Nuclear Magnetic Resonance Spectroscopy

Fundamental principle of NMR and Interpretation of NMR

1. NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY PRESENTED BY DEVISHA TATINENI PHARMACEUTICAL CHEMISTRY M.PHARM 1

2. CONTENTS Introduction Fundamental principles of NMR Interpretation Chemical shift Number of signals Spin-Spin coupling: Splitting of signals Coupling constant Integrals 2

3. NMR Spectroscopy Nuclear Magnetic Resonance is a branch of spectroscopy in which radio frequency waves induce transitions between magnetic energy levels of nuclei of a molecule. 3

4. The Electromagnetic Spectrum High Frequency Longer Wavelength 4

5. The frequency of radio waves lies between 10 7 and 10 8 cps The energy of radio frequency ( rf ) radiation can be calculated by using the equation : E = h v h = Planck ̕ s constant = 6.6 × 10 -27 erg sec v = frequency = 10 7 _ 10 8 cps(cycles per sec). 5

6. E = 6.6 × 10 _27 × 10 7 ( or 10 8 ergs) = 6.6 × 10- 20 ( or 6.6 × 10- 19 ergs ) Energy of rf radiation is very small to vibrate, rotate , or excite an atom or molecule. But this energy is sufficient to affect the nuclear spin of the atoms of a molecule. 6

7. The nuclei of some atoms have a property called “ SPIN ”. NUCLEAR SPIN These nuclei behave as if they were spinning. This is like the spin property of an electron, which can have two spins: +1/2 and -1/2 . Each spin-active nucleus has a number of spins defined by its spin quantum number, I . The number of Spin states = 2I + 1 . 7

8. If the number of neutrons and the number of protons are both even, then the nucleus has NO spin. 12 C, 16 O , 32 S etc. If the number of neutrons plus the number of protons is odd, then the nucleus has a half-integer spin (i.e. 1/2, 3/2, 5/2) 1 H, 19 F, 31 P If the number of neutrons and the number of protons are both odd, then the nucleus has an integer spin (i.e. 1, 2, 3) 2 H, 14 N 8

9. 9

10. Principle NMR spectroscopy is the interaction of magnetic field with spin of nuclei and then absorption of radio frequency. For example, the nucleus of proton 1 H + has two spin rotations : clockwise rotation with a spin quantum number I = +½ and | counterclockwise rotation with a spin quantum number I = - ½ The number of spin sates is 2I+1 which is 2x (1/2) +1 = 2 state 10

11. NUCLEAR SPIN STATES - HYDROGEN NUCLEUS + 1/2 - 1/2 The two states are equivalent in energy in the absence of a magnetic or an electric field. + + The spin of the positively charged nucleus generates a magnetic moment vector, m . m m TWO SPIN STATES 11

12. Without the magnetic field the spin states of nuclei possess the same energy, and energy level transition is not possible. When a magnetic field is applied, the separate levels and radio frequency radiation can cause transitions between these energy levels. 12

13. no difference in absence of magnetic field proportional to strength of external magnetic field Energy Differences Between Nuclear Spin States + +  E  E ' increasing field strength 13

14. Some important relationships in NMR The frequency of absorbed electromagnetic radiation is proportional to the energy difference between two nuclear spin states which is proportional to the applied magnetic field Units Hz kJ/mol (kcal/mol) tesla (T) 14

15. Magnetic properties of nuclei When a charged particle such as a proton spins on its axis, it creates a magnetic field. Thus, the nucleus can be considered to be a tiny bar magnet. 15

16. But when magnetic field is applied, the proton (H) posses spin & their own magnetic field align themselves either or opposite to magnetic field. For e.g. 1H has +1/2 & -1/2 spin state, the proton (H) have +1/2 spin state align themselves with field (Lower energy) and with -1/2 spin state align opposite to field (Higher energy). 16

17. 17

18. Nuclear Spin 18

19. Change in spin state energy separation with increase by applied magnetic field ,B 19

20. External Magnetic Field When placed in an external field, spinning protons act like bar magnets. => 20

21. THE “RESONANCE” PHENOMENON absorption of energy by the spinning nucleus 21

22. N uclear Spin Energy Levels B o +1/2 -1/2 In a strong magnetic field (B o ) the two spin states differ in energy. aligned unaligned N S 22

23. Two Energy States The magnetic fields of the spinning nuclei will align either with the external field, or against the field. A photon with the right amount of energy can be absorbed and cause the spinning proton to flip.

