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Smt. K.R.P.Kanya Mahavidyalaya, Islampur

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Presentation on theme: "Smt. K.R.P.Kanya Mahavidyalaya, Islampur"— Presentation transcript:

1 Smt. K.R.P.Kanya Mahavidyalaya, Islampur
Mass Spectrometry Dr.S.R.Mane M.Sc. B.Ed.Ph.D. Associate Prof.& Head, Dept. of Chemistry, Smt. K.R.P.Kanya Mahavidyalaya, Islampur

2 Chemical Identification of Compound
Comparison of Physical Properties Boiling Point Melting Point Density Optical rotation Appearance Odor Elemental Analysis Burn the compound and measure the amounts of CO2, H2O and other components that are produced to determine the empirical formula Molecular formula

3 Introduction Modern techniques for structure determination of organic compounds include: Mass spectrometry Size and formula of the compound Infrared spectroscopy Functional groups present in the compound Ultraviolet spectroscopy Conjugated p electron system present in the compound Nuclear magnetic resonance spectroscopy Carbon-hydrogen framework of the compound

4 Introduction MS is different than optical spectroscopy(UV-Vis, IR and NMR) as it does not measures the interaction of molecule with light or electromagnetic radiation But since the record obtained resembles very much like optical lines in the spectrum so the name mass spectroscopy is given

5 Theorotical Background
In Mass spectrometer the neutral organic molecules(M) in vapor phase are bombarded with high energy electrons. Due to this, energy of molecule increases and this excitation energy spread through out the molecular structure. This energy is sufficient to knock out an electron of lowest ionization potential from the molecule and the molecular ion (M•+) is formed

6 Organic Molecule 8 to 15 eV Molecular ion / Radical cation/ Parent ion
M (g) e M• e- Organic Molecule 8 to 15 eV Molecular ion / Radical cation/ Parent ion If energy of bombarding electron is increased (50 to 80 eV), the parent ion formed will have excess energy spread through out its molecular structure. But within short time this excess energy collect at a particular bond. Due to this, breaking of bonds at that point takes placeto produce a fragment ions m m2. Cation Radical M•+ Molecular ion m m2 Radical Cation Neutral Molecule

7 Theorotical Background
e.g. Formation of molecular ion from Methane Molecular ion (M+.) m/z = 16

8 Formation of Molecular ion from Ethane
m/z = 15

9 What’s in a Mass Spectrum
Mass-to-charge ratios of a molecule or its fragment are graphed or tabulated according to their relative abundance Fragment Ions Fragment Ions: derived from molecular ion or higher weight fragments

10 Mass Spectrometry Magnetic-Sector Instruments
Electron-impact, magnetic-sector instrument

11 Background Molecular ion (parent ion):
The radical cation corresponding to the mass of the original molecule The molecular ion is usually the highest mass in the spectrum Some exceptions w/specific isotopes Some molecular ion peaks are absent.

12 Background Mass spectrum of ethanol (MW = 46) M+
SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/1/09)

13 Background The cations that are formed are separated by magnetic deflection.

14 Background Only cations are detected. Radicals are “invisible” in MS.
The amount of deflection observed depends on the mass to charge ratio (m/z). Most cations formed have a charge of +1 so the amount of deflection observed is usually dependent on the mass of the ion.

15 Background The resulting mass spectrum is a graph of the mass of each cation vs. its relative abundance. The peaks are assigned an abundance as a percentage of the base peak. the most intense peak in the spectrum The base peak is not necessarily the same as the parent ion peak.

16 The mass spectrum of ethanol
base peak M+ SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/1/09)

17 Background Most elements occur naturally as a mixture of isotopes.
The presence of significant amounts of heavier isotopes leads to small peaks that have masses that are higher than the parent ion peak. M+1 = a peak that is one mass unit higher than M+ M+2 = a peak that is two mass units higher than M+

18 Easily Recognized Elements in MS
Nitrogen: Odd number of N = odd MW SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/2/09)

19 Easily Recognized Elements in MS
Bromine: M+ ~ M+2 (50.5% 79Br/49.5% 81Br) 2-bromopropane M+ ~ M+2 SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/1/09)

20 Easily Recognized Elements in MS
Chlorine: M+2 is ~ 1/3 as large as M+ M+ M+2 SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/2/09)

21 Easily Recognized Elements in MS
Sulfur: M+2 larger than usual (4% of M+) M+ Unusually large M+2 SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/1/09)

