1,1,2-Tribromoethane. Spin-Spin Coupling

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1 NMR Spin oupling Spin-Spin oupling Spectra usually much more complicated than a series of single lines, one for each type of hydrogen. Peaks are often split into a number of smaller peaks, sometimes with complex patterns. Spin-Spin oupling Protons possess magnetic moments. These moments can either be aligned with the external field or against it. 1,1,2-Tribromoethane Nonequivalent protons on adjacent carbons. A second proton that is adjacent to the first proton will feel this magnet moment. If the magnetic moment of the proton is aligned with the field, its neighbor will feel a stronger field. If it is aligned against the field, the neighbor will feel a weaker field. Simple case where proton A has only one other neighboring proton (B), proton A will potentially feel two possible environments one enhanced by proton B and one weakened by proton B. Proton A will therefore show up as two peaks, or a doublet. triplet doublet If proton A signal is split by proton B, then proton B signal must also be split by proton A. 1

2 Doublet: 1 Adjacent Proton Triplet: 2 Adjacent Protons b feel B 0 +x b feel B 0 -x J ab equal amounts of a and doublet J ab a feel B 0 +x a feel B 0 -x triplet a feel B 0 Spin-Spin oupling The N + 1 Rule The coupling of a nuclear spin with other nearby nuclear spins causes multiplet splitting If a signal is split by N equivalent protons, it is split into N + 1 peaks. General rule for splitting: A nucleus coupling with N equivalent nuclei of spin I will have its signal split into (2NI+1) lines. For the most common case of coupling with nuclei of I=½, this simplifies to (N+1) lines. The N+1 Rule 2

3 The Triplet a b J a 1 1 R R' R'' J b A triplet can be envisioned as two successive splittings. The first proton on an adjacent 2 will split the peak into two, and the second proton on the adjacent 2 will split each of these peaks again. The center peaks will overlap to create the triplet with a 1:2:1 peak area ratio. The Quartet 1 a 1 Splitting for Ethyl Groups The relative peak areas for a quartet are 1:3:3:1 and can be easily determined using the same branching method b c 1 3

4 Splitting for Isopropyl Groups Range of Magnetic oupling Equivalent protons do not couple, hence do not split each other. oupling mechanism is via the bonding electrons. Stronger coupling with higher s-hybridization. Protons bonded to the same carbon (geminal) will split each other only if they are not equivalent. Protons on adjacent carbons (vicinal) normally will couple. Protons separated by four or more bonds normally will not couple. Long range coupling possible in conjugated and aromatic systems! The oupling onstant J Distance between adjacent peaks of a multiplet, is the coupling constant J. J must be identical for the two coupling nuclei Units are in z (recall the δ scale!). Notation oupling onstant For two s that are coupled to each other, both will show the same value of J. J is measured in z and is independent of B 0. Ratio ν/j ν/j 10 First rder spectra, normal multiplet intensity patterns 1 J 2 J 3 J 4

5 Doublets As the chemical shifts of the two protons splitting one another get closer together, the pattern deviates from the ideal pattern of each peak being split into two peaks of equal height. Roof effect Peaks that are splitting each other will always slant toward each other like two sides of a roof. This effect can be lessened by moving to a higher field instrument, since the pattern depends on the ratio of ν : J. Second rder Effect This effect is useful for determining which protons are splitting each other (and are therefore adjacent to each other. Spin-Spin oupling In order for two hydrogens to couple to one another, they have to be chemically inequivalent. Equivalent Nuclei hemical equivalence: 2 nuclei have the same chemical environment (i.e. same chemical shift). Magnetic equivalence: must have the same chemical environment and be equally coupled to every other magnetic nucleus in the system. a a 2 N N 2 B A F A F B A, B chemically, magnetically equivalent F A, F B chemically, magnetically equivalent 2 J A,FA = 2 J B,FB = 2 J A,FB = 2 J B,FA 5