24. N S N S Add Energy N S S N Aligned = Low Energy Excited state = High energy N S S N Energy Released Back to low energy ground state 24

25. According to the quantum theory, a spinning nucleus can only have values for the spin angular momentum given by the equation : Spin angular momentum = [I(I+1)] 1/2 h / 2µ I = Spin quantum number h= Planck̕ s constant 25

26. But µ = γ × spin angular momentum µ = magnetic moment of the nucleus γ = gyro magnetic ratio If a nucleus having a magnetic moment is introduced into a magnetic field , H 0, the two energy levels become separate corresponding to m I = -1/2 (anti-parallel to the direction of magnetic field) and m I = +1/2 (parallel to the direction of magnetic field). 26

27. For a nucleus with I = 1/2, the energies E 1 and E 2 for the two states with m I = +1/2 and m I = -1/2 , respectively, are E 1 = -1/2 [ γ h / 2µ ] H E 2 = + 1/2 [ γ h / 2µ ] H 27

28. 28

29. When the nucleus absorbs energy, the nucleus will be promoted from the lower energy state E 1 , to the higher energy state E 2 by absorption of energy , ∆E, equal to the energy difference, E 2 – E 1 . It means that the absorption of energy ∆E changes the magnetic moment from the parallel state m I = +1/2 to the anti-parallel state (m I = -1/2). 29

30. The frequency ν at which energy is absorbed or emitted is given by Bohr̕ s relationship: ν = E 2 – E 1 / h ν = 1/2 [ γ h / 2µ ] H + 1/2 [ γ h / 2µ ] H / h ( or ) ν = γ / 2µ H This is the Larmor equation 30

31. The Larmor Equation!!! γ n = 2 µ H o g is a constant which is different for each atomic nucleus (H, C, N, etc) gyromagnetic ratio g strength of the magnetic field frequency of the incoming radiation that will cause a transition

32. From the Larmor equation : that the frequency absorbed or emitted by a nucleus in moving from one energy level to another is directly proportional to the applied magnetic field. 32

33. When a nucleus is placed in a system where it absorbs energy, it becomes excited. It then loses energy to return to the unexcited state .It absorbs energy and again enters an excited state. This nucleus which alternatively becomes excited and unexcited is said to be in a state of resonance. 33

34. WHEN A SPIN-ACTIVE HYDROGEN ATOM IS PLACED IN A STRONG MAGNETIC FIELD ….. IT BEGINS TO PRECESS A SECOND EFFECT OF A STRONG MAGNETIC FIELD

35. Precessional frequency:- May be defined as revolutions per second made by the magnetic moment vector of the nucleus around the external magnetic field Bo & Alternatively precessional frequency of spinning bar magnet may be defined as equal to the frequency of EMR in megacycles per second necessary to induce a transition from one spin state to another spin state. 35

36. The frequency of precession (proton) is directly proportional to the strength of applied magnetic field. If the precessing nucleus is irradiated with electromagnetic radiation of the same frequency as that frequency of precession nucleus, then Energy is absorbed The nuclear spin is flipped from spin state +1/2 (with the applied field) to -1/2 (against the applied field). 36

37. 37

38. N S w Nuclei precess at frequency w when placed in a strong magnetic field. h n If n = w then energy will be absorbed and the spin will invert. NUCLEAR MAGNETIC RESONANCE NMR RADIOFREQUENCY radiation

39. Interpreting Proton NMR Spectra

40. TYPES OF INFORMATION FROM THE NMR SPECTRUM 1. Each different type of hydrogen gives a peak or group of peaks (multiplet). 3. The integral gives the relative numbers of each type of hydrogen. 2 . The chemical shift ( ᵟ, in ppm ) gives a clue as to the type of hydrogen generating the peak (alkane, alkene, benzene, aldehyde, etc.) 4. Spin-spin splitting gives the number of hydrogens on adjacent carbons. 5. The coupling constant J also gives information about the arrangement of the atoms involved.

41. Chemical Shift (Position of Signals) The utility of NMR is that all protons do not show resonance at same frequency because, it is surrounded by particular no. of valence electrons which vary from atom to atom so, they exist in slightly different electronic environment from one another. Position of signals in spectrum help us to know nature of protons i.e. aromatic, aliphatic, acetylinic, vinylic, adjacent to electron releasing or withdrawing group. Thus they absorb at different field strength. 41

42. 42 The relative energy of resonance of a particular nucleus resulting from its local environment is called chemical shift. NMR spectra show applied field strength increasing from left to right. Left part is down field right part is up field. Nuclei that absorb on up field side are strongly shielded.

