θ-1) B z dθ = 0 r 3 θ=0

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1 207 NMR Spectroscopy and rganic Structure Determination Scott Virgil California Institute of Technology SPIN-SPIN CUPLING IN NMR SPECTRSCPY Interaction of Nuclear Spins in NMR Spectroscopy A. Dipolar Coupling is averaged by tumbling in solution. Nuclear Spins are magnetic dipoles and strong coupling exists θ between nuclei which are in close proximity (~ 50,000 z). r Q: Why don't NMR lines show this coupling? B 0 A: Because molecular motion and tumbling averages the effect to zero for molecules in the solution state. C b µ π B z = (3cos 2 a θ-) B z dθ = 0 r 3 θ=0 B. Spin-Spin J-Coupling arises from interaction of nuclei via the intervening bonds. J values are constant in z. J coupling cannot be averaged out by molecular tumbling. J values represent a discrete quantity of energy (E = hν). J values are reported with the number of intervening bonds as the superscript. The energy associated with J-coupling represents the preference for two spins to be paired (positive J) or unpaired (negative J) relative to each other. The interaction of the NMR nuclei with the paired electrons of the intervening bonds leads to the J-coupling effect. z C e - e - z C θ C e - e - e - e - C. Decoupling of multiplets When we observe a proton in the NMR spectrum, we generally assume that the spin states of neighboring nuclei stay in the same spin state during the collection of data. So, for a spin ½ nucleus, we observe two resonances revealing the two states with different energy. Decoupling is achieved by rapidly changing the spin states of the neighboring nucleus so the observed proton sees an average state. NMR Instrument - Decoupling Experiment: Irradiate at exact frequency of dec. Exchange of acidic protons: e.g., alcohols and amines Quadrupolar Nuclei: Nuclei with I > ½ can average their own nuclear spin states by interaction of the non-spherically symmetrical nuclear charge with the non-spherically symmetrical electric charge environment of the molecule. Generally we see no J coupling to 4 N or in some cases a broadened :: triplet appearance. 3 J Y r a b C observed C Y C no averaging Y C C nucleus observed nucleus I = +/2 I = -/2 Y Y C averaging C C observed observed nucleus observed nucleus I = +/2 I = -/2 observed

2 California Institute of Technology Chemistry 43 NMR Spectroscopy Spin Coupling Theory Important Points to remember: Coupling Constants are always reported in z. (independent of magnetic field strength) bserved spin-spin coupling is a through bond effect. Depends on # of bonds separating the nuclei: C C R* a b C 2 Me C 2 Me J C (+58 z) 2 J (-2 z) 3 J (+6 z) 4 J (-2.5 z) Coupling constants can be + or. For first order spectra, the sign of J has no observed effect. Magnitude of J also affected by:. ybridization J C : ~ 25 z for sp 3, ~ 60 z for sp 2, ~ 250 z for sp 2. Geometry 2 J : bond angle, 3 J : dihedral angle, 4 J : W-coupling 3. Electronegative atoms withdraw e- density. 4. π Conjugating groups facilitate coupling. The Ramsay Theory describes the mechanism of Spin-Spin Coupling: The coupling between two nuclei represents a discrete amount of energy (in z) for the nuclei to be aligned with each other ( ) vs. against each other ( ). Two nuclei couple with each other via the electrons spins of the intervening bonds. In the example below of a -bond C- coupling, the and states are stabilized by the alignment of the nuclei with the intervening electrons, and the and states are destabilized. Transitions between states of the C - spin system of CCl 3. C No Coupling J C = 209 z Effect of Coupling J C = +209 z Energy ω ω ω C ω C ω +J/2 ω -J/2 ω C -J/2 ω C +J/2 ω C +J/2 ω C 3 C ppm J C = +209 z ω ppm ω C -J/2

