1. CHE2202, Chapter 13 Learn, 1 Structure Determination: Nuclear Magnetic Resonance Spectroscopy Chapter 13 Suggested Problems – 1-23,29,34,36-8,44-7,51

2. CHE2202, Chapter 13 Learn, 2 Nuclear Magnetic Resonance Spectroscopy Nuclei are positively charged, have spin, and interact with an external magnetic field, B 0 Magnetic rotation of nuclei is random in the absence of a magnetic field –In the presence of a strong magnet, nuclei adopt specific orientations

3. CHE2202, Chapter 13 Learn, 3 Nuclear Magnetic Resonance Spectroscopy When exposed to a certain frequency of electromagnetic radiation, oriented nuclei absorb energy which causes a spinflip from a state of lower energy to higher energy –Nuclear magnetic resonance - Nuclei are in resonance with applied radiation Frequency that causes resonance depends on: –Strength of external magnetic field –Identity of the nucleus –Electronic environment of the nucleus

4. CHE2202, Chapter 13 Learn, 4 Nuclear Magnetic Resonance Spectroscopy Larmor equation –Relation between Resonance frequency of a nucleus Magnetic field and the magnetogyric ratio of the nucleus

5. CHE2202, Chapter 13 Learn, 5 Worked Example Calculate the amount of energy required to spin-flip a proton in a spectrometer operating at 300 MHz –Analyze if the increase of spectrometer frequency from 200 MHz to 300 MHz increases or decreases the amount of energy necessary for resonance

6. CHE2202, Chapter 13 Learn, 6 Worked Example Solution: –Increasing the spectrometer frequency from 200 MHz to 300 MHz increases the amount of energy needed for resonance

7. CHE2202, Chapter 13 Learn, 7 The Nature of NMR Absorptions Absorption frequencies differ across 1 H and 13 C molecules Shielding: Opposing magnetic field produced by electrons surrounding nuclei to counteract the effects of an external magnetic field –Effect on the nucleus is lesser than the applied magnetic field –B effective = B applied – B local –Individual variances in the electronic environment of nuclei leads to different shielding intensities

8. CHE2202, Chapter 13 Learn, 8 NMR Spectrum of 1 H and 13 C

9. CHE2202, Chapter 13 Learn, 9 Working of an NMR Spectrometer Organic sample dissolved in a suitable solvent is placed in a thin glass tube between the poles of a magnet 1 H and 13 C nuclei respond to the magnetic field by aligning themselves to one of the two possible orientations followed by rf irradiation Varying the strength of the applied field causes each nucleus to resonate at a slightly varied field strength Absorption of rf energy is monitored by a sensitive detector that displays signals as a peak

10. CHE2202, Chapter 13 Learn, 10 Operation of a Basic NMR Spectrometer

11. CHE2202, Chapter 13 Learn, 11 NMR Spectrometer Time taken by IR spectroscopy is about 10 –13 s Time taken by NMR spectroscopy is about 10 –3 s –Provides a blurring effect that is used in the measurement of rates and activation energies of vary fast processes

12. CHE2202, Chapter 13 Learn, 12 Worked Example Explain why 2-chloropropene shows signals for three kinds of protons in its 1 H NMR spectrum Solution: –2-Chloropropene has three kinds of protons Protons b and c differ because one is cis to the chlorine and the other is trans

13. CHE2202, Chapter 13 Learn, 13 The Chemical Shift The left segment of the chart is the downfield –Nuclei absorbing on the downfield have less shielding as they require a lower field for resistance The right segment is the upfield –Nuclei absorbing on the upfield have more shielding as they require a higher field strength for resistance

14. CHE2202, Chapter 13 Learn, 14 The NMR Chart

15. CHE2202, Chapter 13 Learn, 15 The Chemical Shift Chemical shift is the position on the chart at which a nucleus absorbs The delta (δ) scale is used in calibration of the NMR chart –1 δ = 1 part-per-million of the spectrometer operating frequency –The delta scale is used as the units of measurement can be used to compare values across other instruments