43. 43 Let’s consider the just the proton ( 1 H) NMR

44. 44

45. 45 Measuring Chemical Shift Numeric value of chemical shift: difference between strength of magnetic field at which the observed nucleus resonates and field strength for resonance of a reference.

46. For measuring chemical shifts of various protons in a molecule, the signal of TMS is taken as reference. The NMR signal for particular protons will appear at different field strength compared to signal from TMS. This difference in the absorption position of the proton with respect to TMS signal is called as chemical shift ( δ – value). 46

47. TMS shift in Hz tetramethylsilane “TMS” reference compound n Rather than measure the exact resonance position of a peak, we measure how far downfield it is shifted from TMS. Highly shielded protons appear way upfield. Chemists originally thought no other compound would come at a higher field than TMS. downfield

48. TMS (Tetra methyl silane) is most commonly used as IS in NMR spectroscopy. Due to following reasons; It is chemically inert and miscible with a large range of solvents. Its twelve protons are all magnetically equivalent. Its protons are highly shielded and gives a strong peak even small quantity. It is less electronegative than carbon. It is highly volatile and can be easily removed to get back sample. 48

49. chemical shift = d = shift in Hz spectrometer frequency in MHz = ppm This division gives a number independent of the instrument used. parts per million the “chemical shift” in the following way:

50. The alternative system which is generally used for defining the position of resonance relative to the reference is assigned tau (  ) scale.  = 10 -  A small numerically value of  indicates a small downfield shift while large value indicates a large downfield shift. A small value of  represents a low field absorption and a high value indicates a high field absorption.  scale

51. SHIELDING AND DESHIELDING Rotation of electrons ( π ) to nearby nuclei generate field that can either oppose or strong the field on proton. If magnetic field is oppose applied magnetic field on proton, that proton said shielded proton and if field is strong the applied field then, proton feels high magnetic field strength and such proton called as Deshielded proton. 51

52. So, shielded proton shifts absorption signal to right side (upfield) and deshielded proton shifts absorption signal to left side (down field) of spectrum. So, electric environment surrounding proton tells us where proton shows absorption in spectrum. 52

53. The electrons around the proton create a magnetic field that opposes the applied field. Since this reduces the field experienced at the nucleus, the electrons are said to shield the proton.

54. Shielded Protons Magnetic field strength must be increased for a shielded proton to flip at the same frequency. =>

55. The protons are shielded by the electrons that surround them. In an applied magnetic field, the valance electrons of the protons are caused to circulate. This circulation, called a local diamagnetic current, generates a counter magnetic field that opposes the applied magnetic field. This effect, which is called diamagnetic shielding or diamagnetic anisotropy. 55

56. The counter field that shields a nucleus diminishes the net applied magnetic field that the nucleus experiences. As a result, the nucleus precesses at a lower frequency. This means that it also absorbs radiofrequency radiation at this lower frequency. 56

57. When the secondary fields produced by the circulating electrons oppose the applied field at a particular nucleus in the molecule, it means that effective field experienced by the nucleus is less than the applied field. This is known as positive shielding and the resonance position moves up field in NMR spectrum. 57

58. If the secondary field produced by the circulating electrons reinforces the applied field, the position of resonance moves downfield. This is known as negative shielding. 58

59. shielded/upfield: higher electron density requires a stronger field for resonance deshielded/downfield: lower electron density requires a weaker field for resonance (using a constant radiofrequency)

60. Chemical shift depends upon following parameters: Inductive effect Hybridization Anisotropic effect Hydrogen bonding 60

61. Inductive effect 61

62. Electronegativity – chemical shift Compound CH 3 X Element X Electronegativity of X Chemical shift d CH 3 F CH 3 OH CH 3 Cl CH 3 Br CH 3 I CH 4 (CH 3 ) 4 Si F O Cl Br I H Si 4.0 3.5 3.1 2.8 2.5 2.1 1.8 4.26 3.40 3.05 2.68 2.16 0.23 Dependence of the Chemical Shift of CH 3 X on the Element X deshielding increases with the Electronegativity of atom X TMS most deshielded

63. highly shielded protons appear at high field “deshielded“ protons appear at low field deshielding moves proton resonance to lower field C H Cl Chlorine “deshields” the proton, that is, it takes valence electron density away from carbon, which in turn takes more density from hydrogen deshielding the proton. electronegative element DESHIELDING BY AN ELECTRONEGATIVE ELEMENT NMR CHART d - d+ d - d+

64. Substitution Effects on Chemical Shift C H Cl 3 C H 2 Cl 2 C H 3 Cl 7.27 5.30 3.05 ppm -C H 2 -Br -C H 2 -CH 2 Br -C H 2 -CH 2 CH 2 Br 3.30 1.69 1.25 ppm most deshielded most deshielded The effect decreases with increasing distance . The effect increases with greater numbers of electronegative atoms.

