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NMR Spectroscopy CHEM 430 Fall 2011. F IRST O RDER S PECTRA For a spectrum to be 1 st order, the between the chemical shifts of any given pair of nuclei.

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Presentation on theme: "NMR Spectroscopy CHEM 430 Fall 2011. F IRST O RDER S PECTRA For a spectrum to be 1 st order, the between the chemical shifts of any given pair of nuclei."— Presentation transcript:

1 NMR Spectroscopy CHEM 430 Fall 2011

2 F IRST O RDER S PECTRA For a spectrum to be 1 st order, the between the chemical shifts of any given pair of nuclei must be much larger than the value of the coupling constant J between them: J > 10 1 st order spectra exhibit the following characteristics: Spin– spin multiplets are centered on the resonance frequency. Spacings between adjacent components of a spin–spin multiplet = J. Multiplicities that result from coupling exactly reflect the n + 1 rule for I = ½ The intensities of spin–spin multiplets correspond to the coefficients of the binomial expansion given by Pascals triangle for spin- ½ nuclei Nuclei with the same chemical shift do not split each other, even when the coupling constant between them is nonzero. NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy2

3 F IRST O RDER S PECTRA When the chemical shift difference is less than about 10 times J, 2 nd order effects appear in the spectrum, including deviations in intensities from the binomial pattern and other exceptions from the preceding characteristics. Pople notation: Nuclei that have a 1 st order relationship are represented by letters that are far apart in the alphabet (AX) Nuclei that are close in chemical shift and may exhibit a second- order relationship are represented by adjacent letters (AB) Nuclei in the middle of AX are represented as M Higher field spectrometers – > 300 MHz increase (i.e. 1 ppm represents a greater number of Hz) and minimize 2 nd order effects NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy3

4 F IRST O RDER S PECTRA AB Spin System - 1 st to 2 nd order NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy4 /J = 4 /J = 1 /J = 0.4 /J = 15

5 F IRST O RDER S PECTRA AB 2 Spin System – 1 st to 2 nd order NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy5 /J = 4 /J = 1 /J = 0.4 /J = 15

6 C HEMICAL AND M AGNETIC E QUIVALENCE In addition to meeting the requirement that n / J > 10, 1 st order spectra must pass a symmetry test. Any two chemically equivalent nuclei must have the same coupling constant to every other nucleus. Nuclear pairs that fail this test are said to be magnetically nonequivalent, and their spectral appearance is 2 nd order NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy6

7 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Nuclei are chemically equivalent if they can be interchanged by a symmetry operation of the molecule. Thus the two protons in 1,1- difluoroethene or in difluoromethane may be interchanged by a 180° rotation. Nuclei that are interchangeable by rotational symmetry are said to be homotopic. Rotation about CC single bonds is so rapid that the chemist rarely considers the fact that the three methyl protons in CH 3 CH 2 Br are not symmetrically equivalent (dynamic effect) NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy7

8 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Nuclei related by a plane of symmetry are called enantiotopic, provided there is no rotational axis of symmetry. For example, the protons in BrClCH 2 are chemically equivalent and enantiotopic because they are related by the plane of symmetry containing C, Br, and Cl. If the molecule is placed in a chiral environment, (using a solvent composed of an optically active material or by placing the molecule in the active site of an enzyme) as represented by a small hand placed to one side of BrClCH 2 the protons are no longer equivalent because the hand is a chiral object. NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy8

9 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Because the plane of symmetry is lost in a chiral environment, the nuclei are not enantiotopic and have become chemically nonequivalent ( no symmetry operation can interchange them). The term enantiotopic was coined because replacement of one proton of the pair by another atom or group, such as deuterium, produces the enantiomer NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy9

10 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Enantiotopic or homotopic protons need not be on the same carbon. For example the alkenic protons in cyclopropene are homotopic, but those in 3- methylcyclopropene are enantiotopic. Chemically equivalent nuclei (homotopic or enantiotopic) are represented by the same letter in the spectral shorthand of Pople. Cyclopropene is A 2 X 2, as is difluoromethane, since the two fluorine atoms have spins of ½ The ring protons of 3-methylcyclopropene constitute an AX 2 group. NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy10

11 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry To be magnetically equivalent, two nuclei must be chemically equivalent and have the same coupling constant to every other nucleus. This test is more stringent than that for chemical equivalence, because it is necessary to go beyond considering just the overall symmetry of the molecule. For example, in CH 2 F 2 each of the two hydrogens has the same coupling to a specific fluorine atom because both hydrogens have the same spatial relationship to that fluorine. Consequently, the protons are chemically and magnetically equivalent - A 2 X 2. NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy11

12 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Examples that once seemed simple: p-nitrotoluene On first inspection, like CH 2 F 2, it has a plane of symmetry However, on careful inspection the coupling of what would be two chemically identical nuclei (H a and H a) is different to H b so these are magnetically non-equivalent NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy12 J abJ ab

