Page 1 C HAPTER 13 + 14 HW S OLUTIONS : S PECTROSCOPY INFRARED (IR) SPECTROSCOPY 1. A compound containing an NH 2 group should have two peaks in its IR spectrum in the 3100-3500 cm -1 range. Explain why there are two absorbances in this region, and use a drawing to show the type of vibration that produces each peak. The two absorbances are due to two different stretching modes of the two N-H bonds. One is a “symmetric” stretch, where the N-H bonds vibrate “in sync” (they each get longer at the same time), and the other is an “asymmetric stretch” where the vibration is out of sync (as one gets longer, the other gets shorter). 2. The following IR spectra are of the C 5 H 12 O 2 isomers below. Match each spectrum to the correct structure. 3. The following IR spectrum is of 1-hexene. Use arrows on the structure to indicate a possible vibration for each specified wavenumber. Signal at 3105 cm -1 (alkene C-H stretch) Signal at 1650 cm -1 (C=C stretch) O O CH 3 OH O O OH A B C C C H C H H C H H C H H C H H H H H C C H C H H C H H C H H C H H H H H 3105 1650 R N H H symmetric stretch R N H H asymmetric stretch A ~1700 is C=O OH from alcohol OH from carb acid C B
Transcript
Page 1
CHAPTER 13 + 14 HW SOLUTIONS: SPECTROSCOPY
INFRARED (IR) SPECTROSCOPY
1. A compound containing an NH2 group should have two peaks in its IR spectrum in the 3100-3500 cm-1 range. Explain why there are two absorbances in this region, and use a drawing to show the type of vibration that produces each peak.
The two absorbances are due to two different stretching modes of the two N-H bonds. One is a “symmetric” stretch, where the N-H bonds vibrate “in sync” (they each get longer at the same time), and the other is an “asymmetric stretch” where the vibration is out of sync (as one gets longer, the other gets shorter).
2. The following IR spectra are of the C5H12O2 isomers below. Match each spectrum to the correct structure.
3. The following IR spectrum is of 1-hexene. Use arrows on the structure to indicate a possible vibration for each specified wavenumber.
Signal at 3105 cm-1 (alkene C-H stretch)
Signal at 1650 cm-1
(C=C stretch)
O
O
CH3OH
O O OH
A B C
C C
H
C
H
H
C
H
H
C
H
H
C
H
HH
HH C C
H
C
H
H
C
H
H
C
H
H
C
H
HH
HH3105 1650
RN
H
H
symmetric stretchR
NH
H
asymmetric stretch
A
~1700 is C=O
OH from alcohol
OH from carb acid
C B
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4. The following IR spectrum is of 3-pentanone. Use arrows on the structure to indicate a possible vibration for each specified wavenumber.
Signal at 2997 cm-1
(alkane C-H stretch)
Signal at 1712 cm-1
(C=O stretch)
5. Which of the following structures would produce the IR spectrum below?
GENERAL CONCEPTS OF NMR
6. Explain in general what happens inside an NMR instrument when a compound absorbs radiation of the radio frequency.
In the presence of a strong magnetic field (inside the instrument), the nuclei of each spin-active atom (1H, 13C) exist in one of two states (a,b) with a particular energy difference between the states. If the applied radio waves matches that exact energy, the radiation is absorbed by a nucleus in the a state (aligned with the external magnetic field, lower energy), and the nucleus moves to the b state (opposed to the external magnetic field, higher energy). This is a “spin flip.”
7. Briefly explain why deuterated solvents are used in preparing samples for NMR analysis.
The NMR instrument requires dilute samples, and if the solvent contains 1H atoms, the signal from the solvent will overwhelm the spectrum. 2H nuclei (deuterium) have a completely different resonance frequency than 1H nuclei, so if the solvent has deuterium instead of hydrogen atoms, the solvent will appear invisible in the NMR. (Also in modern instruments, the 2H signal is used to “lock” the reference frequency.)
