NMR for Organic Chemistry III

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1 NMR for rganic Chemistry III Lecture 1 Lecture 2 Lecture 3 Lecture 4 Recap of Key Themes from NMR II + Problems CSY + Problems HSQC + Problems HMBC and Solving Structures + Problems 1

2 1. Practical Aspects Glass NMR Tube 5mm Field Spinner Maximum field strength at sample Rf coils The Probe Solution state NMR - sample dissolved in NMR solvent usually CDCl3, CD3D, C6D6 etc 5 mm tube, 700 µl sample, 1 mg Rf inputs and outputs Common NMR instrument MHz 9.4 Tesla Field Strength This means a proton will resonate at 400 MHz 13C would resonate at 100 MHz 2 Linked to a computer

3 1. Practical Aspects Magnet Amplifier Concentric design Superconducting Magnet Cooled in liquid Helium perator Probe Sample tube at cenre Highest Field Strength Console (Rf generator and detector) Modern NMR Spectrometers use superconducting magnets and sophisticated Rf pulse generators. Data is collected and processed automatically. Samples are often loaded by Robot But data interpretation must be done manually. This course is about understanding and interpreting 1H and 13C NMR data 3

4 rigin of Chemical shift (see hv 0 First nucleii B B E ΔE 0 hv 0 Aligned or pposed Single line in spectrum (singlet) Nucleus is shielded from the magnetic field by electrons in bonds. More electrons, more shielding More shielding, smaller ΔE Smaller ΔE, lower frequency In older NMR instruments the magnetic field is varied In this case a higher field strength is required when there is more shielding. B o Chemical Shift, δ, is defined as: δ = frequency of resonance - frequency of standard (Hz) perating frequency of spectrometer (MHz) So 1 ppm = 400 Hz on a 400 MHz instrument Units are ppm (10-6 ) 4

5 Chemical Shift no Coupling to 1 H; no coupling to 12 C Integral / peak intensity not (very) proportional to number of C 5

6 Chemical Shift 1 H - 1 H coupling Integral proportional to number of H 6

7 Aromatic CH Alkene CH C= Aromatic C-C CH-H CH CH2 CH3 7

8 A closer Look at Methyls - 13C Aromatic - Methyl Aromatic - - Methyl Methyl Ester CH-Methyl CH2-Methyl Methyls can be more intense - result of faster relaxation of 13C nucleus 8

9 A closer Look at Methyls - 1H Aromatic - - Methyl Methyl Ester Aromatic - Methyl CH-Methyl CH2-Methyl 9

10 A closer Look at CH2s - 13C 10

11 A closer Look at CH2s - 1H 11

12 A closer Look at CHs - 13C Aromatic and Alkene CH CH-H Epoxide CH CH-CH3 Alkyne CH 12

13 A closer Look at CHs - 1H Alkyne CH Epoxide CH CH-H CH-CH3 Aromatic and Alkene CH 13

14 4.1 Advanced Coupling Multiplets NMe Singlet Magnetically equivalent protons coupling with 0 others H 1.07 N H H δ s CH 2 -CH=CHD NCH 2 -CH= CH 3 -CH 2 CH 3 -CH 2 - d dt Doublet Magnetically equivalent protons coupling with 1 other Triplet Magnetically equivalent protons coupling with 2 others Quartet Magnetically equivalent protons coupling with 3 others Doublet of Triplets Proton coupling with 1 other and 2 other Chem. Inequiv. at the same time. 14

15 4.1 Multiplets 'Pure' Multiplets arising from a single J coupling J J J J Pascal's Triangle H X H M J H A J J J J J δ 1 δ 1 δ δ δ Singlet Doublet Triplet Quartet Pentet etc Composite multiplets arising from two (or more) J couplings e.g. AMX 4 J XY = 1.0 Hz H A H A H Y H X I I H Y H X H A 3 J AX = 6.2 Hz 3 J AY = 10.9 Hz 15

16 4.1 Multiplets Effect of coupling constant and Δδ on appearance of dd H X I I H Y H A 3 J AY = 15.1 Hz 3 J AX = 6.2 Hz H Y H X H A Note 'roofing effect' H X H A H 3 J AX = 7.0 Hz 3 J AY = 7.0 Hz H Y H X I I H Y HA 3 J xy = 7.0 Hz 3 J AX = 7.0 Hz verlap of two doublets appears as a triplet because 3 J AX = 3 J AY 16

17 4.1 Multiplets ther Common Multiplets H X Doublet of Triplets H X H Y H Y H A H A 3 J AX = 6.2 Hz 3 J XY = 15.1 Hz H A 4 J AY = 1.0 Hz H A3 C H M Doublet of Quartets H X H M H X 3 J AM = 6.4 Hz 3 J MX = 15.1 Hz H A 4 J AX = 1.0 Hz 17