3 Y CD 3 CD 2 Jgem 2.28 z Geminal Proton-Proton Coupling For normal molecules, Geminal coupling will only be observed if the >C 2 group is diastereotopic. In order to obtain 2 J values for >C 2 groups of achiral molecules including C 3 groups, deuterium substitution is used. J = γ γ D J D = 6.54 J D : 2 : 3 : 2 : ) Effect of Electronegative Substituents: Electron-withdrawing groups make J more positive (the magnitude of J decreases) For acetone: 2 J = z = -4.9 z C 4 C 3 F C 3 Cl C 3 I C 3 MeC(C 2 C 3 ) 2 C 2 Cl 2 C 2 Br 2 RC 2 C 3 (C 3 ) 4Si b' c c' b b b' a' a a a' Jgem = -2.6 Jaa = -6.0 Jbb = -0.9 Jcc = -2.6 Jaa 2 Jbb = ) Effect of Bond Angle and ybridization: Increasing the -C- bond angle increases the s-character of orbitals making J more positive C 2 C 2 RC=C2 =C2 -N=C2 =C=C2 Jgem = -9.8 Jgem = -5.4 Jgem = -2 to +2 Jgem = +42 Jgem 8.5 Jgem = ) Effect of π-conjugation: C 3 C 2 C 3 CC 3 C 3 CN C 3 N 2 PhC 3 PhC 2 R* PhC 2 CN C 2 (CN) Br Br a,a' b,b' a,a' b,b' Jgem = -2.5 Jaa = -5.5 Jbb = -7.3 Jaa = -0.9 Jbb = -5.3

4 Geminal Coupling Adjacent to Carbonyl Group J gem gets more negative with each additional adjacent π-bond. For a C= group, the effect is sensitive to the dihedral angle (shown at right). -3 z -4 z -5 z -6 z -7 z -8 z Coupling Constants of Vinyl Compounds C2=C (z) -F -Cl -Br -R -Ar -CR -P()(R) 2 -N 2 -NR 2 -CR -CN -CR -R -Ar J gem J geminal J cis J trans J cis J trans Ph Ph Ph Cl (C 3 ) 3 Si Cl (C 3 ) 3 Si Cl C 3 C C N S A B A B A B Cl A B Cl A B Cl A B A B J AB(gem) J AB(trans) J AB(cis)

5 generally positive in sign The Karplus Curve shows how J vicinal varies with the dihedral angle. As shown, J vicinal is attenuated by electron-withdrawing groups and amplified by π- conjugation. Vicinal Coupling φ π-conj. electron withdrawing Vicinal Coupling in Cyclic Systems Simple cyclohexane Derivatives: J ax-ax = 0-3 z J eq-ax = z J eq-eq = z eq ax eq eq ax ax Glucose Anomeric : J ax-ax = 7.9 z J eq-ax = 3.7 z ax ax ax eq Five-membered Rings:!! conformational flexibility makes it difficult to distinguish J trans and J cis. Normally: J cis = 8 - z, J trans = 2-9 z Shown at right is a more rigid case: Bicyclic Systems: J vicinal generally smaller, Especially for bridgehead protons J ac J bd J de J dh 3-4 z. J gem = 8. z J ax-ax = 3.3 z J eq-ax = 7.3 z J ab = 5.6 z J gem = 7.0 z J fg 9.8 z J bd = 2.9 z a J ac = 3.2 z b f c ax eq g ax d CN C 3 J eh = 3.3 z e, h Cl

6 Chemistry 43 NMR and Structure Determination California Institute of Technology NMR of Bicyclic Norbornene Derivative J fg 9.8 z J bd = 2.9 z f g d J eh = 3.3 z J ab = 5.6 z b e, h Cl a J ac = 3.2 z c CN NMR (500 Mz, CDCl 3 )

7 Chemistry 43 NMR and Structure Determination J gem = 0.7 z J trans = 3.3 z J cis = 6.5 z h f g d N J gem =.6 z e C 2 C 2 C3 d, b, c NMR (500 Mz, CDCl 3 ) abc (m, 5) d 2.52 (td,, J = 7.4,.6 z) e 2.34 (td,, J = 7.2,.5 z) f 2.3 (dd,, J = 3.2, 6.5 z) g.90 (dd,, J = 0.7, 3.4 z) h.67 (d,, J = 6.6 z) i.62 (m, 2) j.4 (m, 2) k 0.93 (t, 3, J = 7.5 z)

