Chem 550‐2012 Spring Problem set #2, Answer Key 1. (12 pts) You prepared a sample of naphthalene-d8 (C10D8) to be used as an internal standard
in a GC/MS of polycyclic aromatic compounds in the environment. Proton NMR indicates
incomplete labeling, and you turn to EI mass spectrometry to quantify. At 12 eV of ionizing
energy (achieved by lowering the filament voltage in the ion source), no M - 1 or M - 2 can be
detected in unlabeled naphthalene. The presumably pure C10D8 is then admitted to the mass
spectrometer, and the following abundances were measured by signal averaging. Calculate
the composition (i.e., fraction d8, d7, etc). The isotopic contributions to A + 1 and A + 2 for a
molecule with 10 carbons are 11.0 and 0.54%, respectively.
m/z
133
134
135
136
137
138
Rel. Abund.
16.06
17.83
33.98
100
10.78
0.51
The idea of the isotopic depletion is shown below:
Each peak is contributed by the C10DxH8-x and the isotope (13C, 14C) of species with lower
mass. We know that the relative abundance of A+1(13C) and A+2 (14C) peaks of C10H8 are
11% and 0.54%, respectively. We can remove this pattern (bar graph in same color) to obtain
the “pure” deuterium distribution (box outlined in black).
Relative abundance of
C10D5H3: 16.06
C10D6H2: 17.83 - (16.06 x 0.11) = 16.06
C10D7H1: 33.98 - (16.06 x 0.0054) - (16.06 x 0.11) = 32.13
C10D8: 100 - (16.06 x 0.0054) - (32.13 x 0.11) = 96.38
Peak at m/z of 137 is mainly composed of the 13CC9D8 and 14CC9D7H1, and the peak at m/z of
138 is mainly composed of the 14CC9D8. Now add the peak intensities and renormalize.
The composition of the mixture should be: 10% C10D5H3, 10% C10D6H2, 20% C10D7H1, and
60% C10D8.
1 2. (10 pts) Give a formula and suggest a structure for the compound that gives the following EI
mass spectrum taken at 70 eV and rationalize with a fragmentation map all peaks of rel int >
20%. The first peak in the cluster in the m/z region of 250 is m/z 256. Note, you may not be
able to distinguish isomers.
186
Cl
+.
Cl
Cl
256
221
150
‐HCl
‐Cl
‐Cl
The molecular ion presents specific Cl containing isotopic pattern, and the three major peaks
are the compound with sequential loss of Cl. Therefore, it should be C12H7Cl3 (# of rings
+double bonds = 1 + 12 - (7 x 0.5) - (3 x 0.5) = 8) with chemical structure shown in the figure (It
would be difficult to determine the isomer without the pure reference compound and a
reference spectrum obtained following that of the unknown).
3. (10 pts) Give a formula and suggest a structure for the compound that gives the following 70
eV mass spectrum. Write a fragmentation map that accounts for all the peaks of rel intensity
greater than 15%. Careful measurement reveals that the peak at m/z 312 is 59.2 and that at
m/z 313 is 7.8%
2 152
Br
+.
‐2Br
76
‐C6H4
The isotopic ratio (7.8/59.2/0.011) tells us there are approximately 12 carbons. The molecular
ion presents specific Br-containing isotopic pattern, the base peak indicates loss of 2Br. It
should be C12H8Br2 (# of rings + double bonds = 1 + 12 - (8 x 0.5) - (2 x 0.5) = 8) with chemical
structure shown in the figure. Here too it would be difficult to assign the correct isomer.
4. (16 pts) Could the following spectrum be that of 1-chloro-4-bromothiophene? To test, calculate
the expected isotope pattern for the molecular ion and compare to that shown in the spectrum.
(you might expand the pattern on your computer to measure). You must show your work.