16. CHE2202, Chapter 13 Learn, 16 Worked Example The 1 H NMR peak of CHCl 3 was recorded on a spectrometer operating at 200 MHz providing the value of 1454 Hz –Convert 1454 Hz into δ units Solution:

17. CHE2202, Chapter 13 Learn, 17 Chemical Shifts in 1 H NMR Spectroscopy Chemical shifts are due to the varied electromagnetic fields produced by electrons surrounding nuclei Protons bonded to saturated, sp 3 -hybridized carbons absorb at higher fields Protons bonded to sp 2 -hybridized carbons absorb at lower fields Protons bonded to electronegative atoms absorb at lower fields

18. CHE2202, Chapter 13 Learn, 18 Regions of the 1 H NMR Spectrum

19. CHE2202, Chapter 13 Learn, 19 Representative Chemical Shifts

20. CHE2202, Chapter 13 Learn, 20 Worked Example CH 2 Cl 2 has a single 1 H NMR peak –Determine the location of absorption Solution: For CH 2 Cl 2, δ = 5.30 –The location of absorption are the protons adjacent to the two halogens

21. CHE2202, Chapter 13 Learn, 21 Integration of 1 H NMR Absorptions: Proton Counting In the figure, the peak caused by (CH 3 ) 3 C– protons is larger than the peak caused by – OCH 3 protons Integration of the area under the peak can be used to quantify the different kinds of protons in a molecule

22. CHE2202, Chapter 13 Learn, 22 Worked Example Mention the number of peaks in the 1H NMR spectrum of 1,4-dimethyl-benzene (para- xylene or p-xylene) –Mention the ratio of peak areas possible on integration of the spectrum

23. CHE2202, Chapter 13 Learn, 23 Worked Example Solution: –There are two absorptions in the 1H NMR spectrum of p-xylene –The four ring protons absorb at 7.05 δ and the six methyl-groups absorb at 2.23 δ –The peak ratio of methyl protons:ring protons is 3:2

24. CHE2202, Chapter 13 Learn, 24 Worked Example

25. CHE2202, Chapter 13 Learn, 25 Worked Example

26. CHE2202, Chapter 13 Learn, 26 Worked Example

27. CHE2202, Chapter 13 Learn, 27 Spin-Spin Splitting in 1 H NMR Spectra Multiplet: Absorption of a proton that splits into multiple peaks –The phenomenon is called spin-spin splitting –Caused by coupling of neighboring spins

28. CHE2202, Chapter 13 Learn, 28 Spin-Spin Splitting in 1 H NMR Spectra Alignment of both –CH 2 Br proton spins with the applied field will result in: –Slightly larger total effective field and slight reduction in the applied field to achieve resonance There is no effect if one of the –CH 2 Br proton spins aligns with the applied field and the other aligns against it Alignment of both –CH 2 Br proton spins against the applied field results in: –Smaller effective field and an increased applied field to achieve resonance

29. CHE2202, Chapter 13 Learn, 29 The Origin of Spin-Spin Splitting in Bromoethane

30. CHE2202, Chapter 13 Learn, 30 Spin-Spin Splitting in 1 H NMR Spectra n + 1 rule: Protons that exhibit n + 1 peaks in the NMR spectrum possess –n = number of equivalent neighboring protons Coupling constant is the distance between peaks in a multiplet

31. CHE2202, Chapter 13 Learn, 31 Spin-Spin Splitting in 1 H NMR Spectra It is possible to identify multiplets in a complex NMR that are related –Multiplets that have the same coupling constant can be related –Multiplet-causing protons are situated adjacent to each other in the molecule

32. CHE2202, Chapter 13 Learn, 32 Rules of Spin-Spin Splitting Chemically equivalent protons do not show spin-spin splitting The signal of a proton with n equivalent neighboring protons is split into a multiplet of n + 1 peaks with a coupling constant Two groups of protons coupled together have the same coupling constant, J

33. CHE2202, Chapter 13 Learn, 33 Worked Example The integrated 1 H NMR spectrum of a compound of formula C 4 H 10 O is shown below –Propose a structure