8 Example. Using the solved NMR spectrum for the starting ketone (below), assign the stereochemistry for the two ester products from the Wolff rearrangement. TBS C 3 N 2 TBS C 2 Me C 3 hν C3 TBS Me g f TBS c C 3 (h) d e a Geminal Coupling: J cf = 9. z J eg =.5 z b Vicinal Coupling: J de = 0 z (φ = 90 ) J cd = 0 z (φ = 90 ) J df = 8.0 z J dg = 5.5 z Long-Range Coupling: 4J ad = 2.2 z 4J ah =.5 z (allylic) 4J ce = 3.0 z (W-) (500 Mz, CDCl 3 ) a 5.26 (qd,, J =.5, 2.3 z) b (m, 4) c 2.62 (dd,, J = 3.0, 9.2 z) d 2.48 (ddd,, J = 2.2, 5.4, 7.9) e 2.4 (dd,, J = 3.0,.5 z) f 2.23 (dd,, J = 8.0, 9. z) g 2.07 (dd,, J = 5.5,.5 z) h.69 (d, 3, J =.6 z) i 0.94 (s, 9) j 0.5 (s, 3) k 0.02 (s, 3) J ce = 3.0 z f g TBS d e c C 3 (h) a J ah =.5 z

9

10 Example. Isomer B NMR (500 Mz) spectrum

11 Example. This is a difficult problem because in bicyclic systems we cannot assume that a given J value is vicinal, germinal or long range based on its magnitude. Solve the Spin Systems of Both Spectra. (Match up J values that agree within ~0.2 z of each other) The two spin systems are different (one is linear, the other is branched)! The spin system derived from Spectrum B shows g coupling strongly with both e and d. Therefore, one of the J values is an unusually strong long-range coupling ( 4 J dg = 5.9 z). The W-geometry of long-range coupling secures the assignment. signal ppm mult, J values a d e f g C 3 a d e f g C , dq, J =.5, 2.9 z, d, J = 7.8 z, dt, J = 3.2, 7.0 z, dd, J = 6.8, 8.6 z, d, J = 8.3 z 3, d, J = 2.0 z signal ppm mult, J values , dq, J =.5, 2.9 z, d, J = 5.9 z, t, J = 8. z, dd, J = 2.9, 7.3 z, d, J = 5.9, 8.3 z 3, d, J = 2.0 z Spectrum A spin system: C 3 a e f g 7.8 d Spectrum B spin system: C 3 a f e g 5.9 d 8.4 z 8.3 z 6.8 z f d TBS Me e 7.3 z 7.8 z g TBS g e f d Me C 3 a 4 J = 5.9 z C 3 a In bicyclic systems, some vicinal J values can be ~ 0 z because the angle is locked at 90. d C 2 Me (c) Geminal Coupling: J fg = 8.4 z TBS f e g C 3 (h) a Vicinal Coupling: J eg = 0 z J de = 7.8 z J ef = 6.8 z b Long-Range Coupling: J ae = 2.9 z (c) MeC d Geminal Coupling: J eg = 8.3 z TBS e f g b C 3 (h) a Vicinal Coupling: J df = J fg = 0 z J ef = 7.3 z Long-Range Coupling: J af = 2.9 z J dg = 5.9 z (W-Coupling)

