e. by recording the mass spectrum at 15 eV, 20 eV…..

2. By increasing sample size3. By increasing the time sample spends in the ionization chamber ( decreasing the positive

acceleration potential)4. Increasing the sample pressure.

Meta stable peaks:-

Meta stable peaks are broad and diffused peaks. These are observed at non integral masses.The presence of meta stable peaks in mass spectrum is an evidence for the formation of fragment ion from the molecular ion ( or ) daughter ion by the elimination of neutral species. If M+.

1 is undergoing fragmentation to give rise to M+.2 and a neutral species, the meta stable

peak (m*) is neither observed at the mass of M+1 nor M+

2 . It is actually observed at mass less than both the peaks.

M* = (M+2)2/( M+

1 )

The meta stable ions are formed during the journey of ions from ion source to mass chamber.They have less energy than that, they would have acquired or gained and they formed in the ion source.

Therefore meta stable ions gets recorded at a mass to charge value less than its actual value.

m* +.M

M+2

M+1

m/z valueR

. A

Isotopic peaks:-

Molecular ion peak is the peak observed at highest m/z value except for isotopic peaks of (M+. + 1) (M+. + 2). Normally the mass of M+. Ion is one or two mass units less than the highest mass observed in the mass spectrum. The M+. + 1 and M+. + 2 peaks are known as the isotopic Peaks.

Base peak

M+.1

( molecular ion peak )

+ 2M+.

M+. + 1

( %

)R

elat

ive

abu

nd

ance

The isotopic peaks are observed because certain molecules contains heavier isotopes in placeof normal isotopes. The intensity of the isotopic peaks in relation to the molecular ion peak help in determining molecular formula. The isotopic peaks are observed not only for molecular ion peak, it can also for daughter ion. The intensity of isotopic peaks is related tonatural abundance.

Isotopic peaks

m/z value

Common Isotope100 %

1H12C14N16O32S35Cl79Br

Relative abundance ofheavier isotope (% )

2D (0.016)13C (1.08)

15N ( 0.38)17O ( 0.04), 180 (0.2)33S (0.78), 34S (4.4)37Cl (32.5) i.e. 3:1

81Br (98) i.e. 1:1

19F, 127I are mono isotopic.

Major elements that made contribution to (M+. + 2) are S, Cl, Br. M ajor elements that make contribution to (M+. + 1) are C, S, N. If M+. and (M+. + 2) are in 100:32.5 ratio, then one chlorine atom is present. If M+. and (M+. + 2) are in 1:1 ratio, then one bromine atom is present. If M+. and (M+. + 2) are in 100:4.4 ratio, then one sulphur atom is present.

If more than one hetero atom present, relative abundances of If M+. , M+. + 2, M+. + 4…….can be calculated using binomial expression (a+b)n

Where, a = Relative abundance of lighter isotope of hetero atomb = Relative abundance of heavier isotope of hetero atomn = number of such hetero atoms in that molecule.

Example: 1. CH2Cl2:- a = 35Cl b = 37Cl n = 2Relative Natural 3 : 1

abundance( a + b)2 = a2 + 2ab + b2

= 32 + 2*3*1 + 12

= 9 + 6 + 1i.e. 9 : 6 : 1

Example: 2. CH2Br2:- a = 79 Br b = 81 Br n = 2Relative Natural 1 : 1

abundance( a + b)2 = a2 + 2ab + b2

= 12 + 2*1*1 + 12

= 1 + 2 + 1i.e. 1: 2 : 1

Determination of molecular formula using isotopic peak intensities:-Guide lines :-

1. Identify the molecular ion peak and apply nitrogen rule.

2. If the molecular ion peak is not the base peak ( 100 % ) raise its intensity to 100 % and proportionally recalculate M+. + 1 and M+. + 2.

3. M+. + 2 intensity indicates the presence and absence of Cl/Br/S.

4. The number of carbon atoms is determined by dividing M+. + 1 intensity ( after detecting contribution by nitrogen and sulphur) with 13C abundance (1.1 %).

5. Number of hydrogens in the molecule is obtained by detecting the contribution by C, N, S,halogen from the molecular weight of the molecule. If necessary required oxygen's are

added.

Example:- 1

m/z : 27, 28, 29, 39, 41, 43, 44, 72, 73, 74R. A : 59, 15, 54, 23, 60, 79, 100, 73, 3.3, 0.15

(i) M/Z of molecular ion peak is 72 ( even mass)Therefore zero or even nitrogen's are present.

(ii) M+. = 73/73*100 = 100 %M+. + 1 = 3.3/73*100 = 4.5%M+.+ 2 = 0.15/73*100 = 0.205 %M+.+ 2 intensity is only 0.205 % of molecular ion peak intensity.