34. CHE2202, Chapter 13 Learn, 34 Worked Example Solution: –The molecular formula (C 4 H 10 O) indicates that the compound has no multiple bonds or rings –The 1 H NMR spectrum shows two signals, corresponding to two types of hydrogens in the ratio 1.50:1.00, or 3:2 –Since the unknown contains 10 hydrogens, four protons are of one type and six are of the other type –The upfield signal at 1.22 δ is due to saturated primary protons

35. CHE2202, Chapter 13 Learn, 35 Worked Example –The downfield signal at 3.49 δ is due to protons on carbon adjacent to an electronegative atom - in this case, oxygen –The signal at 1.23 δ is a triplet, indicating two neighboring protons –The signal at 3.49 δ is a quartet, indicating three neighboring protons –This splitting pattern is characteristic of an ethyl group –The compound is diethyl ether, CH 3 CH 2 OCH 2 CH 3

36. CHE2202, Chapter 13 Learn, 36 1 H NMR Spectroscopy and Proton Equivalence Proton NMR is much more sensitive than 13 C and the active nucleus ( 1 H) is nearly 100 % of the natural abundance Proton NMR shows how many kinds of nonequivalent hydrogens are in a compound Theoretical equivalence can be predicted by comparing structures formed by replacing each H with X and determining the number of different compounds Equivalent H’s have the same signal while nonequivalent H’s give different signals –There are degrees of nonequivalence

37. CHE2202, Chapter 13 Learn, 37 1 H NMR Spectroscopy and Proton Equivalence One use of 1 H NMR is to ascertain the number of electronically non-equivalent hydrogens present in a molecule In relatively small molecules, a brief look at the structure can help determine the kinds of protons present and the number of possible NMR absorptions –Equivalence or nonequivalence of two protons can be determined by comparison of structures formed if each hydrogen were replaced by an X group

38. CHE2202, Chapter 13 Learn, 38 1 H NMR Spectroscopy and Proton Equivalence Possibilities –If the protons are chemically unrelated and non-equivalent, the products formed by substitution would be different constitutional isomers

39. CHE2202, Chapter 13 Learn, 39 1 H NMR Spectroscopy and Proton Equivalence –If the protons are chemically identical, the same product would be formed despite the substitution

40. CHE2202, Chapter 13 Learn, 40 1 H NMR Spectroscopy and Proton Equivalence –If the hydrogens are homotopic but not identical, substitution will form a new chirality center Hydrogens that lead to formation of enantiomers upon substitution with X are called enantiotopic

41. CHE2202, Chapter 13 Learn, 41 1 H NMR Spectroscopy and Proton Equivalence –If the hydrogens are neither homotopic nor enantiotopic, substitution of a hydrogen at C3 would form a second chirality center

42. CHE2202, Chapter 13 Learn, 42 Worked Example How many absorptions will (S)-malate, an intermediate in carbohydrate metabolism have in its 1 H NMR spectrum? Explain

43. CHE2202, Chapter 13 Learn, 43 Worked Example Solution: –Because (S)-malate already has a chirality center(starred), the two protons next to it are diastereotopic and absorb at different values –The 1 H NMR spectrum of (S)-malate has four absorptions

44. CHE2202, Chapter 13 Learn, 44 More Complex Spin-Spin Splitting Patterns Some hydrogens in a molecule possess accidentally overlapping signals –In the spectrum of toluene (methylbenzene), the five aromatic ring protons produce a complex, overlapping pattern though they are not equivalent

45. CHE2202, Chapter 13 Learn, 45 More Complex Spin-Spin Splitting Patterns Splitting of a signal by two or more nonequivalent kinds of protons causes a complication in 1 H NMR spectroscopy

46. CHE2202, Chapter 13 Learn, 46 Tree Diagram for the C2 proton of trans-cinnamaldehyde

47. CHE2202, Chapter 13 Learn, 47 Worked Example 3-Bromo-1-phenyl-1-propene shows a complex NMR spectrum in which the vinylic proton at C2 is coupled with both the C1 vinylic proton (J = 16 Hz) and the C3 methylene protons (J = 8 Hz) –Draw a tree diagram for the C2 proton signal and account for the fact that a five-line multiplet is observed

