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Nuclear Magnetic Resonance Spectroscopy

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1 Nuclear Magnetic Resonance Spectroscopy
Chapter 16 Nuclear Magnetic Resonance Spectroscopy Suggested Problems –

2 Intro to NMR Spectroscopy
Nuclear Magnetic Resonance (NMR) spectroscopy is the most powerful method of gaining structural information about organic compounds NMR involves an interaction between electromagnetic radiation (light) and the nucleus of an atom We will focus on C and H nuclei. The structure (connectivity) of a molecule affects how the radiation interacts with each nucleus in the molecule

3 Intro to NMR Spectroscopy
Protons and neutrons in a nucleus behave as if they are spinning If the total number of neutrons and protons is an ODD number, the atoms will have net nuclear spin Examples: The spinning charge in the nucleus creates a magnetic moment A magnetic moment creates a small magnetic field

4 Intro to NMR Spectroscopy
Like a bar magnet, a magnetic moment exists perpendicular to the axis of nuclear spin

5 Intro to NMR Spectroscopy
If the normally disordered magnetic moments of atoms are exposed to an external magnetic field, their magnetic moments will align WHAT if the total number of neutrons and protons is an EVEN number?

6 Intro to NMR Spectroscopy
The aligned magnetic moments can be either with or against the external magnetic field The α and β spin states are not equal in energy.

7 Intro to NMR Spectroscopy
When an atom with an α spin state is exposed to radio waves of just the right energy, it can be promoted to a β spin state The stronger the external magnetic field, the greater the energy gap

8 Intro to NMR Spectroscopy
The magnetic moment of the electrons generally reduces the affect of the external field The more shielded a nucleus is with electron density, the smaller the α  β energy gap.

9 Intro to NMR Spectroscopy
The amount of radio wave energy necessary for the α  β energy transition depends on the electronic environment for the atom When the α spins are flipped to β spins, the atoms are said to be in resonance The use of the term, “resonance” here is totally different from when we are talking about electrons in molecular orbitals

10 Acquiring a 1H NMR Spectrum
NMR requires a strong magnetic field and radio wave energy The strength of the magnetic field affects the energy gap 1.41 tesla corresponds to a 60 MHz NMR 7.04 tesla corresponds to a 300 MHz NMR

11 Acquiring a 1H NMR Spectrum
The strong magnetic field is created when a high current is passed through a superconducting material at extremely low temperature (≈4 Kelvin) The greater the current, the greater the magnetic field In most current NMR instruments, a brief pulse of radio energy (all relevant wavelengths) is used to excite the sample Each of the atoms is excited and then relaxes, emitting energy The emitted energy is recorded as a free induction decay (FID)

12 Acquiring a 1H NMR Spectrum
The FID contains all of the information for each atom A mathematical treatment called a Fourier-transform separates the signals so an individual signal can be observed for each atom Such an instrument is called an FT-NMR Often multiple FIDs are taken and averaged together Before analysis, NMR samples must be prepared neat or in a liquid solution and placed in a small NMR tube The sample is placed into the magnetic field and the tube is spun at a high rate to average magnetic field variations or tube imperfections

13 Acquiring a 1H NMR Spectrum
Solvents are used such as chloroform-d. WHY? The magnet is super-cooled, but the sample is generally at room temp

14 Characteristics of a 1H NMR Spectrum
NMR spectra contain a lot of structural information Number of signals Signal location – shift Signal area – integration Signal shape – splitting pattern

15 Number of Signals Protons with different electronic environments will give different signals Protons that are homotopic will have perfectly overlapping signals Protons are homotopic if the molecule has an axis of rotational symmetry that allows one proton to be rotated onto the other without changing the molecule Find the rotational axis of symmetry in each molecule below

16 Number of Signals Another test for homotopic protons is to replace the protons one at a time with another atom If the resulting compounds are identical, then the protons that you replaced are homotopic

17 Number of Signals Protons that are enantiotopic will also have perfectly overlapping signals Protons are enantiotopic if the molecule has a plane of reflection that makes one proton the mirror image of the other Find the mirror plane that splits each molecule below

18 Number of Signals The replacement test is universal
It will work to identify any equivalent protons whether they are homotopic or enantiotopic If the resulting compounds are enantiomers, then the protons that you replaced are enantiotopic

