15 Problems in Spectroscopy 2010 (1)

8/6/2019 15 Problems in Spectroscopy 2010 (1)

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Fifteen Problems in

Spectroscopy

Sren Brgger Christensen

Department of Medicinal Chemistry

Faculty of Pharmaceutical Sciences

University of Copenhagen

2010

8.0 7.9 7.8 7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.100.96


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2

2009 The University of Copenhagen. All rights reserved. No part of this material may be reproduced

without prior written permission of the publisher.

Preface

The following fifteen problems are a continuation of Ten Problem in Spectroscopy, which was used

as supplementary exercises for a course in spectroscopy at FARMA. The intention is to offer somepossibilities for students to work with problems in spectroscopy either individually or in groups.

The present edition differs from the first edition by adding five problems (Problem 1-5, Part 1) withcompounds, the structures of which can be solved solely by taking advantage of the IR,

1H NMR and

molecular weight as revealed from the mass spectrum. The13

C NMR spectrum and DEPT135 spectrum arealso given. If the signal in the

13C NMR spectrum is assigned a 0 the signal wil l disappear in the DEPT

spectrum, a + indicates that the signal will have a positive intensity and a - indicates a negative intensity.The

13C NMR spectra can be used as a possibility for interpreting such spectra when the students feel

confident with this spectroscopical technique.

In some of the problems two1H NMR spectra are recorded: one spectrum of the compound

dissolved in a deuterated solvent and one spectrum after addition deuterium oxide (D2O). By comparison ofthe spectra you will observe that one or more signals have disappeared from the latter spectrum. In this case

advantage is taken of the possibility of exchanging protons attached to oxygen, nitrogen or sulfur atoms withdeuterium by shaking the solution with deuterium oxide. Consequently it can be concluded that thedisappeared signal originates in a proton attached to either an oxygen, a nitrogen, or a sulfur atom.

In worked Problem 1 an example is given on a strategy for structure elucidation.

The problems 6-15 (Part 2) present molecules with more complex structures. Such problems aresolved using all the information available from MS,

1H NMR, and

13C NMR spectra. Worked Problem 2

presents a strategy for dealing with this kind of problems.

It is always difficult to predict the challenge of problems, especially for someone who knows theanswers. That is why the problems appear at random. I would, however, assume the problems 11, 12, and13 might appear a minor challenge than other problems in Part 2.

In the problems 6 and 7 you will notice that the1H NMR spectrum is recorded at 300 MHz as well as

at 600 MHz. These spectra illustrate how a complex NMR pattern simplifies at higher field. In problem 6 aCOSY spectrum helps interpreting the contiguity of the molecule.

The following spectra are spectra recorded of real compound, meaning that you will experience thefalse signal that might appear e.g. in an IR spectrum, if the potassium bromide used for preparing the disc ishumid.

Before looking up the solution I suggest that you draw the expected molecule in a drawing programthat enables you to predict the spectra (e.g. ChemBioDraw) and ask the program to predict the

1H and

13C

NMR spectra.

The solutions to the problems are given the last page as the CAS numbers of the compounds.

Finally I wish you fun when solving the problems.

Acknowledgement: Nils T. Nyberg has recorded the 600 MHz spectra, Birgitte Simonsen the IR spectra,and Dorthe M. Skytte has performed careful prove reading and given creative inputs.

Copenhagen, October 2009

S. Brgger Christensen


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IndexPreface ........................................................................................................................................................... 2

Part 1 ............................................................................................................................................................. 4

Worked Problem 14

Elucidation of configuration using molecular weight, IR and1H NMR spectra ............................................. 6

Problem 1 .................................................................................................................................................. 8

Problem 2 ................................................................................................................................................ 10

Problem 3 ................................................................................................................................................ 12

Problem 4 ................................................................................................................................................ 15

Problem 5 ................................................................................................................................................ 17

Part 2: Problems 6 15 ................................................................................................................................... 20

Elucidation of the configuration of organic compounds using EIMS, IR,1H NMR and

13C NMR spectra 20

Mass spectrometry for Analysis of Elemental Composition of Organic Compounds.............................. 21

Worked Problem 2....................................................................................................................................... 22