C
H
C
H
H
C
O
C
H
H
C
H
H
H
H
H C
H
C
H
H
C
O
C
H
H
C
H
H
H
H
H
OHO
A B C D
2997 1712
Alkyne C-H
stretch
Alkyne C-C
stretch
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CHEMICAL EQUIVALENCE
8. How many signals should ideally be present in the 1H NMR spectrum of each compound (how many different hydrogen environments are present)? Use labels (a,b,c…) to show which hydrogen atoms should be grouped together.
CHEMICAL SHIFT
9. Use a discussion of shielding to explain why hydrogen “A” absorbs at a higher frequency than hydrogen “B” in an NMR instrument. Use diagrams with your answer.
Hydrogen A is more electron poor than hydrogen B due to its proximity to the electronegative chlorine atoms. This means that the magnetic field generated by the hydrogen’s sigma electrons (which opposes the external magnetic field Bo of the NMR instrument) will be less than in B.
This contrary magnetic field is called “shielding” and causes the hydrogen nucleus to “feel” a smaller magnetic field (net B). Since A has less shielding, the nucleus feels a larger magnetic field, which corresponds to a larger energy gap between the a and b states, and a higher frequency absorption.
10. In each pair, which of the hydrogen atoms nearest the asterisks (*) should produce the furthest left signal in the 1H NMR spectra? Briefly explain each comparison.
The CH3O signal should be further left (higher ppm). These H are next to an electronegative atom so are electron poor, and this shifts the signal to the left.
The asterisked H on the left compound is next to two bromine atoms (right compound is next to one) which makes it more electron poor and should be further left in the NMR spectrum.
Oa
b cde
5 signals
a.O
2 signals
aa
b bb. O
3 signals
a
a a
ab bc cc.
H
O
4 signals
a
a
b c dd.CH3
CH33 signals
a
a
b
bc
ce.
NH2
H2N
3 signals
aa
b
bcc
cc
f.
Cl C
H
C
H
H
HCl
A B
OCH3
vs.a. * * Br Br
Brb.
** vs.
A
Bo
less shielding
net B
B
Bo
more shielding
net B
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11. Using an NMR correlation chart, assign the signals in the 1H NMR spectrum (a-c) to hydrogen atoms in the structure. (Note: signal “a” contains groupings that coincidentally overlap.)
SIGNAL SPLITTING
12. For the carboxylic acid below…
a. What is the expected splitting (singlet, doublet, etc.) for each hydrogen atom?
Signal Ha Hb Hc
Splitting triplet doublet singlet
b. Using the nuclear orientation of the neighbors, explain the origin of the splitting for hydrogen
atoms a +b. Use a diagram with your answer.
Splitting for Hb:
The Hb’s experience the external magnetic field PLUS a contribution from its neighbor Ha.
Since Ha can be either aligned or against the external field (a or b, about 50:50 each), half of the Hb’s “feel” an increased magnetic field (net B), and therefore absorb at a slightly increased frequency. The other half of the Hb’s feel a decreased magnetic field and absorb at a decreased frequency. These two “felt” magnetic fields lead to two signals (doublet).
Splitting for Ha:
Ha senses the magnetic orientation of both Hb neighbors. There are three possibilities for the magnetic spin states of the neighbors: both a (making the “felt” field bigger), both b (making the “felt” field smaller) and one each of a and b (having no effect on the “felt” field). These three situations correspond to the three absorbances of the triplet signal. There are two situations that have no effect on the felt field, giving the 1:2:1 ratio.
O
O
ab
c
aa a
a
02468PPM
a bc
Br C
H
Br
C
H
H
C
O
OH
a b
c
external field (Bo)
Ha (α)Ha (β)
"felt" field (net B)
"felt" field (net B)
external field (Bo)
Hb2 (α)
"felt" field (net B)
"felt" field (net B)
Hb1 (α)
Hb2 (β)
Hb1 (β)
Hb1 (α)
Hb2 (β)
Hb2 (α)
Hb1 (β)
"felt" field (net B)
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13. For the molecule below….
a. What is the expected splitting (singlet, doublet, etc.) for each hydrogen atom?
