Mass Spectrometry

Reading EI Spectra:(1)Recognition of M+

(2)Deduction of Formula(3) Fragmentation of M+

Recognition of Molecular Ion

1. Is highest m/z value even?A. If odd then what?

2. Do highest m/z fragments make sense?A. M-15 present? M-18? M-31? M-29?

B. M-3 M-14 are impossible

3. Even if 1 and 2 are true the highest m/z might not be M+

Nitrogen Rule

Spectrum of an Amine

Spectrum of an Alcohol

M+ present?

m/z0

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Scan No : 521 Ret Time : 7:42.81 TIC : 186674 BP = 72568CPEH02.D

500499499498497496496495494494494493493492492491490490490489488488488487486486486485484482482481481480480479478477476475475474474473472472471471470470469469468468467467466466465464464463462461460460459458457456455455454453452452451451451450449449448447447446445445444443443442442441440440439438438437436435435434434433432431431429428427427426426426425425424423422422421421420420419418417416416416415414413413412412411411411410409409408408407407406406405404403403402402401401400399399398398397396395395394393392391391390390389389388388388387387386386385384384383383382381381381381380380379379378377377376375375374373373372371370370369368368367367366366365364364364363363362362361360360359358358357357356356355355353353352351351350350349349348348346346346345344344343342342341341340340339338337337336336335335334334333333332331331331330330329328328327326325324323323322321320319319319318317316316315315314314313312311311310309309308307306305305304303303303301301300299298298298297296295295294293292291290290289288288287286285285284283283282282281281280280279279278278277277276275274273273272272271271271270269267267266265265265265264264263262262261260259258258257256256255254253253252251250250249248247247246246245244243242242241241241240240239238237237236235234233233232232231231230230229229228227226225225224224223223222222220219217216215214214213213212211210210209208207206205204203202201200200199198

1299883

73

69

55

45

41

<

> 40 60 80 100 120 140 160 180

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Scan No : 309 Ret Time : 5:44.52 TIC : 35545140 BP = 83880963-BROMOCYCLOHEXENE GROUP 7 &8.D

500499498498497496495495494494493493493492492491491490490489489488487486485484484482481481480479479478477477477477476475474474473473472472471470470469469468467467466465465464463463463462461460460459459458457457456456455454453452452451451450450449449448448447447446445444444443442442440439439439438437436434432430428426424424423422422421421420420419418417417417416416415415414413413412412411411410409408407406405404403403402401400400399398397396396395395394393392392391390390389388386385384384381381380380379378377377376376375375374373373372372371370369368367366365364364363362362361360360359358357357355354353352351349348347346345345344344343343343342341340340339339338338338337336335335334333333332332331330330329328328327326325325324323322322321321320319318317316314312311310310309309308307306305304303302301300299299298297296295294294293293292291290289288287286284283282280280279278277277276275275275274274273272271271270269268267266265264263262261260260259258258256256255255254253252251250249248247247247246245244244243242241241240239238236235234233232231230230229228227226225224224223222221220219218217216215214213212212211210209208207206205204203202201200199198

79

7465

535041

<

> 40 60 80 100 120 140 160 180

Deducing molecular formula• Use %m+1 to get # of Cs• Use Nitrogen rule• M-18 peak indicates oxygen• Doublet M+ peaks indicate Cl (3:1) and Br

(1:1)• Peak at 127 = I• Peak at 77 is phenyl group (C6H5)• M+2 peak = 4-5% indicates S• Deduce molecular formula!

Molecular Formula?

Molecular Formula?

Molecular Formula?

m/z %abund

86 20.2

87 1.2

Index of Hydrogen Deficiency

• Units of Unsaturation = r + db• CnHmXxNyOz

r + db = n – m/2 – x/2 + y/2 + 1

Fragmentation of M+

• Draw structure of M+ if possible and then draw possible heterolytic cleavages (curved arrows) or homolytic cleavages (fish hooks)

• Alternatively, show homolytic or heterolytic cleavage cleavage of the molecule and then remove an electron from one of the fragments.

• Fragmentation to give neutral molecule and a radical cation must involve some kind of rearrangement.

General Rules

1. Increase branching decrease M+

2. Add carbons decrease M+

3. Cleave more likely at higher º C

4. Unsubst cyclic have strong M+ (especially aromatic)

5. Alkenes cleave to allylic species

6. R on cycloalkanes α cleave. Cyclohexenes do retro Diels-Alder.

7. R on aromatic rings β-cleave to benzylic

8. Bonds β to heteroatoms cleave

9. Loss of small stable molecules, H2O, NH3, C2H4, HCN, CO, H2S,etc often occur

Probablity of bond cleavage depends on stability of fragments produced. (3º > 2º > 1º > CH3) (oxonium > carbocation) (allylic, benzylic > no resonance)

Alkanes

• Straight chain alkanes – homologous series of ions – smooth curve max at 57 or 71. M-15 absent. “picket fence”

• Discontinuity = branching. Absence of homolog no single cleavage possible

• Cycloalkanes. Larger M+. More even mass fragments (M-C2H4)

Alkenes

• Cleave next to double bond to form allylic fragments. (Double bonds can shift in M+.)

• Homologous series is -2 (27, 41, 55, 69..)• Cyclohexenes do retro Diels Alder – give even

mass fragment (M-C2H4, M-C3H6 etc.)

Aromatics

• No branching = strong M+. Often see M++.• m/z 77, 51 = phenyl (monosubst benzene)• Most prominent peaks are due to beta

cleavage of branches to benzylic (really tropylium.)