(iii) Hence S, Cl, Br are absent.(iv) Assuming nitrogen to be absent in the molecule,

Number of carbon atoms = M+.+ 1 intensity/ % of 13C = 4.5/1.1 = 4 carbonsNumber of hydrogen's = Molecular weight – (weight of total carbon atoms + weight of

hetero atoms)= 72 – 4*12 = 72-48 = 24 hydrogen's.

For four carbons maximum permitted hydrogen's are 10 (CnH2n+2)Hence 24 hydrogen's are not permitted.Assume one oxygen atom is present. Hence 24 – 16 = 8 hydrogen's.Therefore molecular formula = C4H8O

Example :- 2

m/z : 90, 91, 92R.A : 100, 5.1, 4.6

( i ) The base peak formed at m/z of 90, then the mass is 90 ( even mass )Therefore even/zero number of nitrogen's are present.

(ii) M+. and M+. + 2 ratio is 100 : 4.6Therefore sulphur atom is present.

(iii) Assuming nitrogen is absent, then the number of carbon atoms = M+. + 1/1.1 = 5.1/1.1 = 4(iv) Total mass of carbon atoms = 4*12 = 48(v) Mass of sulphur = 1*32 = 32

Therefore remaining mass = 90-80 = 10Hence number of hydrogens = 10Molecular formula = C4H10S

Example : 3m/z : 73, 74, 75R. A : 86, 3.2, 0.18

(i) The mass of the M+. = 73 (odd mass )Therefore molecular ion has odd number of nitrogen's

(ii) M+. = 86/86*100 = 100 %M+. + 1 = 3.2/86*100 = 3.7%M+.+ 2 = 0.18/86*100 = 0.209 %

(iii) Ratio of M+. and M+. + 2 peak intensities are 100 : 0.209Hence oxygen atom is present.

(iv) Assuming one nitrogen is present, Number of carbon atoms = M+.+ 1 intensity/ % of 13C = 3.7/1.3 = 3 carbons

(iv) Mass of carbon atoms = 3*12 = 36Mass of oxygen atom = 16Mass of nitrogen atom = 14Remaining mass = 73-66 = 7. Hence number of hydrogen’s = 7.Molecular formula = C3H7NO

Example : 4 m/z : 66, 64, 51, 49, 38, 37, 36, 35, 29, 28, 15R. A : 23, 69, 2 ,8 ,6, 3, 18, 10, 100, 10, 24

M+.+2 , M+. Are in 3:1 ratio. Hence Chlorine atom is present.Remaining mass = 64-35 = 29 ( may be C2H5 or HCO )

Hence molecular formula = CH3CH2Cl.

Example : 5m/z : 110, 108, 95, 93, 81, 79, 29R. A : 50, 50, 3, 3, 44, 44, 100

M+.+2 , M+. Are in 1:1 ratio. Hence Bromine atom is present.Remaining mass = 108 -79 = 29 ( may be C2H5 or HCO )

Hence Molecular formula = C2H5Br

McLafferty Rearrangement :-Elimination of a neutral molecule from a suitably substituted molecular ion or daughter ionThrough a six membered cyclic transition state involving ϒ-hydrogen transfer is known as McLafferty effect.

C

B

R AD

XH

+.

C

B

R A

H

+.

D

X+

(All homolytic cleavage only)

Neutral molecule

+.

H H

+1.

1- pentene m/z = 42

Retro-Diels alder reaction:

It is reverse of Deils – alder reaction. It is characteristic of cyclic unsaturated molecules. It involves cleavage of the two bonds and formation of 2π bonds. Two fragment ions, mainly diene and dienophile.

Example:

+. +

+

+.

+.cyclohexenem/z = 82

m/z = 54 ( more intense )

m/z = 28 ( less intense )

Double allylic cleavage in a cyclic unsaturated compound is known as Retro Diels-Alder rearrangement.

++.

+.1.

m/z = 82 m/z = 54

+.+

+m/z = 28

m/z = 104

m/z = 132

+.+.

2.

+.

O

+

+

O

O m/z = 28

m/z = 106

m/z = 134

+.+.

3.

Examples:

Ortho effect:Elimination of a neutral species (molecule or free radical ) from a suitable orthosubstituted benzene ( or) cis-olefin under EI through a six membered cyclic transition is known as ortho effect.

Example:

+.

OHOH

O+ H2O

m/z = 106m/z = 124

H

OCH3

O

H

H

CO

+ CH3OH

H

OCH3

O

H

m/z = 100 m/z = 68

Trans isomer

cis methyl crotonate

no m/z at 68

Ortho effect may be used to distinguish ( i) o-disubstituted benzene from corresponding m & p – disubstituted.(ii) cis-olefins from trans olefins.

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