Mass Spectrometry 1

Question 1

Step 4: Detection

Answer 1

• Sensitivity / minimum sample size for analysis: – g routinely
– pg (10-12 g!) and sometimes even better!
• Mainly using electron multiplier:
– 1 ion 105–107 electrons depends on incoming ion… (this is easier to measure than just one ion)
– Detect current

Question 2

Step 4: Detection - ion entering e- multiplier

Answer 2

hits the wall and makes electrons which make more and more

Question 3

Quantification in MS

Answer 3

Quantification not inherent
• Some molecules form ions easier than others
– As you know from solution chemistry!
• Detection ion-dependent

Solution:
• Use internal standard, e.g. same molecule, but isotope labelled (different mass!), at known concentration
• Use relative intensities for quantification
• Method used for doping control (e.g. London 2012)

Question 4

Step 5: Data analysis

Answer 4

Pharmacy (organic chemistry): desired / contaminant
• Confirmation of presence/absence 
• Identify unknowns
– Molecular weight determination 
– Molecular formula determination 
– Structural analysis
– Isotopic incorporation

Also:
• Elemental analysis (major, minor, trace)
• Environmental monitoring (fast, specific, sensitive)
• Time-resolved (fast, specific, sensitive)

Question 5

Data analysis steps

Answer 5

• Molecular ion region 
– Low mass resolution (you can see unit mass, eg difference between 100 and 99) 
– Simple rules (N-rule) 
– Isotopes
• Cl, Br, S
• C (approximate # C)
– (High mass resolution)

• Fragmentation pattern – Fingerprint
– Interpretation

Question 6

Start with the Molecular Ion

Answer 6

• The highest mass peak (excl. noise and isotopes) is hopefully the ionized un-fragmented molecule, the Molecular ion, M+. (or M-).
• The biggest peak (most abundant ion), the Base Peak
– Set to 100%. All other fragments calculated as percentage relative to the base peak
– Base peaks highly characteristic of compounds / classes of compounds

Question 7

base peaks forms..

Answer 7

something stable

Question 8

Molecular Ion (+/-) - advantages

Answer 8

• Knowledge of M+/- or quasi-molecular ion (e.g. [M+H]+, [M+NH4]+) eliminates potential molecules
• Isotope distribution reveals (amount of) Cl, Br, and … C!
• High mass resolution:
  • possibly determine number of C, H, O, N, etc.
  • Not always possible at higher masses as too many possible combinations (biomolecules so not much precision due to the many different combinations)

Question 9

Molecular Ion (+/-) - disadvantages

Answer 9

-Isomers have same masses…
- Other ionisation techniques better at generating more
stable (quasi-) molecular ion:
EI &laquo_space;CI < ESI, MALDI

Question 10

Finding the Molecular Ion

Answer 10

• Start simple: 1 pure compound only
– May require chromatography – Tony’s sessions / HPLC lab
• Where is it?
– Look for highest mass peaks…
• BUT not always detected
– e.g. many molecules with -OH, -COO-, -COOH groups
• A real M+● must be
1. the highest mass excl. isotope satellites
2. for EI: an odd-electron ion (M + e- → M+● + 2e-) 3. capable of forming the high mass fragments

Question 11

Is an ion a potential molecular ion?

Answer 11

• Is it an odd-electron (OE+.) ion?
• The saturation index
= nr of rings + nr of double bonds = R + DB
• ForCxHyNzOn:
R + DB = x – 1⁄2 y + 1⁄2 z + 1
• Note that nr. of O does not matter (why?) - anything that makes two bonds doesn’t count in the formula
• Interchangeable (why?):
– Si, C
– P, N
– S, O (not included)
– F, Cl, Br, H

Question 12

What adds to the z value

Answer 12

Anything that makes 3 bonds adds to the z value

Question 13

The saturation index

Answer 13

• Even electron (EE) ions never have whole number for saturation index

-RULE: Every M+● is an OE ion
– So if it is not an OE ion, it cannot be a mol. ion
– Not all OE ions are molecular ions!
– Rearrangements involving multiple bond breaking can also yield OE ions

Question 14

The Nitrogen Rule

Answer 14

Is M+● even or odd mass?
• Even: nr of N = 0, 2, 4, …
• Odd: nr of N = 1, 3, …
• Why? - if you get an even number then that means you have an even number of N however, if you have an odd number then you’ll have an odd number of N