The isotopic distribution of C4H2SClBr is mainly contributed by C4SClBr in which the isotopic
pattern of S, Cl, and Br can be expected to be dominant over that of carbon. The major
isotopes of S (32S, 34S), Cl (35Cl, 37Cl), and Br (79Br, 81Br) contribute to A, A + 2, A + 4, and A +
6, and we can construct the combination by following “Polynomial Approach” (32S + 34S)(35Cl +
37
Cl)(79Br + 81Br) and generate the table as:
A
32
A+2
A+4
A+6
35 79
34 37 79
S Cl Br
S Cl Br
S Cl Br 34S37Cl81Br
35
79
34
3 32
S37Cl79Br
32 35 81
S Cl Br
32
S37Cl81Br
34 35 81
S Cl Br
The distribution can be obtained:
A: 1003 =1000000
A+2: (4.5x100x100) + (100x31.96x100) + (100x100x97.28)=1337400
A+4: (4.5x31.96x100)+(100x31.96x97.28)+(4.5x100x97.28)=369064.88
A+6: (4.5x31.96x97.28)=13990.8
after normalization this becomes:
A: 74.77
A+2: 100
A+4: 27.59
A+6: 1.04
The contribution of C4 to the pattern is mainly at A+1 where we have 4 × 1.1% = 4.4%,
whereas its contribution to A + 2 is negligible in this simple estimation. Finally, the A+1
contribution of C4 is placed into the “gaps” of the 13C pattern each with 4.4% relative to the
preceding peak:
A: 74.77
A+1: 3.29
A+2: 100
A+3: 4.4
A+4: 27.59
A+5: 1.21
A+6: 1.04
The expected isotopic pattern looks like:
This is the same as the experimental spectrum.
5. (16 pts) You have a sample of an invaluable moon rock and wish to know if it contains zinc.
You take a very small sample, dissolve in acid, and derivatize with acetylacetone to make a
volatile coordination compound (C5H7O2—Zn(II)—C5H7O2 ) that can be analyzed by GC/MS.
The following mass spectrum was obtained.
4 a) Consider the molecular ion region and test whether the isotope pattern agrees with
your calculation. You must show your work to receive credit (8 pts).
b) Explain the formation of ions clustered starting at m/z 205 and 163 (8 pts).
(a) Zn has five stable isotopes: (64Zn: 100%, 66Zn: 57.4%, 67Zn: 8.4%, 68Zn, 38.7%, 70Zn:
1.2%). The isotopic distribution of Zn can be expected to be dominant over that of carbon.
The contribution of C10 to the pattern is mainly at A+1 (10 × 1.1% = 11%), which can be
placed into the gaps of the 13C pattern each with 11% relative to the preceding peak:
A: 100
A+1: 11
A+2: 57.4
A+3: (57.4x0.11)+8.4=14.7
A+4: 38.7
A+5: 4.25
A+6: 1.2
This isotopic pattern agrees with that of the spectrum.
(b) Ion with m/z of 205 is the loss of C3H5O˙ (.O
) from molecular ion.
.O
O
Ion with m/z of 163 is the loss of C5H7O2˙ (
) from molecular ion.
6. (8 pts) Show a calculation that gives the chlorine pattern for any molecule that contains four
chlorines. You must show your calculation to get credit.
The chlorine pattern can be obtained by using “binomial approach”: (a+b)4 =
a4+4a3b+6a2b2+4ab3+b4 ,where a and b are the isotopic abundances of both isotopes (35Cl:
100%, 37Cl: 31.96%). Therefore, the pattern is:
A= 100%
A+2= 4x(100%)3x(31.96%)= 127.84%
A+4= 6x(100%)2x(31.96%)2= 61.28%
A+6= 4x(100%)x(31.96%)3= 13.06%
A+8= (31.96%)4= 1% (You can renormalize this to make the most intense peak 100%)
7. (16 pts) In the following table are the ionization and appearance energies for propionic acid.