Elucidation of configuration: ........................................................................................................................... 24

Problem 6 ................................................................................................................................................ 27

Problem 7 ................................................................................................................................................ 31

Problem 8 ................................................................................................................................................ 35

Problem 9 ................................................................................................................................................ 38

Problem 10 .............................................................................................................................................. 41

Problem 11 .............................................................................................................................................. 45

Problem 12 .............................................................................................................................................. 47

Problem 13 .............................................................................................................................................. 50

Problem 14 .............................................................................................................................................. 53

Problem 15 .............................................................................................................................................. 56

Solutions to the problems (CAS numbers): ............................................................................................. 59


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Part 1

Worked Problem 1

EIMS (75eV)

10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 1000

50

100

15

26

27

29 39

41

42

43

4458

71

86

87

O

IR (Liquid sandwich)

MS+

= 86


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Worked problem 1

1H NMR (300 MHz, CDCl3)

2.5 2.0 1.5 1.0Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.071.953.102.00

13C NMR (75 MHz, CDCl3)

200 180 160 140 120 100 80 60 40 20Chemical Shift (ppm)

0

0.25

0.50

0.75

1.00

NormalizedIntensity

+-

0

+

208.9

1

45.6

5

29.8

7

17.3

213.7

2

-


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Elucidation of the configuration of an organic compound using molecular

weight, IR and1H NMR spectra

The structures of the compounds, in which the spectra given in problems 1 to 5 originate, can be

elucidated solely by interpreting the IR and 1H NMR spectra and taking advantage of the molecular

weight as stated on the EIMS spectrum. The EIMS and13

C NMR spectra are also given. The latterspectra might be interpreted at a later stage.

Molecular weight

The molecular weight is stated to be 86 g/mol.

Functional groups

The IR spectrum reveals a strong absorbance just below 3000 cm-1 indicating the presence of

hydrogen atoms attached to sp3 hybridized carbon atoms. In addition the spectrum reveals a

strong absorbance at 1720 cm-1 indicating the presence of a ketone group. Besides these bands

the spectrum reveals no strong absorbance above 1500 cm-1.

Building blocks1H NMR spectrum

The building blocks of the molecule can be interpreted from the 1H NMR spectrum revealing the

presence of two methyl groups (2 CH3, 2.10 and 0.90 ppm, integral 3H for each) and two

methylene groups (2 CH2, 2.40 and 1.60 ppm, integral 2H for each). A combination of the atoms

present in the groups affords a weight of 215 + 214 = 58. Since the molecular weight was 86 the

weight of the remaining building block must be 28, corresponding to a C=O group. This latter group

already was expected from the IR spectrum revealing that the compound was a ketone.

The contiguity of the molecule

The contiguity of the molecule is revealed from the multiplicities (coupling patterns) of the signals in

the 1H NMR spectrum as listed in the below Table 1.

Intensity Multiplicity J Interpretation

2.40 2 t 6.9 Hz CH2-CH2-CO

2.10 3 s 6.9 Hz CH3-CO

1.60 2 sextet 6.9 Hz CH3-CH2-CH2

0.90 3 t 6.9 Hz CH3-CH2

Table 1: Systematic representation of the signals in the 1H NMR spectrum.

The coupling constants, J, are measured on the spectrum as illustrated below:

1. The distance between 1.0 and 2.0 on the X-axis measured with a ruler is 56 mm.

2. Since the spectrum is recorded at 300 MHz 1 ppm equals 300 Hz. From point 1 it appears

that 56 mm is the distance between 1 and 2 ppm on the X-axis and therefore 56 mm also is

the distance of 300 Hz on the X-axis. Division 300/56 reveals that 1 mm equals 5.3 Hz.


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3. The distances between the peaks of the triplet at 2.4 ppm measured with a ruler is 1.3 mm

corresponding to 1.35.3 = 6.9 Hz.

The signal at 2.40 is a triplet (J6.9 Hz) integrating for two protons. The -value indicates that the

signal is next to the C=O group and the multiplicity indicates that the signal is also next to a CH2-

group.

The signal at 2.10 is a singlet integrating for three protons. The -value indicates that the signal is

next to a C=O group: CH3-(C=O)-.

