Exam 2 Answer

Chemistry 605 (Reich)
SECOND HOUR EXAM
Sat. April 19, 2014
Question/Points
R-13A
/20
R-13B
/15
R-13C
/20
R-13D
/15
R-13E
/20
R-13F
Total
Average
Hi
61
92
/10
Mode
Median
51
64
/100
AB
BC
75
45
Distribution from grade list (average: 61.1; count: 26)
4
Number
3
2
1
0
0.00
10.00
20.00
Name
30.00
40.00
50.00
Grade
60.00
70.00
80.00
90.00
100.00
Answer Key
If you place answers anywhere else except in the spaces provided, (e.g. on the spectra
or on extra pages) clearly indicate this on the answer sheets.
20
Problem R-13A. Assign several of the protons and analyze multiplets of a steroid.
(a) Assign and analyze the following signals (report multiplicity and J values). Use the steroid numbering on the
structure:
H3, tdd, J = 11, 7, 6 Hz
δ 3.55
6
H15, ddd, J = 7, 5, 2 Hz
δ 4.55
H6, dt, J = 5, 2 Hz
δ 5.40
(b) Assign the 2-proton signal between δ 2.4 and δ 2.7. Briefly provide a rationale for this assignment. What kind
AB of ABX
of multiplet is this?
4
Apart from H3, H15 and H6, the H16 protons will be the most downfield. The large 2J (19 Hz) also means the protons
must be next to the keto group
10
Solution 2
JAB
19.1
19.1
JAX
0.6
-14.8
JBX
6.3
21.6
νA
701.2
693.5
νB
683.0
690.7
νAB
18.2
2.9
i14=i15
0.0074
0.47
δA
2.60
2.57
δB
2.53
2.56
1
2
3
4
5
6
7
8
671.99
666.91
691.12
685.71
696.54
695.1
715.5
714.25
7
3
666.9
672.0
4
8
2.65
1 2
2.60
2.55
2.50
B
A
2.45
Sol. 1
A
Sol. 2
679.86
686.15
700.95
701.51
Solution 1 is correct - can't have a
negative 3J, as in Sol 2. Also the Sol. 2
couplings are unreasonably large
685.7
5
6
Solution 1
691.1
c- ± Δνab-/2 = 701.51, 686.15
c+ ± Δνab+/2 = 700.95, 679.86
696.5
695.1
c- = 693.8 Δνab- = 15.3
c+ = 690.4 Δνab+ = 21.2
715.5
714.2
(c) Analyze this multiplet in a mathematically correct fashion (also show a coupling tree), using the frequencies
given. Report coupling constants and chemical shifts. If two solutions are possible present them, and provide a
rationale for choosing one of them.
Problem R-13A
270 MHz 1H NMR Spectrum in CDCl3
Source: Ieva Reich (3/27)
O
12
11
1
H
9
H
HO
14
16
15
2
3
8
H
OH
30
20
10
0
Hz
6
4
H15, ddd, J = 7, 5, 2 Hz
H6, dt, J = 5, 2 Hz
H3, tdd, J = 11, 7, 6 Hz
5.45 5.40 5.35
4.55 4.50
3.6
2.6
3.5
2.4
2.5
2.3
2.2
H11ax
2.1
2.0
1.9
1.8
1.7
1.35 1.30
2.03
1.6
1.5
1.2
1.1
1.00
0.88
0.88
H15
H6
6
5
H3
4
H16
3
ppm
H8
2
1
0
15
Problem R-13B. Compound R-13B is a disubstituted naphthalene. Spectra of just the aromatic region are
provided in two different mixtures of CDCl 3 and C6D6. From an analysis of the spectra, determine the positions of
substitution. The substituents are:
O
O
OMe
Problem R-13B (C16H18O3)
270 MHz 1H NMR Spectrum in C6D6/CDCl3 (4/1)
(Source: Ieva Reich 11/15)
2.85
8.0
H
1.00
O
H 6.95
OMe
O
30 20 10
0
H 7.15
Hz
1.10
H 7.57
H 7.67
H 7.63
8.0
7.9
7.8
7.7
7.6
7.5
7.4
7.3
7.2
7.1
7.0
6.9
ppm
Problem R-13B (C16H18O3)
270 MHz 1H NMR Spectrum in C6D6/CDCl3 (5/2)
(Source: Ieva Reich 11/15)
2.98
30 20 10
0
Hz
1.07
1.00
7.9
7.8
7.7
7.6
7.5
7.4
7.3
7.2
0.97
7.1
7.0
ppm
.(a) Draw a possible structure, and label it with chemical shifts. Use the top spectrum for this, please.