The IE (ionization energy) of the acid is 10.4 eV. The appearance energies for the
fragmentation of the molecule are given in the table below:
Ion
AE (eV) Neutral Products
5 CHO2+
12.84
•C2H5
C2H5+
12.4
•COOH
C3H4O+
10.9
H2O
+
C3H5O
12.20
•OH
C3H5O2+
11.70
•H
a. Calculate the Hf of the molecular radical cation given that the Hf of neutral, gas-phase
propionic acid at RT is -455.8 kJ/mol (8 pts).
1 eV=96.5 KJ/mol
Ionization energy =Hf (ion) - Hf (neutral)
10.4 eV x 96.5 = Hf (ion) + 455.8
Hf (ion)= 547.8 KJ/mol
b. Given that the Hf(H2Og) is -241.8 kJ/mol, what is the Hf(C3H4O+.) ion that forms in the
fragmentation of the propionic acid. Given that the Hf(cyclopropanone radical cation) is
+860 kJ/mol, can the structure of the C3H4O+. be that of the radical cation of
cyclopropanone (i.e., the water loss is 1,3 followed by cyclization)? Explain (8 pts).
C3H6O2+˙
C3H4O+˙ +H2Og
(10.9 - 10.4)eV = Hf (C3H4O+˙) + Hf (H2Og) - Hf (C3H6O2+˙)
48.25 KJ/mol = Hf (C3H4O+˙) -241.8 KJ/mol -547.8 KJ/mol
Hf (C3H4O+˙)= 837.85 KJ/mol
Because Hf (C3H4O+˙) < Hf (cyclopropanone radical cation), the structure of
cyclopropanone radical cation can’t be formed. The structure should be more stable than
that of the cyclic compound. A possibility is •CH2CH2CO+ (a distonic ion)
8. (12 pts) Look up the proton affinities of the following molecules: ammonia, aniline,
ethylenediamine, and 1,4-butanediamine.
a. Explain the trend for ammonia, ethylenediamine, and 1,4-butanediamine (6 pts) .
The proton affinities of ammonia, ethylenediamine, and 1,4-butanediamine are 853.6 KJ/mol,
951.6 KJ/mol, and 1005.6 KJ/mol, respectively. Therefore, the trend of gas-phase basicity
(proton affinity) is: 1,4-butanediamine > ethylenediamine > ammonia. This is because the
methyl groups have a tendency to "push" electrons away from themselves (they are
polarizable and can stabilize charge). The second two compounds can also “chelate” the
proton by allowing it to bond to both amine groups: -H2N—H+—NH2b. Compare the solution Kb values of ammonia and aniline with those in the gas phase,
and explain why aniline is a weaker base in solution but a stronger base than ammonia
in the gas phase (6 pts).
In solution, the pKb of aniline (9.38) is higher than that of ammonia (4.75), which makes aniline
a weaker base in solution compared to ammonia. This is because aniline has a phenyl group,
allowing the lone pair to be delocalized (less available). I think the big effect is that the
6 hydrophobic aniline is not well solvated. Moreover, the interaction of the ammonia ion with
water is more favorable than the interaction with anilinium ion. This is so called the “solvent
effect”.
In the gas phase, without the solvent effect, aniline (gas phase basicity: 850.6 KJ/mol)
becomes stronger base than ammonia (gas phase basicity: 819.0 KJ/mol) because the
positive charge can be delocalized by polarizing the aromatic ring.