48. CHE2202, Chapter 13 Learn, 48 Worked Example Solution: –C2 proton couples with vinylic proton (J = 16) Hz C2 proton’s signal is split into a doublet –C2 proton also couples with the two C3 protons (J = 8 Hz) Each leg of the C2 proton doublet is split into a triplet to produce a total of six lines

49. CHE2202, Chapter 13 Learn, 49 Worked Example

50. CHE2202, Chapter 13 Learn, 50 Uses of 1H NMR Spectroscopy The technique is used to identify likely products in the laboratory quickly and easily –NMR can help prove that hydroboration- oxidation of alkenes occurs with non- Markovnikov regiochemistry to yield the less highly substituted alcohol

51. CHE2202, Chapter 13 Learn, 51 1 H NMR Spectra of Cyclohexylmethanol

52. CHE2202, Chapter 13 Learn, 52 Worked Example Mention how 1 H NMR is used to determine the regiochemistry of electrophilic addition to alkenes –Determine whether addition of HCl to 1- methylcyclohexene yields 1-chloro-1- methylcyclohexane or 1-chloro-2- methylcyclohexane

53. CHE2202, Chapter 13 Learn, 53 Worked Example Solution: –Referring to 1 H NMR methyl group absorption The unsplit methyl group in the left appears as a doublet in the product on the right Bonding of a proton to a carbon that is also bonded to an electronegative atom causes a downfield absorption in the 2.5–4.0 region – 1 H NMR spectrum of the product would confirm the product to be 1-chloro-1-methylcyclohexane

54. CHE2202, Chapter 13 Learn, 54 13 C NMR Spectroscopy: Signal Averaging and FT–NMR Carbon-13 is the only naturally occurring carbon isotope that possesses a nuclear spin, but its natural abundance is 1.1% Signal averaging and Fourier-transform NMR (FT–NMR) help in detecting carbon 13 Due to the excess random electronic background noise present in 13 C NMR, an average is taken from hundreds or thousands of individual NMR spectra

55. CHE2202, Chapter 13 Learn, 55 Carbon-13 NMR Spectra of 1- Pentanol

56. CHE2202, Chapter 13 Learn, 56 13 C NMR Spectroscopy: Signal Averaging and FT–NMR Spin-spin splitting is observed only in 1 H NMR –The low natural abundance of 13 C nucleus is the reason that coupling with adjacent carbons is highly unlikely –Due to the broadband decoupling method used to record 13 C spectra, hydrogen coupling is not seen

57. CHE2202, Chapter 13 Learn, 57 Characteristics of 13 C NMR Spectroscopy 13 C NMR provides a count of the different carbon atoms in a molecule 13 C resonances are 0 to 220 ppm downfield from TMS

58. CHE2202, Chapter 13 Learn, 58 Characteristics of 13 C NMR Spectroscopy General factors that determine chemical shifts –The electronegativity of nearby atoms –The diamagnetic anisotropy of pi systems –The absorption of sp 3 -hybridized carbons and sp 2 carbons

59. CHE2202, Chapter 13 Learn, 59 Carbon-13 Spectra of 2-butanone and para-bromoacetophenone

60. CHE2202, Chapter 13 Learn, 60 Worked Example Classify the resonances in the 13 C spectrum of methyl propanoate, CH 3 CH 2 CO 2 CH 3

61. CHE2202, Chapter 13 Learn, 61 Worked Example Solution: –Methyl propanoate has four unique carbons that individually absorb in specific regions of the 13 C spectrum

62. CHE2202, Chapter 13 Learn, 62 DEPT 13 C NMR Spectroscopy DEPT-NMR (distortionless enhancement by polarization transfer) Stages of a DEPT experiment –Run a broadband-decoupled spectrum –Run a DEPT-90 –Run a DEPT-135 The DEPT experiment manipulates the nuclear spins of carbon nuclei

63. CHE2202, Chapter 13 Learn, 63 DEPT-NMR Spectra for 6-methyl-5- hepten-2-ol

64. CHE2202, Chapter 13 Learn, 64 Uses of 13 C NMR Spectroscopy Helps in determining molecular structures –Provides a count of non-equivalent carbons –Provides information on the electronic environment of each carbon and the number of attached protons Provides answers on molecule structure that IR spectrometry or mass spectrometry cannot provide