There are two patterns of 1,2,4-hydrogens
H 8.0
11
H 6.95
R
H 7.67
H 7.63
X
H 7.15
7.63
6.95
8.0
7.57
7.67
Upfield signals - this
one should have the
electron-donating OMe
group at position X
7.15
O
O
H 7.57
(b) Is another substitution pattern consistent with the spectra? Explain.
4
There is no way to determine the relative position
of the substituents on the two rings with the
information at hand
7.63
7.57
7.15
7.67
O
OMe
O 8.0
OMe
6.95
Problem R-13C. You are given the structure of a pentose thiol triacetate and asked to determine the relative
stereochemistry and conformation from the 270 MHz 1H NMR spectrum presented on the next page.
(a) Analyze the sets of signals and show coupling constants in the standard format. When you have completed
the analysis, assign the individual protons (e.g., H5ax). Use the numbering system given on the structure in part
(b).
2
δ 3.75 dd, J = 12, 9 Hz
This must be H5a from the chemical shift - coupled gem and ax-ax.
Thus H4 must be axial.
2
δ 4.05 ddd, J = 11.5, 4.5, 1 Hz
This must be H5e - coupled gem and eq-ax. There is also a
W-coupling to the equatorial H3
4
δ 4.95 dd, 8, 2.5 - H2
br d, 8 - H1
This is an AB pattern, one half is coupled to an X proton by ca 2.5 Hz,
the other half is broadened.
These must be the H1 and H2 protons, the H1 is broadened by
exchange of the SH (which is a broad singlet at δ 2.17). The 8 Hz
coupling between H1 and H2 means both protons are axial. The small
coupling (2.5 Hz) of H2 to H3 means H3 is equatorial
2
δ 5.07 ddd, J = 8.5, 4.5, 3
This is the proton at H4, coupled to the axial proton at C5 (9 Hz) and
to the eq protons at C5 and C3. Thus H3 must be equatorial.
2
δ 5.55
This is the equatorial H3 proton, coupled twice eq-ax (2.5 Hz) to H4
and H2. There is a small W coupling to H5e.
td, 2.5, 1
(b) Determine the stereochemistry of R-05K. Place the appropriate substituents in each of the boxes on the
structure below.
H
H5e H
4
O
5
O
SH
AcO
5
5
SH
H
1
2
1
3
2
H5a OAc
AcO 4
OAc
3
OAc
H
OAc
(c) Briefly describe how you made the assignment at C-4.
3
H5a and H5e can be assigned from their chemical shifts (most upfield of the ring protons) and coupling
patterns. H5a shows, in addition to the 12 Hz gem coupling, a 9 Hz vicinal coupling, which means H4
must be axial. H4 can be assigned from the three couplings, two of which are to the H5 protons.
Problem R-13C (C11H16O7S)
270 MHz 1H NMR spectrum in CDCl3
(Source: Paul Savage/Gellman 11/32)
O
5
H5a
8.61
dd, J = 12, 9 Hz
H5e
dd, J = 11.5, 4.5, 1 Hz
SH
1
2
AcO
4
OAc
3
OAc
10.00
H3
4.1
dt, 2.5, 1
4.0
3.8
H1
br d, 8
H4
30
20
10
0
3.7
Hz
H2
ddd, J = 8.5, 4.5, 3
dd, 8, 2.5
H
AcO
5
H
3
5.1
5.60 5.55 5.50
2.88
5.0
H
H
4
O
2
1
SH
H AcO
OAc
H
4.9
SH
Curphy-Morrison:
H
SH
X
7.0
1.3
O-Alkyl 2.10
O-Ac
6.5
1.06
1.00
1.02
2.2
2.1
2.0
3.45
6.0
5.5
5.0
4.5
4.0
3.5
ppm
3.0
2.5
2.0
1.5
1.0
0.5
0.0
15
1060.7
1059.0
1057.2
1074.1
1072.3
1070.5
1068.8
1067.5
1085.2
1083.9
1082.1
Problem R-13D. This problem requires you to interpret the partial 500 MHz 1H NMR spectrum of a
1,2-dimethylpiperidine deuterated 94% on the N-methyl group (source: F. A. L. Anet J. Am. Chem. Soc. 1989, 111,
3429).