12 Ch 43 NMR Spectroscopy and rganic Structure Determination AB Spin System Line Positions: Intensities: ν = m + 2 Δν 2 +J J i = ( - ν2 = m + 2 Δν 2 +J 2-2 J i 2 = ( + ν3 = m - 2 Δν 2 +J J i 3 = ( + ν4 = m - 2 Δν 2 +J 2-2 J i 4 = ( - Let 2C = Δν 2 +J 2 J ) Δν 2 +J 2 J ) Δν 2 +J 2 J ) Δν 2 +J 2 J ) Δν 2 +J 2 ν J ν A 2C ν A + ν m = B 2 Δν = 8 z J = 6 z 2C = Δν 2 +J 2 - J ν 2 ν = 9.0 z J ν B ν 4. J = ν2-ν = ν4-ν3 Some Useful Points spacing of 2. = 2C = doublet centers Δν2 +J 2 Δν suppose you know where all peaks are, what is Δν? it may be shown that: Δν = 3. i 2 i = i 3 i 4 = ν -ν4 ν2-ν3 = 2C+J 2C-J (ν-ν4)(ν2-ν3) suppose you only know ν,ν2,i,i 2, where is the other doublet? it may be shown: ν2-ν3 = 2C-J = 2J a- using eq. 2: Δν = 2J a a- where a = i 2 i Example Spectra (with J = 6 z).5.5 4J 20J 8J 2 2 2J 3 3 indistinguishable from a simple quartet peak height raio (:a) 2C-J Δν :. 20J 20.98J :.25 8J 8.94J :.5 4J 4.9J :2 2J 2.82J :3 J.73J :5 0.5J.2J :0 0.22J 0.70J : J 0.202J

13 California Institute of Technology Chemistry 43 NMR and Structure Determination The NMR of Limonin displays multiple diasterotopic C 2 s which display the leaning features of the AB system. d 4.72 (d,, J =. z) e 4.43 (d,, J =. z) h 2.92 (dd,, J = 3.0, 4.8 z) i 2.82 (t,, J = 2.2 z) j 2.65 (dd,, J = 0.8, 4.8 z) k 2.52 (dd, J =.5, 0.2 z) l 2.4 (dd, J = 2.7, 2.7 z) m 2.20 (dd,, J = 2.8,.5 z) h j f C 3 C 3 d e k m l b C 3 a C 3 g c a' Limonin NMR (250 Mz, CDCl 3 ) i

14 Ch 43 NMR Spectroscopy and rganic Structure Determination AB Spin System Br A Δν = 5 z J AB = -2 z J A = 5 z J AB = -2 z B J A + J B = 22 z J B = 7 z ΔJ = J A J B = 8 z 2 J AB AB portion 4 2D + J AB 3 ν A + ν B m = 2 JAB 2' ' 4' J AB 3' 2D + = 2D - = portion J A + J B 2D - ν 2D + - 2D - (5+8/2) 2 +(-2) 2 = 9.2 z (5-8/2) 2 +(-2) 2 = 2.2 z 5 5' 6 6' 7 7' 2D + + 2D - Define: D (Δν + ΔJ/2) 2 + J 2 2 AB (Δν - ΔJ/2) 2 + J 2 + = and D - = 2 AB where ΔJ = J A -J B A Transitions: ν = m + D + + ½ J AB + ¼ (J A + J B ) i = ( J AB / 2D + ) ν = m + D + ½ J AB ¼ (J A + J B ) i = ( J AB / 2D ) ν 2 = m + D + ½ J AB + ¼ (J A + J B ) i 2 = ( + J AB / 2D + ) ν 2 = m + D ½ J AB ¼ (J A + J B ) i 2 = ( + J AB / 2D ) B Transitions: ν 3 = m D + + ½ J AB + ¼ (J A + J B ) i 3 = i 2 ν 3 = m D + ½ J AB ¼ (J A + J B ) i 3 = i 2 ν 4 = m D + ½ J AB + ¼ (J A + J B ) i 4 = i ν 4 = m D ½ J AB ¼ (J A + J B ) i 4 = i Transitions (lines 7,7 are called combination lines): ν 5 = ν + ½ (J A + J B ) i 5 = ν 5 = ν ½ (J A + J B ) i 5 = ν 6 = ν + D + D ν 6 = ν D + + D ν 7 = ν + D + + D ν 7 = ν D + D i 6 = i 6' = i 7 = i 7' = Δν 2 - (ΔJ/2) 2 + J AB 2 (2D + )(2D - ) Δν 2 - (ΔJ/2) 2 + J AB 2 (2D + )(2D - ) These line positions have been loaded into an Excel Spreadsheet to Simulate AB spectra.

