NAME ID # INTERPRETATION OF ORGANIC SPECTRA …
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NAME _______________________ ID # _______________________ INTERPRETATION OF ORGANIC SPECTRA (4361/8361) 1:30 – 3:30 pm, December 15, 2012 Final Exam This exam is open book and open note. You are permitted to use any written materials you have brought as aids on this exam. You may also use a simple calculator. Other than this, please do not use any other electronic devices (cell phones, computers, recording devices, etc.) during the exam. You may use pen or pencil. However, re-grades will be considered only for exams completed in pen. Please write your answers in the boxes/spaces provided. If your answer is not in the appropriate space (say, for example, it’s on the back of the page), draw us an arrow and/or note telling us where to look. Feel free to remove the corner staple if this helps you analyze the spectra; you will have the opportunity to re-staple your exam at the end. You will be given 2 hours total to finish the test. This exam contains two main problems, which are split into parts. Many of these parts can be answered independently. Do not get stuck on one part and then assume that you will be unable to answer the rest of the question—move on. In addition, partial credit will be given for incorrect but still plausible answers, so guess on problems you cannot answer perfectly. At the end of the 2 hour exam period you will be asked to return your exam to the proctor. Please do not take any part of the exam packet with you when you are done; everything will be returned to you after the exams are graded. This packet should contain 22 pages, including this one. Please check to make sure that your packet contains 22 pages before beginning your exam.
Transcript of NAME ID # INTERPRETATION OF ORGANIC SPECTRA …
NAME _______________________
ID # _______________________
INTERPRETATION OF ORGANIC SPECTRA (4361/8361)
1:30 – 3:30 pm, December 15, 2012
Final Exam This exam is open book and open note. You are permitted to use any written materials you have brought as aids on this exam. You may also use a simple calculator. Other than this, please do not use any other electronic devices (cell phones, computers, recording devices, etc.) during the exam. You may use pen or pencil. However, re-grades will be considered only for exams completed in pen. Please write your answers in the boxes/spaces provided. If your answer is not in the appropriate space (say, for example, it’s on the back of the page), draw us an arrow and/or note telling us where to look. Feel free to remove the corner staple if this helps you analyze the spectra; you will have the opportunity to re-staple your exam at the end. You will be given 2 hours total to finish the test. This exam contains two main problems, which are split into parts. Many of these parts can be answered independently. Do not get stuck on one part and then assume that you will be unable to answer the rest of the question—move on. In addition, partial credit will be given for incorrect but still plausible answers, so guess on problems you cannot answer perfectly. At the end of the 2 hour exam period you will be asked to return your exam to the proctor. Please do not take any part of the exam packet with you when you are done; everything will be returned to you after the exams are graded. This packet should contain 22 pages, including this one. Please check to make sure that your packet contains 22 pages before beginning your exam.
NAME _______________________
Scoring: 1. _________ / 5 5. _________ / 39 9. _________ / 4 2. _________ / 5 6. _________ / 5 10. ________ / 10 3. _________ / 48 7. _________ / 7 11. ________ / 18 4. _________ / 16 8. _________ / 15 12. ________ / 18 Total Score: _________ / 200 Melatonin is a human hormone that varies regularly in concentration over the course of a day, and has been implicated in the regulation (via “circadian rhythms”) of a variety of biological processes. Melatonin is a substituted indole, with the incompletely determined structure shown at right. In this structure, the -OCH3 group could be attached to any available carbon on the indole benzene ring; and the alkylamide group could be attached to either free carbon of the indole pyrrole ring. In the first part of this exam, you will use 1H, 13C, 1H-1H COSY, 1H-13C HSQC, and 1H-13C HMBC spectra (attached to the back of this exam) to determine the correct structure of melatonin. 1. All of the NMR spectra were collected on samples of melatonin dissolved in
deuterated DMSO-d6, which has molecular formula C2D6SO. Deuterium atoms are invisible to 1H NMR, and yet we still see a multiplet at = 2.5 ppm corresponding to the NMR solvent. Why?
2. The 1H resonance at = 8.0 ppm corresponds to a proton attached to a nitrogen
atom. This resonance looks sharper, better resolved, than we would expect for an N-H proton. What must be true of the NMR sample to achieve this unusually sharp N-H resonance? (Answer on the next page.)