65. CHE2202, Chapter 13 Learn, 65 13 C NMR Spectrum of 1- methylcyclohexane

66. CHE2202, Chapter 13 Learn, 66 Worked Example Propose a structure for an aromatic hydrocarbon, C 11 H 16, that has the following 13 C NMR spectral data: –Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, 139.8 δ –DEPT-90: 125.5, 127.5, 130.3 δ –DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 δ; negative peak at 50.2 δ

67. CHE2202, Chapter 13 Learn, 67 Worked Example Solution: –Calculate the degree of unsaturation of the unknown compound C 11 H 16 has 4 degrees of unsaturation –Look for elements of symmetry 7 peaks appearing in the 13 C NMR spectrum indicate a plane of symmetry –According to the DEPT-90 spectrum, 3 of the kinds of carbons in the aromatic ring are CH carbons

68. CHE2202, Chapter 13 Learn, 68 Worked Example –The unknown structure is a monosubstituted benzene ring with a substituent containing CH 2 and CH 3 carbons

69. CHE2202, Chapter 13 Learn, 69 Nuclear Magnetic Resonance (NMR) Spectroscopy NMR spectroscopy identifies the carbon–hydrogen framework of an organic compound. Certain nuclei, such as 1 H, 13 C, 15 N, 19 F, and 31 P, have a nonzero value for their spin quantum number; this property allows them to be studied by NMR.

70. CHE2202, Chapter 13 Learn, 70 The Spin State of a Nucleus is Affected by an Applied Magnetic Field

71. CHE2202, Chapter 13 Learn, 71 The ∆ Energy Between the Two Spin States Depends on the Strength of the Applied Magnetic Field (B o )

72. CHE2202, Chapter 13 Learn, 72 The Operating Frequency of the Spectrometer Depends on the Strength of the Magnetic Field

73. CHE2202, Chapter 13 Learn, 73 All the Hydrogens in a Compound Do Not Experience the Same Magnetic Field The electrons surrounding the nucleus decrease the effective applied magnetic field sensed by the nucleus.

74. CHE2202, Chapter 13 Learn, 74 Where Protons Show a Signal

75. CHE2202, Chapter 13 Learn, 75 Chemically Equivalent Protons (protons in the same environment)

76. CHE2202, Chapter 13 Learn, 76 Each set of chemically equivalent protons give a signal in the 1H NMR spectrum. How Many Signals?

77. CHE2202, Chapter 13 Learn, 77 How Many Signals?

78. CHE2202, Chapter 13 Learn, 78 How Many Signals?

79. CHE2202, Chapter 13 Learn, 79 The Reference Compound TMS is the reference compound (it appears at = 0 ppm).

80. CHE2202, Chapter 13 Learn, 80 An 1 H NMR Spectrum The greater the chemical shift, the higher the frequency.

81. CHE2202, Chapter 13 Learn, 81 Terms to Remember

82. CHE2202, Chapter 13 Learn, 82 Relative Positions of the Signals Protons in electron-poor environments show signals at high frequencies. Electron withdrawal causes NMR signals to appear at a higher frequency (at a larger  value).

83. CHE2202, Chapter 13 Learn, 83 Relative Positions of the Signals The closer the electronegative the atom (or group), the more it deshields the protons.

85. CHE2202, Chapter 13 Learn, 85 Where 1 H NMR Signals Appear

86. CHE2202, Chapter 13 Learn, 86 Where They Show a Signal Methine protons appear at higher frequency than methylene protons, which appear at a higher frequency than methyl protons.

87. CHE2202, Chapter 13 Learn, 87 The Relative Positions of Signals In the same environment, a methine proton appears at a higher frequency than methylene protons, which appear at a higher frequency than methyl protons.

88. CHE2202, Chapter 13 Learn, 88 Protons Attached to sp 2 Carbons The chemical shift of protons attached to sp 2 carbons appear at higher frequencies than one would predict.

89. CHE2202, Chapter 13 Learn, 89 Diamagnetic Anisotropy (Benzene Ring Protons) The protons show signals at higher frequencies because they sense a larger effective magnetic field.