9. (20 pts) Given the importance of mass spectrometry in the world of proteomics, it is important
that one know how to use efficiently a web-based database. For the following problem, use
the Protein Prospector website http://prospector.ucsf.edu. The database to be searched is
SwissProt.2011.10.10, and the cysteines are unmodified; use the default setting unless
otherwise indicated.
a. You produce a protein, digest it with trypsin, and acquire a MALDI-ToF spectrum of the
resulting peptides: m/z 1034.5299, 2502.1639, 2673.1919, 2759.3127, 3689.7541,
3929.8282, and 3950.8236. You input this data set into the MS Fit program on the
prospector website in hopes of identifying your protein. What do you find? (4 pts)
The search gave no protein hit.
b. The results are discouraging, so you take your problem to senior grad student for some
advice. He points out that you have only large peptides and that perhaps your trypsin
digestion didn’t go to completion. So you head back to the MS Fit program and allow
for up to 4 missed cleavages. What is a “missed cleavage”? (2 pts) With this
adjustment, what do you find? (2 pts)
Missed cleavage means an incomplete digestion caused by the protease (trypsin) missed
cleavage on several amino residues (K, R).
With the adjustment, the search gave two protein hits: Troponin C and Cyclin-T.
c. Later, you realize your MALDI-ToF can provide mass accuracy within 10 ppm (better
than the default), so you head back to the MS Fit program again and adjust the mass
tolerance (Tol) to 10 ppm. What do you find? (2 pts) Explain! (2 pts)
By narrowing the tolerance window, the search gave only one hit: Troponin C.
Database contains a large number of proteins, making possible a large number of tryptic
peptides upon digestion “in silico”. Some peptides from different proteins may have similar
masses. If the mass differences are within the mass tolerance, the database will consider all
these protein as possible hits. Once we have an accurate mass measurement (< 5-10 ppm),
the mass tolerance can be narrowed and the confidence of the database search increased.
d. The results are better, but you really want to be sure, and so you try the experiment
again; this time with a more complete trypsin digestion. Your MALDI-ToF analysis gives
7 the following ions of m/z 805.4236, 849.4135, 875.4655, 990.4235, 1026.5142,
1049.4680. You use MS Fit program with the default settings. What do you find? (2 pts)
Explain! (2 pts)
The search gave only one hit: Troponin C.
A more complete digestion generates smaller peptides and reduces the possibility of having
different combination of peptide sequences with similar masses. Therefore, it increases the
confidence level of database search and provides a more certain result.
e. To complete your protein identification, you decide that a MS/MS scan is required. You
choose the ion of m/z 1049.4680 as your precursor. Your product-ion spectrum includes
the following ions: m/z 233.0954, 246.1561, 348.1224, 375.1987, 446.2358, 476.1810,
574.2944, 604.2395, 675.2767, 702.3529, 804.3192, 817.3799, 875.3564, and
918.4275. Explain whether the MS Tag program gives you a more certain identification?
(2 pts) Why? (2 pts)
Yes. All product ions are matched to one unique peptide (MTDQQAEAR) of Troponin C. The
peptide sequence is determined by its sequential fragment ions rather by its mass, this,
indeed, increases the level of confidence.
10. (16 pts) A biochemist has observed a biomarker protein in patients having Alzheimer’s disease
syndrome. He/she purified it and needs to know the mass of the protein. The following is an
ESI+ mass spectrum.
a. Deduce the molecular mass three times by using different combinations of data. Report
an average and standard deviation. You must show work to obtain full credit.
We can follow the equation:
mi = (M + nH)/n, where mi is the m/z of one peak, M is the molecular mass, n is the number of
hydrogen (H) absorbed on the molecular.
Therefore, we can calculate the molecular mass as:
882.59 = (M + n)/n
860.55 = (M + n + 1)/(n + 1)
839.59 = (M + n + 2)/(n + 2)
We first solve and obtain n = 39
The M can be calculated for these three peaks to be: 34382.01, 34382, and 34382.19.
And the average mass of this protein is 34382.07 with standard deviation of 0.1.
8 b. Consider the small peaks between the major peaks, especially in the low m/z part of the
spectrum. Your colleague in the lab says they are Na+ adducts and you need not
worry. Is this correct? If not, suggest an origin for these peaks.
No. The Na+ adducts should have smaller difference of m/z relative to each peak. The small
peaks represent species with mass ~350 Da higher than the protein. This might be an isoform
of the protein, which was co-expressed with the main form.
9