chem 2423. chapter 13. nuclear magnetic spectroscopy

1 CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W

CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W

Short Answer

1. For a nucleus to exhibit the nuclear magnetic resonance phenomenon, it must be magnetic. Magnetic nuclei

include:

a. all nuclei with even numbers of neutrons and protons

b. all nuclei with odd numbers of protons

c. all nuclei with odd numbers of neutrons

d. both b and c

2. Nuclear magnetic resonance spectroscopy provides information about a molecule's:

a. conjugated pi electron system.

b. size and formula.

c. carbon-hydrogen framework.

d. functional groups.

3. Explain why all protons in a molecule do not absorb rf energy at the same frequency.

Exhibit 13-1

The following question(s) pertain to the charting of NMR spectra. MATCH a term to each description below.

Place the letter of the term in the blank to the left of the description.

a. TMS e. low-field or downfield side

b. high-field or upfield side f. chemical shift

c. MHz g. specific absorption

d. delta ()

4. _____ When looking at an NMR chart the right-hand part of the chart is the _____.

5. _____ The exact place on the chart at which a nucleus absorbs is called its _____.

6. _____ The calibration standard for

1H and

13C NMR is _____.

7. _____ The NMR charts are calibrated using an arbitrary scale that is divided into _____ units.

Exhibit 13-2

For each of the compounds below tell how many signals you would expect the molecule to have in its normal,

broadband decoupled 13

C NMR spectra.

8.


9.

10.

11.

12.

13.

14.

15.

Exhibit 13-3

Identify the indicated sets of protons as unrelated, homotopic, enantiotopic, or diastereotopic.

16.


17.

18.

19.

Exhibit 13-4

For each compound below tell how many types of nonequivalent protons there are.

20.

21.

22.

23.

24.


25.

26.

27.

Exhibit 13-5

Predict the splitting patterns you would expect for each proton in the molecules below.

28.

29.

30.

31. The

1H NMR spectrum of styrene oxide shows that protons 1, 2, and 3 all have different chemical shift values.

Proton 1 is coupled to both proton 2 (J = 5.8 Hz) and proton 3 (J = 2.5 Hz). Draw a tree diagram for the

proton 1 signal.


Exhibit 13-6

Refer to the structure of 3-methyl-2-butanone below to answer the following question(s).

32. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled a.?

a. septet

b. quartet

c. doublet

d. singlet

33. Refer to Exhibit 13-6. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled b.?

a. septet

b. quartet

c. doublet

d. singlet

34. Refer to Exhibit 13-6. The carbonyl-carbon resonance of 3-methyl-2-butanone occurs at 208.7 ppm downfield

from TMS. How many hertz downfield from TMS would this carbonyl-carbon absorb if the spectrometer

used to measure this absorption were operating at 200 MHz?

35. Treatment of tert-butyl alcohol with hydrogen chloride yields a mixture of tert-butyl chloride (SN1 product)

and 2-methylpropene (E1 product). After chromatographic separation, how would you use 1H NMR to help

you decide which was which?

Exhibit 13-8

Below are three isomeric chlorobutanes and their 13

C NMR spectral data. MATCH the spectral data to the

correct structures by placing the letter of the spectrum in the blank to the left of the corresponding structure.

a = , 55.4, 36.2, 19.3

b = , 56.2, 35.3

c = , 56.0, 36.0, 24.2, 8.2


36. _____

37. _____

38. _____

Exhibit 13-8

Propose structures for compounds that fit the following 1H NMR data.

39. C8H9Br

3H doublet at 2.0 , J = 7 Hz

1H quartet at 5.0 , J = 7 Hz

5H singlet at 7.3

40. C7H14O

6H triplet at 0.9 , J = 7 Hz

4H sextet at 1.6 , J = 7 Hz


41. C3H6Br2

2H quintet at 2.4 , J = 6 Hz


42. C10H14

6H doublet at 1.2 , J = 7 Hz

3H singlet at 2.3

1H septet at 2.9 , J = 7 Hz

4H singlet at 7.0

43. C6H14

12H doublet at 0.8

2H septet at 1.4

Exhibit 13-9

To answer the following question(s), consider the data and 1H NMR spectrum below:

The mass spectrum of this compound shows a molecular ion at m/z = 113, the IR spectrum has characteristic

absorptions at 2270 and 1735 cm1

, and the 13

C NMR spectrum has five signals.


Spectrum obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/

44. Refer to Exhibit 13-9. Based on the mass spectral data and the IR data, what functional groups are present in

this compound?

45. Refer to Exhibit 13-9. How many types of nonequivalent protons are there in this molecule?

46. Refer to Exhibit 13-9. Describe the signal at 3.5 in terms of its integration, splitting pattern and chemical

shift.

47. Refer to Exhibit 13-9. Describe the signals at 4.35 and 1.3 in terms of their integration, splitting and

chemical shift.

48. Refer to Exhibit 13-9. What is the significance of the

13C NMR data?

49. Refer to Exhibit 13-9. Propose a structure for this compound.

50. How would you use

1H and

13C NMR to help you distinguish between these two isomeric structures?