NH
HN CH3
O
H3CO
melatonin
3. In the chart below, assign chemical shift values (to within 0.05 ppm) to every proton
in melatonin, using the numbering scheme on the right. I have provided a box for every possible location of a proton on the melatonin skeleton, but remember that two of these locations will be occupied by other groups (the -OCH3 and alkylamide groups). As a result, you should leave two of the boxes empty.
proton (ppm) proton (ppm) proton (ppm)
H1 H5 H11
H2 H6 H12
H3 H7 H14
H4 H10 H15
4. In the boxes on the right,
name four unique coupling constants that you observe in the 1H NMR spectrum of melatonin, and give the J value for each (to within 1 Hz).
coupling constant name “J(H#,H#)”
J (Hz)
NH
HN CH3
O
H3CO
1413
1211
10
2
3
1
45
67
8
9
15
5. In the chart below, assign chemical shift values (to within 0.1 ppm) to every carbon atom in melatonin, using the same numbering scheme as on the previous page.
proton (ppm) proton (ppm) proton (ppm)
C2 C7 C11
C3 C8 C13
C4 C9 C14
C5 C10 C15
C6
6. In principle, 13C-13C INADEQUATE NMR could be used to determine the carbon-
carbon connectivity of melatonin. But INADEQUATE is an NMR technique that is almost never used. Why not?
7. Melatonin absorbs UV light, with a fairly large extinction coefficient at = 275 nm.
Both the aromatic indole and the amide group contribute to absorption at this wavelength. Which would you expect contributes more?
NH
HN CH3
O
H3CO
1413
1211
10
2
3
1
45
67
8
9
15
indole amide both contribute
equally or ? (Circle one.)
Assuming that the contributions of these two parts of the molecule to the extinction coefficient at 275 nm (275) are additive, what would you estimate the extinction coefficient to be closest to?
8. The IR spectrum of melatonin is shown below. What functional groups are
responsible for each of the peaks labeled on the spectrum? 9. There are a lot of unanswered questions about how melatonin is distributed in
tissues, and how that distribution may be affected by the daily cycle of sunlight and darkness. It would be great to be able to image melatonin in cells and tissues. Unfortunately, melatonin is not fluorescent, so this distribution cannot be imaged by fluorescence microscopy. Propose an alternate technique to image the distribution of melatonin in a tissue sample. You do not need to describe the technique—just a name is enough.
= 3308 cm-1
functional group
= 3012 cm-1
= 1623 cm-1
10 100 or ? (Circle one.) 1000 10000 100000
1000 1500 2000 2500 3000 3500
Wavenumbers (cm-1)
Tra
nsm
issi
on (
%T
)
3308
3012
1623
Bo Zhou (Xing group, Med. Chem.) conducted a small-scale, Pd-catalyzed hydrogenolysis of starting material 1, with the hope of removing its methoxybenzoyl protecting group and obtaining product 2. Bo analyzed the crude reaction mixture by electrospray-ionization mass spectrometry (ESI-MS), in positive-ion mode. He found that the starting material had indeed been consumed in the reaction, but that the parent mass of the primary product was not consistent with structure 2. He suspected, however, that the product was probably structurally related to 2. In the second part of this exam, you will analyze Bo’s MS data to determine the structure of the actual product obtained in this reaction. The ESI-MS spectra of pure 1 and the major reaction product are shown below.
10. What is the most likely chemical formula for the ions represented by each of the
peak masses listed on the next page? In each formula, give not only the atomic symbol for each element, but also the isotope number for any element that is not >90% abundant in nature. Make sure to indicate the charge state of the ion.
O
O
O
O
O
Br
NH
OCH3
1 (C24H26BrNO6)
Pd-C, H2 CH3OH
O
O
O
O
O
Br
NH2
2 (C16H18BrNO5, not observed)
Example of answer format:
m/z = 42: C2(13C)1H5
+
11. Assuming that the m/z = 306 peak in the ESI-MS spectrum of the reaction product
corresponds to an unfragmented parent ion, what is the structure of this m/z = 306 ion? And how does it fragment to yield a m/z = 218 daughter ion? In the box below, draw both ion structures, and a mechanism (using “electron pushing”) that connects them.