15 Chemistry 43 NMR and Structure Determination California Institute of Technology AB Spin System Vinyl Bromide in C 6 D 6 with added CDCl 3 NMR (500 Mz) l k j l 55% CDCl 3 Δν/J = -2.9 k 53% CDCl 3 Δν/J = -.3 j 52% CDCl 3 Δν/J = 0 i 5% CDCl 3 Δν/J = 0.6 h 50% CDCl 3 Δν/J =.5 g 49% CDCl 3 Δν/J =.8 f 48% CDCl 3 Δν/J = 2.4 e 47% CDCl 3 Δν/J = 3 d 45% CDCl 3 Δν/J = 4 c 43% CDCl 3 Δν/J = 5.7 b 39% CDCl 3 Δν/J = 9 a 27% CDCl 3 Δν/J = 7 i h g f e d c b a

16 Example 2. Understanding the AB portion of an AB spectrum. The proton decoupled 40.5 Mz 3 P NMR spectrum of square-planar Rh(py)Cl[P(para-tolyl) 3 ] 2 is shown below relative to the standard 85% 3 P 4 at 0 ppm. (py = pyridine) Drago, et. al. JACS 983, 05, (a) What is the abundance of 03 Rh and its spin? (b) Are the phosphine ligands cisor trans- in this complex? Give the Pople (e.g. A 2 or AB) notation for the complex. (c) Assign the lines in the spectrum. m Rh = +/2 ν PB ν PA J(Rh-P B )=98 z J(Rh-P A )=67 z m Rh = -/2 m Rh = +/2 m Rh = -/2 Solution: The coupling of two inequivalent phosphine ligands P A and P B to the I=/2 03 Rh nucleus results in an AB spin system of which we are observing the AB portion. If the complex were of cis- geometry, we would only see a doublet for the equivalent 3 P nuclei of an A 2 spin system. 2 J(P A -P B )=47.3 z (p-tolyl) 3 P (p-tolyl) 3 P Rh py (m = +/2) Cl AB pattern for Rhodium m = + /2 (p-tolyl) 3 P py Rh (m = -/2) (p-tolyl) 3 P Cl AB pattern for Rhodium m = - /2 The intensities of the AB portion can be understood by considering the two trans-bis(phosphine) species having Rh in the m = +/2 and -/2 spin states. The first state (non-dashed box) with the Rh in the m = +/2 spin state causes each phosphorus to resonate at a higher frequency [by J/2 = (+67/2) z for P A and J/2 = (+98/2) z for P B ]. The two phosphorus nuclei of this species couple to each other with the intensities of an AB system. The species with the m = -/2 Rh spin state leads to the other AB system with expected intensities.

17 Chemistry 43 NMR and Structure Determination California Institute of Technology Special Effects of Strong Coupling Virtual Coupling in an AB system B When A and B are strongly Et coupled (δν/j 2), the line A separations in the portion of the spectrum do not individually 3 J AB = 6.62 z correspond to J A and J B. Even 3 J A = 7.4 z when J B 0, a strongly coupled 4 J B = 0.04 z system may show a familiar dd δν AB 6 z pattern. This has been termed virtual coupling which can be somewhat misleading. Although the full spread of the multiplet is = J A + J B, there is nearly zero coupling between B and in the example at right. The virtual Coupling effect is also seen for the methyl resonances of straight-chain aliphatic compounds. Although the methyl group only has significant J coupling with the adjacent methylene group, each successive C2 group resonates at nearly the same frequency. This causes the methyl pattern to be distorted from a true triplet. ften, the methyl triplet appears to lean towards the upfield direction. See: Musher, J. I. and Corey, E. J. Tetrahedron 962, 8, ~.34 ppm 0.9 ppm C 2 C 2 C 3 R C 2 C 2 ~.34 ppm.34 ppm NMR (300 Mz, CDCl 3 )

