chm 424l organic laboratory, dr. laurie s. starkey introduction...

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CHM 424L Organic Laboratory, Dr. Laurie S. Starkey Introduction to Mass Spectrometry Mass spectrometry is used to determine a sample's molecular mass and molecular formula. Some structural information can also be determined by mass spec. This technique is especially valuable when used in conjuction with gas chromatography (called GC-MS); the GC separates the components of a mixture and then the mass spec analyzes each component. To obtain a mass spectrum, a sample is vaporized and then bombarded with a high energy beam of electrons (this technique is called electron-impact ionization, or EI). This ionizes the sample by knocking an electron out of the structure. A nonbonded or bonded electron is removed, resulting in radical cation species. Although it is now missing an electron, the intact molecule still has the same molecular mass and is known as the molecular ion (M + ), or the parent ion. In addition, this high-energy environment causes the molecular ion to fragment and the various pieces can be analyzed to learn something about the original structure. All charged species are separated by mass as they are passed through a magnetic field and a detector records how many pieces of each mass have been formed. Note that because this technique does not involve the absorption or emission of energy, it is not called spectroscopy. sample of a neutral molecule is vaporized blast with electron beam to ionize e - e - e - loss of an electron results in a radical cation (no change in mass) + + + fragmentation occurs + + mixture of charged and uncharged particles magnetic deflection separates charged particles by mass-to-charge ratio, m/z (charge is usually +1) + + lower mass higher mass (M + ) detector records the frequency of each mass; the taller the peak, the more abundant the fragment base peak (most abundant) defined as 100% to set scale molecular ion, M + (~highest m/z) gives molecular mass (in amu) spectra have been reproduced from the NIST Chemistry WebBook http://webbook.nist.gov/chemistry + + Spectral Analysis In this spectrum, the sample has a molecular mass (M) of 100 amu. The most abundant fragment, called the base peak, has a m/z of 43 amu, or (M–57). This fragment was formed when a 57 amu segment of the molecule broke off. Many other fragments were formed; not all will be analyzed. note that a small peak is observed at (M+1)

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CHM 424L Organic Laboratory, Dr. Laurie S. StarkeyIntroduction to Mass Spectrometry

Mass spectrometry is used to determine a sample's molecular mass and molecular formula. Somestructural information can also be determined by mass spec. This technique is especially valuable when usedin conjuction with gas chromatography (called GC-MS); the GC separates the components of a mixture andthen the mass spec analyzes each component. To obtain a mass spectrum, a sample is vaporized and thenbombarded with a high energy beam of electrons (this technique is called electron-impact ionization, or EI).This ionizes the sample by knocking an electron out of the structure. A nonbonded or bonded electron isremoved, resulting in radical cation species. Although it is now missing an electron, the intact molecule stillhas the same molecular mass and is known as the molecular ion (M+), or the parent ion. In addition, thishigh-energy environment causes the molecular ion to fragment and the various pieces can be analyzed to learn something about the original structure. All charged species are separated by mass as they are passed througha magnetic field and a detector records how many pieces of each mass have been formed. Note that becausethis technique does not involve the absorption or emission of energy, it is not called spectroscopy.

sample of a neutral molecule

is vaporized

blast with electron beam

to ionize

e-e-

e-

loss of an electron results in a radical cation

(no change in mass)

+

+

+

fragmentationoccurs +

+

mixture of charged and uncharged

particles

magnetic deflection separates charged

particles by mass-to-charge

ratio, m/z(charge is

usually +1)

++

lower mass higher mass(M+)

detector records the frequency of each mass; the taller the peak,

the more abundant the fragment

base peak(most abundant)defined as 100%

to set scalemolecular ion, M+

(~highest m/z)gives molecular mass (in amu)

spectra have been reproduced from the NIST Chemistry WebBook http://webbook.nist.gov/chemistry

+

+

Spectral AnalysisIn this spectrum, the

sample has a molecular mass (M) of 100 amu.

The most abundant fragment, called the base

peak, has a m/z of 43 amu, or (M–57). This fragment was formed

when a 57 amu segment of the molecule broke

off. Many other fragments were formed; not all will be analyzed.

note that a small peak is observed at (M+1)

What is the mass of a single molecule?

CH4 Mass = C + 4(H) =

but ~1.1% of carbon atoms exist as 13C isotope!

13CH4 Mass = 13C + 4(H) =

What ratio is expected for the molecular ion peaks of C2H6?

the number of carbon atoms affects the relative height of the M+1 peak

Other isotopes of high abundance

Cl (35.45 amu) atoms are 76% 35Cl and 24% 37Cl

Br (79.90 amu) atoms are 51% 79Br and 49% 81Br

What ratios are expected for the molecular ion peaks of CH2Br2?

possible masses? CH2 Br Br total amu

fragments containing Cl or Br have significant M+2 peaks

mass spec of chloroethane (M+ 64) mass spec of bromobutane (M+ 137) mass spec of dibromomethane (M+ 172)

Determining molecular formula from high-resolution mass spectrometry (high-res mass spec)exact masses of various elements:

1H 1.007825 amu12C 12.000000 amu14N 14.003050 amu16O 15.994914 amu

these molecules all have a molecular mass of 98 g/mol

C4H6N2O 98.0480 amuC5H6O2 98.0368 amuC5H10N2 98.0845 amuC6H10O 98.0732 amuC7H14 98.1096 amu

the molecular formula can be determined from precise molecular mass

Fragmentation of the molecular ion

Which is more stable, a carbocation C+ or a radical R ?

