Mass spectrometry of Fragmentation extra information and example question

Question 1

In some cases, electrons and their mass is not important, in what circumstance is this important to consider?

Answer 1

High resolution mass spec is when electrons are important (mass of an electron is 1/2000 of a proton and is negligible but can sometimes be important)

Question 2

What do isotopes have a different number of?

Answer 2

Isotopes are different number of neutrons

Question 3

Electrons are hybridised, what does this mean?

Answer 3

They like to go round in pairs

Question 4

Note:Positive charge: M (neutral molecule)  M+. (+ve charge and dot represent loss of an electron as now have more protons than electrons)

Question 5

Elemental composition of amphetamine:

Answer 5

Count number of carbons, count number of hydrogens, count number of nitrogen (9x12=108, 13x1= 13 and 1x14= 14, add them all up is 135) if can find molecular ion could be because its fragmented or made a mistake in the maths- if the molecular weight is odd it is because there is an odd number of nitrogen’s and following the nitrogen rule means that the molecular weight being odd is right

Question 6

Looking at an isotope:

Answer 6

**o Carbon (working at a carbon-12 scale) **
 C12 isotope= 12.000 (6 protons, 6 electrons and 6 neutrons)
 C13 isotope= 13.003355 (6 protons, 6 electrons and 7 neutrons)
 1.1% is the proportion of C13/C12
 The more carbons there are the greater the chance of one of those carbons being C13 i.e., with amphetamine there are 9 carbons and 9x1.1%= 9.9% and this means there is almost a 10% chance that one of those 9 carbons is C13

**o Chlorine **
 Cl35 and Cl37
 Relative abundance is 3:1 of these isotopes
 Get two peaks of 3:1 sizing apart for any compound which contains chlorine

Question 7

What can odd electron species lose?

Answer 7

A radical or a neutral
Radicals can be H., CH3. Or .OH and neutrals can be CO, H2O and HCl

Question 8

What is meant by tandem mass spectrometry?

Answer 8

Tandem mass spectrometry: MS-MS is another term for this, and two mass spectrometers are working in tandem together, ionise molecule, only looking at that one and not another, select precursor ion and then collide with gas to make it fragment. Can adjust this collision by changing energy (most of time don’t want this to occur as after fragmentation don’t want this to collide. Only want molecule to collide with gas, not another molecule, so all done under controlled conditions)

Question 9

What can fragment ions lose?

Answer 9

Radicals or neutral molecules

Question 10

What won’t even electrons fragment lose?

Answer 10

Radicals, else it wont be stable

Question 11

Whats makes a radical more stable?

Answer 11

Larger radicals are more stable than smaller radicals

Question 12

Exam example question

Answer 12

**o Starting with mebendazole **
 186 m/z (100% relative abundance) is the base peak
 187 therefore is the carbon 13 isotope at 10%- this suggests there are 9 carbons
 9x12= 108
 186 m/z of that drug is what of the molecule…
 77m/z is the benzene ring at 52% abundance
 295 m/z is ….
 The molecule has 16 carbons
 The molecule has 13 hydrogens
 The molecular has 3 Nitrogen’s
 The molecule has 3 oxygens
 3 nitrogen’s in molecule is odd m/z
 C16H13N3O3 of the molecule
 C16 192
H13 13
N3 42
O3 48
Total= 295 which is the M+. m/z 295
 m/z 296 is C13
C16 16 x 1.1% x 41 (% relative abundance of 295 m/z)= 7.2% which is similar to the 7% of the 296 m/z of the molecule
 m/z of 263 and 264
295-263= 32
Look for 32 in molecule… OCH3 + H that is nearby makes CH3OH which is a neutral
C16H13N3O3 (odd electron) – CH4O (odd electron?) = [C15H9N3O2].+

m/z 264 is 7% relative abundance
m/z 263 is 40% relative abundance
same principle for molecular ion
look at elemental composition
C15 15 x 1.1 x 40= 6.6% which is close to 7%
 m/z 218 and 219
subtract 218 from molecular ion (always start with molecular ion)
295 – 218 = 77
Benzene is m/z 77
Loss as a radical is the fragment
C16H13N3O3 .+ – C6H5 . = C10H8N3O3 +
218 is at 54%
Working out the isotope at m/z 219 C13 10 x 1.1 x 54 (always multiple by C12 % to work out the C13 %)= 5.94% which is roughly 6% and this is what was found
 m/z 186 and 187
295 – 186= 109 m/z
m/z of 109 can be 109-32= 77 (could be from a fragment, a second loss)
C15H9N3O2 .+ – C6H5 . = C9H4N3O2 + Is 186 m/z
C13 isotope for m/z 187 9 x 1.1 x 100= 9.9% which is roughly 10%
 m/z of 105 and 106
C7H5O+
m/z is 105
C13 7 x 1.1% x 37% = 2.8% which is close to 3% which is the abundance of the 106 m/z
 m/z of 77 and 78
77 m/z is C6H5. At 52%
m/z 78 is C13 6 x 1.1 x 52%= 3.4% which is roughly 3%, however 78 m/z is found at 5%….
the extra % could be from:
78 could be C13 + C5H5+ with C12=5 to be C6H5+ m/z 78
78 could also be C6H6 .+
 the m/z 31 at relative abundance of 14%
CH3O+ of the end of the molecule

o Proguanil
 There are 5 nitrogens and that means theres an odd m/z
 Chlorine isotopes have 3:1 ratio
 The m/z 43 is the C3H7 which is the base peak at 100%
 m/z 195, 196 and 197 is the Cl35, the C13 and the Cl37 respectively
 the patterning helps you know what side of the molecule the cl is on and therefore what part of the molecule the ions are forming from
 or approach as did for other question, start from top and work way through