Alkanes undergo extensive fragmentationAlkanes undergo extensive fragmentation

m/zm/z

DecaneDecane

142142

4343

5757

7171

8585

9999

CHCH33—CH—CH22—CH—CH22—CH—CH22—CH—CH22—CH—CH22—CH—CH22—CH—CH22—CH—CH22—CH—CH33

Relative Relative intensityintensity

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– More stable carbocations will be more abundant.

CH3CH2CH2CHCH3

CH3

m/z 86

+•

CH3CH2CH2CHCH3

CH3

m/z 86

+•

Facile

Facile

More difficult

CH3CH2CH2CHCH3

CH3

+• +

m/z 43

CH3 CH3CH2CH2CH

CH3

CH3CH2CH2CH

CH3

m/z 71

+• +

+• +

m/z 57

CH3CH2 CH2CHCH3

CH3

+CH3 m/e = 15

CH3

|+C - CH3 m/e = 57 | CH3

CH3 - CH2 +m/e = 29

CH3 |

CH3 - CH2 -- C -- CH3 m/e = 71 +

CH3

|+C - H m/e = 43 | CH3

– MS of 2,2,4-trimethylpentane

• Branched alkanes have small or absent M+

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Relative Relative intensityintensity

120120

9191 CHCH22—CH—CH22CHCH33

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+ +91

Propylbenzene fragments mostlyat the benzylic positionPropylbenzene fragments mostlyat the benzylic position

CycloalkanesCycloalkanes

– Loss of side chain

– Loss of ethylene fragments

.Ionization followed by fragmentation

Splitting out of ethylene.

Ejection of electron

Radical/cation

fragmentation

– MS of methylcyclopentane:MS of methylcyclopentane:

Alkene FragmentationAlkene Fragmentation– Fairly prominent M+

– Fragment ions of CnH2n+ and CnH2n-1+

– Terminal alkenes lose allyl cation if possible to form resonance-stabilized allylic cations

R - R

[CH2=CHCH2CH2CH3] CH2=CHCH2+ + •CH2CH3

+•

CycloalkenesCycloalkenes

– Prominent molecular ion

– Retro Diels-Alder cleavage

• Typically you see both.• More stable cation will predominate• Also works for hetero-substituted (e.g. make enol)• Both EI (shown) and in CI. (protonated molecular ion, cleave, then reprotonation)

+

+•

+•

+

+•

Observed!

Observed!

– Cyclohexenes give a 1,3-diene and an alkene, a process that is the reverse of a Diels-Alder reaction.

A radical cation (m/z 68)

+•

•+

+

CH3

CH3C CH2

CCH2H3C

CH3

Limonene(m/z 136)

A neutral diene(m/z 68)

4-terpineol(MW 154)

OH

OHO

mz 86+

+

mz 68

EI Mass Spectrum

+

Mass spectrum of 4-terpineol as a good example for Retro Diels Alder fragmentation

Alkyne FragmentationAlkyne Fragmentation– Molecular ion readily visible

– Terminal alkynes readily lose hydrogen atom

– Terminal alkynes lose propargyl cation if possible(resonance-stabilized propargyl cation or a substituted propargyl cation).

HC C=CH2HC C-CH2+ +3-Propynyl cation

(Propargyl cation)

Aromatic Hydrocarbon FragmentationAromatic Hydrocarbon Fragmentation– Molecular ion usually strong– Alkylbenzenes cleave at benzylic carbon tropylium ion formation

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m/zm/z

m/zm/z = 78= 78

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Relative Relative intensityintensity

– Some molecules undergo very little Some molecules undergo very little fragmentationfragmentation– Benzene is an example. Benzene is an example.

The major peak corresponds to the molecular ion.The major peak corresponds to the molecular ion.

+

+

+

-H-CO

-H2

m/z 108m/z 107

m/z 79[C6H7]

+ m/z 77[C 6 H5 ]+

CH2OH OH

HHH

H

H

H

H

+

7 memberd ring

Benzonium ionPhenyl ion

m/z 91

+ ++

H

H - CH3

CH3

CH3

CH3

Benzene Compounds

Tropylium ion

Tropylium ion

Mass Spectrum of n-Octylbenzene

Br

Bromine pattern

Tropylium ion

Mass Spectrum of benzyl bromide

•

Molecular ion(a radical cation)

A radical

••

• ++O HC R

••

••

+R'-C O H

A resonance-stabilized oxonium ion

R

R"

R'

R"R'-C=O-H

R"

+

– One of the most common fragmentation patterns of alcohols is loss of H2O to give a peak which corresponds to M-18 and no M peak.