6 Equivalent Nuclei Spin-Spin oupling A B F A F B A, B chemically equivalent but NT magnetically equivalent Two coupled methylenes will therefore give a pair of triplets. A 2 2 B 3 J AFA 3 J BFA and 3 J AFB 3 J BFB ( J trans > J cis ) B A 3-2 Br Fast rotation around single bonds normally leads to magnetic equivalence Br 2 igher rder Splitting 2 3 If an atom has more than one type of adjacent to it, as in propyl bromide, the will be split by BT of its neighbors. The central 2 (methylene) group is split by a methyl on one side and a methylene on the other side, here all with the same J-value (coupling constant). The couplings need not be equal! 60 Mz propyl bromide 300 Mz propyl bromide Br Spin-Spin oupling If 3 J 2,3 > 3 J 2,2 then would see a quartet of triplets If 3 J 2,2 > 3 J 2,3 then would see a triplet of quartets! 6

7 ommon Alkyl Groups For alkyl chains, hydrogens on adjacent carbons will typically split each other with J-values of 6-8 z. ommon alkyl groups show splitting patterns that are characteristic and recognizable. Going through the branching exercise is not necessary at every step. Example: 1-hloropropane l n-propyl triplet - sextet - triplet Isolated methyl groups will always give a singlet that integrates to 3 hydrogens. Ethyl groups give the familiar quartet-triplet pair that integrate in a 2:3 ratio. X Integration ratio of 2:2:3 Isopropyl Group Example: 2-Methylpropanenitrile (Isovaleronitrile) 3 N 3 X iso-propyl septet - doublet Integration Ratio of 1:6 7

8 Example: 1-Bromobutane Br 3 n-butyl: triplet-quintet-sextet-triplet Integration: 2:2:2:3 Beyond Simple Alkyls Acyclic alkyls are simple because the J values tend to always be the very similar (~same). A step up in complexity are alkenes, where geminal splitting is also possible, and trans and cis alkenes show different J-values. The influence of the geminal-relationship is over the shortest distance X The magnetic influence of the transrelationship is over the longest distance The cis-relationship, is over an intermediate distance Branching Diagrams Still Work! Example: Styrene Three different J-values are observed in this system. a would appear as a doublet of doublets. c 2 J gem = 0 1 z The observed multiplet for a is a doublet of doublets A B B Note the strong anisotropic effect of the ring current! X a b 3 J trans = z 3 J cis = 6-15 z 3 J A 3 J AB 3 J AB A Geminal couplings are very small (0-1 z) and are usually not observable unless very high field instruments are used. Normally not seen in standard 200 to 300 Mz spectra! 8

9 Styrene Styrene A B J A J AB A B A signal J A J AB doublet of doublets Aromatics Protons on benzene rings can split each other regardless of how they are positioned. They can even split a hydrogen as far away as the para-position. The J-value (coupling constant) gets smaller as the distance increases. J para X para ortho meta J ortho J meta In low-field 1 NMR the signal for this proton would be split into a doublet by the proton ortho to it. n a high field instrument one finds this J ortho as well as a J meta and a J para from the effect of the protons meta and para to it Typically: J ortho = 7-10 z J meta = 1-3 z J para = 0-1 z Example: Nitrobenzene a b N 2 a c b Nitrobenzene has three different types of hydrogens, labelled a, b, and c. We can assign the peaks on the left to these hydrogens using their chemical shifts, integration, and splitting patterns. 9

10 Example: Nitrobenzene Example: Nitrobenzene b? a b N 2 a b c a b N 2 a b c c Firstly, nitro is a strong EWG via resonance. Expect it to deshield a and c more so than b. A good initial guess might be that the rightmost peak is b. Next look at the integration. Expect the peak areas corresponding to these hydrogens to have the ratio of a : b : c = 2:2:1. For the spectrum on the left, the center peak has an integration of 1, so it must be c. Example: Nitrobenzene Nitrobenzene Expected Splittings N 2 a N 2 a c a a a b b b a b b c b Definitive assignment of a and b is straightforward using the splitting patterns. J AB = 8 z J Ac = 1.2 z c J BA = 8 z J B = 8 z a will be split by b with J = 7-10 z and by c by 1-3 z. b will be split by a with J = 7-10 z and by c by 7-10 z. 10