30
20
10
0
Hz
AB q of 1:1:1 triplets
Normal
Spectrum
CH3
N
CH3
CH2D
CH3
JHD = 1.8 Hz
1059.0
94%
1085.3
1084.0
6%
2
1072.6
1070.8
N
AB q
JAB = 11.6 Hz
Deuterium
decoupled
νAB =
2
JHH
(25.0) (1.8) = 6.71 Hz
νA = 1071.7 + 6.71/2 = 1075.1 δ 2.15
νB = 1071.7 - 6.71/2 = 1068.3
δ 2.17
2.18
c = 1071.7
2.16
2.14
ppm
2.12
δ 2.14
2.10
Identify all of the peaks in both the top and bottom spectra. Briefly explain the pattern, extract all coupling
constants and determine exact chemical shifts, and report them in the standard format (e .g., δ 3.44, t, nJxy = 3.5
Hz).
3
The signals we are looking at are the N-CH3 protons (δ 2.17) and the N-CH2-D protons (2.11-2.17)
3
In the deuterium decoupled spectrum the singlet at δ 2.17 is N-CH3
9
In D-decoupled the AB quartet centered at δ 2.14 is the N-CH2-D group. The CH2 group is diastereotopic (there
is an asymmetric center at C-2) and so forms an AB quartet.
2
JAB = 11.7 Hz, δA = 2.15, δB = 2.13
3
Normal spectrum: each peak of the AB quartet is split into a 1:1:1 triplet by coupling to D. 2JHD = 1.7 Hz
20
Problem R-13E (C18H25O2PSi). In this problem you are given the structure of a phosphorus compound and
asked to interpret parts of the NMR spectrum. For each multiplet or set of multiplets report the pattern in the
standard format: δ 0.00, triplet of pentets, nJXY = 0.0 Hz. You may use first order analysis.
O
SiMe3
P
(a) Assign and interpret the three signals centered at δ -0.05.
O-CH2-CH3
Me3Si signal
The two small peaks are
4
29
Si satellites
2
JH-Si = 7Hz
(b) Assign and interpret the multiplet at δ 0.75. Draw and label a coupling tree.
B
This is the P-CH2-SiMe3 group - the protons are diastereotopic (P is an asymmetric
center), so this is an ABX pattern X = 31P
JAB = 14 Hz (2JHH)
δA = 0.73
7
30
20
10
0
A
2
JHH
Hz
JAX = 16 Hz (2JH-P)
δB = 0.78
2
2
JH-P
JBX
JBX = 18 Hz (2JH-P)
(this is "AMX" treatment which always corresponds to
Solution1, not a proper ABX analysis, so if this happens to be a
Solution 2 situation, then the J's would be very wrong - they do
look OK though, and so Solution 1 is probably fine)
JH-P
JAX
0.85 0.80 0.75 0.70 0.65
(c) Assign and interpret the multiplets at δ 3.7 - 4.2. Draw
and label a coupling tree.
ABM3X
Each pattern is a doublet of pentets, J = 10, 7
Hz for the diastereotopic O-CH2-CH3 group.
The pentet arises because 3JHH to the CH3
group and 3JHP are nearly the same
6
3
JHH = 7 Hz
3
JHP = 7 Hz
2
JHH = 10 Hz
We know the 10 Hz coupling is the
gem JHH because of the size, and
leaning effects.
4.20 4.15 4.10 4.05 4.00
3.85 3.80 3.75 3.70
(d) Assign and interpret the multiplet at δ 8.2.