18 Chemistry 43 NMR and Structure Determination California Institute of Technology Consider the series of,3-dibromopropane (a),,4-dibromobutane (b) and,5-dibromopentane (c). Although spectra (a) and (c) each give a triplet for the -C 2 Br resonance, spectrum (b) shows a seriously distorted peak. The -C 2 Br group of,4-dibromobutane couples to the adjacent C 2 group, and since the next C 2 group is equivalent by symmetry, a higher order coupling pattern results. Tetrahydrofuran shares a similarly characteristic coupling pattern. a: Br Br (500 Mz, CDCl 3 ) b: Br Br (500 Mz, CDCl 3 ) c: Br Br (500 Mz, CDCl 3 )

19 Chemistry 43 NMR and Structure Determination California Institute of Technology If two chemically equivalent nuclei share the same J coupling to each other nucleus in the molecule, they are also magnetically equivalent. Examples of this are cyclopropane and difluoromethane. These molecules display a simple A 2 2 spectrum consisting of triplets for both the A and resonances. If two chemically equivalent nuclei have different J coupling constants to another nucleus in the molecule, they are magnetically inequivalent. The AA spin system can lead to a more complex multiplet pattern for the A and resonances. A 2 2 : AA'': J A A A A Cl J A J A A' Cl ' J A' J AA' A A' J A J A' ' J ' AA'' Spin System ν5 = νa + 2 K + 2 K 2 +L 2 i 5 = 2 ( - K K 2 +L 2 ) ν6 = νa - 2 K + 2 K 2 +L 2 i 6 = 2 ( + K K 2 +L 2 ) ν7 = νa + 2 K - 2 K 2 +L 2 i 7 = i 6 Define K,L,M,N: 2 Lines at each νa,ν K = J AA' + J ' ν = νa + N 2 i = L = J A - J A' ν2 = νa + N 2 i 2 = M = J AA' - J ' ν3 = νa - N 2 i 3 = N = J A + J A' ν4 = νa - N 2 i 4 = ν8 = νa - 2 K - 2 K 2 +L 2 i 8 = i 5 ν9 = νa + 2 M + 2 M 2 +L 2 i 9 = 2 ( - M M 2 +L 2 ) ν0 = νa - 2 M + 2 M 2 +L 2 i 0 = 2 ( + M M 2 +L 2 ) ν = νa + 2 M - 2 M 2 +L 2 i = i 0 ν2 = νa - 2 M - 2 M 2 +L 2 i 2 = i 9 all i 5 - i 2 satisfy: i # = L 2 L 2 +4x 2 where x = ν# - νa,2 3,4 J AA' = -2 z J ' = -8 z J A = 0 z J A' = 2 z 6 K = -20 L = 8 M = -4 N = y = L 2 L 2 +4x 2 7 same peak observed at ν x K ν A N K ν M M K 2 +L 2 - K M 2 +L 2 - M

20 Some Common AA'' Systems para-disubstituted Aromatic (J J AA' J ', J A' 0 ) A,2,9,0 3,4,,2 J AA' (small) A J A' <.5 Y J ' (small) J AA' = 2.3 z K = 4.6 J ' = 2.3 z L = 8.3 J A = 8.3 z M = 0 N = 8.3 J A' = 0 z para-anisaldehyde. For para-disubstituted aromatic molecules, the coupling constant between para protons is less than 0.5 z. So L = N = J A. Also, J AA' J ' = J meta 2 z. Therefore, M = 0 and lines 9 through 2 will add to the peaks at νa ± /2 J A. Four smaller peaks will be observed for lines 5-8. Diphenylcyclobutanedicarboxylic esters. A similar analysis may be applied to the head-to-head cyclobutane isomer shown A' ' 5 J AA' +J ' 6 ν A J A z A C 3 C 2.4 z 2.4 z A' ' Ph C 2 Me Ph C 2Me