Signal Ha Hb Hc Hd He
Splitting triplet sextet triplet quartet triplet
b. Using the nuclear orientation of the neighbors, explain the origin of the splitting for hydrogen “d.”
Just as in the previous problem, Hd has 3 neighbors (He), each of which can be a or b. This leads to four different net magnetic orientations (as summarized below), corresponding to four absorptions in the quartet signal (with a ratio of 1:3:3:1).
aaa aab abb bbb
aba bab
baa bba
c. Sketch a drawing of the splitting pattern for the following hydrogen atoms as accurately as
possible. In each, state the relative heights of the lines in each signal (for example the lines of a triplet will be in the ratio of 1:2:1).
Signal Ha Hb Hd
Splitting drawing
Use Pascal’s triangle:
ab
c de
O
1 2 1 1 510 10 51 1 3 3 1
11 1
1 21 3 3 1
1
1 4 6 41 5 10 10
15 1
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DRAWING 1H NMR SPECTRUM FROM STRUCTURE
14. Draw the expected 1H NMR spectrum for each compound, paying attention to chemical shift and splitting. Mark the integration of each peak, then assign each peak in the NMR to hydrogen atoms in the structure.
8 7 6 5 4 3 2 1 0 ppm
H3CC
CH2CH3
Oa.
2H, q
3H, s
3H, t
a
b ca b
c doesn't matter if a or b is further left, both need to
be in 2-3 range.
8 7 6 5 4 3 2 1 0 ppm
b.
H3CC
O
H3C CH3
CH2 CH3
2H, q
9H, s
3H, t
a
b ca
a ab
cdoesn't matter if a or c is further left, both need to
be near 1.
8 7 6 5 4 3 2 1 0 ppm
c.
H3C C
H
CH3
Br
1H, sept.
6H, d
b
aa a
b
8 7 6 5 4 3 2 1 0 ppm
d.
BrCH
CH2
Br
CH2 Br
2H, q
2H, t
1H, t
aab b
cc
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DEGREES OF UNSATURATION
15. Calculate the degrees of unsaturation associated with each chemical formula. Then circle the structures beside it that are possible for the formula (which match the degrees of unsaturation).
DETERMINING STRUCTURE FROM AN 1H NMR SPECTRUM
16. Draw the structure that matches the molecular formula and would produce the 1H NMR spectrum for each problem below. Then assign all peaks in the NMR spectrum (use labels a,b,c) to hydrogen atoms in the structure.
a. C5H10
b. C6H10O2
OH
OHO
OHO
OH3CO
O
c. C7H6Br2 Br
Br
Br
Br BrBr
CH3
Br
Br
d. C5H9NOH2N
O
O
NH2
NH2
OH
O
NH2
012345678PPM
1H, m
6H, d
5H total
C9H12
c
b
a
4 d.u.
012345678910PPM
2H, quint.
3H, t
1H, t
C3H6O
a b
c
1 d.u.
[2(5)+2] – 10 = 1 d.u. 2
[2(6)+2] – 10 = 2 d.u. 2
[2(7)+2] – 8 = 4 d.u. 2
[2(6)+2] – 10 = 2 d.u. 2
a
b c
ca
a
a a
H
O
a bc
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17. The following four 1H NMR spectra correspond to four different alcohols with a molecular formula of C5H12O. Determine which alcohol corresponds to which spectrum, then assign the peaks in the spectrum to hydrogen atoms in each structure.
012345PPM
2H, m
3H, t
1H, s
2H, t
4H, m OH
e
d
d
c
ba
b e
c
da
"coincidental overlap"
012345PPM
1H, s1H, sext.