18 4.1 Multiplets ther Common Multiplets H A3 C H M CH A3 Heptet CH A3 H M Triplet of Quartets A B C 3 J BC = 8.2 Hz J AB = 6.6 Hz H C H B H A

19 4.2 Multiplets Measuring J For a spectrum obtained at X MHz, then 1 ppm = X Hz Remeber that J is constant, so multiplets appear differently at different field strengths: H A δ = 1.29, 3 J AX = 8.0 Hz H A H A H A H A H A 60 MHz 90 MHz 270 MHz 400 MHz 600 MHz Common 1 H NMR frequencies if multiplet is annotated in Hz then simple substraction gives J if multiplet is annotated in ppm then need to know Frequency of Expt. Here 400 Mhz 61.3 Hz = ppm = Hz 3 J = ppm 3 J = 0.02 x 400 = 8.0 Hz 19

20 4.2 Multiplets Measuring J at different field strengths Br Cl H 3H 1.79 H H H H 3.04; ; verlapping Multiplets Very hard to interpret 4H 90 MHz 2H 3H 2 J = 12.4 Hz 3 J = 7.0 Hz dd dd dd dd 3H 600 MHz 2H 2H 1H 1H 20 3 J = 6.8 Hz

21 4.3 Multiplets Diastereomeric Protons H X Br H M1 H M2 Me H M1 H M2 H X C()Me Br H M1 and H M2 are NT chemically equivalent. H M1 is gauche to Br, but H M2 is anti to Br ddq Thought Experiment H X Br D H M2 H X Br H M1 D Replacing one hydrogen, R the other hydrogen with D results in the creation of Diastereomers. 3H, d Diastereomers are not identical - thus we expect different δ values for H M1 and H M2. They will also couple to each other, and differently to other protons. Such protons are described as diastereotopic. H X 1H, dd 1H, dd 3 J XMe = 6.8 Hz 1H, ddq 2 J M1M2 = 12.8 Hz H 3 X J XM1 = 7.0 H M1 H 3 M2 J XM2 = 7.1 Note: 6.8, 7.0 and 7.1 Hz J values are effectively identical, so multiplets overlap 21

22 4.3 Multiplets Diastereomeric Protons ther examples: H A and H B are diastereotopic 5.91 H A 6.21 H B 2.27 H C H B H A dd dd dd 2 J AB = 2.1 Hz 3 J AC = 10.0 Hz 3 J BC = 16.8 Hz H C

23 Using Coupling to solve Connectivity 23

24 Using Chemical Shift to Solve Functional Groups Methyls -CH2 Alkene C= 24

25 Now: Problems Next Time: Adding CSY to give more information 25

26 NMR for rganic Chemistry III Problems 1 1

27 1A. Assign each 1 H and 13 C resonance to the corresponding atoms. Explain the answer. 2

28 1B. Assign each 1 H and 13 C resonance to the corresponding atoms. Explain the answer. 3

29 1C. Assign each 1 H and 13 C resonance to the corresponding atoms. Explain the answer. 4

30 1D. Assign each 1 H and 13 C resonance to the corresponding atoms. Explain the answer. 5

31 2. Match each compound to its NMR spectra. Assign each resonance to its H or C atom(s) and then answer the questions below. H H H A B 4 PPM PPM PPM PPM 50 0 C D 10 PPM PPM PPM PPM 0 2a. What is the 1 H chemical shift of a CH2-C= group? 2b. What is the 1 H chemical shift of a CH2-H group? 2c. What is the 1 H chemical shift of a CH2-Ac group? 2d. What is the 1 H chemical shift of a CH2-CH2-CH2 group? Draw up an ordered list of electron withdrawing ability based on these observations 2e. What is the 13 C chemical shift range for an ester or acid carbonyl carbon? 2f. What is the 13 C chemical shift of an aldehyde carbon? 2g. Predict the 13 C chemical shift of a ketone carbonyl carbon. 6

32 3. Match each compound to its NMR spectra. Assign each resonance to its H or C atom(s) and then answer the questions below. N A C 3 2 PPM PPM PPM PPM 20 0 B D 4 2 PPM PPM PPM PPM a. What is the 13 C chemical shift of an amide carbonyl? 3b. Explain the trend in 13 C chemical shifts of carbonyls 3c. The spectra for dimethylacetamide are wrong - why? 3d. Compare the CH2 resonances in the 5-membered and 3-membered rings - explain why they differ. 7

33 4. Identify the compound of formula C 9 H 10 2 which has the following NMR spectra: 8

34 5. Identify the compound of formula C 9 H 10 2 which has the following NMR spectra: 9

35 6. Identify the compound of formula C 9 H 10 2 which has the following NMR spectra: 10

36 7. Identify the compound which has the following NMR spectra: 11

37 8. Identify the compound which has the following NMR spectra: 12

38 9. Identify the compound which has the following NMR spectra: 13

39 10. Identify the compound of formula C 7 H 7 N 2 which has the following NMR spectra: 14

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