65. Electron withdrawing groups– reduces electron density around the proton Deshielding. Examples: F, Cl, Br, I, OH, NH 2, NO 2 Electron releasing groups- increases electron density around the proton Shielding. Examples: Alkyl groups 66

66. Hybridization sp 3 Hydrogens ( no electro negative elements and Π bonded groups ) 3 > 2 > 1 > cyclopropane 67

67. sp 2 Hydrogens: In an sp 2 C-H bond, the carbon atom has more s character (33% s), which effectively renders it more electronegative than an sp 3 carbon (25% s). If the sp 2 carbon atom holds its electrons more tightly, this results in less shielding for the H nucleus than in an sp 3 bond. Another effect anisotropy. 68

68. sp Hydrogens: On the basis of hybridization, acetylenic proton to have a chemical shift greater than that of vinyl proton. But chemical shift of acetylenic proton is less than that of vinyl proton. Finally sp 2 > sp > sp 3 . 69

69. Anisotropic Effect Due to the presence of pi bonds 70

70. Anisotropic effects constitute shielding and deshielding effects on the proton because of induced magnetic fields in other parts of the molecule which operate through space. For example if a magnetic field is applied to a molecule having п electrons, these electrons begin to circulate at right angles to the direction of the applied field thereby producing induced magnetic field. 71

71. The effect of this field on the nearby proton has been found to depend upon the orientation of the proton with respect to the Π bond producing the induced field. 72

72. fields add together

73. C=C H H H H B o ANISOTROPIC FIELD IN AN ALKENE protons are deshielded shifted downfield secondary magnetic (anisotropic) field lines Deshielded fields add

74. B o secondary magnetic (anisotropic) field H H C C ANISOTROPIC FIELD FOR AN ALKYNE hydrogens are shielded Shielded fields subtract

75. Protons attached to sp 2 hybridized carbon are less shielded than those attached to sp 3 hybridized carbon. H H H H H H C C H H H H CH 3 CH 3  7.3 ppm  5.3 ppm  0.9 ppm

76. Acidic Hydrogens & Hydrogen bonding 77

77. Acidic hydrogens: Both resonance and the Electronegativity effect of oxygen withdraw electrons from the acid proton. 78

78. HYDROGEN BONDING DESHIELDS PROTONS The chemical shift depends on how much hydrogen bonding is taking place . ( in concentrated solution) Hydrogen bonding lengthens the O-H bond and reduces the valence electron density around the proton - it is deshielded and shifted downfield in the NMR spectrum.

79. Carboxylic acids have strong hydrogen bonding - they form dimers. In methyl salicylate, which has strong internal hydrogen bonding, Notice that a 6-membered ring is formed. SOME MORE EXTREME EXAMPLES

80. APPROXIMATE CHEMICAL SHIFT RANGES (ppm) FOR SELECTED TYPES OF PROTONS R-C H 3 0.7 - 1.3 R-C=C-C- H 1.6 - 2.6 R-C-C- H 2.1 - 2.4 O O RO-C-C- H 2.1 - 2.5 O HO-C-C- H 2.1 - 2.5 N C-C- H 2.1 - 3.0 R-C C -C- H 2.1 - 3.0 C -H 2.3 - 2.7 R-N-C- H 2.2 - 2.9 R-S-C- H 2.0 - 3.0 I-C- H 2.0 - 4.0 Br-C- H 2.7 - 4.1 Cl-C- H 3.1 - 4.1 RO-C- H 3.2 - 3.8 HO-C- H 3.2 - 3.8 R-C-O-C- H 3.5 - 4.8 O R-C=C- H H 6.5 - 8.0 R-C- H O 9.0 - 10.0 R-C-O- H O 11.0 - 12.0 O 2 N-C- H 4.1 - 4.3 F-C- H 4.2 - 4.8 R 3 C H 1.4 - 1.7 R-C H 2 - R 1.2 - 1.4 4.5 - 6.5 R-N- H 0.5 - 4.0 Ar-N -H 3.0 - 5.0 R-S -H R-O -H 0.5 - 5.0 Ar-O- H 4.0 - 7.0 R-C-N- H O 5.0 - 9.0 1.0 - 4.0 R-C C- H 1.7 - 2.7

