Chemistry 605 (Reich)
SECOND HOUR EXAM
Thur. April 12, 2012
Question/Points
R-11H /30
R-11J,K /15
R-11L,M /30
Name
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.
Total /100
R-11N /17
R-11O /8
Average 69
Hi 97 (2)
Median 71
AB 78
BC 50
CD 40
Distribution from grade list (average: 69.1; count: 27)
0 10 20 30 40 50 60 70 80 90 100
Grade
0
1
2
3
Number
Problem R-11H (C12H16O3). You are provided the 1
H and 13
C NMR spectra of a compound. Interpret the
spectra, and determine the structure or structures. Note that the signal at δ 6.5 disappeared when D2O was
added.
(a) DBE
(b) Analyze the multiplets below. Identify the patterns (e.g., ABXYZ - underline the observed nuclei). If they are
first order, report them in the standard format (δ 0.00, dqt, J = 0.0, 0.0, 0.0, 2H). Provide part structure(s) defined by
these protons.
(c) Identify at least 3 signals in the 13
C NMR spectrum which provide significant structural information, and
describe the part structures obtained from them.
(d) Draw the structure of R-11H below. If more than one structure fits the data, draw them, but circle your best
choice.
δ 1.7-2.1
δ 3.9-4.1
δ 6.7-7.2
2.1 2.0 1.9 1.8 1.7
4.2 4.1 4.0
7.2 7.1 7.0 6.9 6.8
0
Hz
102030
CD
F
HI
ABX3
ABM3Y
δ 2.05, d quintets, J = 13.4, 7.5
δ 1.75, d quintets, J = 13.5, 7.5
CH3
HAHB
H
δ 4.13, dq, J = 11, 7
δ 4.08, dq, J = 11, 7
AA'XX' (or AA'BB')
CH3
HBHA
O
R1
R2
HA
HB
HA'
HB'
R1, R2 must be electron donating
δ 175.2 - This is an ester carbonyl carbon
δ 155.1 - This is likely an sp2
carbon (aromatic) bearing an electronegative group C-O
δ 60.9 - sp3
carbon, probably a CH2-O group
HO
OO
*
A
B
C,D
E
F
G
H
I
δ 52.7 - sp3
CH
O
O
HO
10
8
O
OHO
8
O
CO2H
5
3 or 4 other structures at 2-3 points
10
3
5
5
5
2
5
30
There arealso two CH3 carbons (neither one on O) and an aliphatic CH (52.7) , also probably not on O
O
OH
EtO
8
Calc:
3.85
Calc:
4.00
δ
δ
δ
10 9 8 7 6 5 4 3 2 1 0
ppm
1.98 1.96
0.99
1.98
0.99 1.00 0.98
2.92 2.95
1.25 1.20 1.15
0.90 0.85
3.45 3.40 3.35
7.2 7.1 7.0 6.9 6.8
4.2 4.1 4.0
2.1 2.0 1.9 1.8 1.7
Problem R-11H (C12H16O3)
300 MHz 1
H NMR spectrum in CDCl3
Source: Chris Marvin/Charlie Fry 09/32
0
Hz
102030
ExchangeswithD2O
x
ABCDEFGHI
ABX3
ABX4 (actually ABM3X)
AA'XX' - p-disubstituted benzene
240 220 200 180 160 140 120 100 80 60 40 20 0 -20
ppm
12.0
14.0
26.5
52.7
60.9
76.6
77.0
77.4
115.4
129.0
130.7
155.1
175.2
DEPT-90
DEPT-135
Problem R-11H (C12H16O3)
75 MHz 13
C NMR spectrum in CDCl3
Source: Chris Marvin/Charlie Fry 09/32
Normal {i
H}
CH2CH CH3 CH3CH2OCC
O
-O C
Ph
H
O
Ph
CH3
H
Ph
H
O
Ph
H
CH3H H
Ph
H
OPh
CH3
H
Ph
H
OPh
H
CH3
H H
1 2
7 8
Ph
H
O
Ph
CH3
H
Ph
H
O
Ph
H
CH3H H
Ph
H
OPh
CH3
H
Ph
H
OPh
H
CH3
H H
3 4
5 6
Problem R-11J and K (C19H18O). You are given 200 MHz 1
H NMR spectra of two stereoisomers of a
compound which differ at one stereocenter only (i. e., 1 and 2, or 6 and 8), the possibilities are 1 to 8 below. Your
task is to make both a structural and a stereochemical assignment. Explain the basis of your assignment below,
taking care to clearly identify the signals you are using.
(a) What spectral features allow you to distinguish the two structural
types (1 to 4 versus 5 to 8)?
(b) What spectral features allow you to distinguish the pair of
isomers? Write the spectrum number (R-11J or R-11K) in the
appropriate blank.
15
(c) Just to show you understand the spectra, give just the chemical shifts of the two protons at the CH2 group
(marked with an arrow in 1).