13 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Examples that once seemed simple: 1,1-difluroethene Like the previous example, the molecule has a plane of symmetry Here, the fluorines are spin-active ( ± 1/2), so each hydrogen is coupled differently to F a so these are magnetically non-equivalent NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy13 J aFJ bF

14 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Why is this important? Any spin system that contains nuclei that are chemically equivalent but magnetically nonequivalent is, by definition, 2 nd order. Moreover, raising the magnetic field cannot alter basic structural relationships between nuclei, so that the spectrum remains second order at the highest accessible fields. Nuclei that do not have the same chemical shift (anisochronous) also are magnetically nonequivalent because they resonate at different resonance frequencies (chemical shift criterion). NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy14

15 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Isochronous nuclei that are magnetically non-equivalent by having unequal couplings to another nucleus are said to fail the coupling constant criterion. Nuclei that are chemically equivalent but not magnetically equivalent are given the same letter in the Pople notation, but one is denoted by a prime Thus 1,1-difluoroethene is a AAXX system NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy15

16 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Spectra for these systems are complex – 1 H spectrum of 1,1-difluoroethene: NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy16

17 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Another example – 1 H spectrum of 1,2-dichlorobenzene: NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy17

18 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry Even open- chain systems such as ClCH 2 CH 2 OH contain magnetically nonequivalent spin systems: NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy18 A 1 st order spectrum would have comprised two 1: 2: 1 triplets (not case here) and instead of three peaks in each resonance, there are four (look to sides of center resonance)

19 C HEMICAL AND M AGNETIC E QUIVALENCE Symmetry NMR - The Coupling Constant 4-2 CHEM 430 – NMR Spectroscopy19

20 S IGNS AND M ECHANISMS Fermi Contact Interaction Spin–spin coupling arises because information about nuclear spin is transferred from nucleus to nucleus via the electrons. Both nuclei and electrons are magnetic dipoles, whose mutual interactions normally are described by the point–dipole approximation (as used by McConnell to describe diamagnetic anisotropy) Fermi found that this approximation breaks down when dipoles are very close (comparable to the radius of a proton). Under these circumstances, when the nucleus and electron in essence are in contact, their interaction is described by a new mechanism, the Fermi contact term. NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy20

21 S IGNS AND M ECHANISMS Fermi Contact Interaction The energy of the interaction is proportional to the gyromagnetic ratios of the nucleus and of the electron, the scalar (dot) product of their spins (I for a nucleus, S for an electron), and the probability that the electron is at the nucleus (the square of the electronic wave function evaluated with zero distance from the nucleus):. E FC – n e I · S 2 (0) NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy21

22 S IGNS AND M ECHANISMS Fermi Contact Interaction Because the nuclear and electronic gyromagnetic ratios have opposite signs, the more stable arrangement is when the nucleus and the electron are antiparallel (spins paired): NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy22 Nuclear spinelectron spin Energy Nuclear spinelectron spin

23 S IGNS AND M ECHANISMS Fermi Contact Interaction Here, a single bond (two electrons) joins two spin-active nuclei: 13 C- 1 H The bonding electrons will tend to avoid one another, if one is near the 13 C nucleus (in this example) the other will be near the 1 H nucleus By the Pauli principle, these electrons must be opposite in spin The Fermi model then predicts that the most stable condition between the two nuclei must be one in which they too are opposite in spin: NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy23 13 C spin 1 H spin electrons opposite in spin

24 S IGNS AND M ECHANISMS Fermi Contact Interaction When one spin slightly polarizes another spin oppositely the coupling constant J between the spins has a positive sign by convention. A negative coupling occurs when spins polarize each other in the same (parallel) direction. Qualitative models indicate that coupling over two bonds, as in HCH, is negative, while coupling over three bonds, as in HCCH is positive. There are numerous exceptions to this qualitative model, but it is useful in understanding that J has sign as well as magnitude. NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy24

25 S IGNS AND M ECHANISMS Fermi Contact Interaction There are many variations of the subscripts and superscripts associated with J constants In general, the superscript numeral to the left of J is the number of intervening bonds through which the coupling is taking place 2 J is a coupling constant operating through two bonds - geminal 3 J is a coupling constant operating through three bonds – vicinal 4 J is a coupling constant operating through four bonds – long range Subscripts to the right of J can be used to show the type of coupling, such as HH for homonuclear between protons or HC for heteronuclear between a carbon and proton Often, this subscript will be used to order the various J-constants within a complex multiplet: J 1, J 2, J 3, etc. or J AB, J BC, J AC NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy25

26 S IGNS AND M ECHANISMS Fermi Contact Interaction High resolution NMR spectra normally are not dependent on the absolute sign of coupling constants. Simultaneous reversal of the sign of every coupling constant in a spin system results in an identical spectrum. Many spectra, however, depend on the relative signs of component couplings. For example, the general ABX spectrum is determined in part by three couplings, J AB, J AX, and J BX Different spectra can be obtained when J AX and J BX have the same sign from when they have opposite signs even when the magnitudes are the same. NMR - The Coupling Constant 4-3 CHEM 430 – NMR Spectroscopy26