3H, t
3H, d
4H, m
OH
d e
c
cba b e
d
c
a
012345PPM
1H, s
6H, s
2H, q 3H, td
b
c
ca b
c
d
OH a
012345PPM
1H, s
2H, t
1H, m
6H, d
2H, q HO
b
d
c
e
eab
cd
ea
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18. Draw the structure that matches the molecular formula and would produce the 1H NMR spectrum for each problem below. Then assign all peaks in the NMR spectrum (use labels a,b,c) to hydrogen atoms in the structure.
012345PPM
2H
3H
3H
C4H8O2
O
Ob
a
c
a
b
c
012345678PPM
2H 2H
3HC7H7Cl
CH3
Cl
a
bb
a
c
a
c
b
01234PPM
2H, t 2H, t
9H, sC7H13BrO
O
Br
ba
ca
bc
c c
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18 continued
02468PPM
5H, m2H, t
2H, t1H, s
C9H13NHN
CH3
3H, s
ba
c
d ed
eca
a
aa
a
b
0123PPM
2H
3H
2H1H
3H
C4H11N NH2
edac
a bc
d
e
b
OH
Br
02468PPM
2H, d 2H, d
1H, s
3H, d
1H, q
C8H9BrO
Br
HO
a b
c
d
e This would be OK too:a
a
b
b
c
d e
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CARBON-13 NMR SPECTROSCOPY
19. List two differences between 13C NMR and 1H NMR spectroscopy, or the spectra they produce. • Different resonance frequencies: 1H absorbs at 300 MHz, and 13C at 75 MHz (on a 300 MHz
magnet). • Different ranges of ppm: 1H spans 0-12 ppm, and 13C spans 0-220 ppm. • Integrations correlate with the number of hydrogens in 1H NMR, but don’t correlate to the
number of carbons in 13C NMR. • Signals are often split in 1H NMR, but all signals are singlets in 13C NMR.
20. Determine the total number of signals present in the 13C spectrum of each compound. Then use a
correlation chart to indicate the approximate frequency in ppm of each carbon pointed to with an arrow.
21. An unknown is narrowed down to be one of the three structures below. How can the 13C NMR
spectrum of the unknown be used to determine the compound’s structure?
Due to differences in symmetry, each compound would have a different number of signals (2, 4, 3) in the 13C NMR. Counting the number of signals can determine the unknown’s structure.
22. Assign all peaks in the 13C NMR spectrum (use labels A,B,C) to carbon atoms in the structure.
Cl
Cl
O
O
H3CO
OH
HO
OH
OH
OH
OH
B C
D
A
OCH3
O
AB
C
DD
D
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COMBINED SPECTROSCOPY PROBLEMS
23. Below are 5 possible structures with a formula of C3H6O2. Identify which of the molecules A-E “fit” each piece of data. In many cases more than one compound matches each piece of data.
Data Which match?
The IR spectrum contains a large, broad peak around 3300 cm-1. (OH) A, B, C
The 1H NMR spectrum has a signal at 9.5 ppm. (Aldehyde H) B, C, E
The IR spectrum contains a strong peak at 1745 cm-1. (C=O of ester) (Assume this signal represents exactly what it suggests from the correlation chart.)
D, E
All signals in the 1H NMR spectrum are singlets. A, D
The 1H NMR spectrum contains a quartet integrating to 2H at 4.0 ppm. E
24. Which compound from the previous problem (A-E) would have the following IR signals and 1H
NMR spectrum? After identifying the compound, assign the IR signals to bonds in the molecule (fill in the table), and assign each peak in the 1H NMR spectrum to hydrogen atoms in the structure. Key IR signals:
Frequency (cm-1) Intensity Bonds represented
3210 to 3420 strong +broad O-H
2955 strong Alkane C-H
1725 strong C=O
O
OHH
O
H
O O
OCH3 H
O
O
OH
OHA B C D E
012345678910PPM
1H, t 2H, qa b
c
da
b
c
d
2H, t
H
O
OH
1H
Page 13
25. Use the chemical formula and given spectra (1H NMR, 13C NMR, and/or IR spectra) to determine the structure of each compound. Then assign each peak in the 1H NMR spectrum to hydrogen atoms in the structure.