R-11J ,
R-11K ,
Each spectrum has an isolated AB pattern at 2.7 and
3.1, and an ABMX3 pattern, which is as expected for
the compounds 5 - 8. In compounds 1 - 4 the coupling
pattern would be more complicated, since one of the
cyclopropane protons would be coupled to the CH2.
The AB coupling of the cyclopropane protons is identical in
both, so stereochemistry has to be the same (both cis). The
other coupling constants are quite similar, and do not
provide much insight. However, there are some large
chemical shift differences - in G the Me doublet is unusually
upfield (δ 0.2), and the CHMe multiplet normal (δ 2.5),
whereas in H the methyl is normal (δ 0.9) , and the multiplet
unusually upfield (δ 1.2). This can only be explained by
large anisotropy effects of the phenyl group in compounds
5 and 6.
R-11K R-11J
H
O
Ph
CH3
H
H
O
Ph
H
CH3H H
5 (R-11K) 6 (R-11J)
Endo protons are both upfield
The relatively large vicinal coupling of the two cyclopropane
protons (10 Hz) shows a cis relationship between them.
0.3
1.95
1.1
0.95 Normal
2.0 2.7
2.0 2.7
Both 3
J
same
2
3
2
10
2 2
1 1
5 5
2 2
Problem R-11J (C19H18O)
200 MHz 1
H NMR Spectrum in CDCl3
(Source: Drew Weber/Zimmerman 11/18)
Problem R-11K (C19H18O)
200 MHz 1
H NMR Spectrum in CDCl3
(Source: Drew Weber/Zimmerman 11/18)
3.0 2.5 2.0 1.5 1.0 0.5 0.0ppm
1.00
2.98
1.05
2.90
H2O
H2O
3.0 2.5 2.0 1.5 1.0 0.5 0.0ppm
1.00
2.08
0.94 0.97
3.05
Hz
020406080
Hz
020406080
Hz
020406080
Ph
H
O
Ph
H
CH3H
R-11K
Ph
H
O
Ph
CH3
HH
R-11J
Problem R-11L and R-11M. From the 270 MHz 1
H NMR spectra of two stereoisomeric bromo pentaacetoxy
cyclohexanes assign stereochemistry and conformation ("interpret" means give δ, J and multiplicity).
OAc
Br
AcO
(a) Interpret the signal at δ 4.5 in R-11L. Suggest possible part structures. Circle the proton at δ 4.5
(c) Identify other significant multiplets in the expansion (δ 5.3 -
5.8) on the right (R-11L), draw coupling trees, and identify part
structures. HINT: these are all first order
OAc
AcO
AcO
(b) Interpret the signal at δ 5.2 in R-11L. Suggest possible part structures. Circle the proton at δ 5.2.
(d) Complete the structure for R-11L below by placing bromo and acetoxy groups with the appropriate
stereochemistry on the structure.
R-11L
Br
H
5.7 5.6 5.5 5.4
ppm
30
δ 4.5, t, J = 4 Hz - the chemical shift
suggests a CHBr proton, the small
couplings must be ee or ae:
AcO
H
Br
OAc
Less likely, but possible
δ 5.2, dd, J = 10, 4
- must be an axial proton (10 Hz J) with one neighboring eq and one ax
- probably coupled to 4.5 proton (the one at 5.7 has a smaller
J, and leaning suggests it is coupled to the t at 5.59)
Br
H
H
OAc
OAc
4.5
5.2
δ 5.43, t (J = 10), axial proton, axial on both sides
δ 5.56, t (J = 10), axial proton, axial on both sides
δ 5.59, t (J = 3.5), equatorial proton
δ 5.70, dd (J = 10, 2.5), axial on one side, equatorial on other
AcO
OAc
Br
OAc
OAc
AcO H
H
H
H
H
H
The two adjacent triplets (5.43 and 5.56) means that there
must be four adjacent axial hydrogents. The other two protons
must be equatorial (triplets with J <4 Hz)
4
4
6
4
4.5
5.6
5.7
5.5
5.4
5.43
5.56
5.595.70
5.2
5.43 and 5.56 are coupled to each other - hence adjacent
H
Any stereochem here
4.5
5.65.7
5.5
5.4 5.2
4
4
4
10
10
10
e
ea
a
a a
(e) What do the signals at δ 2 tell you about the structure of R-11M (compare them to the δ 2 signals of R-11L).
(f) Assign and interpret the signal at δ 4.0 in R-11M. Suggest possible part structures.
(g) Analyze the rest of the NMR spectrum of R-11M. Point out significant features of the spectrum which can be
used to assign stereochemistry. HINT: there are some second-order effects in the multiplet δ 5.1 - 5.5.
(h) Complete the structure of R-11M below by placing bromo and acetoxy groups with the appropriate
stereochemistry on the structure.