27 C OUPLINGS O VER O NE B OND The one- bond coupling between 13 C and 1 H is readily measured from the 13 C spectrum when the decoupler is turned off Although it complicates routine 13 C interpretation, this coupling can provide useful information and illustrates several important principles. Because a p orbital has a node at the nucleus, only electrons in s orbitals can contribute to the Fermi coupling mechanism (s orbitals have a maximum in electron density at the nucleus). For protons, all electrons reside in the 1s orbital, but, for other nuclei, only that proportion of the orbital that has s character can contribute to coupling. NMR - The Coupling Constant 4-4 CHEM 430 – NMR Spectroscopy27

28 C OUPLINGS O VER O NE B OND When a proton is attached to an sp 3 carbon atom (25% s character), 1 J HC is about half as large as that for a proton attached to an sp carbon atom ( 50% s character). These numbers define a linear relationship between the %-s character of the carbon orbital and the one- bond coupling: %-s(CH) = 0.2 J( 13 CH) 1 J HC ethane = 125 Hz (sp 3 ) 1 J HC ethene = 156 Hz (sp 2 ) 1 J HC ethyne = 249 Hz (sp) The zero intercept of this equation indicates that there is no coupling when the s character is zero, in agreement with the Fermi contact model. NMR - The Coupling Constant 4-4 CHEM 430 – NMR Spectroscopy28

29 C OUPLINGS O VER O NE B OND The 1 J CH coupling ranges from about 100 to 320 Hz, and may be interpreted in terms of the J– s relationship The coupling constant in cyclopropane (162 Hz) demonstrates that the carbon orbital to hydrogen is approximately sp 2 hybridized! Other examples include tricyclopentane (144 Hz, 29% s, sp 2.4 ), cubane (160 Hz, 32% s, sp 2 ) and quadricyclane (179 Hz, 36% s, sp 1.8 ). Although the J–s relationship works well for hydrocarbons not as applicable to polar bonds. NMR - The Coupling Constant 4-4 CHEM 430 – NMR Spectroscopy29

30 C OUPLINGS O VER O NE B OND For other nuclei variations in the effective nuclear charge and hybridization effects, may alter the coupling constants. Just as the resonance frequency of a nucleus is proportional to its gyromagnetic ratio, the coupling constant between two nuclei, as noted above, is proportional to the product of both gyromagnetic ratios. Nuclei with very small gyromagnetic ratios, such as 15 N, tend to have correspondingly small couplings. For 15 N has a negative sign, whereas 1 H and 13 C are positive. Therefore 1 J NH between 15 N and hydrogen have a negative sign One bond couplings have been studied for other nuclei but are more complex NMR - The Coupling Constant 4-4 CHEM 430 – NMR Spectroscopy30

31 C OUPLINGS O VER O NE B OND NMR - The Coupling Constant 4-4 CHEM 430 – NMR Spectroscopy31

32 G EMINAL C OUPLINGS By the Fermi model for geminal coupling (HCH) 2 J is usually negative Geminal coupling (H–C–H) cay be measured directly from the spectrum when the coupled nuclei are chemically nonequivalent, (the AB or AM part of an ABX, AMX, ABX 3 … spectrum). If the relationship is 1 st order (AM), the coupling may be measured by inspection. In 2 nd order cases (AB), the spectrum must be simulated computationally, unless the two spins are isolated (a two- spin system) When nuclei are chemically equivalent but magnetically nonequivalent (as in the AA' part of an AA'XX' spectrum) their coupling constant is accessible by computational methods NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy32

33 G EMINAL C OUPLINGS Here, a intervening atom (usually a spin-inactive 12 C) communicates spin information between the two interacting nuclei The Fermi model then predicts that the most stable condition between the these two geminal nuclei must be one in which they are parallel in spin: NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy33 1 H spin 12 C is spin inactive 13 H spin 1 H spin 12 C is spin inactive

34 G EMINAL C OUPLINGS IMPORTANT Splittings are not observed between coupled nuclei when they are magnetically equivalent, but the coupling constant may be measured by replacing one of the nuclei with deuterium. For example, in CH 2 Cl 2 - the geminal H– C– D coupling is seen as the spacing between the components of the 1: 1: 1 triplet (deuterium has a spin of 1). Since coupling constants are proportional to the product of the gyro-magnetic ratios of the coupled nuclei, J( HCH) may be calculated from J(HCD): NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy34

35 G EMINAL C OUPLINGS As the H-C-H decreases, the amount of electronic interaction between the two orbitals increases, the electronic spin correlations also increase and J becomes larger (more negative) NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy35 H-C-H 109 o 2 J HH = Hz H-C-H 118 o 2 J HH = -4.3 Hz H-C-H 120 o 2 J HH = +0-3 Hz In general: 2 J HH

36 G EMINAL C OUPLINGS Variations in J also result from hybridization and ring size As ring size decreases, C-C-C decreases, along with p-character; the resulting H-C-H increases, along with the corresponding s -character – J becomes smaller NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy36 2 J HH (Hz) = to -15