The signal at 1.60 is a sextet (J6.9 Hz) integrating for two protons, indicating that the signal is

only weakly affected by elctronegative groups and 5 neighboring protons are affecting the signal.

The neighboring groups have to be a methyl and a methylene group.

The signal at 0.90 is a triplet (J6.9 Hz) integrating for three protons, indicating a methyl group

affected by a neighboring methylene group.

All this information is given in the table.

Based on this knowledge the molecule has to be: CH3-(C=O)-CH2-CH2-CH3.

Prediction of spectrum

Figure 1 reveals the spectrum predicted by ChemBioDraw (Draw the compound in the drawing

program, mark the compound, go into structure, and choose predict 1H-NMR shifts):

Figure 1:1H NMR spectrum predicted by ChemBioDraw. As often is the case with simple molecules the

prediction is very good.


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Problems 1-5

Problem 1

EIMS (75 eV)

10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85 90 95 1000

50

100

26

27

28

29

30 39 41 4356

57

58

86

87

O

IR (Liquid sandwich, the peak at 1600 cm -1 is an internal reference)

MS+

= 86


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Problem 1


3.5 3.0 2.5 2.0 1.5 1.0 0.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.002.06


220 200 180 160 140 120 100 80 60 40 20 0 -20Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

-

0

211.9

5

35.4

5

7.94


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Problem 2

EIMS (75 eV)

10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 780

50

100

18

27

28

29

30

31

41

42

43

45

46 55 57

59

73

HO

IR (Liquid sandwich, the peak at 1600 cm -1 is an internal reference)

MS+

= 74


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Problem 2


4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.933.032.021.201.00

1H NMR (300 MHz, CDCl3 added D2O)

3.5 3.0 2.5 2.0 1.5 1.0Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.953.162.081.00

3.85 3.80 3.75 3.70 3.65 3.60 3.55 3.50 3.45Chemical Shift (ppm)

0

0.05

0.10

0.15

NormalizedIntensity

1.00


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.933.032.02


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Problem 2


85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10 5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

+

-

+

69.4

2

32.0

12

2.8

9

10.0

3

Problem 3

EIMS (75 eV)

10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 780

50

100

15 26

27

28

29

30

31

33

38

39

40

41

42

43

4450 53

5557

5973

74

OH

MS+

= 74


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Problem 3



4.0 3.5 3.0 2.5 2.0 1.5 1.0

Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.181.001.132.15

2.00 1.95 1.90 1.85 1.80 1.75 1.70 1.65 1.60 1.55i l i

0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

Normalized

Intensity

1.00


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Problem 3


3.5 3.0 2.5 2.0 1.5 1.0 0.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

5.950.932.00


85 80 75 70 65 60 55 50 45 40 35 30 25 20 15 10Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

+

-

69.6

4

30.8

1

18.9

1


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Problem 4

EIMS (75 eV)

10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 780

50

100

14

15

19

26

27

28

29

31

32

33

38

39

41

42

43

45

55

56

5773

OH

IR spectrum (Liquid sandwich, the peak at 1600 cm-1 is an internal reference)

