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Year 2 Chem > Mass Spectrometry > Flashcards

Mass Spectrometry Flashcards

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1
Q

How does Electron ionisation work?

A
  • Gas phase molecules are converted to ions by bombardment with a beam of electrons
  • An electron passing close to the molecule extracts an electron (an electron pair departs, overall the molecule has lost one electron)
  • Molecule is converted to a cation with an unpaired electron (radical cation)
  • The molecule gains additional energy from the electron beam
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2
Q

What is the molecular ion?

A
  • The peak with the highest m/z, ignoring isotopes

- The molecular ion is always a radical cation

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3
Q

How do you tell whether the peak with the highest m/z is a molecular ion or an isotope?

A

There is usually a cluster of fragments at the highest m/z value. The molecular ion will have a much higher intensity than the isotopes around it with similar m/z, the isotope peaks are very small.

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4
Q

What does the molecular ion tell you about the molecule?

A
  • The m/z of the molecular ion is the same as the molecular mass of the molecule
  • The intensity of the molecular ion tells us how stable the molecule is
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5
Q

What do fragment ions tell you?

A
  • The mass of the ion that it has formed

- How easily the ion is formed, if the intensity is large then the ion forms more easily

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6
Q

What happens when the energy supplied equals the ionisation energy of a molecule?

A

M•+ is formed when ionising energy populates vibrational states within the energy well. Ions of various internal energies are formed. Bond length increases on ionisation

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7
Q

How does fragmentation occur?

A

During ionisation, transfer of energy in excess of the enthalpy of the highest vibrational state leads to bond rupture, this produces fragment ions

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8
Q

Which bonds are most susceptible to break in fragmentation?

A

The weakest bonds in the molecule

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9
Q

What is the appearance energy?

A
  • The energy that fragment ions start to be produced at
  • At around 70eV, the abundance of molecular ions start to decrease as the energy is too high and molecular ions are dissociating faster than they’re produced
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10
Q

What energy are mass spectra typically recorded at?

A

Approx. 35eV, this gives enough energy to produce both fragment ions and molecular ions

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11
Q

How are even electron ions produced?

A

A proton or a sodium ion is added to the molecule and an adduct ion is formed, [M + H]+ or [M + Na]+

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12
Q

Why is EI mass spec not suitable for all molecules?

A
  • Many molecules show absent or weak molecular ions in their EI spectra as some functional groups cause spontaneous fragmentation
  • The solution to this is to use soft ionisation or protect the groups that cause spontaneous fragmentation, this ensures that molecular ions show up
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13
Q

What are the examples of soft ionisation techniques?

A

ESI, MALDI, CI and APCI

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14
Q

What are the characteristics of ESI and MALDI?

A
  • Produces ions from molecules that are not suited to EI (involatile or unstable)
  • Generates ions that have low amounts of residual energy
  • This leads to little fragmentation and prominent molecular adduct ions eg. [M + H]+ or [M + Na]+
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15
Q

What are the disadvantages of using ESI/MALDI?

A

There is a loss of information on sub structures as this comes from fragmentation

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16
Q

Can soft ionisation be used for molecules with weak bonds?

A

Because of the low energy transfer which results in formation of low energy ions, molecules with weak bonds can be ionised and still produce molecular ions

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17
Q

How does Chemical Ionisation (CI) work?

A
  • A reagent gas is ionised in the chemical ionisation source, this produces an ionising agent
  • The ionising agent reacts with the analyte to form an even electron ion
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18
Q

What are the different reagent gases that can be used in chemical ionisation?

A

CH4, NH3, 2-methylpropane (isobutane)

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19
Q

Why does proton transfer occur in CI?

A

Proton transfer occurs if the proton affinity of the analyte is higher than the proton affinity of the reagent gas

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20
Q

When is ΔE largest in CI?

A
  • When the difference in proton affinity is largest, there is a larger energy transfer and more chance of fragmentation of the analyte
  • When there is a small difference between proton transfer there is a greater chance of a molecular adduct ion
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21
Q

Is proton transfer in CI exothermic?

A

Proton transfer is exothermic, when energy is transferred to the analyte, the overall system loses energy because the analyte is more stable attached to a proton than when the reagent gas has a proton attached

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22
Q

How does ACPI work?

