1. Nuclear Magnetic Resonance Spectroscopy, cont. Dr. Todebush Chemistry 2412L

2. So far… Determine the number of sets of equivalent hydrogen atoms  Number of signals on spectrum Determine the number of hydrogen atoms in each set  Integration line Determine general information about adjacent groups  Chemical shift (  )

3. Next… Determine specific information about adjacent groups In particular, how many H atoms on the adjacent atoms  Signal splitting

4. Spin-Spin Splitting Nonequivalent protons on adjacent carbons have magnetic fields that may align with or oppose the external field This magnetic coupling causes the proton to absorb slightly downfield when the external field is reinforced and slightly upfield when the external field is opposed All possibilities exist, so signal is split into multiple peaks

5. 1,1,2-Tribromoethane Nonequivalent protons on adjacent carbons

6. Doublet: 1 Adjacent Proton

7. Triplet: 2 Adjacent Protons

8. The (n + 1) Rule If a signal is split by n equivalent protons, it is split into (n + 1) peaks ◦ n = neighboring

9. Range of Magnetic Coupling Equivalent protons do not split each other Protons bonded to the same carbon will split each other only if they are not equivalent Protons on adjacent carbons normally will couple Protons separated by four or more bonds will not couple

10. Ethyl Iodide SignalAB Chemical shift  1.2-1.9  3-4 Integration3H2H n23 n + 134 Splittingtq AB

12. Splitting for Ethyl Groups

13. 3-Methyl-2-butanone SignalIntegrationSplitting Chemical Shift (  ) A2.0-2.4 B0.7-1.3 C2.0-2.4 ACB

15. Splitting for Isopropyl Groups

16. 3-Methyl-2-butanone How many signals in the 13 C-NMR?

18. Which isomer best fits this spectrum? or

19. Which isomer best fits this spectrum? or

20. Splitting of Hydroxyl Proton Ultrapure samples of ethanol show splitting Ethanol with a small amount of acidic or basic impurities will not show splitting

21. N-H Proton Moderate rate of proton transfer Peak may be broad

22. Coupling Constants (J) Distance between the peaks of a split signal Measured in Hz (usually 0-18) Not dependent on strength of the external field Gives info on type of H ◦ Multiplets with the same coupling constants may come from adjacent groups of protons that split each other ◦ Structural features

23. Values for Coupling Constants

24. Stereochemical Nonequivalence Usually, two protons on the same C are equivalent and do not split each other If the replacement of each of the protons of a - CH 2 group with an imaginary “Z” gives stereoisomers, then the protons are non- equivalent and will split each other

25. Some Nonequivalent Protons

26. How do 1 H- and 13 C-NMR differ? Signals in 13 C-NMR are weak ◦ Noisy baseline, many scans are necessary Peak areas in 13 C-NMR are not proportional to number of carbons ◦ No integration Signals in 13 C-NMR are not split ◦ C-C splitting is negligable ◦ C-H splitting too difficult to analyze; H-decoupled mode

27. Interpreting 13 C-NMR The number of different signals indicates the number of different kinds of carbon The chemical shift indicates the functional group Use to support 1 H-NMR analysis

28. Solving NMR Problems Given: ◦ Molecular formula ◦ 1 H-NMR ◦ IR (sometimes) ◦ 13 C-NMR (sometimes) Goal: ◦ Determine structure First step: ◦ Determine the Index of Hydrogen Deficiency  Short cut to number of rings or pi bonds

29. IHD = C – ½(H + X) + ½N + 1 Sum of number of rings +  bonds Remember: alkynes have 2  bonds C 6 H 12 IHD = 6 – ½(12) + 1 = 1 C 5 H 3 N 2 O 2 ClIHD = 5 – ½(3+1) + ½(2) + 1 = 5 Anytime IHD > 4 and C ≥ 6, think benzene C 6 H 6 IHD = 6 – ½(6) + 1 = 4

30. In-class 1 H-NMR Problem: What is the structure of the compound with the following 1 H-NMR spectrum? a b c d e