81. YOU DO NOT NEED TO MEMORIZE THE PREVIOUS CHART IT IS USUALLY SUFFICIENT TO KNOW WHAT TYPES OF HYDROGENS COME IN SELECTED AREAS OF THE NMR CHART aliphatic C-H CH on C next to pi bonds C-H where C is attached to an electronegative atom alkene =C-H benzene CH aldehyde CHO acid COOH 2 3 4 6 7 9 10 12 X-C-H X=C-C-H MOST SPECTRA CAN BE INTERPRETED WITH A KNOWLEDGE OF WHAT IS SHOWN HERE

82. Number of signals In a given molecule the protons with different environments absorb at different applied field strengths whereas the protons with identical environments absorb at the same field strength. A set of protons of identical environments are known as equivalent protons while the protons with different environments are known as non-equivalent protons. The number of signals in a PMR spectrum tells us how many kinds of protons are present in a given molecule. 83

83. Prediction of different kinds of protons: suppose each hydrogen or proton in the molecule to be substituted by some other atom Z. If the substitution of either of two protons by Z is expected to furnish the same product or enantiomeric products (i.e. mirror images), the two protons would be chemically and magnetically equivalent, otherwise not. 84

84. Ethyl chloride CH3-CH2-Cl Substitution of a methyl proton CH2Z-CH2-Cl Substitution of a methylene proton CH3-CHZ-Cl So methyl protons are not equivalent to methylene protons. 85

85. Prediction of Signal Number

86. Acetone (one signal)

87. Benzene (one signal)

88. p- xylene (Two NMR signals)

89. Methyl Acetate (Two NMR signals)

90. Ethyl benzene (Three NMR signals)

91. Propane2-ol (3 NMR signals)

92. Wrong Methyl cyclopropane (Three signals)

93. (4 signals)

94. 2- chloropropene Three signal

95. 2- chloropropene having three methyl and two vinylic protons. The methyl protons are non-equivalent to vinylic protons but equivalent amongst themselves. In case of the vinylic protons, replacement of either of the two would yield one of a pair of diastereomeric products. 96

96. 97

97. One signal 98

98. Two signals 99

99. Two signals 100

100. Two signals 101

101. Three signals 102

102. Three signals 103

103. Three signals 104

104. Four signals 105

105. INTEGRATION

106. The area under a peak is proportional to the number of hydrogens that generate the peak. Integration = determination of the area under a peak INTEGRATION OF A PEAK Not only does each different type of hydrogen give a distinct peak in the NMR spectrum, but we can also tell the relative numbers of each type of hydrogen by a process called integration.

107. 55 : 22 : 33 = 5 : 2 : 3 The integral line rises an amount proportional to the number of H in each peak METHOD 1 integral line integral line simplest ratio of the heights Benzyl Acetate

108. Benzyl Acetate (FT-NMR) assume CH 3 33.929 / 3 = 11.3 33.929 / 11.3 = 3.00 21.215 / 11.3 = 1.90 58.117 / 11.3 = 5.14 Actually : 5 2 3 METHOD 2 digital integration Modern instruments report the integral as a number. Integrals are good to about 10% accuracy.

109. Each different type of proton comes at a different place . NMR Spectrum of Phenyl acetone

110. SPIN-SPIN SPLITTING

111. Often a group of hydrogens will appear as a multiplet rather than as a single peak. SPIN-SPIN SPLITTING Multiplets are named as follows: Singlet Quintet Doublet Septet Triplet Octet Quartet Nonet This happens because of interaction with neighboring Hydrogens.

112. 1, 1, 2-Trichloroethane 113

113. The sub peaks are due to spin-spin splitting and are predicted by the n+1 rule. Each type of proton senses the number of equivalent protons (n) on the carbon atom(s) next to the one to which it is bonded, and its resonance peak is split into (n+1) components. 114

114. n + 1 RULE

115. two neighbors n+1 = 3 triplet one neighbor n+1 = 2 doublet singlet doublet triplet quartet quintet sextet septet MULTIPLETS this hydrogen’s peak is split by its two neighbors these hydrogens are split by their single neighbor

116. EXCEPTIONS TO THE N+1 RULE IMPORTANT ! Protons that are equivalent by symmetry usually do not split one another no splitting if x=y no splitting if x=y 1) 2) Protons in the same group usually do not split one another or more detail later