A
  • Analyte introduced into the source in solution at atmospheric pressure
  • Warm N2 gas converts solvent into a mist, the heater vaporises both the solvent and analyte
  • The corona discharge needle, held at a potential of around 5kV emits electrons which ionise gas molecules
  • Secondary reactions produce reagent ions (protonated water and water clusters)
  • Reagent ions transfer proton to analytes having higher PA
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23
Q

List the series of reactions that produce the reagent ions in ACPI

A
  • N2 → N2•+ → N4•+
  • N4•+ + H2O → H2O•+ + 2N2
  • H2O•+ + H2O → H3O+ + OH•
  • H3O+ + H2O → (H2O)nH+
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24
Q

What are the disadvantages of ACPI?

A

Soft ionisation meaning that fragmentation isn’t possible, we can only determine the mass of the molecular ion

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25
Q

Why is CI used over EI?

A

CI can be used to determine the mass of the molecular ion of an unstable compound that would instantly fragment in EI meaning there is no molecular ion peak. A carrier gas that has a similar proton affinity to the analyte has to be used to avoid too much fragmentation and ensure the molecular ion peak still appears.

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26
Q

What is the mono-isotopic mass of a molecule? (Sometimes called exact mass)

A

The molecule in the form where each atom is in its most table isotope (isotope with the lowest mass). This means the mono-isotopic mass of the molecule is the peak in the molecular ion cluster with the lowest mass

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27
Q

What are isotopologues?

A

Molecules that differ only in the isotopic composition of one or more of their atoms

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28
Q

What is the nominal mass of a molecule?

A

Value calculated from the most abundant isotope of each element rounded to the nearest integer

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29
Q

What is the mass defect

A

Difference between the nominal mass and monoisotopic mass

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30
Q

How is resolution calculated?

A
  • R = (m/z) / (Δm/z)
  • m/z is the mass of the peak in profile mode
  • Δm/z is the width of the peak at 50% of the height of the peak
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31
Q

What is unit resolution?

A

The resolution at which Δm/z is approx. 1

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32
Q

What happens to Δm/z at high resolutions?

A

It is much smaller as the analyser can pick up much smaller mass differences meaning that isobars (molecules with the same nominal mass but different exact mass) can be distinguished

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33
Q

What is an isobar?

A

Atomic or molecular species with the same nominal mass but different exact mass

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34
Q

What do isotope profiles tell us?

A
  • Isotope clusters recorded in low resolution reflect the presence and number of particular elements
  • Isotopologues give peaks at integer m/z value separations
  • Peak intensities reflect the natural relative abundances of the isotopes in the molecule
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35
Q

What are the notations for elements with isotopes?

A
  • A = only one isotope so one possible mass value
  • A + 1 = Two isotopes so two possible mass values, differ by 1 Da
  • A + 2 = At least two isotopes, at least one difference of 2 Da
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36
Q

What is the ratio of peaks for Cl2?

A
  • 9:6:1 for 70 72 and 74 respectively

- 70 has two 35Cl, 72 has one 35Cl and one 37Cl, 74 has two 37Cl

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37
Q

What is the ratio of peaks for a molecule with 5 Chlorines?

A

4:6:4:3

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38
Q

What is the rule for working out how many A+2 elements in structure?

A

The number of peaks separated by 2u, -1 tells you the number of A+2 elements in the structure

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39
Q

What is the ratio of peaks for a molecule with 3 Bromines?

A

1:4:4:1

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40
Q

What is the ratio of peaks for a molecule with 2 Bromines?

A

1:2:1

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41
Q

How can the statistical distribution of isotopologues be used to estimate molecular formulae for carbon?

A
  • One additional isotope of 13C has a natural abundance of 1.1%
  • For a molecule with 3 carbons, there is a 1.1% chance of there being an isotope at any of the 3 carbons
  • This makes the overall probability of a 13C appearing in the molecule 3 x 1.1% = 3.3%
  • If we calculate the probability of 2 carbons being replaced by 13C in the molecule, we do (1.1%)^2 x No of carbon atoms. However, this number is negligible for molecules with a small number of carbons
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42
Q

When calculating the number of carbons in a molecule, what do the M peak and M+1 peak need to be scaled to?

A
  • The M peak needs to be normalised to 100%, this is done by (M+1/M) x 100
  • This gives you a value that if you divide by 1.1 tells you the number of carbons in the molecule
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43
Q

What error estimate do we apply to the answer of the abundance of carbons in a molecule?

A
  • +-10%
  • If your answer is 8.82 take +-10% = 7.94 to 9.70
  • We calculated there to be 8 carbons, 8.8 falls in that range, whereas 7.7 and 9.9 fall outside the range meaning there must be 8 carbons in the molecule
44
Q

When determining the number of carbons in of a molecule that has an A+2 element, what correction do you need to make before calculating?