19 Number of Signals If the protons are neither homotopic nor enantiotopic, then they are NOT chemically equivalent Perform the replacement test on the protons shown in the molecule below How would you describe the relationship between the protons shown? These protons are diastereotopic Practice with SkillBuilder 16.1

20 Number of Signals There are some shortcuts you can take to identify how many signals you should see in the 1H NMR The 2 protons on a CH2 group will be equivalent if there are NO chirality centers in the molecule The 2 protons on a CH2 group will NOT be equivalent if there is a chirality center in the molecule

21 Number of Signals There are some shortcuts you can take to identify how many signals you should see in the 1H NMR The 3 protons on any methyl group will always be equivalent to each other Multiple protons are equivalent if they can be interchanged through either a rotation or mirror plane Practice with SkillBuilder 16.2

22 Number of Signals Identify all the groups of equivalent protons in the molecules below and describe their relationships Hc and Hf are diastereotopic

23 Number of Signals Recall that cyclohexane chairs have 6 equatorial and 6 axial protons Do the axial and equatorial protons have different electronic environments? Try the replacement test for an axial/equitorial pair How many signals should you see for the molecule in the 1H NMR?

24 Number of Signals At room temperature, the chair interconversion occurs rapidly The NMR is not fast enough to see the individual structures, so the average is observed (1 signal) What might you expect to see if the temperature of the NMR sample were brought down to -100 C? 2 signals

25 Chemical Shifts Tetramethylsilane (TMS) is used as the standard for NMR chemical shift In many NMR solvents, 1% TMS is added as an internal standard The shift for a proton signal is calculated as a comparison to TMS For benzene on a 300 MHz instrument = 7.27 ppm

26 Chemical Shifts

27 Chemical Shifts The shift for a proton signal is calculated as a comparison to TMS For benzene on a 60 MHz instrument The Hz of the signal is different in different instruments, but the shift relative to TMS (δ) is constant The ppm value is independent of the instrument.

28 Chemical Shifts The shift for a proton signal is calculated as a comparison to TMS The shift relative to TMS (δ) is a dimensionless number, because the Hz units cancel out Units for δ are often given as ppm (parts per million), which simply indicates that signals are reported as a fraction of the operating frequency of the spectrometer Most 1H signals appear between 0-10 ppm

29 Chemical Shifts Early NMRs analyzed samples at a constant energy over a range of magnetic field strengths from low field strength = downfield to high field strength = upfield Shielded protons required a stronger external magnetic field to be excited at the same energy as deshielded protons. WHY?

30 Chemical Shifts Current NMRs analyze samples at a constant magnetic field strength over a range of energies Shielded protons have a smaller magnetic force acting on them, so they have smaller energy gaps and absorb lower energy radio waves Higher Energy Lower Energy

31 Chemical Shifts Alkane protons generally give signals around 1-2 ppm
Protons can be shifted downfield when nearby electronegative atoms cause deshielding. HOW? Electronegative atoms pull electrons away and decrease the shielding of the nucleus

32 Chemical Shifts To predict chemical shifts, start with the standard ppm for the type of proton (methyl, methylene, or methine) Use table 16.1 to adjust the ppm depending on proximity to certain function groups

33 Chemical Shifts

34 Chemical Shifts Handbooks can be used for functional groups beyond table 16.1 Practice with SkillBuilder 16.3

35 Chemical Shifts Predict chemical shifts for all of the protons in the molecule below Ha is a methyl group (0.9 ppm) + slight deshielding from the carbonyl and the –OH to around probably between 1-2 ppm. Hb and Hd are diastereotopic, but their signals would likely be similar in shift (1.2 ppm) deshielded 1.0 by the carbonyl and probably 1-2 ppm by the CCl3 to probably between Hc is a methine (1.7 ppm) deshielded 1.0 by the carbonyl and 2.5 by the –OH to around 5.2 ppm

36 Chemical Shifts When the electrons in a pi system are subjected to an external magnetic field, they circulate a great deal causing diamagnetic anisotropy Less energy is required to flip benzene protons that are outside the ring. The are deshielded and their signals appear downfield.