Exhibit 13-10

Answer the question(s) for the compound whose 1H NMR spectra is shown below.

C4H8O


51. Refer to Exhibit 13-10. Describe each signal in terms of its integration, splitting and chemical shift.


Exhibit 13-11 Answer the question(s) for the compound whose

1H NMR spectra is shown below.

C5H12O



53. Refer to Exhibit 13-11. Calculate the degree of unsaturation for this compound.

54. Refer to Exhibit 13-11. Describe each signal in the

1H NMR in terms of its integration, splitting and chemical

shift.


Exhibit 13-12

Answer the question(s) for the compound whose 1H NMR spectra is shown below.

C8H7ClO



56. Refer to Exhibit 13-12. Calculate the degree of unsaturation in this compound.

57. Refer to Exhibit 13-12. Describe the signals that occur between 7 and 8 in terms of integration, splitting and

chemical shift.

58. Refer to Exhibit 13-12. Describe the signal at 2.6 in terms of integration, splitting and chemical shift.


60. Propose a structure for a compound, C6H14O, with the following

13C NMR spectral data:

Broadband decoupled 13

C NMR: 23.0, 68.4

DEPT-90: 68.4

DEPT-135: positive peaks at 23.0, 68.4 ; no negative peaks

61. Propose a structure for Compound X, which has M

+ = 120 and (M + 2) = 122 of approximately equal

intensity in its mass spectrum and has the following 13

C NMR spectral data:


C NMR: 32.6, 118.8, 134.3

DEPT-90: 134.3

DEPT-135: positive peaks at 134.3 ; negative peaks at 32.6, 118.8


62. Propose a structure for Compound Z, which has the following spectroscopic properties:

MS: M+ = 88

IR: 3380 cm1

1H NMR: 0.85 (6H doublet); 1.40 (3H multiplet); 2.68 (1H singlet); 3.55 (2H

triplet)


C NMR: 22.7, 25.0, 41.8, 60.5

DEPT-90: 25.0

DEPT-135: positive peaks at 22.7, 25.0 ; negative peaks at 41.8, 60.5

63. Describe how you could differentiate the following compounds using a DEPT NMR experiment.

64. Which structure of molecular formula C4H8Cl2 fits both the

1H NMR and

13C NMR spectra shown below?

a.

b.

c.

d.


Spectra obtained from: SDBSWeb: http://www.aist.go.jp/RIODB/SDBS/


CHEM 2423. Chapter 13. Nuclear Magnetic Spectroscopy (Homework) W

Answer Section

SHORT ANSWER

1. ANS:

d

PTS: 1

2. ANS:

c

PTS: 1

3. ANS:

All nuclei in molecules are surrounded by electron clouds. When a uniform external magnetic field is applied

to a molecule, the circulating electron clouds set up tiny local magnetic fields of their own. These local

magnetic fields act in opposition to the applied field, so that the effective field actually felt by a nucleus is a

bit smaller than the applied field.

Beffective = Bapplied Blocal

This effect is termed shielding. Each nucleus is shielded to a slightly different extent, so each unique kind of

proton in a molecule resonates at a slightly different frequency and gives rise to a unique NMR signal.

PTS: 1

4. ANS:

b

PTS: 1

5. ANS:

f

PTS: 1

6. ANS:

a

PTS: 1

7. ANS:

d

PTS: 1

8. ANS:


three

PTS: 1

9. ANS:

two

PTS: 1

10. ANS:

five

PTS: 1

11. ANS:

six

PTS: 1

12. ANS:

five

PTS: 1

13. ANS:


three

PTS: 1

14. ANS:

five

PTS: 1

15. ANS:

three

PTS: 1

16. ANS:

homotopic

PTS: 1

17. ANS:

enantiotopic

PTS: 1

18. ANS:

diastereotopic

PTS: 1

19. ANS:

unrelated

PTS: 1

20. ANS:


four

PTS: 1

21. ANS:

one

PTS: 1

22. ANS:

four

PTS: 1

23. ANS:

four

PTS: 1

24. ANS:


four

PTS: 1

25. ANS:

two

PTS: 1

26. ANS:

five

PTS: 1

27. ANS:

four

PTS: 1

28. ANS:

No. of adjacent

Proton Protons Splitting

1

1 doublet

2

6 septet

PTS: 1

29. ANS:


No. of adjacent


1

0 singlet

2

3 quartet

3

2 triplet

PTS: 1

30. ANS:

No. of adjacent


1

0 singlet

2

1 doublet

3

1 doublet

PTS: 1

31. ANS:

PTS: 1

32. ANS:

d

PTS: 1

33. ANS:

c

PTS: 1

34. ANS:


PTS: 1

35. ANS:

2-Methylpropene has two kinds of hydrogens. It will have a vinylic absorption (4.56.5 ) representing two

hydrogens and an unsplit signal (1.01.5 ) due to the six equivalent methyl hydrogens.

tert-Butyl chloride has only one kind of hydrogen, which results in one unsplit signal.