A
  • The M+1 peak will have contribution from the A+2 elements A+1 form if it is abundant enough
  • For example 32S has 2 isotopes 33S and 34S, the 33S will contribute to the M+1 peak in this case
  • Its abundance its 0.79 so you take this away from the abundance of the M+1 peak, this ensures the peak only has carbon characteristics and you can accurately calculate the number of carbons in the molecule
45
Q

What is the equation for degrees of unsaturation?

A
  • Degrees of unsaturation = (No of Tetravalent atoms) - 0.5(No of Monovalent atoms) + 0.5(No of Trivalent atoms) + 1
  • If you have the structure of the molecule you can just add up the number of rings and double bonds to get the same answer
46
Q

What is the nitrogen rule?

A
  • Compounds with odd numbers of Nitrogen atoms give M•+ at odd m/z
  • Compounds with even numbers of Nitrogen atoms give M•+ at even m/z
47
Q

Which electron is removed when forming a molecular ion?

A
  • The electron with the lowest ionisation energy is removed

- Electrons in non-bonding orbitals have the lowest ionisation energy followed by π-orbitals followed by σ-orbitals

48
Q

Which ions can OE ions give rise to?

A

EE ions by cleavage of one bond and OE ions by cleavage of two bonds (by rearrangement or ring cleavage)

49
Q

Which ions can EE ions give rise to?

A

EE ions typically only give rise only to new EE ions (separation of electron pair to form two radicals is unfavourable)

50
Q

For reactions proceeding with charge retention which product ion is favoured?

A

The product ion with lower IE is favoured

51
Q

Why does charge migration occur in certain cleavages?

A

When the products are more stable when the charge migrates

52
Q

How do you determine the enthalpy of reaction of a simple bond cleavage?

A

The heats of the formation of the products which is equal to the bonds cleaved in the neutral counterpart of the reactant + the IE of the ion product of the reaction

53
Q

What is Stevenson’s rule?

A

Where cleavage of a single bond generates more than one ion and radical pair, the fragment with the higher ionisation energy retains the unpaired electron, the ion is generated from the other fragment. (Both pathways can occur but Stephenson’s rule tells us which is favoured

54
Q

What is an alkane ion series?

A

Series of EE ions differing by 14 units, formula of molecules in series: CnH2n+1

55
Q

How does the cleavage work in an ion series?

A

Primary fragmentation occurs on the chain to produce a cation, secondary fragmentation then occurs so that a stable ethene molecule is lost. Secondary fragmentation occurs at a difference of 28 so the ion series has gaps of 28. However, because the primary fragmentation can occur at any carbon it means that the difference of 28 can occur in between the original difference of 28, this means there is representation at 14 units apart

56
Q

Why are the ions at C3 to C6 more abundant in the ion series of alkanes?

A

The ions are more stable as secondary fragmentation is less favourable meaning the ions are more abundant

57
Q

When fragmentation produces ions of similar stability, what is the determining factor for which product is favoured?

A

The loss of the largest alkyl radical is favoured if the ions produced have similar stability, this can lead to a preference for the least stable ion because it has lost the largest alkyl radical

58
Q

Which compound class is the exception from the 14 Da difference between ions in its ion series?

A

Aromatics, the pattern of ions is: 38, 39, 50-52, 63-65, 75-78

59
Q

What is the main factor influencing ion abundance?

A

Product ion stability:

  • Tertiary cation more stable than primary, therefore tertiary cations more abundant
  • If the ion can form resonance structures, this also makes it more abundant as its more stable
60
Q

What are the most favoured products from cleavage of a single bond?

A
  • Hydrocarbons: Allyl, benzyl and tropylium ions
  • O-containing: Oxonium ions, acylium ions
  • N-containing: Immonium ions and imminium ions
61
Q

When is the stability of ions negligible?

A

When the cations are all of the same order eg. All secondary, this is when the loss of the largest alkyl determines abundance

62
Q

Which neutral molecules can be lost from EE ions only?

A

H2O, NH3 and HCl

63
Q

What is Fields rule?

A

EE decompositions forming the same product ion eject the neutral molecule with the lowest proton affinity

64
Q

What determines the site of ionisation in a fragmentation reaction?