– Molecular ion strength depends on substitutionprimary alcohol weak M+

secondary alcohol VERY weak M+

tertiary alcohol M+ usually absent

– Another common pattern is loss of an alkyl group from the carbon bearing the OH (-Cleavage) to give a resonance-stabilized oxonium ion and an alkyl radical (largest R group lost as radical).

Alcohol FragmentationAlcohol Fragmentation

Elimination of water.

74 – 56 = 18 (water).Elimination of propyl radical.

74 – 31 = 43 (C3H7)

MS of 1-butanol is an example demonstrating both MS of 1-butanol is an example demonstrating both processes.processes.

Mass Spectrum of 3-methyl-1-butanol

Carbonyl Compounds Carbonyl Compounds FragmentationFragmentation

Dominant fragmentation pathways:

•-cleavage

•-cleavage•McLafferty rearrangement

m/z 128

+•-cleavage

m/z 43+

•

+

+

m/z 113CH3

O

O

•

O

+

Aldehydes and KetonesAldehydes and Ketones

Characteristic fragmentation patterns are:

– Cleavage of a bond to the carbonyl group (-cleavage).

C O

R

R'

C OR'

C OR'

+ R

Acylium ion

Note that an cleavage of an aldehyde could produce a peak at M – 1 by eliminating H atom.

This is useful in distinguishing between aldehydes and ketones.

•+

m/z 58

McLaffertyrearrangement

Molecular ionm/z 114

•+O

HO

H+

– McLafferty rearrangement.

Splitting out an alkene (neutral molecule) and producing a new radical/cation.

Mass Spec of 2-octanone displays both cleavage and McLafferty

CH3CO+

resulting from cleavage.

CH3CH2CH2CH2CH2CH2CO+ resulting from cleavage.

Carboxylic AcidsCarboxylic Acids

Characteristic fragmentation patterns are:

–-cleavage to give the ion [CO2H]+ of m/z 45.

– McLafferty rearrangement if H present.

-cleavage• +

m/z 45

O=C-O-HOH

O

Molecular ionm/z 88

+

•+•+

+

McLaffertyrearrangement

m/z 60

OH

OH

OH

OHMolecular ion

m/z 88

– Loss of water, especially in CI– Loss of 44 is the loss of CO2

Fatty Acids

+

+

•+•+ m/z 71

m/z 59

Molecular ionm/z 102

-cleavage

OCH3

OCH3

O

O

OCH3

O+

m/z 74

McLaffertyrearrangement +

+•+•

O

OCH3

H OH

OCH3

Molecular ionm/z 102

EstersEsters

-cleavage and McLafferty rearrangement:

Ether FragmentationEther Fragmentation– -cleavage– C-O cleavage (requires stable cation to lose)– Ion rearrangement

Ms of 2-Chloroethylphenyl etherMs of 2-Chloroethylphenyl ether

Inte

nsi

ty

20

40

60

80

100

20 40 60 80 100 120 140

94

156107

160

94

107

O

Cl H

m/z 30

++•

-cleavageCH3 CH3

CH3-CH-CH2-CH2-NH2 CH3-CH-CH2 CH2=NH2

AminesAmines–The most characteristic fragmentation pattern of 1°, 2°, and 3° aliphatic amines is -cleavage

Halide FragmentationHalide Fragmentation– Loss of halogen atom– Elimination of HX– -cleavage

– Fortunately there is an [M+2]+ of 4% for the natural abundance of 34S. This is diagnostic for S vs 2x16O

– Aliphatic thiols can split out H2S, [M-34]

– -cleavage at carbon bearing the sulfur in thiols, thioethers, similar to ethers, etc.

Sulfur CompoundsSulfur Compounds

S+

S+R

•

-R•

N+

O•

OO

+

Loss of •N=O

Loss of CO

m/z = 93(this can form from lots of different origins)

CH+Aromatic!

m/z=65

Good test for aryloxy

NitroaromaticsNitroaromatics

– Spaced by unit mass

– Each peak is for the same molecular formula

– Different peaks because there are some molecules with 13C, 2H etc.