37 G EMINAL C OUPLINGS Electron withdrawal by induction tends to make the coupling constant more positive - for alkanes the negative coupling thus decreases in absolute value (becoming less negative) CH to CH 3 OH Hz CH 3 I -9.2 to CH 2 Br Hz Electron donation makes the coupling more negative CH 4, to TMS Hz Analogous substitution on sp 2 carbon changes the coupling profoundly: NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy37

38 G EMINAL C OUPLINGS These effects of withdrawal or donation of electrons through the -bonds (induction) can be augmented or diminished by -effects such as hyperconjugation. Lone pairs of electrons can donate electron density and make 2 J more positive, whereas the orbitals of double or triple bonds can withdraw electrons and make 2 J less positive (or more negative). The above-mentioned large increase in the geminal coupling of imines or formaldehyde compared with ethene results from reinforcement of the effects of withdrawal and donation NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy38

39 G EMINAL C OUPLINGS These effect of -withdrawal occurs for carbonyl, nitrile, and aromatic groups as in the values for acetone (14.9 Hz), acetonitrile (– 16.9 Hz), and dicyanomethane (– 20.4 Hz). The effect is some-what reduced by free rotation in open-chain systems, so that particularly large effects are created by constraints of rings: -donation by lone pairs makes J more positive. This effect also explains the difference in the geminal couplings of three-membered rings: cyclopropane and oxirane NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy39

40 G EMINAL C OUPLINGS Remember splittings are not observed for magnetically equivalent nuclei (like the 4 hydrogens on CH 4 ) ; the 2 J values in this table are used as reference values and generated by observing the 2 J HD for the deuterated analog of these compounds (top page 92 in text). NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy40

41 G EMINAL C OUPLINGS Geminal couplings between protons and other nuclei also have been studied. The H–C– 13 C coupling responds to substituents in much the same way as does the H–C–H coupling; values are smaller, due to the smaller of 13 C. Unlike the H–C–H case, the H–C– 13 C geminal coupling pathway can include a double or triple bond; such couplings can be useful to determine stereochemistry: NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy41

42 G EMINAL C OUPLINGS The 2 J HCN between hydrogen and 15 N strongly depends on the presence and orientation of the nitrogen lone pair. 2 J HCN is a useful structural diagnostic for syn–anti isomerism in imines, oximes, and related compounds as the H–C– 15 N coupling in imines is larger and negative when the proton is cis to the lone pair but smaller and positive for a proton trans to the lone pair: The cis relationship between the nitrogen lone pair and hydrogen also is found in heterocycles such as pyridine 2 J HCN In saturated amines with rapid bond rotation values typically are quite small and negative (CH 3 NH 2, -1.0). NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy42

43 G EMINAL C OUPLINGS 2 J between 15 N and 13 C follow a similar pattern and also can be used for structural and stereochemical assignments. The carbon on the same side as the lone pair (syn) in imines again has a larger, negative coupling ( Hz). The anti-isomer has a 2 J CCN of 1.0 Hz. Likewise, the two indicated carbons in quinoline have couplings differentiated by their geometry - as one is syn and the other anti to the nitrogen lone pair. NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy43

44 G EMINAL C OUPLINGS 2 J HCP between 31 P and hydrogen also have been exploited stereochemically. The maximum positive value of 2 J HCP is observed when the HC bond and the phosphorus lone pair are eclipsed (syn), and the maximum negative value when they are orthogonal or anti. The situation is similar to that for couplings between hydrogen and 15 N, but signs are reversed as a result of the opposite signs of the gyromagnetic ratios of 15 N and 31 P. The coupling also is structurally dependent, as it is larger for P(III) than for P(V): 27 Hz for (CH 3 ) 3 P and 13.4 Hz for (CH 3 ) 3 P=O. NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy44

45 G EMINAL C OUPLINGS Geminal H– C– F couplings are usually close to for an sp 3 carbon (47.5 Hz for CH 3 CH 2 F) and for an sp 2 carbon (84.7 Hz for CH 2 =CHF). Geminal F–C–F couplings are quite large for saturated carbons (240 Hz for 1,1-difluorocyclohexane), but less than 100 Hz for unsaturated carbons (35.6 Hz for CH 2 =CF 2 ). NMR - The Coupling Constant 4-5 CHEM 430 – NMR Spectroscopy45

46 Coupling constants between protons over three bonds have provided the most important early stereochemical application of NMR spectroscopy - vicinal coupling As with geminal coupling, the overall lowest energy spin state is one where the 1 H nuclei and electron spins are paired ( 12 C is spin inactive) Observe the two possible spin interactions: V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy46

47 Observe that the orbitals must overlap for this communication to take place – weaker J-constants To communicate spin information, one additional flip must take place, and the J-values are usually positive The magnitude of the interaction, it can readily be observed, is greatest when the orbitals are at angles of 0 o and 180 o to one another: V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy47 0 o dihedral angle 180 o dihedral angle