MS+

= 74


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Problem 4



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.002.062.061.082.07



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.002.041.922.03

1.60 1.55 1.50 1.45 1.40 1.35 1.30 1.25 1.20 1.15

0

0.05

0.10

0.15

0.20

NormalizedIntensity

2.062.06


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Problem 4



-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+--

-

62.6

0

34.8

2

18.9

4

13.9

1

Problem 5

EIMS (75 eV)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 2300

50

100

29

39 4550

65 76 93 105 121

132

149

177

222

O

O

O

O

MS+

= 222


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Problem 5



7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.002.050.99

0.93


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Problem 5



0

0.25

0.50

0.75

1.00

NormalizedIntensity +

-

+

+

0

167.4

1

132.0

7

130.7

9

128.6

9

61.6

0

14.1

7

0


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Part 2: Problems 6 15

Elucidation of the configuration of an organic compound using EIMS, IR, 1H

NMR and13C NMR spectra

On the following pages you will find the EIMS, IR, 1H NMR, and 13CNMR spectra of ten organiccompounds. The challenge is to elucidate the configurations of the compounds. A possible strategyis firstly to interpret the MS to deduce the molecular weight. Inspection of the pattern at M+,(M+1)+, and (M+2)+ or the similar pattern at dominant peaks in the mass spectrum can be used forestimating the elemental composition of the molecule (cf. the following paragraph on massspectrometry). The functional group(s) in the molecule might be deduced from the IR spectrumand, if the functional group contains a carbon atom, from the 13C NMR spectrum. The DEPT-135combined with the 13C NMR spectrum reveals the number of quaternary carbon atoms, the numberof methine, methylene and methyl groups in the molecule. The DEPT-135 spectrum is indicated inthe 13C NMR spectra by annotation of 0 if the signal disappears, + if the signal has a positiveintensity, and - if the signal has a negative intensity in the DEPT-135 spectrum. Use theintegration curves in the 1H NMR spectrum to confirm the number of hydrogen atoms in themolecule. By adding the weight of the revealed groups an estimate of the molecular weight can bemade. If the estimated weight corresponds to the weight as revealed from the MS the buildingblocks involved in the molecule are known. Finally the contiguity of the molecule to give the finalconfiguration can be obtained by interpreting the coupling pattern of the signals in the 1H NMRspectrum.

In Worked Problem 2 you will find an example of structure elucidation using the mentionedstrategy. In the following paragraph you will find see how the signals in the mass spectra with m/zvalues a few units higher than the value of the molecular weight might be used for estimation of themolecular formula.


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Mass spectrometry for Analysis of Elemental Composition of Organic

Compounds

Almost all elements are a mixture of more isotopes. In general the major isotope possesses thesmallest atomic mass (an exception is borine). The molecular mass is defined as the mass of the

molecule containing the lightest isotopes. A mass spectrum thus will display a peak for themolecular ion plus a number of peaks with larger values for m/z. The pattern of these peaks willdepend on the elemental composition of the molecule and might thus be helpful for estimating themolecular formula. The most common elements in organic molecules are listed in Table 1:

Table 2: Isotopes of common elements in organic compounds:

Element

A A+1 A+2

Type of elementAtomic weight % Atomic weight % Atomic weight %H 1 100 2 0.015 A

C 12 100 13 1.1 A+1

N 14 100 15 0.37 A+1

O 16 100 17 0.04 18 0.20 A+2

F 19 100 A

Si 28 100 29 5.1 30 3.4 A+2P 31 100 A

S 32 100 33 0.79 34 4.4 A+2

Cl 35 100 37 32.0 A+2

Br 79 100 81 97.3 A+2

I 127 100 A

The elements might be grouped into three types of elements: Type A, Type A+1, and Type A+2. Ifthe intensity of the peak two units higher than the molecular ion is less than three per cent of thatof the molecular ion it may be concluded that none of the elements Si, S, Cl or Br is present. Somecharacteristic patterns for molecules containing chlorine and bromine molecules are given in (Table2).

Table 3: Abundance of isotope peaks in chlorine and bromine containing ions.

A A+2 A+4 A+6

Cl 100 32

Cl2 100 64 10

Br 100 97

Cl3 100 96 31 3

BrCl 77 100 24

BrCl2 62 100 45 6

Br2 51 100 49

Some drawing software like ChemDraw shows the pattern above the molecular peak in theanalysis window (in ChemBioDraw mark the molecule, choose the View drop down menu, choose

Analysis Window).

The number of carbon atoms in the molecule can be calculated according to the below equation:

in which 1.1 is the percentage of naturally occurring 13C, n is the number of carbon atoms in themolecule, I(M+1) is the intensity of the peak at (M+1), I(M) is the intensity of the molecular peak.


