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Structure Determination: Nuclear Magnetic Resonance Spectroscopy


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Structure Determination: Nuclear Magnetic Resonance Spectroscopy

Structure Determination:

Nuclear Magnetic Resonance Spectroscopy

1a. Gjør kort rede for NMR spektroskopi.

      Læreboken side 424-429.

1b. Gjør rede for skjerming av kjerner og om hvordan nærliggende elektronnegative kjerner vil pavirke skjermingen av kjerner.

      Læreboken side 426-427

1c. Grei ut om spin-spin splitting i 1H NMR spektre.

      Læreboken side 443-447.

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.

Answer:  c

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

Answer:  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.

The following questions 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

b.   high-field or upfield side

c.   MHz

d.   delta

e.   low-field or downfield side

f.    chemical shift

g.   specific absorption

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

Answer:  b

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

Answer:  f

6.              The calibration standard for  and  NMR is:

Answer:  a

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

Answer:  d

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


Answer:  three


Answer:  two


Answer:  five


Answer:  six


Answer:  five


Answer:  three


Answer:  five


Answer:  three

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


Answer:   homotopic


Answer:   enantiotopic


Answer:   diastereotopic


Answer:   unrelated

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


Answer:  four


Answer:  one


Answer:  four


Answer:  four


Answer:  four


Answer:  two


Answer:  five


Answer:  four

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







31. The  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  and proton 3 .  Draw a tree diagram for the proton 2 signal.


Refer to the structure of 3-methyl-2-butanone below to answer the following questions.

32. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled a.?

a.   septet

b.   quartet

c.   doublet

d.   singlet

Answer:  d

33. What is the splitting pattern for the hydrogens in 3-methyl-2-butanone labeled b.?

a.   septet

b.   quartet

c.   doublet

d.   singlet

Answer:  c

34. 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 is 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  NMR to help you decide which was which?

Answer:  2-Methylpropene has two kinds of hydrogens.  It will have a vinylic absorption (4.5–6.5 ) representing two hydrogens and an unsplit signal (1.0–1.5 ) due to the six equivalent methyl hydrogens.

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

Below are three isomeric chlorobutanes and their  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


Answer:  b


Answer:  c


Answer:  a

Propose structures for compounds that fit the following  NMR data:


      3H doublet at 2.0 ,

      1H quartet at 5.0 ,

      5H singlet at 7.3



      6H triplet at 0.9 ,

      4H sextet at 1.6 ,

      4H triplet at 2.4 ,



      2H quintet at 2.4 ,

      4H triplet at 3.5 ,



      6H doublet at 1.2 ,

      3H singlet at 2.3

      1H septet at 2.9 ,

      4H singlet at 7.0



      12H doublet at 0.8

      2H septet at 1.4


To answer the following questions, consider the data and NMR spectrum below:





3 H


The mass spectrum of this compound shows a molecular ion at m/z = 113, the IR spectrum has characteristic absorptions at 2270 and 1735 cm-1, and the 13C NMR spectrum has five signals.

Spectrum obtained from:  SDBSWeb:

44. Based on the mass spectral data and the IR data, what functional groups are present in this compound?

Answer:  A molecular ion of mass 113 indicates an odd number of nitrogen atoms in the compound.  Coupled with the IR absorption at 2270 cm-1, a nitrile functional group is indicated.  The IR absorption at 1735 cm-1 indicates a carbonyl group, possibly an ester.

45. How many types of nonequivalent protons are there in this molecule?

Answer:  three

46. Describe the signal at 3.5 d in terms of its integration, splitting pattern and chemical shift.

Answer:  The integration of the signal at 3.5 d 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 d indicates that this -CH2- group has at least one electronegative atom or group bonded to it, possibly a carbonyl group


47. Describe the signals at 4.35 d and 1.3 d in terms of their integration, splitting and chemical shift.

Answer:  The signal at 4.35 d 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 d 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-

48. What is the significance of the 13C NMR data?

Answer:  The 13C NMR has five signals, which means that there five differenct kinds of carbon in this compound.

49. Propose a structure for this compound.

Answer:  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:

50. How would you use  and  NMR to help you distinguish between these two isomeric structures?


# peaks

Distinguishing Absorptions



methyl doublet at 1.5 (overlaps other methylene signals), no vinyl protons

one carbonyl carbon, no vinyl carbons



split vinylic peak; rel. area 2

methyl singlet between 3.5–4.0

one vinylic carbon, no carbonyl carbon

Answer the questions for each of the compounds whose  NMR spectra are shown below.








Spectrum obtained from:  SDBSWeb:

51. Describe each signal in terms of its integration, splitting and chemical shift.

Answer:  The signals at 1.05 and 2.45 d are characteristic, in their integration and slitting (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 d, a 3H singlet, is characteristic for a CH3 attached to a carbonyl group.

52. Propose a structure for this compound.

Answer:  2-butanone,





Spectrum obtained from:  SDBSWeb:

53. Calculate the degree of unsaturation for this compound.

Answer:  The base formula for this compound is C5H12, which is corresponds to a saturated compound, so there is no unsaturation in this compound.         

54. Describe each signal in the 1H NMR in terms of its integration, splitting and chemical shift.

Answer:  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 d 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 d is a 3H singlet which is characteristic of a methyl group bonded to an oxygen.

55. Propose a structure for this compound.

Answer:  tert-butyl methyl ether, (CH3)3COCH3






Spectrum obtained from:  SDBSWeb:

56. Calculate the degree of unsaturation in this compound.

Answer:  The base formula for this compound is C8H8; the saturated formula is C8H18; the degree of unsaturation is five [(18 - 8) ¸ 2].

57. Describe the signals that occur between 7 and 8 d in terms of integration, splitting and chemical shift.

Answer:  The chemical shift of the signals between 7 and 8 d 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.

58. Describe the signal at 2.6 d in terms of integration, splitting and chemical shift.

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

59. Propose a structure for this compound.

Answer:  4-chloroacetophenone

64. Which structure of molecular formula C4H8Cl2 fits both the 1H NMR and 13C NMR spectra shown below?


Spectra obtained from:  SDBSWeb:

Answer:  B.



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