117. 3) The n+1 rule applies principally to protons in aliphatic (saturated) chains or on saturated rings. EXCEPTIONS TO THE N+1 RULE or but does not apply (in the simple way shown here) to protons on double bonds or on benzene rings. NO NO YES YES

118. SOME COMMON PATTERNS

119. SOME COMMON SPLITTING PATTERNS ( x = y ) ( x = y )

120. SOME EXAMPLE SPECTRA WITH SPLITTING

121. NMR Spectrum of Bromoethane

122. NMR Spectrum of 2-Nitropropane 1:6:15:20:16:6:1 in higher multiplets the outer peaks are often nearly lost in the baseline

123. NMR Spectrum of Acetaldehyde offset = 2.0 ppm

124. INTENSITIES OF MULTIPLET PEAKS PASCAL’S TRIANGLE

125. PASCAL’S TRIANGLE Intensities of Multiplet peaks

126. THE ORIGIN OF SPIN-SPIN SPLITTING HOW IT HAPPENS

127. 129

128. 130

129. C C H H C C H H A A upfield downfield B o THE CHEMICAL SHIFT OF PROTON H A IS AFFECTED BY THE SPIN OF ITS NEIGHBORS 50 % of molecules 50 % of molecules At any given time about half of the molecules in solution will have spin +1/2 and the other half will have spin -1/2. aligned with B o opposed to B o neighbor aligned neighbor opposed +1/2 -1/2

130. C C H H C C H H one neighbor n+1 = 2 doublet one neighbor n+1 = 2 doublet SPIN ARRANGEMENTS The resonance positions (splitting) of a given hydrogen is affected by the possible spins of its neighbor.

131. two neighbors n+1 = 3 triplet one neighbor n+1 = 2 doublet SPIN ARRANGEMENTS methylene spins methine spins

132. three neighbors n+1 = 4 quartet two neighbors n+1 = 3 triplet SPIN ARRANGEMENTS C C H H H H H C C H H H H H methyl spins methylene spins

133. THE COUPLING CONSTANT

134. J J J J J THE COUPLING CONSTANT The coupling constant is the distance J (measured in Hz) between the peaks in a simple multiplet . J is a measure of the amount of interaction between the two sets of hydrogens creating the multiplet. J

135. The coupling constant is a measure of how strongly a nucleus is affected by the spin states of its neighbor. Coupling constant is expressed in Hertz (Hz). 137

136. 100 MHz 200 MHz 1 2 3 4 5 6 1 2 3 100 Hz 200 Hz 200 Hz 400 Hz J = 7.5 Hz J = 7.5 Hz 7.5 Hz 7.5 Hz Coupling constants are constant - they do not change at different field strengths The shift is dependant on the field ppm FIELD COMPARISON Separation is larger

137. NOTATION FOR COUPLING CONSTANTS The most commonly encountered type of coupling is between hydrogens on adjacent carbon atoms. This is sometimes called vicinal coupling. It is designated 3 J since three bonds intervene between the two hydrogens. Another type of coupling that can also occur in special cases is 2 J or geminal coupling Geminal coupling does not occur when the two hydrogens are equivalent due to rotations around the other two bonds. ( most often 2 J = 0 ) 3 J 2 J

138. Couplings larger than 2 J or 3 J also exist, but operate only in special situations. Couplings larger than 3 J (e.g., 4 J, 5 J, etc) are usually called “long-range coupling.” C C C H H 4 J , for instance, occurs mainly when the hydrogens are forced to adopt this “W” conformation (as in bicyclic compounds). LONG RANGE COUPLINGS

139. 6 to 8 Hz 11 to 18 Hz 6 to 15 Hz 0 to 5 Hz three bond 3 J two bond 2 J three bond 3 J three bond 3 J SOME REPRESENTATIVE COUPLING CONSTANTS H ax ,H ax = 8 to 14 H ax ,H eq = 0 to 7 H eq ,H eq = 0 to 5 three bond 3 J trans cis geminal vicinal

140. 4 to 10 Hz 0 to 3 Hz four bond 4 J three bond 3 J 0 to 3 Hz four bond 4 J cis trans 6 to 12 Hz 4 to 8 Hz three bond 3 J Couplings that occur at distances greater than three bonds are called long-range couplings and they are usually small (<3 Hz) and frequently nonexistent (0 Hz).

141. Interpretation of some functional groups 143

142. Alkanes Alkanes have three types of hydrogens – methyl, methylene and methyne. 144

143. 145

144. Alkenes 146 Alkenes have two types of hydrogens – vinyl (those attached directly to the double bond) and allylic hydrogens (those attached to the α - carbon, the carbon atom attached to the double bond).