Note: this only works for molecular ions – So first do previous confirmation

Question 15

Natural Isotopic Abundances

Answer 15

m/z differs for (naturally occurring) (stable) isotopes:
• C: 12C & 13C (& 14C…) – 98.9% 12C
– 1.1% 13C
• “A+2” elements
- Br : 79 (main mass), 81 (2nd mass), 79.9 (avg)
- Cl : 35 (main mass), 37 (2nd mass), 35.5 (avg)

Question 16

Natural Isotopic Abundances 2

Answer 16

• No important isotopes: H,F,P,I
• Natural abundance:
– 35/37Cl: doublet: 75.77 / 24.23 % -> ~ 3:1
– 79/81Br: doublet: 50.69 / 49.31 % -> ~ 1:1

– 12/13C: “doublet”: 98.90 / 1.10 %
– 14/15N: doublet: 99.63% / 0.37%

– 32/33/34/36S: quadruplet: 95.02 / 0.75 / 4.21 / 0.02 %
ignoring 36: triplet ≈ 125 : 1 : 6
– 16/17/18O: triplet: 99.762 / 0.038 / 0.200%
ignoring 17: doublet ≈ 500 : 1

Question 17

Isotopes

Answer 17

• Mass / charge detected differs for isotopes
• Natural abundance:
– 35/37Cl: 75.77 / 24.23%
– 79/81Br: 50.69 / 49.31%
– 12/13C: 98.90 / 1.10%
– 16/17/18O: 99.762 / 0.038 / 0.200%
32/33/34/36S: 95.02 / 0.75 / 4.21 / 0.02%

Question 18

“A+2” Elements

Answer 18

• Check for isotope distribution M+2:
– Determine ratio of peaks
– Check for A+4 etc!
– Compare expected for “A+2” elements (1, 2, …)
– Also check if fragments contain this element
– Tricky if more than a single “A+2” element…

Question 19

Use 12C and 13C to determine the Molecular Formula

Answer 19

• A+1: 12C & 13C
– Determine ratio of peaks (A+1) / A
– Divide by 0.011 to indicate nr. of C
(Because 13C = 1.1.% of total C)

Example: peaks at 16 & 17
• Measure heights: 100% & 1.6%
• Calculate ratio of (M+1)/M: 1.6 / 100 = 0.016
• Divide by 0.011 : 1.4
• At least 1 C and at most 2 C (impossible here!)

Question 20

Use 12C and 13C to determine the Molecular Formula

Example 2: butane

Answer 20

Example 2: butane
• peaks at 58 & 59m/z
• Intensity ratio M+1/M:0.042
• 0.042/0.011=3.8
• 3 ≤ nr of C ≤ 5
• 5 C = 60 so not possible → must be 3 or 4 C

Note: in principle should consider 15N
• but odd/even rule normally used.
• 15N only 0.37% so difficult to detect

Question 21

Isotopic labelling

Answer 21

• NMR: D, 13C
• MS: D, 13C, 15N, 18O

Why?
• Aid identification
• Study processes – Reaction
– Metabolic pathway
– Bio-medical applications:
• Cellular level: Does molecule (e.g. drug) enter cell? How? Where does it partition to? What happens to it?
• Organism level – similar to above

Question 22

General Procedures for Interpretation

Answer 22

• Often several techniques are required to determine an unknown (IR, NMR, UV/VIS, pyrolysis, …)
• Consider all data incl. everything that may be present and happen in a reaction (contaminants, side-products
Mass spectrum:
• Does spectrum look complex or not?
• Bonds broken easily / stable ions

Question 23

General Procedures for Interpretation - steps

Answer 23

1. General inspection
2. Look at patterns identifying “A + 2” elements
3. Identify highest m/z (take isotopes into account!)
   Odd / even → nr. of N
4. Use above & find all potential molecular formulae
5. Check if formulae can explain OE+.
6. (Identify other significant OE)
   (Note: if mol. ion even, fragment OE also even)
7. (Calculate R + DB for significant OE fragments)
8. Derive structure using all information incl. tables

Question 24

EI: General Guidelines

Molecular ion abundances - table

Answer 24

Strong : Medium (Conjugated alkenes) :
Weak / absent

Aromatic HC (ArH)
Ar~Br
Long-chain aliphatic compounds

ArF
Ar~I
Branched alkanes

ArCl
ArCO~R
Tert. Aliphatic Alcohols

ArCN
ArCH2~R
Tert. Aliph. Bromides

ArNH2
ArCH2~Cl
Tert. Aliph. Iodides

~ = weak bond