37 Chemical Shifts Diamagnetic anisotropy means that different regions in space will have different magnetic strengths Some regions will be more shielded than others

38 Chemical Shifts The result of the diamagnetic anisotropy effect is similar to deshielding for aromatic protons. What about the other protons? Why does it appear that only one signal is given for all of the aromatic protons?

39 Chemical Shifts The result of the diamagnetic anisotropy effect is similar to shielding for protons that extend into the pi system Some of the protons in [14] Annulene appear at 8 ppm while others appear at -1 ppm. Which are which?

40 Chemical Shifts Explain the chemical shifts in table 16.2
Practice conceptual checkpoints 16.10

41 Integration The integration or area under a peak quantifies the relative number of protons giving rise to that signal A computer will calculate the area of each peak representing that area with a step-curve The curve height represents the integration

42 Integration The computer operator sets one of the peaks to a whole number to let it represent a number of protons The computer uses the integration ratios to set the values for the other peaks 1.48 1.56 1.00 1.05

43 Integration Integrations represent numbers of protons, so you must adjust the values to whole numbers If the integration of the first peak is doubled, the computer will adjust the others according to the ratio 2.96 3.12 2.00 2.10

44 Integration The integrations are relative quantities rather than an absolute count of the number of protons Predict the 1H shifts and integrations for tert-butyl methyl ether Symmetry can also affect integrations Predict the 1H shifts and integrations for 3-pentanone Practice with SkillBuilder 16.4

45 Multiplicity When a signal is observed in the 1H NMR, often it is split into multiple peaks Multiplicity or a splitting patterns results

46 Multiplicity Multiplicity results from magnetic affects that protons have on each other Consider protons Ha and Hb We already saw that protons align with or against the external magnetic field Hb will be aligned with the magnetic field in some molecules. Other molecules in the sample will have Hb aligned against the magnetic field Some Hb atoms have a slight shielding affect on Ha and others have a slight deshielding affect

47 Multiplicity The resulting multiplicity or splitting pattern for Ha is a doublet A doublet generally results when a proton is split by only one other proton on an adjacent carbon

48 Multiplicity Consider an example where there are two protons on the adjacent carbon There are three possible affects the Hb protons have on Ha

49 Multiplicity Half of the Ha atoms will not experience a signal shift. WHY? ¼ of the Ha atoms will be shielded and ¼ deshielded

50 Multiplicity Ha appears as a triplet
The three peaks in the triplet have an integration ratio of 1:2:1 WHY?

51 Multiplicity Consider a scenario where Ha has three equivalent Hb atoms splitting it Explain how the magnetic fields cause shielding or deshielding

52 Multiplicity Ha appears as a quartet
What should the integration ratios be for the 4 peaks of the quartet?

53 Multiplicity By analyzing the splitting pattern of a signal in the 1H NMR, you can determine the number of equivalent protons on adjacent carbons

54 Multiplicity The trend in table 16.3 also allows us to predict splitting patterns Explain how the n+1 rule is used

55 Multiplicity Remember three key rules
Equivalent protons can not split one another Predict the splitting patterns observed for 1,2-dichloroethane To split each other, protons must be within a 2 or 3 bond distance

56 Multiplicity Remember three key rules
The n+1 rule only applies to protons that are all equivalent The splitting pattern observed for the proton shown below will be more complex than a simple triplet Complex splitting will be discussed later in this section Practice with SkillBuilder 16.5

57 Multiplicity Predict splitting patterns for all of the protons in the molecule below Ha is a doublet Hb is a septet Hc is a doublet Hd is a triplet He is a doublet Hf is a doublet

58 Multiplicity The degree to which a neighboring proton will shield or deshield its neighbor is called a coupling constant The coupling constant or J value is the distance between peaks of a splitting pattern measured in units of Hz When protons split each other, their coupling constants will be equal Jab = Jba

59 Multiplicity The coupling constant will be constant even if an NMR instrument with a stronger or weaker magnetic field is used Higher field strength instruments will give better resolution between peaks, because the coupling constant is a smaller percentage of the overall Hz available

60 Multiplicity Sometimes recognizable splitting patterns will stand out in a spectrum An isolated ethyl group gives a triplet and a quartet Note the integrations The triplet and quartet must have the same coupling constant if they are splitting each other