145. 147

146. Aromatic compounds Aromatic compounds have two types of hydrogens – aromatic ring hydrogens (benzene ring hydrogens) and benzylic hydrogens (those attached to an adjacent carbon atom). 148

147. 149

148. MONOSUBSTITUTED RINGS

149. R = alkyl (only) Apparently the ring current equalizes the electron density at all the carbons of the ring and, therefore, at all of the hydrogen atoms. In monosubstituted rings with an alkyl substituent all ring hydrogens come at the same place in the NMR spectrum. ALKYL-SUBSTITUTED RINGS

150. NMR Spectrum of Toluene 5 3

151. X = OH, OR, + + + : .. : - - - : : : : .. Electron-donating groups shield the o- , p- positions due to resonance (see below). .. .. .. .. .. .. .. .. unshared pair ester Electronegative elements with unshared pairs shield the o- and p- ring positions, separating the hydrogens into two groups. NH 2 , NR 2, -O(CO)CH 3 SUBSTITUENTS WITH UNSHARED PAIRS

152. Anisole (400 MHz) 2 3 shielded The ring protons in toluene come at about 7.2 ppm at the red line. Compare:

153. Only the o- protons are in range for this effect. The same effect is sometimes seen with C=C bonds. When a carbonyl group is attached to the ring the o- protons are deshielded by the anisotropic field of C=O THE EFFECT OF CARBONYL SUBSTITUENTS

154. Acetophenone (90 MHz) 2 3 3 deshielded The ring protons in toluene come at about 7.2 ppm at the red line. Compare:

155. para -DISUBSTITUTED RINGS

156. 1,4-Disubstituted benzene rings will show a pair of doublets, when the two groups on the ring are very different an example: 1-iodo-4-methoxybenzene para -Disubstitution

157. NMR Spectrum of 1-iodo-4-methoxybenzene CHCl 3 impurity 2 2 3

158. NMR Spectrum of 1-bromo-4-ethoxybenzene 4 2 3

159. X = Y X ~ X’ X = X THE p-DISUBSTITUTED PATTERN CHANGES AS THE TWO GROUPS BECOME MORE AND MORE SIMILAR all H equivalent All peaks move closer. Outer peaks get smaller …………………..… and finally disappear. Inner peaks get taller…………………………. and finally merge. same groups

160. NMR Spectrum of 1-amino-4-ethoxybenzene 4 2 2 3

161. NMR Spectrum of p - Xylene (1,4-dimethylbenzene) 4 6

162. 3/27/2013 164

163. NMR 3/27/2013 165 m o/p The Amino group is Electron Donating and Activates the ring. Increases electron density around Ortho & Para protons relative to Meta . Chemical Shift, , of ring protons is up field, decreased ppm o m m o p 2 Amino Protons Aniline (C 6 H 7 N)

164. 3/27/2013 166 Nitro group is electron withdrawing and deactivates the ring. Protons in ring are deshielded moving Chemical Shift downfield . Magnetic Anisotropy causes the Ortho protons to be more deshielded than the Para & Meta protons. p m o o m m p o Nitrobenzene (C 6 H 5 NO 2 )

165. 3/27/2013 167 Ha Ha’ Hb Hb ’ Ha&Ha ’ Hb&Hb ’ 2 Amino Protons Para Di-Substituted Benzene ring Ha & Ha’ have same Chemical Shift Hb & Hb’ have same Chemical Shift Ha is split into doublet by Hb Hb is split into doublet by Ha Two sets of peaks produced by relative Electronegativity of Amino & Cl groups P- Chloroaniline (C 6 H 6 ClN)

166. Alkynes Alkynes have two types of hydrogens – acetylenic hydrogen and α hydrogens found on carbon atoms next to the triple bond. 168

167. 169

168. Alkyl halides Alkyl halides have one type of hydrogen – α hydrogen (the one attached to the same carbon as the halogen). 170

169. 171

170. Alcohols Alcohols have two types of hydrogens – hydroxyl proton and the α hydrogens (those on the same carbon as the hydroxyl group). 172

171. 173

172. Ethers Ethers have one type of hydrogen – the α hydrogens (those attached to the α carbon, which is the carbon atom attached to the oxygen). 174

173. 175

174. Amines Amines have two types of hydrogens – those attached to the nitrogen (the hydrogens of the amino group) and those attached to the α carbon (the same carbon to which the amino group is attached). 176