90. CHE2202, Chapter 13 Learn, 90 Diamagnetic Anisotropy (Alkenes and Aldehydes) The protons show signals at higher frequencies because they sense a larger effective magnetic field.

91. CHE2202, Chapter 13 Learn, 91 A Hydrogen Bonded to an sp Carbon The chemical shift of a hydrogen bonded to an sp carbon appears at a lower frequency than it would if the π electrons did not induce a magnetic field.

92. CHE2202, Chapter 13 Learn, 92 Integration Identifies the Relative Number of Protons The area under each signal is proportional to the number of protons giving rise to the signal.

93. CHE2202, Chapter 13 Learn, 93 These Compounds Will Show the Same Integration

94. CHE2202, Chapter 13 Learn, 94 Splitting (N + 1) N is the number of equivalent protons on adjacent carbons that are not equivalent to the protons that produce the signal.

95. CHE2202, Chapter 13 Learn, 95 Splitting a is a triplet b is a quartet c is a singlet

96. CHE2202, Chapter 13 Learn, 96 a is a triplet b is a multiplet c is a triplet d is a singlet Splitting

97. CHE2202, Chapter 13 Learn, 97 Equivalent Protons Do Not Split Each Other’s Signals

98. CHE2202, Chapter 13 Learn, 98 What Causes Splitting?

99. CHE2202, Chapter 13 Learn, 99 What Causes Splitting?

100. CHE2202, Chapter 13 Learn, 100 Why a Quartet? Why 1:3:3:1? the ways in which the magnetic fields of three protons can be aligned

101. CHE2202, Chapter 13 Learn, 101 Splitting is Observed if the Protons are Separated by No More Than Three Bonds Splitting may occur through four bonds if one is a double bond.

102. CHE2202, Chapter 13 Learn, 102 The 1 H NMR Spectrum of 1,3-Dibromopropane Triplet two neighboring protons Quintet four neighboring protons

103. CHE2202, Chapter 13 Learn, 103 An 1 H NMR Spectrum

104. CHE2202, Chapter 13 Learn, 104 An 1 H NMR Spectrum of Allyl Bromide

105. CHE2202, Chapter 13 Learn, 105 A Quartet versus a Doublet of Doublets

106. CHE2202, Chapter 13 Learn, 106 The signals for the H c, H d, and H e protons overlap because the electronic effect of an ethyl substituent is similar to that of a hydrogen. An 1 H NMR Spectrum of Ethylbenzene

107. CHE2202, Chapter 13 Learn, 107 An 1 H NMR Spectrum of Nitrobenzene The signals for the H a, H b, and H c protons do not overlap because the nitro group is strongly electron withdrawing.

108. CHE2202, Chapter 13 Learn, 108 Coupling Constants Coupled protons have the same coupling constant. The coupling constant (J) is the distance between two adjacent peaks of a split NMR signal in hertz.

109. CHE2202, Chapter 13 Learn, 109 Coupling Constants

110. CHE2202, Chapter 13 Learn, 110 Summary The number of signals tells us the number of sets of equivalent protons in the compound. The value of the chemical shifts tells us the nature of the chemical environment: alkyl, alkene, benzene, etc. The integration values tells us the relative number of protons. The splitting tells us the number of neighboring protons. The coupling constants identifies coupled protons.

111. CHE2202, Chapter 13 Learn, 111 The Coupling Constant is Greater for Trans Protons Than for Cis Protons

112. CHE2202, Chapter 13 Learn, 112 A Splitting Diagram The number of peaks observed depends on the relative magnitudes of the coupling constants.

113. CHE2202, Chapter 13 Learn, 113 Why a Quintet Rather Than a Quartet?

114. CHE2202, Chapter 13 Learn, 114 J ab = J ac

115. CHE2202, Chapter 13 Learn, 115 Adjacent Protons: Different or the Same? When two different sets of protons split a signal, the N + 1 rule is used separately for each set if the coupling constants for each set are different. When the coupling constants are similar, the N + 1 rule is used for the two sets of protons as though they were equivalent.