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Worked Problem 2

EIMS (75 ev)

10 14 18 22 26 30 34 38 42 46 50 54 58 62 66 70 74 780

50

100

15

26

27

28

29

31

37

38

39

40

41

43 49

51

52

53

54

6365

67

68



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1H NMR Spectrum (300 MHz, CDCl3)

2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.162.121.002.06

13C NMR Spectrum (75 MHz, CDCl3)

85 80 75 70 65 60 55 50 45 40 35 30 25 20 15Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedInt

ensity

+-

-+

0

84.4

6

68.1

5

22.0

0

20.4

413.4

2


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Elucidation of configuration of worked problem 2:

Molecular weight

EIMS indicates a molecular weight of 68

Building blocks13C NMR and DEPT-135 shows

1 quaternary C

2 CH2

1 CH and 1 CH3, or 2 CH, or 2 CH3.

The chemical shifts of the signals indicate 1 CH and 1 CH3. This is confirmed by inspection of the1H NMR spectrum in which one signal integrates for 1H and one signal for 3H.

Addition of the weights of these building blocks gives 12 + 214 + 13 + 15 = 68. All the buildingblocks of the molecule have been found.

Molecular formula: C5H8 DBE: (10+2-8) = 2

Functional groups

IR spectrum:

A strong band at 3311 cm-1 and 2117 cm-1 shows the presence of a terminal triple bond. Noadditional functional groups appear to be present.

The presence of a triple bond is confirmed by signals at 84 and 68 ppm in the 13C NMR spectrum.

The contiguity of the molecule1H NMR spectrum

The contiguity of the molecule is revealed from the coupling patterns of the signals of the protonsin the molecule (Table 4).

Intensity Multiplicity J Interpretation

2.15 2 dt 3 and 7 Hz H- - CH2-CH2

1.93 1 t 3 Hz H- - CH2

1.55 2 sextet 7 Hz CH2-CH2-CH3

0.90 3 t 7 Hz CH3-CH2

Table 4: Systematic representation of the signals in the 1H NMR spectrum.

The small size of the coupling constant (3 Hz) of the signals at 2.15 ppm and 1.93 ppm indicatethat this is a long range coupling (larger than 3-bonds coupling, indicated by using two hyphensbetween the H and the C in the Table). All the other coupling constants (7 Hz) are as expected forvicinal couplings (3-bonds couplings).


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The coupling constants are measured as described in Worked Problem 1

The splitting pattern of the terminal CH3 (t) reveals that it must be bound to a CH2 group: CH3-CH2-.

The splitting pattern of the CH2 group bound to the CH3 (sextet) reveals that it must be bound to anadditional CH2: CH3-CH2-CH2-.

Remaining building blocks are C and CH. Since the IR and 13C NMR spectrum indicated a terminaltriple bond we can conclude that these two carbons must be a CC-H group yielding the molecule:

CH3-CH2-CH2-CC-H.

The presence of a coupling constant of 3 Hz in only in the two signals at 2.15 and 1.93 proves thatthe protons, in which these signals origin, must couple with each other.

The coupling pattern of the CH2-group next to the triple bond can be explained by the couplingdiagram (Figure 2):


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.06

Figure 2: Coupling diagram of the CH2 group next to the triple bond.

Prediction of the spectra

If the molecule is drawn in ChemBioDraw and the program is prompted to predict the 1H NMRspectrum (mark the molecule, choose the drop down menu at Structure and ask for the predictionof 1H NMR) the below spectrum is predicted (Figure 3):

Figure 3: 1H NMR spectrum predicted by ChemBioDraw for 1-pentyne.


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Notice that the signal originating in the CC-H is predicted to appear at 2.7 ppm but actuallyappears at 1.93 ppm. The other signals are reasonably well predicted. Notice also the goodprediction of the coupling patterns.

The predicted 13C NMR spectrum is shown below (Figure 4):

Figure 4: 13C NMR of 1-pentyne predicted by ChemBioDraw.

Interpretation of the spectra (Figure 5):1H NMR 13C NMR

Figure 5: Interpretation of the NMR spectra

Interpretation of the MS

In the MS spectrum the peak at m/z53 must origin form M+-15, the peak at m/z67 must originform M+-1, and the peak at m/z40 equals the loss of an ethylene molecule, which could be causedby at McLafferty rearrangement (Figure 6):

m/z= 68 m/z= 40

Figure 6: Possible McLafferty fragmentation of 1-pentyne.