CHH

HCH

H

Hvs.

fragmentation is more likely if it gives more stable carbocations and radicals

CH3 CH2 CH2 CH2 CH2

Alkanes

CH3

hexane

remove1 e-

CH3 CH2 CH2 CH2 CH2 CH3

a radical cation (M+) m/z 86

Mass Spec Features of Various Functional Groups

CH3 CH2 CH2 CH2 CH2 CH3

fragment

break a bond to give two fragments: one cation and one radical

CH3 CH2 CH2 CH2 CH2 CH3

CH3 CH2 CH2 CH2 CH2 CH3

m/z 57 (M-29)

m/z 29 (M–57)

CH3 CH2 CH2 CH2 CH2 CH3

fragmentCH3 CH2 CH2 CH2 CH2 CH3

m/z 43 (M–43)

note: hexane has peaks at m/z 15, 29, 43, 57, 71, 86

CH3 CH2 CH2 CH2 CH2 CH3

fragmentCH3

m/z 71 (M-15)

m/z 15 (M–71)

CH3 CH2 CH2 CH2 CH2

CH3CH3 CH2 CH2 CH2 CH2

Why are the peaks at m/z 15 and 71 so small?

Branched Alkanes

CH3 CH2 CH2 CH2 CH CH3

CH3

Explain why the base peak of 2-methylhexane is at m/z 43 (M-57).

CH3 CH2 CH2 CH CH

Alkenes

CH3

remove1 e-

CH3 CH2 CH2 CH CH CH3

CH3 CH2 CH2 CH CH CH3

high-energy pi electron is most likely removed

CH3 CH2 CH2 CH CH CH3

allylic cleavage gives stable carbocation

CH2 CH CH CH3fragment

allylic carbocation is resonance-stabilized

Aromatic

CH3 CH2 CH2

tropylium cationm/z 91

CH3 CH2

Alcohols

CH3 CH2 CH2 OH α cleavage

CH3 CH2 CH2 OH

high-energy nonbonded electron is most likely removed

CH2 OH

also, loss of H2O (M–18) may be observed (multistep mechanism first involves an intramolecular proton transfer)Note: loss of molecular piece results in retention of EVEN mass number

CH3 CH CH2

OHH

Ethers

R CH2 CH2 Oα cleavage

also, loss of RO may be observed

CH2 CH2 OR -OR cleavage

CH2 CH2 R'

CH2 O CH2 CH2 R'

R CH2 CH2 O CH2

CH3 CH2 CH2

Aminesα cleavageCH3 CH2 CH2 NH2

Nitrogen Rule: M+ is odd if there are an odd number of nitrogens!odd molecular mass indicates one nitrogen (or 3 or 5 or 7 nitrogens...)even molecular mass indicates zero nitrogens (or 2 or 4 or 6 nitrogens...)

Aldehydes & Ketones

R C R'O α cleavage R C R'

O

acylium ion is resonance-stabilized

m/z (M–R)

loss of larger group more likely(more stable R )

R CCH2

OCH2

CHR'H

McLafferty rearrangement is a common fragmentation for molecules containing pi bonds (e.g., C=O)Note: loss of molecular piece results in retention of EVEN mass number

favorable 6-membered ring in transition state

McLafferty rearrangement

R CCH2

O

CH2

CH R'H

molecular ion cleaves between α and β carbons

(note transfer of H for calc.)

Esters

R C OR'Oesters can undergo α cleavage, McLafferty

rearrangement, or loss of RO

Mass Spectrometry Discussion Questions

m/z (M–OR)

Note: seemingly illogical fragmentations are common in mass spectrometry. Most amines have a peak at m/z 30 from the very stable cation shown. N

H

?!CH2 NH2

Mass Spectrometry Discussion Questions

For the given molecule (M=57), do you expect the more abundant peak to be m/z 15 or m/z 43? Explain.

CH3 CH2 CH2 CH3

For the given molecule (M=74), which peak do you expect to be most abundant: m/z 31, m/z 45 or m/z 59? Explain.

HO CH2 CH2 CH2 CH3

Explain why the mass spectra of methyl ketones typically have a peak at m/z 43. Provide the structure of this fragment.

How could you use mass spectrometry to distinguish between the following two compounds (M=73)? Provide structures (and m/z values) for the significant fragments expected.

CH3 CH2 CH2 CH2 NH2 CH3 CH2 CH2 NH CH3and

What would be the m/z ratio for the fragment resulting from a McLafferty Rearrangement for the following molecule (M=114)? What fragment accounts for its base peak at m/z 57?

CH3 CH CH2 C CH2

O

CH3

O CH2 CH2 CH2 CH3CH3

In the mass spectrum of the given molecule (M=88), account for the peaks at m/z 45 and m/z 57.

Mass spec problems (Wade 7th or 8th ed.) Chapter 12: 8-11, 18, 20.

CH3

1. Explain the indicated MS peaks for methyl butyrate - provide structures (and m/z values) of the fragments and show how they were formed.

Mass Spectrometry Problems

CH3O

O

CH2CH2CH3

2. How could you use mass spectrometry to distinguish between the following pairs of isomers? Provide structures (and m/z values) for the significant fragments expected.

3. (PBtext) From the mass spec and NMR data given, determine the number of Cl and/or Br atoms present, the chemical formula, the degrees of unsaturation (DU) and the number of rings in the structure.

m/z = 43

m/z = 59

m/z = 71

m/z = 74

M+ 102

O O

M+ 287 100%

M+1 288 12.2%

M+2 289 97.3%

M+3 290 11.9%

1H NMR shows14 H atoms

13C NMR shows 3 π bonds

m/z abundance