22 Easily Recognized Elements in MS
Iodine I+ at 127 Large gap M+ Large gap I+ SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/2/09)

23 Fragmentation process
There are 3 type of fragmentations: Cleavage of s bond + . ---- C + ---- C – C ---- + . C ---- At heteroatom + . ---- C + ---- C – Z ---- + . Z ---- a to heteroatom + . C=Z + ---- C - C – Z ---- + ---- C . + . Z + . ---- C - C – Z ---- + ---- C = C

24 Fragmentation process
Cleavage of 2 s bond (rearrangements) + . ---- C=C + ---- HC – C – Z ---- + HZ Retro Diels-alder + + . McLafferty + . + .

25 Fragmentation process
There are 3 type of fragmentations: Cleavage of Complex rearrangements

26 Fragmentation rules in MS
Intensity of M.+ is Larger for linear chain than for branched compound Intensity of M.+ decrease with Increasing M.W. (fatty acid is an exception) Cleavage is favored at branching  reflecting the Increased stability of the ion Stability order: CH3+ < R-CH2+ < R R CH+ < C+ R R R R’ R CH R” Loss of Largest Subst. Is most favored

27 Illustration of first 3 rules (large MW)

28 Branched alkanes M.+ is absent with heavy branching
MW=170 M.+ is absent with heavy branching Fragmentation occur at branching: largest fragment loss

29 Illustration of first 3 rules (Linear alkane with Smaller MW)
Molecular ion is stronger than in previous sample

30 Illustration of first 3 rules (Branched alkane with Smaller MW)
43 Molecular ion smaller than linear alkane Cleavage at branching is favored

31 Cleavage Favored at branching
Rule 3 Alkanes Cleavage Favored at branching Loss of Largest substituent Favored Rule1: intensity of M.+ is smaller with branching

32 Fragmentation rules in MS
Aromatic Rings, Double bond, Cyclic structures stabilize M.+ Double bond favor Allylic Cleavage  Resonance – Stabilized Cation

33 Aromatic ring has stable M.+

34 Cycloalkane ring has stable M.+

35 Fragmentation rules in MS
a) Saturated Rings lose a Alkyl Chain (case of branching) Unsaturated Rings  Retro-Diels-Alder + . + -R. + + .

36 Retro Diels-alder + . + .

37 Fragmentation rules in MS
Aromatic Compounds Cleave in b  Resonance Stabilized Tropylium -R. Tropylium ion m/z 91

38 Tropylium ion

39 Fragmentation rules in MS
C-C Next to Heteroatom cleave leaving the charge on the Heteroatom - [RCH2] - [R2] larger

40 Fragmentation rules in MS
Cleavage of small neutral molecules (CO2, CO, olefins, H2O ….) Result often from rearrangement McLafferty - CH2=CH2 Ion Stabilized by resonance Y  H, R, OH, NR2

41 Alkenes CH2 CH Et Me -Et +CH2 CH Et Me CH2 CH CH + Et Me -29
Most intense peaks are often: m/z 41, 55, 69 Rule 4: Double Bond Stabilize M+ Rule 5: Double Bond favor Allylic cleavage CH2 CH + Et Me -Et +CH2 CH Et Me CH2 CH CH + Et Me -29 M+ = 112 m/z = 83

42 Alkenes

43 Aromatic compound

44 Alcohols

45 Hydroxy compounds - R3 If R1=H m/z 45, 59, 73 …
Loss of largest group If R1=alkyl m/z 59, 73, 87 … – (H2O) M – (H2O) - H2O - CHR=CHR M – (H2O) – (C1=C2) Alkene

46 Phenol

47

48 Aromatic Ether - R - CH2=CH2 Molecular ion is prominent -CO 1) C5H5
m/z 65 m/z 93 Cleavage in b of aromatic ring 2) - CH2=CH2 Rearrangement m/z 94

49 Aliphatic Ether B CH3 —CH2 —O—CH2 —CH2 —CH2 —CH3 CH3 —CH2 —O+ =CH2
Cleavage of C-C next ot Oxygen Loss of biggest fragment • + CH3 —CH2 —O—CH2 —CH2 —CH2 —CH3 CH3 —CH2 —O+ =CH2 m/z 59 CH3 —CH2 —O —CH2+

50 Ether Rearrangement CH =O+ —CH2 CH3 H— CH3 CH3 —CH2—CH —O —CH2 —CH3
m/z 45 B m/z 73 1- Cleavage of C-C next to Oxygen CH =O+ —CH2 CH3 H— CH3 Box rearr. CH3 —CH2—CH —O —CH2 —CH3 CH3 M·+ m/z 73 CH =O+ H CH3 2- Cleavage of C-O bond: charge on alkyl m/z 45 Index MS MS-fragmentation-2

51 Fragmentation Patterns
The impact of the stream of high energy electrons often breaks the molecule into fragments, commonly a cation and a radical. Bonds break to give the most stable cation. Stability of the radical is less important.