48 In 1961, Karplus derived a mathematical relationship between 3 J HCCH and dihedral HCCH. The cos 2 relationship results from strong coupling when orbitals are parallel. They can overlap at the syn-periplanar or anti-periplanar geometries. When orbitals are staggered or orthogonal, coupling is weak. A and C are empirically determined constants; C and C usually are neglected, as they are thought to be less than 0.3 Hz while A and A imply that J is different at the syn and the anti-maximum V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy48

49 Unfortunately, these multiplicative constants vary from system to system in the range 8–14 Hz and quantitative applications cannot be transferred easily from one structure to another. In general: V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy J (Hz)

50 In chair cyclohexane J aa is large as aa is close to 180° J ee (0– 5 Hz) and J ae (1– 6 Hz) are small as ee and ae are close to 60° When cyclohexane rings are flipping between two chair forms, J aa is averaged with J ee to give a J trans in the range 4– 9 Hz, and J ae is averaged with J ea to give a smaller J cis, still in the range 1– 6 Hz. In conformationally locked systems (no ring flip) the effect can be used to assign stereochemistry V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy50

51 Further examples: V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy51 3 J aa = Hz J ee = 4-5 Hz 60 3 J ae = 4-5 Hz 60

52 For alkenes 3 J trans ( = 180 o ) is always larger than 3 J cis ( = 0 o ) 3 J trans > 3 J cis > 2 J gem allows assignment of the vinyl system (AMX, ABX or ABC spectrum) trivial V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy52 3 J trans = Hz J cis = 6-15 Hz 0

53 For cyclic alkenes internal bond angles may affect 3 J cis V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy53 3 J trans = Hz J cis = 6-15 Hz o angle in H-C-C bond reduces overlap 3 J cis = Hz

54 Despite the potentially general application of the Karplus equation to dihedral angle problems, there are quantitative limitations. The 3 J H–C–C–H depends on the C–C bond length or bond order, the H–C–C valence angle, the electronegativity and orientation of substituents on the carbon atoms in addition to the H–C–C–H dihedral angles. A properly controlled calibration series of molecules must be rigid (mono- conformational) and have unvarying bond lengths and valence angles. Three approaches have been developed to take the only remaining factor, substituent electronegativity, into account: 1. Derive the mathematical dependence of 3 J on electronegativity. 2. Empirical allowance by the use of chemical shifts that depend on electronegativity in a similar fashion as 3 J. 3. Eliminate the problem through the use of the ratio (the R value) of two 3 J coupling constants that respond to the same or related dihedral angles and that have the same multiplicative dependence on substituent electronegativity, which divides out in R. V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy54

55 These more sophisticated versions of the Karplus method have been used quite successfully to obtain reliable quantitative results. The existence of factors other than the dihedral angle results in ranges of vicinal coupling constants at constant even in structurally analogous systems. Saturated hydro-carbon chains (H–C–C H) exhibit vicinal couplings in the range 3– 9 Hz, depending on substituent electronegativity and rotamer mixes and 8.90 Hz for. Higher substituent electronegativity always lowers the vicinal coupling constant. In small rings, the variation is almost entirely the result of substituent electronegativity, with cis ranges of 7– 13 Hz and trans ranges of 4– 10 Hz in cyclopropanes. V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy55

56 These more sophisticated versions of the Karplus method have been used quite successfully to obtain reliable quantitative results. The existence of factors other than the dihedral angle results in ranges of vicinal coupling constants at constant even in structurally analogous systems. Saturated hydro-carbon chains (H–C–C–H) exhibit vicinal couplings in the range 3–9 Hz, depending on substituent electronegativity and rotamer mixes. Higher substituent electronegativity always lowers the vicinal coupling constant. In small rings, the variation is almost entirely the result of substituent electronegativity, with cis ranges of 7– 13 Hz and trans ranges of 4– 10 Hz in cyclopropanes. V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy56

57 In small rings, the variation is almost entirely the result of substituent electronegativity, with cis ranges of 7– 13 Hz and trans ranges of 4– 10 Hz in cyclopropanes. Coupling constants in oxiranes ( epoxides) are smaller because of the effect of the electronegative oxygen atom. 3 J is proportional to the overall bond order, as in benzene and in naphthalene. The ortho-coupling in benzene derivatives varies over the relatively small range of 6.7– 8.5 Hz, depending on the resonance and polar effects of the substituents. The presence of heteroatoms in the ring expands the range at the lower end down to 2 Hz, because of the effects of electronegativity ( pyridines) and of smaller rings (furans, pyrroles). V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy57