PTS: 1

36. ANS:

b

PTS: 1

37. ANS:

c

PTS: 1

38. ANS:

a

PTS: 1

39. ANS:

PTS: 1

40. ANS:

PTS: 1

41. ANS:

BrCH2CH2CH2Br

PTS: 1

42. ANS:


PTS: 1

43. ANS:

PTS: 1

44. ANS:

A molecular ion of mass 113 indicates an odd number of nitrogen atoms in the compound. Coupled with the

IR absorption at 2270 cm1

, a nitrile functional group is indicated. The IR absorption at 1735 cm1

indicates a

carbonyl group, possibly an ester.

PTS: 1

45. ANS:

three

PTS: 1

46. ANS:

The integration of the signal at 3.5 indicates that it is due to two equivalent hydrogens, probably a CH2

group. Since the signal is a singlet (not split) there are no nonequivalent hydrogens attached to the atoms

adjacent to the carbon to which these two hydrogens are bonded. The chemical shift to 3.5 indicates that this

CH2 group has at least one electronegative atom or group bonded to it, possibly a carbonyl group.

PTS: 1

47. ANS:

The signal at 4.35 is owing to two equivalent hydrogens split by three adjacent hydrogens or a CH2 next

to a CH3. It is shifted by attachment to an electronegative atom like oxygen. The signal at 1.3 is three

equivalent hydrogens split by two adjacent hydrogens of a CH3 next to a CH2. It is shifted slightly

downfield by the presence of an electronegative atom bonded to the adjacent CH2.

PTS: 1

48. ANS:

The 13

C NMR has five signals, which means that there are five different kinds of carbon in this compound.

PTS: 1

49. ANS:


By compiling all the information deduced from the data provided, we note that this compound is a) a nitrile

and an ester [two carbons accounted for], and b) has two CH2 groups and one CH3 group [three more carbons

accounted for]. One of the CH2 groups is attached to the oxygen of the ester and one is attached to the

carbonyl of the ester. The singlet CH2 is also bonded to the nitrile. Based on the chemical shifts, the singlet

CH2 must be bonded to the carbonyl and the quartet CH2 must be bonded to the oxygen of the ester. So the

compound is ethyl cyanoacetate:

PTS: 1

50. ANS:

# peaks Distinguishing Absorptions

1H 5 methyl doublet at 1.5 (overlaps other

methylene signals), no vinyl protons

13

C 6 one carbonyl carbon, no vinyl carbons

1H 4 split vinylic peak; rel. area 2

methyl singlet between 3.54.0

13

C 4 one vinylic carbon, no carbonyl carbon

PTS: 1

51. ANS:

The signals at 1.05 and 2.45 are characteristic, in their integration and splitting (a 2H quartet and a 3H

triplet), for an ethyl group. The shift of the CH2 to 2.45 indicates that the ethyl group is attached to a carbonyl

group. The signal at 2.1 , a 3H singlet, is characteristic for a CH3 attached to a carbonyl group.

PTS: 1

52. ANS:

2-butanone, CH3COCH2CH3

PTS: 1

53. ANS:

The base formula for this compound is C5H12, which corresponds to a saturated compound, so there is no

unsaturation in this compound.

PTS: 1

54. ANS:


In looking at the spectrum, it is important to note that the relative number of hydrogens for each signal adds

up to four. The formula of the compound indicates that there are a total of 12 hydrogens. Therefore, the

integration of 3:1 corresponds to 9H:3H. The signal at 1.2 is a 9H singlet, which is characteristic for a

tert-butyl group. The slight downfield shifts indicates that this group is attached to an electronegative atom, in

this case, oxygen. The signal at 3.2 is a 3H singlet which is characteristic of a methyl group bonded to an

oxygen.

PTS: 1

55. ANS:

tert-butyl methyl ether, (CH3)3COCH3

PTS: 1

56. ANS:

The base formula for this compound is C8H8; the saturated formula is C8H18; the degree of unsaturation is five

[(18 8) 2].

PTS: 1

57. ANS:

The chemical shift of the signals between 7 and 8 indicates that these hydrogens are aromatic. The

integration of four hydrogens and the splitting pattern of two doublets is characteristic of a para disubstituted

aromatic compound.

PTS: 1

58. ANS:

This signal is a 3H singlet shifted by attachment to a carbonyl group; a methyl ketone, probably.

PTS: 1

59. ANS:

4-chloroacetophenone

PTS: 1

60. ANS:

PTS: 1

61. ANS:

PTS: 1

62. ANS:

(CH3)2CHCH2CH2OH

PTS: 1


63. ANS:

Both compounds contain two different kinds of carbon, so an ordinary broadband decoupled 13

C NMR

spectrum of either compound will show two peaks. The DEPT-90 spectrum, however, will show no peaks for

1,4-dichlorobutane while the DEPT-90 spectrum of 2,3-dichlorobutane will show one peak. The DEPT-135

spectrum for 1,4-dichlorobutane will show two negative peaks, while the DEPT-135 spectrum of

2,3-dichlorobutane will show two positive peaks.

PTS: 1

64. ANS:

b

PTS: 1

chem 2423. chapter 13. nuclear magnetic spectroscopy

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