61 Multiplicity A peak with an integration equal to 9 suggests the presence of a tert-butyl group An isolated isopropyl group gives a doublet and a septet Note the integrations Practice with conceptual checkpoint 16.17

62 Multiplicity Complex splitting results when a proton is split by NONequivalent neighboring protons If Jab is much greater than Jbc, the signal will appear as a quartet of triplets In the molecule shown, Hb is split into a quartet by Ha and into a triplet by Hc

63 Multiplicity Complex splitting results when a proton is split by NONequivalent neighboring protons If Jbc is much greater than Jab, the signal will appear as a triplet of quartets If Jbc is similar to Jab, the signal will appear as a multiplet

64 A Splitting Diagram The number of peaks observed depends on
the relative magnitudes of the coupling constants.

65 Multiplicity Complex splitting results when a proton is split by NONequivalent neighboring protons Predict the splitting patterns for (S)-pent-2-en-4-ol If Jbc is equal to Jab, what type of patterns will be observed? Practice with conceptual checkpoint 16.18

66 Multiplicity Splitting is not observed for some protons. Consider ethanol The protons bonded to carbon split each other, but the hydroxyl proton is not split

67 Multiplicity The hydroxyl proton and other labile or exchangeable protons undergo rapid exchange with trace amounts of acid. Such exchange blurs the shielding/deshielding effect of the neighboring protons giving a singlet that is often broadened If ethanol is rigorously purified to remove traces of acid, then hydroxyl proton splitting is generally observed Aldehyde protons also often appear as singlet because their coupling constants are sometimes too small to cause observable splitting

68 Multiplicity Signals for exchangeable protons such as those shown below disappear completely when the 1H NMR sample is prepared for analysis in a deuterated solvent such as chloroform-d. WHY? Protic compounds have exchangeable protons

69 Predicting Expected 1H Spectra for a Compound
Predict the chemical shift, integration, and splitting patterns for all of the protons in the following molecule Draw a spectrum for the molecule Practice with SkillBuilder 16.6

70 Predicting Expected 1H Spectra for a Compound
Ha is a singlet with an integration of 3 Hb is a doublet with an integration of 2 Hc is a singlet with an integration of 2 Hd is a triplet with an integration of 1 He and Hf are both doublets with integrations of 1 Ha Hc Hb Hd Hf He The chemical shifts above are estimates

71 Using 1H Spectra to Distinguish Between Compounds
The three molecules below might be difficult to distinguish by IR of MS. WHY? Explain how 1H NMR could distinguish between them Practice with SkillBuilder 16.7

72 Using 1H Spectra to Distinguish Between Compounds
Explain how 1H NMR could be used to distinguish between the two molecules below Would you see a quartet with either of these?

73 Analyzing a 1H NMR Spectrum
With a given formula and 1H NMR spectrum, you can determine a molecule’s structure by a 4-step process Calculate the degree or unsaturation or hydrogen deficiency index (HDI). What does the HDI tell you? Consider the number of NMR signals and integration to look for symmetry in the molecule Analyze each signal, and draw molecular fragments that match the shift, integration, and multiplicity Assemble the fragments into a complete structure like puzzle pieces Practice with SkillBuilder 16.8

74 Analyzing a 1H NMR Spectrum
Consider the data below, and propose a structure for the molecule The formula is C7H13Cl 1H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d 3H); 1.0 (s 6H) Calculate degree of unsaturation Consider the number of NMR signals and integration to look for symmetry in the molecule = 1

75 Analyzing a 1H NMR Spectrum
Consider the data below, and propose a structure for the molecule The formula is C7H13Cl 1H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d 3H); 1.0 (s 6H) Analyze each signal, and draw molecular fragments that match the shift, integration, and multiplicity Assemble the fragments into a complete structure like puzzle pieces

76 Acquiring a 13C NMR Spectrum
Because 1H is by far the most abundant isotope of hydrogen, 1H NMR signals are generally strong 13C only accounts for about 1% of carbon atoms in nature, so a sensitive receiver coil and/or concentrated NMR sample is needed In 1H NMR, shift, splitting, and integration are important In 13C NMR, only the number of signals and the shift will be considered

77 Acquiring a 13C NMR Spectrum
In 13C NMR, the 1H-13C splitting is often so complex that the spectrum is unreadable To elucidate the 13C spectrum and make it easier to determine the total number of 13C signals, 13C NMR are generally decoupled In the vast majority of 13C spectra, all of the signals are singlets Decoupling means that the proton frequencies are irradiated eliminating coupling of protons with carbons.