58 V ICINAL C OUPLINGS NMR - The Coupling Constant 4-6 CHEM 430 – NMR Spectroscopy58

59 F IRST O RDER S PECTRA LONG- RANGE COUPLINGS Coupling between protons over more than three bonds is said to be long range. Sometimes coupling between 13C and protons over more than one bond also is called long range, but the term is inappropriate for 2J( CCH) and 3J( CCCH). Long- range cou-pling constants between protons normally are less than 1 Hz and frequently are unob-servably small. In at least two structural circumstances, however, such couplings com- monly become significant. Overlap. Interactions of bonds with electrons of double and triple bonds and aromatic rings along the coupling pathway often increase the magnitude of the coupling constant. One such case is the four bond allylic coupling,, with a range of about to and typical values close to. Larger values are ob-served when the saturated C ¬ Ha bond is parallel to the p orbitals ( 4- 31). This s– p Hz - 1 Hz HC C CH S– P C ¬ H( s) p NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy59

60 L ONG R ANGE C OUPLINGS As can be deduced from the reduced J values for vicinal coupling and the Karplus relationship, the greater the number of intervening bonds the less opportunity for orbital overlap over long range (> 3 bond) Long- range coupling constants between protons normally are less than 1 Hz and frequently are unobservably small. In at least two structural circumstances, however, such couplings commonly become significant. Allylic and homoallylic coupling W-coupling In cases where a rigid structural feature preserves these overlaps, however, long range couplings are observed – especially where C-H -bonds interact with adjacent -systems NMR - The Coupling Constant 4-7 CHEM 430 – NMR Spectroscopy60

61 L ONG R ANGE C OUPLINGS Allylic systems are the simplest example of a 4 J coupling Here, if the allyl C-H a bond is orthogonal to the system, 4 J = 0 Hz; if this bond is parallel to the vinyl C-H a bond, 4 J = 3 Hz In acyclic systems, the dihedral angle is averaged over both favorable and unfavorable arrangements, so an average 4 J is found, as in 2- methylacryloin Ring constraints can freeze bonds into the favorable arrangement, as in indene or in an exactly parallel arrangement as in allene (right) NMR - The Coupling Constant 4-7 CHEM 430 – NMR Spectroscopy61

62 L ONG R ANGE C OUPLINGS When this type of coupling is extended over five bonds, it is referred to as homoallylic coupling Examples include the meta- and para- protons to the observed proton on an aromatic ring and acetylenic systems: NMR - The Coupling Constant 4-7 CHEM 430 – NMR Spectroscopy62 5 J = J J 0-1 Hz

63 L ONG R ANGE C OUPLINGS Rigid aliphatic ring systems exhibit a specialized case of long range coupling – W-coupling – 4 J W The more heavily strained the ring system, the less flexing can occur, and the ability to transmit spin information is preserved NMR - The Coupling Constant 4-7 CHEM 430 – NMR Spectroscopy63 4 J = J = 3 4 J = 7 Hz

64 L ONG R ANGE C OUPLINGS Although coupling information always is passed via electron-mediated pathways, in some cases part of the through- bond pathway may be skipped, and effect known as through-space coupling Two nuclei that are within van der Waals contact in space can interchange spin information if at least one of the nuclei possesses lone pair electrons - found most commonly, but not exclusively, in H– F and F– F pairs. The six- bond H--F coupling is negligible on the left (2.84 Å) but is 8.3 Hz on the right ( 1.44 Å) ( the sum of the H and F van der Waals radii is 2.55 Å). This is likely is important in the geminal F– C– F coupling, which is unusually large: 2 J FCF for sp 3 CF 2 (~200 Hz, o ) compared to sp 2 CF 2 (~50 Hz, 120 o ) NMR - The Coupling Constant 4-7 CHEM 430 – NMR Spectroscopy64

65 S PECTRAL A NALYSIS Analysis of coupling constants in first-order spectra The general scheme for the interpretation and analysis of 1 st order multiplets: Some helpful constraints: For every signal split into a multiplet, the component J-value(s) must match some other multiplet in the spectrum The distance (Hz) between the two outermost peaks in a multiplet is equal to the sum of each of the coupling constants The smallest J-value is typically given by the difference between the first and second peaks in the multiplet First order multiplets are symmetrically distributed about the center NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy65

66 S PECTRAL A NALYSIS Analysis of coupling constants in first-order spectra Method: 1. Your book suggests letting the spectrometer do the work, however this method still requires spectral common sense 2. Our method is adapted from: Hoye, T. R.; Hanson, P. R. Vyvyan, J. R. J. Org. Chem. 1994, 59, and basically builds the same tree analysis from the branches on in 3. This paper is a well loved and regarded classic by 17 years of grateful graduate students NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy66

67 S PECTRAL A NALYSIS Lets start with a simple spectrum: Crotonic acid (trans-2-butenoic acid) NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy67

68 S PECTRAL A NALYSIS Assuming we analyze the spectrum as we have before, we should get a structure similar (or isomeric) to the actual one: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy68 Simple spectrum: 13 C NMR: 4 unique carbons (too few for aromatic) 1 C=O; 2-alkenyl, 1 alkyl 1 H NMR: 2 multiplets in the alkene region 1 multiplet in the alkyl region In CDCl 3, the acidic proton appears at d 12.4 (acid) Integrals are 1(a):1:1:3

69 S PECTRAL A NALYSIS Without much effort, and any more detailed analysis of chemical shifts, several possibilities arise: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy69