This must be the proton ortho to the phosphonate group approximately a ddd, J = 12, 6, 2 Hz. There are some
second-order effects, so couplings are suspect
3
3
JHH = 8 Hz (ortho coupling)
4
H
O
P
SiMe3
O-CH2-CH3
JHH = 2 Hz (meta coupling)
3
JHP = 12 Hz (ortho 31P coupling)
8.25 8.20 8.15
Problem R-13E (C18H25O2PSi)
300.1 MHz 1H NMR Spectrum in CDCl3.
Source:Olafs Daugulis/Vedejs 08/24
30
20
10
0
O
Hz
B
O
2
SiMe3
P
SiMe3
P
X
A
A
HB H
JHH
OEt
2
2
7.6
7.5
7.4
3
O
P
d pentets
(actually dqd)
HB HA
O
d pentets
(actually dqd)
CH3
3
JHH = 7 Hz
3
JHP = 7 Hz
2
JHH = 10 Hz
4.20 4.15 4.10 4.05 4.00
8.25 8.20 8.15
1.30 1.25
3.85 3.80 3.75 3.70
0.00 -0.05
3.05
ddd, J = 12, 6, 2 Hz
JHH = 8 Hz (ortho coupling)
4
JHH = 2 Hz (meta coupling)
3
JHP = 12 Hz (ortho 31P coupling)
2.03
0.96
10
JH-P
JAX
0.85 0.80 0.75 0.70 0.65
7.3
7.97
Benzene Shifts
O
0.48
P OMe
0.16
OMe
0.24
JH-P
JBX
9
1.00
8
7
6
5
ppm
4
0.90
3
2
1
0
10
Problem R-13F. Here are the two parts of quiz 217 we didn't get to finish during my first lecture.
Put your analysis (δ, J, multiplicity) in the space provide, and suggest some possible structures from
the quiz answer sheet.
Below are proton multiplets in different molecules. Analyze the pattern and extract the coupling
constants. From the chemical shift given, what kind of protons are these likely to be? What part structures
are suggested by this pattern? Find one or more molecules on the "NMR Quiz Answers" sheet that fit the
chemical shift and coupling pattern.
40
30
20
Hz
(b) Integration: 2H
5
HD
NH2
HC
Y
HB
HB
7.40
7.35
ppm
0
These are two aromatic protons (HA and HB), coupled to each
other with J = 8 Hz (ortho). The upfield proton HB has a
second ortho coupling to HC. Both protons are meta-coupled
also, HA to HC, and HB to HD
HB td, 8, 1.5
dd, J = 8, 2 HA
10
X
7.30
O
Br
HA
17
HA
8 9 10 25 also fit
22
ddd, δ 3.45, J = 8.5, 8, 3.3 (almost a dt)
976.7
986.2
985.6
983.5
995.3
993.0
992.4
1002.0
1030.2
1026.9
1038.3
1035.3
(c) Integration: 2H
1047.1
1044.0
5
less well because J in pyridine are more varied
ddd, δ 3.30, J = 9.0, 8.5, 6.8
9.0
8.5
8.5
8.0
3.3
Most likely structure:
HA HX
HA and HB of an ABXY pattern O C C
HB HY
J / Hz
6.8
O
3.50
3.45
3.40
3.35
ppm
Coupling constants are
consistent for those in a
five-membered ring. They don't
fit very well for chair
cyclohexane
3.30
3.25
O
Ph
N
HB
HY
N+
HA
HX
18
19
HB
HA
NMR Quiz Answers
Hans Reich
Chem 605
CF3
OEt
O
Cl
O
OH
S
CF3
2
S
CF3
N-H
OH
S
O
O
O
CF3
O
OEt
1
CF3
CF3
4
3
5
O
H
O
Br
H
N
O
H
6
OH
OH
7
8
O
9
SeCH2Ph
10
O
H
HO
O
SiMe3
H
S
Cl
S
H
SeCH2Ph
H
11
12
14
13
15
16
O
NH2
Br
O
Ph
O
N
18
S
19
23
27
SiMe3
O
S
H
OH
O
O
25
24
O
O
OEt
O
CO2tBu
21
Ph
O
O
O
OMe
22
O
H
20
HO
Ph
Ph
H
O
N
N
EtO
Ph
P
H
17
O
O
N+
Br
O
26
OH
O
O
H
28
29
30
31

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