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Problem 6

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 2400

50

100

27

41

55

60

69

73 83

87

97

101 115

125

135

143

163

179 193 205 223

HO

Br

O

IR (KBr)

MS+

= 222


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Problem 6


12 11 10 9 8 7 6 5 4 3 2 1Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

8.042.032.042.000.68

3.5 3.0 2.5 2.0 1.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

8.042.032.042.00


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Problem 6



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

180.30

77.4

2

77.0

0

76.5

8

33.9

9

33.8

832.6

7

28.8

1

28.3

7

27.9

2

24.5

0

The signal at 180 will disappear in a DEPT-135 experiment and all the other peaks will obtain a

negative intensity.

1H NMR spectrum (600 MHz, CDCl3)

12 11 10 9 8 7 6 5 4 3 2 1Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

4.022.012.082.000.80

The signal at 11 ppm disappears after addition of D2O.

36 34 32 30 28 26 24 22 20Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

33.9

9

33.8

8 32.6

7

28.8

1

28.37

27.92

24.5

0


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30

Problem 6


3.7 3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.082.00


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

4.022.022.072.01


31/59

31

Problem 7

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 1900

50

100

15

29

39

41 45

51

53 63

65

68

74

79

81

92

96

107

122

137

149

153

167

182

OHO

O

O

IR (KBr)


32/59

32

Problem 7


12 11 10 9 8 7 6 5 4Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

N

ormalizedIntensity

3.182.151.000.69

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 6.6Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.100.96

11.5 11.0 10.5 10.0 9.5 9.0ChemicalShift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedInt

ensity

After addition of D2O


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33



0

0.25

0.50

0.75

1.00

NormalizedIntensity

+

+

+0

+

0

0

0

165.3

3

151.8

9

148.0

0

124.8

21

23.7

0

121.9

9

117.3

3

62.1

9

56.1

7

+


34/59

34

Problem 7

1H NMR spectrum (600 MHz, CDCl3)

8.0 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.272.030.90

7.85 7.80 7.75 7.70 7.65 7.60 7.55 7.50 7.45 7.40 7.35 7.30 7.25 7.20 7.15 7.10 7.05

Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Normalized

Intensity

2.030.90


35/59

35

Problem 8

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 1800

50

100

1529

31

39

41

43

51

55

63

65

69

76

90

95

104

107

112

122 133

148

166

O

O O

IR (KBr)


36/59

36

Problem 8



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.892.141.060.90

7.7 7.6 7.5 7.4 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.5 6.4 6.3 6.2Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedI

ntensity

2.141.06


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37

Problem 8



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

+

0

+

0

+

189.1

7

161.9

6

135.7

7

114.1

4

103.7

2

77.4

3

77.0

0

76.5

8

56.0

4


38/59

38

Problem 9

EIMS (75 ev)

40 50 60 70 80 90 100 110 120 130 140 1500

50

100

4251

5563

66 69

81

109

138

HO

OH

O

IR (KBr)


39/59

39

Problem 9

1H NMR Spectrum (300 MHz, CD3OD)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

0.981.10

7.10 7.05 7.00 6.95 6.90 6.85 6.80 6.75 6.70 6.65 6.60 6.55 6.50 6.45 6.40 6.35 6.30 6.25Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

0.982.08


40/59

40

Problem 9

13C NMR (75 MHz, CD3OD)

192 184 176 168 160 152 144 136 128 120 112 104 96 88Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

+

0

0

+

194.0

6

160.4

2

140.0

3

109.8

2

108.7

6


41/59

41

Problem 10

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 1800

50

100

1527

39

41

51

53

55

63

65

75

77

81

91

94

103

115

121

131

137

149

164

OH

O

IR (liquid sandwich)


42/59

42

Problem 10



0

0.05

0.10

0.15

0.20

0.25

0.30

0.35

0.40

0.45

0.50

0.55

NormalizedIntensity

2.183.172.050.900.952.001.02

7.05 7.00 6.95 6.90 6.85 6.80 6.75 6.70 6.65 6.60 6.55 6.50 6.45 6.40Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.001.02

6.2 6.1 6.0 5.9 5.8 5.7 5.6 5.5 5.4 5.3 5.2 5.1 5.0ChemicalShift (ppm)