70 S PECTRAL A NALYSIS From analysis of J-values and knowledge of which protons are coupled by the various J-values can finish the analysis: Inspection of the three multiplets shows the following Hz values: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy70 H a H b H c

71 S PECTRAL A NALYSIS Start with the simplest multiplet: H c is an apparent doublet of doublets (dd) Step 1: The distance between the first two lines always represents the smallest J value If the ratio of these two lines (integral) is 1:1, this J is unique; if it is 1:2, 1:3, etc. there are two or more identical smallest J s Label this J small – = 1.80 Hz NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy71 H c

72 S PECTRAL A NALYSIS Step 2: (most difficult step for complex multiplets) Find the full set of pairs within the multiplet that are separated by J small Each pair will have a reflected partner through the center of the multiplet For pairs where one of the lines has a relative intensity >1, that line will be part of more than one pair – = 1.80 Hz – = 5.25 Hz – = 1.64 Hz – = 1.80 Hz NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy72 H c While these two may not seem equal, they must be matched if 1 st order

73 S PECTRAL A NALYSIS Step 3: Find the centers of each of the pairs generated in step 2 These will collectively represent a new pattern (as if the J small was selectively decoupled) As with step 1, the spacing between the first two lines of this multiplet represent the next smallest J Label this J as med-small, etc. as necessary NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy73 H c = 6.97

74 S PECTRAL A NALYSIS Step 4: Find the midpoint(s) of this new pair(s), and repeat step 3 Step 5: Repeat as necessary until all J-values have been found Remember, it must be internally consistent and all the J values must add up to the difference between the outer peaks of the multiplet. We check: The two J s we determined are 1.80 (1.64) and 6.97 ( )/ = 8.69 The difference between the outermost peaks of the multiplet: – = 8.69 NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy74

75 S PECTRAL A NALYSIS This effectively generates the tree-diagram discussed in the text: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy75 J small = 1.72 Hz J med = 6.97 Hz This proton is coupled to two other protons, with coupling constants of 1.72 and 6.97; (dd, 6.97, 1.72)

76 S PECTRAL A NALYSIS Repeat the analysis with the next most complex multiplet, look for possible J- values of 1.72 and Analyzing the H b proton; apparent doublet of quartets (dq): NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy76 H b Hz 1.65 Hz 1.80 Hz 1.64 Hz Step 1&2Step 3& Hz Step Hz symmetrical Hz Check: = – = 20.51

77 S PECTRAL A NALYSIS The tree diagram for H b : NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy77 J small = 1.64 Hz We see three generational iterations of the 1.64 similar to ~1.7 as found for H c We also see a new constant of J small = 1.64 Hz J large = Hz

78 S PECTRAL A NALYSIS Lastly, repeat the process for H a ; looking for 1.64, or 6.97: Analyzing the H a proton; apparent doublet of quartets (dq): NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy Hz 7.06 Hz 6.89 Hz 1.64 Hz Step 1&2Step 3& Hz 6.89 Hz Step Hz symmetrical Hz Check: = – = H a Hz 6.89 Hz

79 S PECTRAL A NALYSIS The tree diagram for H a : NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy79 J small = 6.89 Hz J med = Hz We see three generational iterations of the 6.89 similar to H c We also see a new constant of a similar constant to 15.51, at J small = 6.89 Hz

80 S PECTRAL A NALYSIS From analysis of J-values we can finish the analysis: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy80 H a has coupling constants of and 6.89 H b has coupling constants of 1.64 and H c has coupling constants of 1.72 and 6.97 Conclusion: H a is coupled to H b and H c H b is coupled to H a and H c H c is coupled to H b and H a Wow.

81 S PECTRAL A NALYSIS Conclusion: On such a small system, the conclusion that all three of these proton families are coupled to one another is trivial; on more complex systems, this is a powerful tool to determine the position and relationship of protons on chains or rings In this system, it helps us deduce which isomer we are observing by analysis of the magnitude of the coupling constants that were generated For each isomer, hypothesize as to what Js would be observed: NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy81 J trans 1 Hs = J allyl = 1-3 J cis 1 Hs = 6-15 Hz J allyl = 1-3 J geminal 1 Hs = 1-3 Hz No allylic coupling J trans = 15.5

82 S PECTRAL A NALYSIS Analysis of coupling constants in first-order spectra Another Method: 1. From our adapted method: Hoye, T. R.; Hanson, P. R. Vyvyan, J. R. J. Org. Chem. 1994, 59, There is a second method for determining coupling constants – graphical analysis 2. For first-order spectra, there are a finite number of combinations of an observed 1 H with a maximum number of interactions 3. Theoretically, the maximum 3 J interactions would be nine: 4. In synthetically interesting systems (middle of chains, rings, etc.) it is typically 2-4 couplings NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy82