175. 177

176. Nitriles Nitriles have one type of hydrogen – the α hydrogens (those attached to the same carbon as the cyano group) 178

177. 179

178. Aldehydes Aldehydes have two types of hydrogens – aldehyde hydrogen and the α hydrogens (those attached to the same carbon as the aldehyde group). 180

179. 181

180. Ketones Ketones have one type of hydrogens – hydrogens attached to the α carbon. 182

181. 183

182. Esters Esters have two types of hydrogens – those on the carbon atom attached to the oxygen atom in the alcohol part of the ester and those on the α carbon in the acid part of the ester ( that is, those attached to the carbon next to the C=O group). 184

183. 185

184. Carboxylic acids Carboxylic acids have two types of hydrogens – acid proton (the one attached to the –COOH group) and the α hydrogens (those attached to the same carbon as the carboxyl group). 186

185. NMR Spectrum of 2-Chloropropanoic Acid offset = 4.00 ppm COOH 1 1 3 ~12 ppm

186. Amides Amides have three types of hydrogens – hydrogens attached to nitrogen, α hydrogens attached to the carbon atom on the carbonyl side of the amide group, and hydrogens attached to a carbon atom that is also attached to the nitrogen atom. 188

187. 189

188. Nitroalkanes Nitroalkanes have one type of hydrogens – α hydrogens, those hydrogen atoms that are attached to the same carbon atom to which the nitro group is attached. 190

189. 191

190. EXAMPLE SPECTRA FOR DISCUSSION PART ONE

191. NMR 07 Methyl Ethyl Ketone 1

192. NMR 08 Ethyl Acetate 2 Compare the methylene shift to that of Methyl Ethyl Ketone (previous slide).

193. a - Chloropropionic Acid NMR 04 3

194. t- Butyl Methyl Ketone NMR 03 4 (3,3-dimethyl-2-butanone)

195. NMR 05 1-Nitropropane 5

196. NMR 17 1,3-Dichloropropane 6

197. EXAMPLE SPECTRA FOR DISCUSSION PART TWO

198. NMR 09 a -Bromobutyric Acid 7

199. NMR 14 Phenylethyl Acetate 8

200. NMR 18 Ethyl Succinate 9

201. NMR 19 Diethyl Maleate 10

202. NMR 11 Ethanol 11

203. NMR 12 Benzyl Alcohol 12

204. 13 NMR 13 n-Propyl Alcohol

205. 2-Nitropropane

206. 2-phenylethylethanoate

207. Ibuprofen

208. Paracetamol

209. Phenacetin (C 10 H 13 NO 2 ) 3/27/2013 211 n+1=0+1=1 (singlet) n+1=3+1=4 (quartet) 2H 1H 3H P-Disubstitution n+1=2+1=3 (triplet) 3H 4H

210. SPECTRUM OF ETHYL IODIDE (CH 3 CH 2 I)--

211. Example : para -methoxypropiophenone Each of two types of equivalent Ar- H protons is split by its neighbor (n = 1): a doublet The -OC H 3 protons are unsplit (n = 0): a singlet The 2 equivalent O=CC H 2 protons are split by 3 –C H 3 protons (n = 3): a quartet and vice-versa, a triplet

212. One triplet 1.2 d One quartet 3.5 d C H 3 C H 2 OC H 2 C H 3

213. Phenyl acetone 3/27/2013 215

214. Benzyl acetate Integrals (Signal Area) ( Con’t ) NMR Spectrum – Benzylacetate ( 3/27/2013 216 h 1 h 2 h 3 (a) (b) (c)

215. Four slides demonstrating a process for interpreting an NMR Spectra 3/27/2013 217 3H  Chemical Shift (  ) in PPM Note: Magnetic Field (H o ) increases 

216. Sextet Quintet 2 protons see 4 protons  5 peaks (quintet) produced 3 protons see 2 protons  3 peaks (triplet) produced 1 proton sees 5 protons  6 peaks (sextet) produced 3 protons see 1 proton  2 peaks (doublet) produced 3H Mono-substituted Benzene Ring Doublet Triplet From chemical shifts, peak integration values, and splitting patterns, develop substructures for each signal.

217. 3/27/2013 219 Solve the Puzzle

218. 3/27/2013 220 2-Phenylbutane

219. References Instrumental Methods of Chemical Analysis by Gurdeep R. Chatwal, Sham K. Anand. Spectroscopy by Pavia, Lampman, Kriz, Vyvyan. www.google.com 221

220. THANK YOU