78 Chemical Shifts in 13C NMR Spectra
Compared to 1H, 13C atoms require a different frequency of energy to excite (resonate) Compared to the standard TMS, 13C NMR signals generally appear between 220 and 0 ppm Each signal on the 13C spectra represents a carbon with a unique electronic environment

79 Chemical Shifts in 13C NMR Spectra
Note how symmetry affects the number of signals for the molecules above How many 13C signals should be observed for the molecule below

80 Chemical Shifts in 13C NMR Spectra
Like 1H signals, chemical shifts for 13C signals are affected by shielding or deshielding Practice with SkillBuilder 16.9

81 Chemical Shifts in 13C NMR Spectra
Predict the number of signals and chemical shifts in the 13C NMR spectrum for the molecule below 8 total signals Estimated shifts C1 (20-30 ppm) C2 and C3 (40-60 ppm) C4, C5, and C6 ( ppm) C7 and C8 ( ppm)

82 DEPT 13C NMR Spectra 13C spectra generally give singlets that do not provide information about the number of hydrogen atoms attached to each carbon Distortionless Enhancement by Polarization Transfer (DEPT) 13C NMR provides information the number of hydrogen atoms attached to each carbon Full decoupled 13C spectrum: shows all carbon peaks DEPT-90: Only CH signals appear DEPT-135: CH3 and CH give (+) signals, and CH2 give (-) signals

83 DEPT 13C NMR Spectra Full decoupled 13C spectrum: shows all carbon peaks DEPT-90: Only CH signals appear DEPT-135: CH3 and CH = (+) signals, CH2 = (-) signals Practice with SkillBuilder 16.10

84 DEPT 13C NMR Spectra Explain how DEPT 13C spectra could be used to distinguish between the two molecules below The peak shifted farthest downfield in the first molecule will be a methylene (CH2). The peak most deshielded in the second molecule will represent a methine (CH).

85 Medically Speaking MRI (magnetic resonance imaging) instruments are essentially 1H NMR spectrometers The body is analyzed rather than a sample in an NMR tube Different tissues have different concentrations of protons. WHY? The MRI gives a 3D image of different tissues.

86 Additional Practice Problems
Explain why a deuterated solvent is used in NMR experiments rather than a protonated solvent. Is such a solvent necessary when analyzing 13C NMR? If a protonated solvent is used, then the protons attached to the solvent molecules will give signals in the 1H NMR, and because the solvent is usually present in a much greater quantity than the solute, the signals for the solute can be dwarfed in comparison to the point where they are no greater than the noise. Because D has an even mass, it will not give an NMR signal.

87 Additional Practice Problems
Identify all the groups of equivalent protons in the molecule below and explain WHY enantiotopic protons are equivalent while diastereotopic protons are not. Diastereotopic protons have different chemical environments making them magnetically nonequivalent. Consider protons d and e. Because the molecule is chiral, the Cl atom has a different spatial relationship with Hd than with He.

88 Additional Practice Problems
Predict the chemical shift, integration, and splitting patterns for all of the protons in the following molecule Ha – singlet at 3-4 ppm, integration = 3 Hb – doublet of doublets or multiplet at ppm, integration = 4 Hc – triplet at 4-6 ppm, integration = 2 Hd – quintet at ppm, integration = 1

89 Additional Practice Problems
Predict the number of signals and chemical shifts in the 13C NMR spectrum and DEPT spectrum for the molecule below C1 – ppm DEPT 90 – no signal, DEPT 135 – signal up C2 – ppm further downfield than C1 DEPT 90 – signal up, DEPT 135 – signal up C3 – 0-50 ppm DEPT 90 – no signal, DEPT 135 – signal down C4 – ppm DEPT 90 – signal up, DEPT 135 signal up


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