A
  • The electron is most easily removed from the site having the lowest ionisation energy
  • The site of ionisation directs fragmentation by either homolytic or heterolytic cleavages
65
Q

What is the difference between homolytic and heterolytic cleavage?

A
  • Homolytic cleavage is a one electron process, a radical is lost
  • Heterolytic cleavage is a two electron process in which a negative ion is lost
66
Q

How can we tell which mechanism is favoured for fragmentation of a molecule?

A
  • We can tell if a molecule favours fragmentation by i-cleavage (charge site initiation) or α-cleavage (radical site initiation) by the abundance of the ion formed by each process
  • If the ion formed by α-cleavage has a higher abundance then we know that the molecule favours fragmentation by radical site initiation
  • If the ion formed by i-cleavage has higher abundance then we know the molecule favours fragmentation by charge site initiation
67
Q

How does radical site initiation work?

A
  • An odd electron is donated to form a new bond to adjacent α-atom, this results in cleavage of a different bond to the α-atom
  • RCR2Y•+ → R• + CR2=Y+
68
Q

What is the α-position in an odd electron ion?

A

The carbon adjacent to the radical

69
Q

What is the order of tendency for alpha cleavage?

A
  • N > S, O, π, R• > Cl > Br > H
  • Tendency for radical site initiation follows the tendency of the radical site to donate electrons
  • The tendency is affected by the environment in the molecule, eg. inductive effects and resonance effects
70
Q

What determines which bond to the alpha carbon is cleaved in an aliphatic alcohol?

A

Stability is negligible so the loss of the largest radical is favoured

71
Q

What do the intensities of ions formed by alpha cleavage reflect?

A
  • The favourability of the different alpha cleavages, if there are many alpha bonds that that can be cleaved then the intensity of the ion will be larger
72
Q

Can alpha cleavage contribute to the ion series of a molecule?

A

Yes, ions produced by alpha cleavage correspond to members of the ion series

73
Q

When there is an option between the largest alkyl being lost and a smaller alkyl being lost but there are two bonds that can be cleaved to form the same ion, which ion is likely to have a higher abundance?

A

The ion that can be formed by cleaving two different bonds, the intensity of this ion is very large as there is a high chance of forming it because you can cleave 2 possible bonds

74
Q

How does charge site initiation (i-cleavage) work?

A
  • The positive charge in the molecule attracts an electron pair from a bond adjacent to the positive charge
  • Can occur with both OE and EE ions
75
Q

Which is more favourable out of alpha cleavage and i-cleavage?

A

Generally, alpha cleavage is more favourable as there is no charge site migration

76
Q

How does i-cleavage of R2C=Y•+ occur?

A
  • An electron pair from the double bond moves onto the Y•+ to form R2C+—Y•
  • An electron pair from the bond between one of the R groups and the C moves onto the C to form RC—Y•
  • The free electron on the Y forms a bond with one of the electrons from the electron pair on the carbon, this forms R—C•=Y
77
Q

What does the tendency to form R+ from RY depend on?

A
  • How electronegative Y is, the more electronegative the element, the more likely i-cleavage is to take place
  • The more electronegative Y is, the less stable it is to have a positive charge on the Y, charge site migration becomes more important as Y becomes more electronegative
  • Hal > O, S&raquo_space; N, C
78
Q

For OE ions with oxygen as the heteroatom, which mechanism of fragmentation is favoured?

A
  • Supports both alpha cleavage and i-cleavage
  • Different bonds are broken by each mechanism
  • We can see which mechanism is favoured by the abundance of the ions
79
Q

For OE ions with C-Y bonds, which heteroatoms favour inductive cleavage?

A
  • Heteroatoms that form polarised bonds with C, eg. the characteristic of the bond is already partially ionic making it more favourable for heterolytic cleavage
  • NO2, I, Br, Cl
80
Q

What does alpha cleavage favour when forming a cation?

A

Loss of the largest radical

81
Q

What does inductive cleavage favour when forming a cation?

A

Formation of the most stable cation

82
Q

How do you determine which part of the molecule retains the charge in the charge site initiation of an EE ion?

A

Fields rule states that the structure with the highest proton affinity will retain the charge

83
Q

How does further fragmentation of EE ions occur?

A

To be energetically favourable, further fragmentation often requires charge remote fragmentation (cleavage several bonds away from charge site) or rearrangement

84
Q

How do cycloalkanes decompose to form ions for mass spectrometry?

A
  • One bond in the structure is cleaved, this causes no change in mass
  • A distonic radical is formed
  • Alpha cleavage produces an OE ion and eliminates a neutral molecule
85
Q

What is a distonic radical?