– Especially significant for Cl, Br

m/z

Clusters of IonsClusters of Ions

The Nominal mass is m/z of the lowest member of the cluster. This is the isotopomer that has all the C’s as 12C, all protons as 1H, all N’s as 14N, etc.

– Mass spectrometry “sees” all the isotopomers as distinct ions

– An ion with all 12C is one mass unit different from an ion with one 13C and the rest 12C

– Since the isotope distribution in nature is known for all the elements (13C is 1.1%), the anticipated range and ratios of ions for a given formula can be predicted and calculated

Analyzing Ion Clusters: Analyzing Ion Clusters: a way to rule candidate structuresa way to rule candidate structures

HH

HH HH

HHHH

HH

HH

HH HH

HHHH

HH

HH

HH HH

HHHH

HH

all H are all H are 11H and all H and all C are C are 1212CC

one C isone C is 1313CC one H isone H is 22HH

7878 7979 7979

93.4%93.4% 6.5%6.5% 0.1%0.1%

Isotopic ClustersIsotopic Clusters

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m/zm/z

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Relative Relative intensityintensity

112112

114114

visible in peaks visible in peaks for molecular ionfor molecular ion

3535ClCl 3737ClCl Isotopic Clusters in ChlorobenzeneIsotopic Clusters in Chlorobenzene

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m/zm/z

Relative Relative

intensityintensity

7777

HH

HH

HH

HH

HH ++

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Isotopic Clusters in ChlorobenzeneIsotopic Clusters in Chlorobenzene

no no mm//zz 77, 79 pair; therefore ion 77, 79 pair; therefore ion responsible for responsible for mm//zz 77 peak 77 peak does not contain Cldoes not contain Cl

81Br

Isotopic AbundanceIsotopic Abundance

– Isotopes: present in their usual abundance.– Hydrocarbons contain 1.1% C-13, so there

will be a small M+1 peak.– If Br is present, M+2 is equal to M+.– If Cl is present, M+2 is one-third of M+.– If iodine is present, peak at 127, large gap.– If N is present, M+ will be an odd number.– If S is present, M+2 will be 4% of M+.

Molecules with HeteroatomsMolecules with Heteroatoms

Molecular FormulaMolecular Formulaas aas a

Clue to StructureClue to Structure

– One of the first pieces of One of the first pieces of information we try to obtain when information we try to obtain when determining a molecular structure determining a molecular structure is the molecular formula.is the molecular formula.

– We can gain some information We can gain some information about molecular formula from the about molecular formula from the molecular weight. molecular weight.

– Mass spectrometry makes it Mass spectrometry makes it relatively easy to determine relatively easy to determine molecular weights.molecular weights.

Molecular WeightsMolecular Weights

CHCH33(CH(CH22))55CHCH33

HeptaneHeptane

CHCH33COCO

OO Cyclopropyl acetateCyclopropyl acetate

Molecular formulaMolecular formula

Molecular weightMolecular weight

CC77HH1616 CC55HH88OO22

100100 100100

Exact massExact mass 100.1253100.1253 100.0524100.0524

– Mass spectrometry can measure exact masses.– Therefore, mass spectrometry can be used to

distinguish between molecular formulas.

Exact Molecular WeightsExact Molecular Weights

– Molecules containing atoms limited to C,H,O,N,S,X,P of even-numbered molecular weight contain either NO nitrogen or an even number of N.

– This is true as well for radicals as well.

– Not true for pre-charged, e.g. quats, (rule inverts) or radical cations.

– In the case of Chemical Ionization, where [M+H]+ is observed, need to subtract 1, then apply nitrogen rule.

– Example, if we know a compound is free of nitrogen and gives an ion at m/z=201, then that peak cannot be the molecular ion.

The “Nitrogen Rule”The “Nitrogen Rule”

– A molecule with an A molecule with an odd number of odd number of nitrogens has an odd nitrogens has an odd molecular weight.molecular weight.

– A molecule that A molecule that contains only C, H, contains only C, H, and O or which has an and O or which has an even number of even number of nitrogens has an even nitrogens has an even molecular weight.molecular weight.