R-11M
H
OAc
H
Br
H
OAc
The CH3-C
O
signals are in a 2:2:1 ratio, so there is likely a symmetry element in the molecule (there are 5
different ones in R-11L)
δ 4.0, t (J = 10)
From the shift, this must the CHBr proton. The 10 Hz triplet means
the proton is axial, and so are both flanking protons
δ 5.15, t (J = 9 Hz)
δ 5.35, probably a dd J = 8, 9 Hz)
It looks as if all protons are of the axial type, although there is a bit of a mess
at δ 5.3, proably some second order effects (or an impurity?). Can't be due
to one equatorial proton, because that will cause 2 other protons to have one
small coupling.
AcO
H
OAc
H
H
OAc
AcO OAc
Br
H
H
H
This has a plane of symmetry, so there are only 4 kinds of
CH protons and 3 kinds of acetate Me
3
3
3
3
4.05.15
5.355.27
Problem R-11L (C16H21BrO10)
270 MHz 1
H NMR Spectrum in CDCl3
(Source: Ieva Reich 11/15)
AcO
OAc
Br
OAc
OAc
AcO H
H
H
H
H
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 0.5 0.0
0.84
3.82
15.78
1.00
5.2 4.45 4.40
2.2 2.1 2.0
5.7 5.6 5.5 5.4
ppm
1456.8
1466.4
1476.0
1492.1
1501.8
1506.0
1509.4
1511.3
1512.6
1530.9
1534.3
1541.1
1544.3
Problem R-11M (C16H21BrO10)
270 MHz 1
H NMR Spectrum in CDCl3
(Source: Ieva Reich 11/15)
AcO
H
OAc
H
H
OAc
AcO OAc
Br
H
H
H
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 0.5 0.0
1.00
5.23
15.94
5.4 5.3 5.2 5.1 4.1 4.0 3.9 2.1 2.0
Problem R-11N (C9H16ClN). In this problem you are required to determine the position of a Cl substituent in a
1-aza-bicyclo[2.2.2]heptane from the 1
H NMR spectra. You are given the 1
H NMR spectra of the compound and the
7,7-dideuterated analog.
(a) Analyze the coupling system of R-11N and report your results below. For each position either give the
multiplicity, J and δ values, or enter Cl if that is where you think it is. NOTES: 1 .You may use first order analysis there
are no significant second order effects. 2. There are no effects detectable due to coupling between H and D.
2x
2n
3x
3n
6x
6n
7s
7a
(b) Briefly describe how you decided on the location of the chlorine
(d) Briefly describe how you distinguished the x and n signals at carbons 2 and 3
(c) Briefly describe specifically how you distinguished proton(s) at 2 from those at 3.
17
δ 3.1, ddd, J = 13, 10, 4.5
δ 2.6, dddd, J = 10, 9, 7, 3
δ 1.3, ddd, J = 12, 10, 6
δ 1.7, dddd, J = 10, 9, 5, 2
Cl
δ 4.6, d, J = 2
δ 2.15, dd, J = 10,2
δ 2.95, dt, J = 10, 2
The four protons at C-2 and C-3 are all coupled to each other (and the endo protons are coupled
to 7s), if the Cl was on one of these they would be much simpler.
The proton at δ 4.6 has to be the CHCl proton. It is coupled only to one of the C-7 protons (7a). So
the Cl must be in the exo position
Simple chemical shift argument - α to N is downfield. Couplings are nearly the same
Only the endo (2n, 3n) can have a long range W-coupling to the C-7 protons - that coupling
disappears in the D compound
3
2
2
10
3
There is also the absence of a W-coupling between H2x and H6x, which would be expected to be
significant in any other isomer
H7aH7s
H2x
H3x
H3n
H2n
H6x
H6n
N
2
3
6
5
7
One proton replaced by Cl
4 3 2 1
ppm
1.00
3.02
2.05
0.98
1.02
2.18
2.08
0
Hz
102030
One of the H atoms at positions 2, 3, 5 or 6 is
substituted by a Cl. Which one?
DD
H2x
H3x
H3n
H2n
H6x
H6n
N
2
3
6
5
7
H7aH7s
H2x
H3x
H3n
H2n
H6x
H6n
N
2
3
6
5
7
Problem R-11N (C9H16ClN)
100 MHz 1
H NMR Spectrum in CDCl3
(Source: JACS 1968, 90, 13551 4/45)
H2x
H3xH3n
H2n
H6n
H7a
H7s
2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0
ppm
HS
SH
2.8 2.6 2.4 2.2 2.0 1.8 1.6 1.4 1.2 1.0
ppm
SH
Problem R-11O. Identify the SH protons in the two 300 MHz 1
H NMR spectra (CDCl3) below, and explain the
difference in their appearance (Source: Aldrich NMR Library).
8
SH
SH
The spin system is AA'BB'B''B''' - so decidely second order. Specifically, the B protons (CH2CH2) are stongly
coupled, so the A protons (SH) are second order - an example of virtual coupling.
In the top spectrum the SH proton is coupled to the CH2 protons, which are well separated from their other coupling
partner, the second CH2 group. So the SH proton is a simple triplet.
2
2
4