52 Fragmentation Patterns
Alkanes Fragmentation often splits off simple alkyl groups: Loss of methyl M+ - 15 Loss of ethyl M+ - 29 Loss of propyl M+ - 43 Loss of butyl M+ - 57 Branched alkanes tend to fragment forming the most stable carbocations.

53 Fragmentation Patterns
Mass spectrum of 2-methylpentane

54 Fragmentation Patterns
Alkenes: Fragmentation typically forms resonance stabilized allylic carbocations

55 Fragmentation Patterns
Aromatics: Fragment at the benzylic carbon, forming a resonance stabilized benzylic carbocation (which rearranges to the tropylium ion) M+

56 Fragmentation Patterns
Aromatics may also have a peak at m/z = 77 for the benzene ring. 77 M+ = 123

57 Fragmentation Patterns
Alcohols Fragment easily resulting in very small or missing parent ion peak May lose hydroxyl radical or water M or M+ - 18 Commonly 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+

58 Fragmentation Patterns
MS for 1-propanol M+-18 M+ SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09)

59 Fragmentation Patterns
Amines Odd M+ (assuming an odd number of nitrogens are present) a-cleavage dominates forming an iminium ion

60 Fragmentation Patterns

61 Fragmentation Patterns
Ethers a-cleavage forming oxonium ion Loss of alkyl group forming oxonium ion Loss of alkyl group forming a carbocation

62 Fragmentation Patterns
MS of diethylether (CH3CH2OCH2CH3)

63 Fragmentation Patterns
Aldehydes (RCHO) Fragmentation may form acylium ion Common fragments: M+ - 1 for M for

64 Fragmentation Patterns
MS for hydrocinnamaldehyde 91 M+ = 134 105 SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09)

65 Fragmentation Patterns
Ketones Fragmentation leads to formation of acylium ion: Loss of R forming Loss of R’ forming

66 Fragmentation Patterns
MS for 2-pentanone M+ SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09)

67 Fragmentation Patterns
Esters (RCO2R’) Common fragmentation patterns include: Loss of OR’ peak at M+ - OR’ Loss of R’ peak at M+ - R’

68 Frgamentation Patterns
105 77 M+ = 136 SDBSWeb : (National Institute of Advanced Industrial Science and Technology, 11/28/09)

69 Rule of Thirteen The “Rule of Thirteen” can be used to identify possible molecular formulas for an unknown hydrocarbon, CnHm. Step 1: n = M+/13 (integer only, use remainder in step 2) Step 2: m = n + remainder from step 1

70 Rule of Thirteen Example: The formula for a hydrocarbon with M+ =106 can be found: Step 1: n = 106/13 = 8 (R = 2) Step 2: m = = 10 Formula: C8H10

71 Rule of Thirteen If a heteroatom is present,
Subtract the mass of each heteroatom from the MW Calculate the formula for the corresponding hydrocarbon Add the heteroatoms to the formula

72 Rule of Thirteen Example: A compound with a molecular ion peak at m/z = 102 has a strong peak at 1739 cm-1 in its IR spectrum. Determine its molecular formula.

73 GC-Mass Spec: Experiment 23
Mass Spec can be combined with gas chromatography to analyze mixtures of compounds. GC separates the components of the mixture. Each component is analyzed by the Mass Spectrometer.

74 GC-Mass Spec: Experiment 23
Assignment: Observe the GC-mass spec experiment Record experimental conditions Analyze the mass spectrum of each component of your mixture: Parent ion peak? Heteroatoms apparent from spectrum? A minimum of 1 or two significant fragments and their structures

75 GC-Mass Spec: Experiment 23
Assignment (cont.): Using the Mass Spec data, retention times, and boiling points, identify the components of your mixture. Write three paragraphs (one per compound) summarizing and interpreting all data. See your data sheet for more details.


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