NNHH22 9393

138138

NNHH22OO22NN

183183

NNHH22OO22NN

NNOO22

The Nitrogen RuleThe Nitrogen Rule

– Relates molecular formulas to multiple bonds Relates molecular formulas to multiple bonds and ringsand rings

– For a molecular formula, CFor a molecular formula, CccHHhhNNnnOOooXXxx, the degree of unsaturation can be calculated by:

index of hydrogen deficiencyindex of hydrogen deficiency = ½ (2c + 2 - h - x + n)

Index of Hydrogen Index of Hydrogen Deficiency Deficiency

Degree of UnsaturationDegree of Unsaturation

– Knowing that the molecular formula of a substance is Knowing that the molecular formula of a substance is CC77HH1616 tells us immediately that is an alkane because it tells us immediately that is an alkane because it corresponds to Ccorresponds to CnnHH22nn+2+2

– CC77HH1414 lacks two hydrogens of an alkane, therefore lacks two hydrogens of an alkane, therefore contains either a ring or a double bond contains either a ring or a double bond

Molecular FormulasMolecular Formulas

Index of Hydrogen DeficiencyIndex of Hydrogen Deficiency

index of hydrogen deficiency = index of hydrogen deficiency =

11

22((molecular formula of alkanemolecular formula of alkane – – molecular formula of molecular formula of

compoundcompound))

Index of hydrogen deficiency Index of hydrogen deficiency

1122

((molecular formula of alkanemolecular formula of alkane – – molecular formula of molecular formula of

compoundcompound))

C7H14C7H14

1122

(C(C77HH1616 – C – C77HH1414))

==

==

1122

(2) = 1(2) = 1==

Therefore, Therefore, one ring or one double bond.one ring or one double bond.

Example 1Example 1

C7H12C7H12

1122

(C(C77HH1616 – C – C77HH1212))==

1122

(4) = 2(4) = 2==

ThereforeTherefore, two rings, one triple bond,, two rings, one triple bond,two double bonds, or one double bond + one ring.two double bonds, or one double bond + one ring.

Example 2Example 2

CHCH33(CH(CH22))55CHCH22OH (1-heptanol, COH (1-heptanol, C77HH1616O) has same O) has same number of H atoms as heptanenumber of H atoms as heptane

Index of hydrogen deficiency = Index of hydrogen deficiency =

1122

((CC77HH1616 –– CC77HH1616OO)) = 0= 0

No rings or double bondsNo rings or double bonds

Oxygen has no effectOxygen has no effect

Index of hydrogen deficiency = Index of hydrogen deficiency =

1122

((CC55HH1212 –– CC55HH88OO22)) = 2= 2

One ring plus one double bondOne ring plus one double bond

CHCH33COCO

OO Cyclopropyl acetateCyclopropyl acetate

Treat a halogen as if it were hydrogen.Treat a halogen as if it were hydrogen.

CC CC

CHCH33

ClClHH

HH

CC33HH55ClCl

same index of hydrogensame index of hydrogendeficiency as for Cdeficiency as for C33HH66

If halogen is presentIf halogen is present

– Index of hydrogen deficiency tells us the sum Index of hydrogen deficiency tells us the sum ofofrings plus multiple bonds.rings plus multiple bonds.

– Using catalytic hydrogenation, the number ofUsing catalytic hydrogenation, the number ofmultiple bonds can be determined.multiple bonds can be determined.

Rings versus Multiple BondsRings versus Multiple Bonds

Select a candidate peak for the molecular ion (M+)

Examine spectrum for peak clusters of characteristic isotopic patterns

Test (M+) peak candidate by searching for other peaks correspond to reasonable losses

Look for characteristic low-mass fragment ions

Compare spectrum to reference spectra

Interpretation of Mass Interpretation of Mass SpectraSpectra

Isotopic Abundances

• Atomic weights are not integers (except 12C)– 14N = 14.0031 amu; 11B = 11.0093 amu; 1H = 1.0078 amu– 16O = 15.9949 amu; 19F = 18.9984 amu; 56Fe = 55.9349 amu

• Sum of the mass defects depends on composition– Hydrogen increases mass defect, oxygen decreases it

• Accurate mass measurements narrow down the possible formulae for a particular molecular weight– 301 entries (150 formulae) in NIST’02 with nominal MW = 321– 4 compounds within 0.0016 Da (5 ppm) of 321.1000.

• Mass spectrum and user info complete the picture– Isotope distributions indicate/eliminate elements (e.g. Cl, Br, Cu)– User-supplied info eliminates others (e.g. no F, Co, I in reaction)

• Isomers are not distinguished in this analysis

Formula Matching BasicsFormula Matching Basics