A

A molecule where there is a distance between the radical and the positive charge

86
Q

How is a methyl radical lost from a cycloalkane?

A

A rearrangement of the molecule occurs prior to alpha cleavage, a methyl radical can then be eliminated from the molecule

87
Q

What is the Diels-Alder reaction?

A
  • A diene is added to an alkene to form a a cycloalkene

- Reaction occurs through a cyclic 6-membered transition state

88
Q

What is the Retro-Diels-Alder reaction?

A
  • The reverse of the Diels-Alder reaction, a cyclocalkene is broken down into a diene and an alkene
  • There are two pathways in which the cycloalkene can be broken down, α-cleavage and i-cleavage
  • α-cleavage forms an ionised diene whereas i-cleavage forms an ionised alkene
  • The product with the lowest ionisation energy is the ionised diene, this means that α-cleavage is favoured
89
Q

How can charge stabilisation influence preferred fragmentation mechanism

A
  • If the charge is conjugated with a double bond or a benzene ring for example, the ionisation energy of the product ion is much lower
  • Lower ionisation energy means that product is more favourable and therefore the mechanism to produce that ion is favoured
90
Q

If α-cleavage and i-cleavage are competing, what determines which fragmentation pathway is favoured?

A

The mechanism that forms the cation with the lowest ionisation energy is favoured

91
Q

Why do rearrangements of atoms or groups in ionised molecules usually not tell us anything about the structure of the molecule?

A

Because ionisation occurs in many different places in the molecule such as in σ- and sometimes in π- ionisation, this means that rearrangements are extensive and you can’t determine anything about the structure as it is happening in many sites

92
Q

When are rearrangements beneficial to determining the structure of the molecule?

A

When the rearrangements are specific, they have considerable diagnostic value

93
Q

How are rearrangements initiated?

A

By charge site or radical site

94
Q

What molecule is often responsible for a peak at m/z 58?

A

A methyl ketone

95
Q

How does the McLafferty rearrangement work?

A
  • A γ-H rearrangement to an unsaturated group by beta cleavage occurs
  • Formation of a 6-membered transition state occurs and then either inductive cleavage or alpha cleavage occurs depending on which one produces the more stable ion
  • If the R group can stabilise the charge then inductive cleavage will be favoured, if not then alpha cleavage is favoured
96
Q

How long does an alkyl chain have to be for a McLafferty rearrangement to be able to occur?

A

When the alkyl chain is 3 or more atoms long

97
Q

Can radical sites on a saturated heteroatom accept H in rearrangement?

A

Yes, the radical forms a bond with one electron from the original bond that the H was involved in, the other electron goes onto the carbon that the H was connected to

98
Q

What are rearrangements with displacement (rd)?

A

Formation of a bond between the radical site and a distant atom leads to cyclisation with displacement of the radical

99
Q

What are charge site rearrangements?

A
  • Stabilisation by resonance can lead to charge site rearrangement being favourable
  • If the ion produced by charge site rearrangement has resonance, then it is favourable
  • The charge site remains unchanged but the ion is a fragment of the original molecule
100
Q

What is the isotope method?

A

Double bonds can be identified by adding deuterium to your analyte, 2D will be added over each double bond and 3D will be added over each triple bond

101
Q

Other than the isotope method, which other method is used to identify multiple bonds positions?

A

React the double bond I2/dimethylsulphide to form 1,2 dithiane

102
Q

How does tandem mass spec work?

A
  • Specialised mass spectrometers can perform sequential stages of mass spectrometry
  • The first and last quadrupoles perform mass analysis, the central quadrupole acts as an ion guide
  • Q2 is a collision cell where collisions generate smaller ions
103
Q

When scanning for a product ion in tandem mass spec, what should you set up each quadrupole to do?

A
  • Q1, fix on single m/z value
  • Q2, CID (collision induced dissociation)
  • Q3, scan m/z range
104
Q

When scanning for a neutral loss in tandem mass spec, what should you set up each quadrupole to do?

A
  • Q1, scan m/z range
  • Q2, CID (collision induced dissociation)
  • Q3, offset scan m/z range (offset corresponds to neutral)
105
Q

When scanning for a precursor ion in tandem mass spec, what should you set up each quadrupole to do?

A
  • Q1, scan m/z range
  • Q2, CID (collision induced dissociation)
  • Q3, fix on single m/z value

Year 2 Chem (5 decks)

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