11 Example: Nitrobenzene Para-disubstituted Rings N 2 a a a c b b c b an now successfully assign all three hydrogen types. Para disubstituted rings show a characteristic pattern that consists of a pair of doublets. N 2 ur initial assignment of b based only on chemical shifts was correct. Mono-substituted benzene indicated by TREE multiplets (but might be overlapping!) Meta Disubstituted Benzene Rings Meta Disubstituted Rings a c b Br a 3 a N 2 d d c b d b c 11

12 Example: meta-dinitrobenzene rtho-disubstituted: Methyl Salicylate a 2 N N 2 b c 3 a d c d b Example: Indoline Alkyl Ring Systems When cyclic alkanes are present, rotation about the - bonds is hindered or impossible. This may result in the two hydrogens of a methylene group becoming chemically inequivalent. 3 a b N In the above example, a and b are inequivalent since a is cis to the methyl group and b is trans to the methyl group. a and b will be split by the methine () with different J values and they will split each other. This is an example of geminal coupling. 12

13 a yclic Alkyls c b Predicting J values: Geminal ouplings As the bond angle -- increases, J becomes smaller o 2 J = z o 2 J = 5 z In general: J o 2 J = 0-3 z Geminal oupling Geminal coupling constants get larger as the ring size increases. Geminal coupling is NT observed if the two hydrogens are chemically equivalent. 2 J (z) = ~2 ~4 ~9 ~11 ~13 9 to 15 Vicinal oupling The vicinal coupling constant is strongly dependent on the dihedral angle between the hydrogens that are splitting one another. 3 J diaxial = z 3 J diequitorial = 4-5 z α = 180 ο α = 60 ο 3 J axial-eq. = 4-5 z α = 60 ο 13

14 Karplus Equation This observation was quantified by Martin Karplus, who determined that the experimental data best fit the following equation: 3 J = A + B cos α + cos 2α Where A, B, and are empirically determined constants a yclic Alkyls c b J (z) α ο Styrene xide J Values Simulated Spectrum (Splitting nly) Dihedral Dihedral J a - b z a b a b c a - c z c b - c n/a 5.5 z 14

15 Diastereotopic Protons If a methylene is adjacent to or nearby a chiral center, the two hydrogens in that methylene can be chemically inequivalent without the need for a cyclic system. These protons can have different chemical shifts and they will split each other. They are called a pair of diastereotopic hydrogens. Example: Diastereotopic Protons All peaks appear as expected except there are two peaks in the 4-5 region, and each is a doublet! This molecule would be expected to show 9 in the aromatic region ( ppm), an acid peak (1) around 12 ppm, two methyl singlets (6) in the ppm region, and a methylene singlet in the 4-5 ppm region (deshielded by electronegative oxygen). Diastereotopic Protons A closer look at the structure. The benzylic carbon has 4 different groups attached, and so is a chiral center. Because the methylene is next to a chiral center, its protons are diastereotopic. Example: 2-phenyl-2-butanol 3 Me Ar * Ac The molecule is redrawn emphasizing the methylene position. As shown in this orientation, one of the methylene s is cis to the aromatic group, and the other is cis to the methyl group. Therefore they experience different chemical environments and they will have different chemical shifts. In addition, they will split each other. 15

16 Simplification of omplex Spectra Simplification of omplex Spectra The 1 spectral range is very narrow. Normally 0 to 10 δ. For most molecules of any size, the spectra quickly become congested. Due to many different chemical shifts. Due to overlap of multiplet splittings. Recall definition of δ scale For close/overlapping δ peaks, can increase B 0 value. This separates the peaks but leaves them at the same δ values. Recall that coupling constants (J) are independent of B 0. Thus use higher field magnets to separate overlapping multiplets. Double Resonance 3.4 δ 80 Mz Menthol z 200 Mz 680 z Mz 1224 z 600 Mz 2040 z Selectively radiate a particular type of with high power. auses rapid transitions between upper and lower spin states, causing saturation. The signal for that nucleus disappears from the spectrum ALL coupling to the irradiated nucleus also disappears from other peaks in the spectrum. spin decoupling 16

17 Double Resonance Double Resonance Allows simplification of complex or overlapping multiplets. Allows identification of which s are coupled. Double Resonance eteronuclear spin decoupling Allyl bromide - 3 decoupled Aromatic decoupled 2 Br Radiate at 2 Br frequency A series of doublets 19 F spectrum ALL decoupled 68 17

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