CH9080
UNIVERSITY OF WARWICK
MASTER OF SCIENCE EXAMINATIONS:
MARCH 2010
MASS SPECTROMETRY
Time allowed 2 hours
Read carefully the instructions on the answer book and make sure that the particulars
required are entered on each answer book.
Percentages in square brackets are intende, as a guide to the time candidates should
spend in answering the corresponding part of the question.
ANSWER ANY TWO QUESTIONS
1
Question 1
[50%, 15% part a, 35% part b]
a. (15 pts) What is the structure associated with this positive ion electron impact mass
spectrum?
Notes: isotope pattern indicates chlorine. 35 is just Cl+. 47 has one
chlorine, so it must be CCl+. 82 has two chlorines, must be CCl2+. 117 has
three chlorines, must be CCl3+. 118 intensity indicates 1 carbon. 117
cannot be the molecular ion – it’s an even electron ion. So, it’s probably
carbon tetrachloride. It cannot be HCCl3 – which would have a molecular
ion at 118. It cannot be BrCCl3 (or any other halogen substitution), which
would generate a mixed spectrum.
Answer: carbon tetrachloride.
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B. (35 pts) What is the structure associated with this positive ion electron impact
spectrum? Assign structures to each peak whose mass is noted on the spectrum.
Draw 'fishhook' diagrams for each structure to explain its formation.
Notes: alkane series suggests a 4-carbon alkyl group – it goes up to 56
(butyl) and no higher – probably linear n-butyl because no particularly
intense peak and lack of a -15 Da loss from the precursor (though in reality,
isomer determination is hard here). 77 indicates a benzyl group. Complete
lack of 91 indicates that benzyl group has no alkyl derivatives. 105
suggests a carbonyl attached to the benzyl (+28 Da). 123 is 55 below 178,
suggesting loss of the whole butyl group (as •C4H7) – this is a pseudo
complementary ion to 56. 123-105 = 18, suggesting a bridging ether.
Answer: n-butyl benzoate
3
Question 2
A. (15 pts) Draw a diagram of a MALDI-TOF mass spectrometer, label the
appropriate parts and describe their function. What is the difference between "linear"
mode and "reflector" mode?
In linear mode, ions hit the first detector. Resolution is determined by the
kinetic energy distribution in the source, and by the flight tube length. In
reflector mode, ions are bounced off an electrostatic mirror to the second
detector. Resolution is higher because the mirror partially corrects the kinetic
energy distribution by making faster (more energetic) ions penetrate the mirror
slightly farther - giving them a longer path length to the detector.
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B. (10 pts) In a linear mode MALDI-TOF experiment, calculate the time-of-flight of
singly charged ions using the following parameters: Ion source potential, 20 kV; drift
tube grounded. Drift tube length: 1.5 meters. Mass of the ion: 1348.6 Da.
1 2 1 d2
K  mv  m 2 ; solve for t.
2
2 t
thus: t 
md 2
(1348.6)(1.67e  27 /1.6e  19)(1.5)2

 28.1 s
2K
2* 20000
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C. (15 pts) Draw a diagram of an electrospray quadrupole ion trap mass
spectrometer. Explain how the ion trap works, and explain how ions can be
"scanned" out into a detector.
The ion trap works by trapping ions in an oscillating electric field between the
endcaps and ring electrodes. If the amplitude and frequency are correct, ion
trajectories inside the trap are stable. Ions can be scanned out to the detector in two
ways. First, the RF amplitude can be ramped, causing lighter ions to become
successively unstable in the trap and fall out. Second, a resonant RF pulse can be
added on top of the main trapping field and scanned to dump the ions out of the trap.
Either way, a plot of amplitude or pulse frequency versus ion current is the resulting
mass spectrum.
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D. (10 pts) In a quadrupole ion trap, using the equation below calculate the low mass
cut off (LMCO) for the trap if it is run at 500 Vrf amplitude, 850 kHz, a pole-to-pole
(z - axis) minimum inner distance of 12 mm, and a z/r aspect ratio of 1.1.
LMCO is given by qz  0.908 
8eV
m (r 2  2 z 2 )
2
Low mass cut off (LMCO) is given by qz  0.908 
8eV
,so
m (r 2  2 z 2 )
2
8V
, with V = 500 Volts, ω =
0.908 (r 2  1.12 r 2 )
2*pi*850 kHz, and r = 12/2 mm, z = 1.1r. So: LMCO = 186 Da.
rearranging, we get m / z 
2
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Question 3
[50%, 25% part a, 25% part b]
a. (25 pts) Given the sequence of the peptide (VFDKDGDGYISAAELR), assign all
the peaks in the spectrum which are labeled with a mass (if possible). This is a
CID/CAD MS/MS spectrum of the 2+ ion, at m/z 878.5. Hint: this is the same sequence
as in part b, so the results should be consistent.
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Notes: students will need to calculate this table by hand, and then the
assignment (marked above) is trivial.
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b. (25 pts) Given the sequence of the peptide (VFDKDGDGYISAAELR), assign all
the peaks in the spectrum which are labeled with a mass (if possible). This is an ETD
MS/MS spectrum of the 2+ ion, at m/z 878.5. Hint: this is the same sequence as in part a,
so the results should be consistent
10
Notes: students will need to calculate this table by hand, and then the
assignment (marked above) is trivial.
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