21 C2 C 2 Y Molecules J AA' (gem) A C A' J A J A' C ' Y J ' (gem) J AA' = -0 z J ' = -4 z J A = 0 z J A' = 6 z,2 K = -24 L = 4 M = 4 N = 6 5,8 3,4 BrC2C2CN. For freely rotating systems, JA and JA' will be averaged over all conformations leading to a first order pattern. TMSC2C2C2Na. When and Y are larger, the anticonformation will predominate and JA JA'. Since JAA' and J' are both geminal couplings (-0 to -8 z) and JA and JA' are both vicinal and probably only differ by -8 z, approximate K 2 >> L 2. Lines 5 and 8 will both appear at ν5 = ν 8 = ν A with i5 + i8 = 2 and lines 6 and 7 will nearly disappear (i6 = i7 0). N-Methylmorpholine. In a cyclic molecule, the C2C2Y system is restricted to the eclipsed and gauche conformations. Again, K 2 >> L 2. The upfield pattern is further broadened by coupling to 4 N. 3 J = 6.8 z Br C 2 C 2 CN C 3 C 3 Si C 2 C 2 M 2 +L 2 C()R ν A A 2 C ' 2 3 AA'' Me 3 Si - M M N A J A J A' M C()R A' C 3 N C 2 C 2 J A A J A' ' A' C 2 C 2 AA'' NMe C 2 C 2

22 ne Molecule that Contains the A4, AA BB and AA Spin Systems. The 400 Mz spectrum of a cyclophane containing the A4, AA BB, and AA spin systems. 3 2 b b a a 2. Try to assign the peaks in this example

23 Chemistry 43 NMR and Structure Determination California Institute of Technology A First rder Analysis of a Simplified AA Spin System What happens when we bring two equivalent A spin systems together? J C = 25 z Two separated C groups: Two equivalent C groups with 2 J CC : many bonds J CC = 0 z 3 C 3 C Two fully separated 3 C- subunits would just give a doublet with line separation = J C J C = 25 z J C = 25 z 2 J CC > 0 z 3 C Y 3 J C = 0 z 3 C 4 J = 0 z J C = 25 z What will the 3 C NMR look like? When the equivalent C groups are brought together with J CC = 30 z, we can analyze the three separate species having, and spin states. For the species with the proton spin states, the two carbons would resonate at the same frequency they are equivalent and would give one line at the downfield frequency. Same for the spin state giving the upfield line. But for the species with the spin states, one 3 C would resonate at the downfield frequency and the other would resonate upfield frequency. They are inequivalent and would couple to each other as an AB spin system! 3 C A 3 C A 3 C A Y Y Y 3 C B 3 C B 3 C B Sum of above three species: ( 3 C NMR) m( A, B ) = +/2 m( A ) = +/2 ν C J(C)=25 z ν C J(C)=25 z Δν= J(C)=25 z 2C = Δν 2 + J 2 = 28 z m( A, B ) = -/2 m( B ) = -/2 2 J(C A -C B ) = 30 z As an Example, we can see the anionic species below gives the pattern with J C = 39 z and 2 J CC = 34 z (offmann, et. Al rganometallics 200, 20, ): ipr MgCl CI 3 - ipr-i I 3 C I 3 C I I I - + Mg Cl Normally in organic compounds, 2 J CC are less than 5 z. The remarkably large 2 J CC = 34 z is ascribed to the heavy atom effect of the iodine J C = 39 z 2 J CC = 34 z C NMR (00 Mz) of [I 2 C-I-CI 2 ] - without decoupling.

24 Chemistry 43 NMR and Structure Determination California Institute of Technology Use of Excel spreadsheets to calculate AB spectra The 2 AB transitions are calculated in the Excel spreadsheet. Just by manipulating the values for Δν (z), one can reproduce the spectra from the Spin Coupling Lecture. ne can also start with this spreadsheet and modify J AB, J A and J B to generate the simulated spectra for other molecules. Vinyl Bromide: Rh(py)Cl[P(para-tolyl) 3 ] 2 AB portion:

25 Chemistry 43 NMR and Structure Determination California Institute of Technology Use of Excel spreadsheets to calculate AA spectra