12. Bo didn’t even try analyzing his reaction via electron-ionization (EI) mass
spectrometry, because he expected that both his starting material and any products would fragment too easily. But what if he had performed an EI-MS on his starting material—what fragments would we expect? In the boxes on the next page, draw mechanisms that show how electron-ionized molecule 1 might undergo two common fragmentation pathways in EI-MS: -cleavage, and the McLafferty rearrangement. In each case, make sure you draw unpaired electrons and formal charges, and illustrate each mechanism using “electron pushing”.
m/z = 504:
m/z = 505:
m/z = 306 m/z = 218
-cleavage
McLafferty rearrangement
11.5
11.0
10.5
10.0
9.5
9.0
8.5
8.0
7.5
7.0
6.5
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
Che
mic
al S
hift
(ppm
)
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
Normalized Intensity
2.7
2.0
2.0
3.0
1.1
1.1
1.0
1.1
1.1
1.0
1.8134
3.7601
1 H N
MR
, 500
MH
z, in
DM
SO
-d6
10.7
510
.70
10.6
510
.60
10.5
5C
hem
ical
Shi
ft (p
pm)
0
0.05
0.10
0.15
0.20
0.25
0.30
Normalized Intensity
10.6554
1 H N
MR
, 500
MH
z, in
DM
SO
-d6
(clo
seup
s)
8.10
8.05
8.00
7.95
7.90
7.85
7.95017.9613
7.9722
10.6529
10.6570
reso
lutio
n-en
hanc
ed
7.30
7.25
7.20
7.15
7.10
7.05
7.00
6.95
6.90
6.85
6.80
6.75
6.70
6.65
Che
mic
al S
hift
(ppm
)
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
Normalized Intensity
6.70666.71146.72396.7287
7.02127.0260
7.10407.1085
7.2183
7.2360
1 H N
MR
, 500
MH
z, in
DM
SO
-d6
(clo
seup
)
3.45
3.40
3.35
3.30
3.25
3.20
Che
mic
al S
hift
(ppm
)
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
Normalized Intensity
3.2897
3.30483.3167
3.3311
1 H N
MR
, 500
MH
z, in
DM
SO
-d6
(clo
seup
s)
2.90
2.85
2.80
2.75
2.70
2.65
2.7654
2.7802
2.7953
168 160 152 144 136 128 120 112 104 96 88 80 72 64 56 48 40 32 24 16Chemical Shift (ppm)
-0.05
0
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
Nor
mal
ized
Inte
nsity
22.7
477
25.3
095
55.3
230
100.
0898
111.
0575
111.
7018
112.
0182
123.
2756
127.
5758
131.
3804
152.
9727
169.
0564
13C NMR, 125 MHz, in DMSO-d6
40.5 40.0 39.5 39.0 38.5
39.4
643
10 9 8 7 6 5 4 3 2F2 Chemical Shift (ppm)
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
6.0
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
F1 C
hem
ical
Shi
ft (p
pm)
1H-1H gCOSY, 500 MHz, in DMSO-d6
see closeupnext page
7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6F2 Chemical Shift (ppm)
6.5
7.0
7.5
8.0
8.5
9.0
9.5
10.0
10.5
F1 C
hem
ical
Shi
ft (p
pm)
1H-1H gCOSY, 500 MHz, in DMSO-d6
(closeup)
10 9 8 7 6 5 4 3 2F2 Chemical Shift (ppm)
24
32
40
48
56
64
72
80
88
96
104
112
120
128
136
144
152
160
168
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C HSQC, 500/125 MHz, in DMSO-d6
see closeupnext page
7.30 7.25 7.20 7.15 7.10 7.05 7.00 6.95 6.90 6.85 6.80 6.75 6.70F2 Chemical Shift (ppm)
100
102
104
106
108
110
112
114
116
118
120
122
124
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C HSQC, 500/125 MHz, in DMSO-d6
(closeup)
10 9 8 7 6 5 4 3 2F2 Chemical Shift (ppm)
24
32
40
48
56
64
72
80
88
96
104
112
120
128
136
144
152
160
168
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C gHMBC, 500/125 MHz, in DMSO-d6
see closeupsfollowing pages
8.0 7.5 7.0 6.5F2 Chemical Shift (ppm)
100
105
110
115
120
125
130
135
140
145
150
155
160
165
170
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C gHMBC, 500/125 MHz, in DMSO-d6
(closeup)
7.25 7.20 7.15 7.10 7.05 7.00 6.95 6.90 6.85 6.80 6.75 6.70F2 Chemical Shift (ppm)
108.5
109.0
109.5
110.0
110.5
111.0
111.5
112.0
112.5
113.0
113.5
114.0
114.5
115.0
115.5
116.0
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C gHMBC, 500/125 MHz, in DMSO-d6
(closeup)
8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6F2 Chemical Shift (ppm)
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C gHMBC, 500/125 MHz, in DMSO-d6
(closeup)
3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6F2 Chemical Shift (ppm)
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
F1 C
hem
ical
Shi
ft (p
pm)
1H-13C gHMBC, 500/125 MHz, in DMSO-d6
(closeup)