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.10

0.11

0.12

0.13

NormalizedIntensity

After addition of D2O


43/59

43

Problem 10



0

0.05

0.10

0.15

NormalizedIntensity

2.050.900.95


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.183.17


44/59

44

Problem 10



-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

-

+

+

+

+

0

+

0

0

146.2

3 143.6

9

137.6

6

131.7

5

121.0

2

115.4

01

14.1

1

110.9

6

77.4

2

77.0

0

76.5

8

55.8

3

39.9

2

COSY spectrum (300 MHz, CDCl3)

7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0F2 Chemical Shift (ppm)

3

4

5

6

7

F1ChemicalShift(ppm)

-


45/59

45

Problem 11

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 2100

50

100

28

30

38

50

63

75

8092

117 129

143

155

171

185

201

N

O O

Br

IR (KBr)


46/59

46

Problem 11

1H NMR (CDCl3, 300 MHz)

8.5 8.0 7.5 7.0 6.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.002.01

13C NMR (CDCl3, 75 MHz)

165 160 155 150 145 140 135 130 125 120 115 110 105Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

+

0

+

0

14

6.8

0

132.4

6

129.8

4

124.8

6


47/59

47

Problem 12

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 1300

50

100

15 27

29

41

43

55

58

71

8199 114

O


MS+

= 114


48/59

48

Problem 12


2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.003.062.962.07


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

3.062.962.07

1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.003.06


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49

Problem 12


220 200 180 160 140 120 100 80 60 40 20 0 -20Chemical Shift (ppm)

0

0.25

0.50

0.75

1.00

NormalizedIntensity

+

+

+

-

-

0

209.2

5

41.8

5

32.6

6

27.7

0

22.3

8


50/59

50

Problem 13

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 1100

50

100

15

27

28

29

31

41

43

44

45

52

54

55

56

57

59

60 70

84

98

O

N


M+

= 99


51/59

51

Problem 13



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.802.004.02

3.5 3.0 2.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.004.02


52/59

52

Problem 13


2.5 2.0 1.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.802.00


120 112 104 96 88 80 72 64 56 48 40 32 24 16Chemical Shift (ppm)

0

0.25

0.50

0.75

1.00

NormalizedIntens

ity

+

--

-

0

118.1

9

67.0

5

65.3

1

19.3

0

15.3

6


53/59

53

Problem 14

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 1400

50

100

1518

28

30

39

41

44

46 57 70

74

86

NH2

OH

O

IR (KBr)

M+

= 131


54/59

54

Problem 14

1H NMR (D2O, 300 MHz)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.003.180.98

M e C N


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

0.98

MeCN


55/59

55

Problem 14

1H NMR (D2O, 300 MHz)


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.003.18

M e C N

13C NMR (D2O, 75 MHz)


0

0.25

0.50

0.75

1.00

NormalizedIntensity

+

+

-+

0

175.5

4

53.5

9 40.0

2

24.4

0

22.2

8

21.1

2

+


56/59

56

Problem 15

EIMS (75 ev)

10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 2700

50

100

18

29

41

57

73

87

101

111

129

143

147

156

185

O

O

O

O


M+

= 258


57/59

57

Problem 15



0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

6.002.170.972.132.07


0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

NormalizedIntensity

2.132.07

2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9 0.8 0.7 0.6 0.5Chemical Shift (ppm)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

Norm

alizedIntensity

6.002.170.97


58/59

58

Problem 15



0

0.25

0.50

0.75

1.00

NormalizedIntensity

+

-

+

--

0

173.2

2

70.4

8

34.0

2

27.7

62

4.5

5

19.1

7


59/59

Solutions to the problems (CAS numbers):

Problem 1 96-22-0

Problem 2 78-92-2

Problem 3 78-83-1

Problem 4 71-36-3

Problem 5 84-66-2

Problem 6 17696-11-6

Problem 7 1521-38-6

Problem 8 3392-97-0

Problem 9 26153-38-8

Problem 10 97-53-0

Problem 11 586-78-7

Problem 12 110-12-3

Problem 13 2141-62-0

Problem 14 61-90-5

Problem 15 141-04-8

Documents

15 Problems in Spectroscopy 2010 (1)