83 S PECTRAL A NALYSIS Another Method: 5. The authors of the article suggest the method of visual pattern recognition to save some of the tedium of the more analytical analysis we just covered 6. In the paper, they generated a series of tables that systematically cover the most commonly encountered spin systems 7. For each table, an example compound is used to show where such a spin pattern would be observed 8. It is suggested that you go through the paper, in conjunction with your text (as a reference for representative J-values) to see how each of the tables apply to the example the authors used NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy83

84 S PECTRAL A NALYSIS Example: In each table, one coupling constant is varied (J z ) while all others are held constant As a consequence for any table you see a symmetrical distribution of the multiplet which converge as Jz approaches zero at the bottom of each table 1Hs that are exemplified are highlighted by a circle or square Note, as we have discussed the Js is equal to the distance between the outermost peaks in a multiplet (Hz scale below each graphical analysis) NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy84

85 S PECTRAL A NALYSIS Example: It is important to note, that as additional coupling decreases, or certain couplings become equal, the patterns simplify to what we recognize from the n+1 rule Therefore, it should start to be clear that the n+1 rule is a highly coincidental case where all possible couplings ( 3 J s) become equal In other words, the n+1 rule is the exception, and the first order relationships we have just discussed are the rule NMR - The Coupling Constant 4-8 CHEM 430 – NMR Spectroscopy85

86 S ECOND O RDER S PECTRA Second- order spectra are characterized by peak spacings that do not correspond to coupling constants, by nonbinomial intensities, by chemical shifts that are not at resonance midpoints, or by resonance multiplicities that do not follow the n + 1 rule Even when the spectrum has the appearance of being first order, it may not be. Lines can coincide in such a way that the spectrum assumes a simpler appearance than seems consistent with the actual spectral parameters ( a situation termed deceptive simplicity). NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy86

87 S ECOND O RDER S PECTRA For example, in the ABX spectrum, the X nucleus is coupled to two nuclei ( A and B) that are closely coupled ( AB /J <<<10). Under these circumstances, the A and B spin states are fully mixed, and X responds as if the nuclei were equivalent. Thus the ABX spectrum resembles an A 2 X spectrum, as if J AX = J BX. In the ABX spectra at the right (a) AB = 3.0 Hz, in (b) AB = 8.0 Hz NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy87

88 S ECOND O RDER S PECTRA The AA'XX' spectrum often is observed as a deceptively simple pair of triplets, resembling A 2 X 2. In this case, it is the A and A' nuclei that are closely coupled ( and J AA' is large). Such deceptive simplicity is not eliminated by raising the field because A and A' are chemically equivalent. The chemist should beware of the pair of triplets that falsely suggests magnetic equivalence ( A 2 X 2 ) and equal couplings, when the molecular structure suggests AA'XX'. Sometimes the couplings between A and X may be observed by lowering the field to turn the AA'XX' spectrum into AA'BB' with a larger number of peaks that may permit a complete analysis. NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy88

89 S ECOND O RDER S PECTRA Another example of second-order complexity occurs in the ABX spectrum (or, more generally, A x B y X z ) when A and B are very closely coupled, J AX is large, and J BX is zero. With no coupling to B, the X spectrum should be a simple doublet from coupling to A. Since A and B are closely coupled, however, the spin states of A and B are mixed, and the X spectrum is perturbed by the B spins (the phenomenon has been termed virtual coupling, which is something of a misnomer, since B is not coupled to X). NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy89

90 S ECOND O RDER S PECTRA NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy90 Methyl resonance at 60 MHz

91 S ECOND O RDER S PECTRA Sometimes proton spectra are second order even at 500 MHz or higher ( aside from the AA' case). Some institutions still have access only to iron core, 60 MHz spectrometers, which produce largely second- order proton spectra. In each case, these spectra may be clarified somewhat by the use of paramagnetic shift reagents. These molecules contain unpaired spins and form Lewis acid– base complexes with dissolved substrates. The unpaired spin exerts a strong paramagnetic shielding effect (downfield) on nuclei close to it. The effect drops off rapidly with distance, so that those nuclei in the substrate that are closest to the site of acid–base binding are affected more. Consequently, the shift to higher frequency varies through the substrate and hence leads to greater separation of peaks. NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy91

92 S ECOND O RDER S PECTRA Two common shift reagents contain lanthanides: tris( dipivalomethanato) europium(III) · 2( tris(dipivalomethanato)europium(III)·2( pyridine) [called Eu(dpm) 3 without pyridine] and 1,1,1,2,2,3,3-heptafluoro-7,7-dimethyl- octanedionatoeuropium( III) [or Eu( fod) 3 ]. Shift reagents are available with numerous rare earths as well as other ele- ments -- Almost all organic functional groups that are Lewis bases have been found to respond to these reagents. When the shift reagent is chiral, it can complex with enantiomers and generate separate resonances from which enantiomeric ratios may be obtained. NMR - The Coupling Constant 4-9 CHEM 430 – NMR Spectroscopy92

93 F IRST O RDER S PECTRA NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy93

94 F IRST O RDER S PECTRA NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy94

95 F IRST O RDER S PECTRA NMR - The Coupling Constant 4-1 CHEM 430 – NMR Spectroscopy95


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