116. CHE2202, Chapter 13 Learn, 116 Enantiotopic Hydrogens Enantiotopic hydrogens are chemically equivalent. Replacing one of the enantiotopic hydrogens with a deuterium (or any other atom or group other than CH 3 or OH) forms an asymmetric center.

117. CHE2202, Chapter 13 Learn, 117 A prochiral carbon will become an asymmetric center if one of the Hs to which it is attached is replace by a deuterium. If the pro-R hydrogen is replaced by deuterium, the asymmetric center will have the R configuration. If the pro-S hydrogen is replaced by deuterium, the asymmetric center will have the S configuration. Prochiral, Pro-R, Pro-S

118. CHE2202, Chapter 13 Learn, 118 Diastereotopic hydrogens are not chemically equivalent. Diastereotopic Hydrogens Replacing each of the diastereotopic hydrogens in turn by a deuterium forms a pair of diastereomers.

119. CHE2202, Chapter 13 Learn, 119 Diastereotopic Hydrogens are Not Chemically Equivalent Diastereotopic hydrogens react with achiral reagents at different rates.

120. CHE2202, Chapter 13 Learn, 120 Equivalent Hydrogens The three methyl hydrogens are in different environments because of rotation about the C—C bond; but on the NMR time scale, they are in the same environment.

121. CHE2202, Chapter 13 Learn, 121 The axial and equatorial hydrogens of cyclohexand-d 11 are equivalent and show one sharp signal. The rate of chair–chair interconversion is temperature dependent: as the temperature decreases, the signal broadens and eventually two signals are observed. A Signal Represents an Average of the Proton’s Environment

122. CHE2202, Chapter 13 Learn, 122 Protons Bonded to O or N pure ethanol ethanol with acid

123. CHE2202, Chapter 13 Learn, 123 Acid-Catalyzed Proton Exchange The chemical shift depends on the extent of hydrogen bonding. They generally appear as broad signals.

124. CHE2202, Chapter 13 Learn, 124 Deuterium Signals are Not Seen in an 1 H NMR Spectrum

125. CHE2202, Chapter 13 Learn, 125 60-MHz Versus 300-MHz a 60-MHz 1 H NMR spectrum a 300-MHz 1 H NMR spectrum

126. CHE2202, Chapter 13 Learn, 126 To see separate signals with clean splitting patterns, the difference in the chemical shifts of the two coupled protons must be 10 times the value of the coupling constant. The Resolution of 1 H NMR Spectra

127. CHE2202, Chapter 13 Learn, 127 13 C NMR Spectroscopy The number of signals reflects the number of different kinds of carbons in a compound. The chemical shift ranges over 220 ppm. The reference compound is TMS.

129. CHE2202, Chapter 13 Learn, 129 Where 13 C NMR Signals Appear

130. CHE2202, Chapter 13 Learn, 130 The 13 C NMR Spectrum of 2-Butanol

131. CHE2202, Chapter 13 Learn, 131 The Proton-Coupled 13 C NMR Spectrum of 2-Butanol

132. CHE2202, Chapter 13 Learn, 132 A 13 C NMR Spectrum Carbons that are not attached to hydrogens give very small signals.

133. CHE2202, Chapter 13 Learn, 133 A DEPT 13 C NMR Spectrum (Distinguishes CH 3, CH 2, and CH Groups)

134. CHE2202, Chapter 13 Learn, 134 Two-Dimensional NMR Spectroscopy (A COSY Spectrum) Cross peaks indicate pairs of protons that are coupled.

135. CHE2202, Chapter 13 Learn, 135 A COSY Spectrum of 1-Nitropropane

136. CHE2202, Chapter 13 Learn, 136 A HETCOR Spectrum of Ethyl Isopropyl Ketone A HETCOR spectrum of indicates coupling between protons and the carbon to which they are attached.

137. CHE2202, Chapter 13 Learn, 137 Nuclear Magnetic Resonance (NMR) Magnetic Resonance Imaging (MRI) an MRI scanner

138. CHE2202, Chapter 13 Learn, 138 An MRI The white region is a brain lesion.The spectrum indicates an abscess.

139. CHE2202, Chapter 13 Learn, 139 An X-Ray Diffraction Pattern