1. 1 CHEM 212 – NMR Spectroscopy Spring 2014

2. 2 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR 1.NMR is rarely used in a vacuum to do a “forensic” analysis of an unknown 2.NMR (all nuclei) is usually used: –To analyze the product of a chemical reaction along with IR –To elucidate the structures of natural products (like the spice lab compounds in CHEM 213) in conjunction with mass spectrometry (which gives molecular weights and formulas), IR and UV 3.In this course, you will be given one of three pieces of data with an 1 H NMR for consideration: –A molecular formula –An IR spectrum –The first part of a chemical reaction – for example:

3. 3 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 1: Do a quick assessment of the information you are given Molecular formula – one of the most important pieces of information Use the index of hydrogen deficiency (HDI) to determine the possible number of rings, double and triple bonds in the molecule: For a chemical formula: C x H y N z O (halogens count as Hs) HDI = x – ½ Y + ½ Z + 1 Example: C 4 H 8 O -- HDI = 4 – ½ (8) + ½ (0) + 1 HDI = 1 This compound contains either one double bond or one ring

4. 4 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 2: Do a quick assessment of the 1 H NMR you are given –Is the molecule simple or complex? –Is the molecule aromatic, aliphatic or both? –What are the total number of resonances that you observe Be careful with overly simple spectra – remember a large molecule may appear to be small and simple if it is highly symmetrical Consider: Durene, C 10 H 14 1 H NMR spectrum consists of two singlets!

5. 5 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Example: 1 H NMR for C 4 H 8 (which has an HDI of 1)

6. 6 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 3: Use the integration along with the molecular formula to make sure you can find all of the 1 Hs (this is 1 H NMR after all!) If you do not have a molecular formula, use the integration to attempt to tabulate the number of 1 Hs in the formula (does it make sense?) If one hydrogen appears to be “missing”, you may suspect it is acidic or exchangeable with the dueterated NMR solvent Remember, integration gives you the least common denominator of the total number of protons of each type Keep in mind organic molecules contain –C-H’s, -CH 2 - ’s, -CH 3 ’s and multiples of chemically equivalent ones!

7. 7 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis Continue our example: For C 4 H 8 O we need to find 8 protons The integral for the quartet at  2.5 measures 30 mm, the singlet at  2.2 measures 44 mm, the triplet at  1.1 measures 43 mm The ratio: 30:43:42 is roughly 2:3:3 2 + 3 + 3 = 8 We found all 8 protons 30 44 43

8. 8 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 4: Classify each of the proton resonances by using the general correlation table Reconsider the number of rings, double or triple bonds that are possible given the HDI, and reconcile this data with what the 1H chemical shifts are telling you Some hints: If you calculate an HDI > 4, you probably have an aromatic ring, and this should show on the spectrum If you calculate an HDI of 1 or 2 and see no protons that are part of an alkene and alkyne – suspect rings if no oxygen's are present, carbonyls if (C=O) they are

9. 9 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis Continue our example: For C 4 H 8 O we have 3 families of chemically equivalent protons in a ratio of 2:3:3 or –CH 2 -, -CH 3, -CH 3 Both the  2.2 and 2.5 resonance correspond to protons on carbons next to electron withdrawing groups The  1.1 resonance corresponds to protons on carbons bound to other aliphatic carbons

10. 10 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis Continue our example: We found –CH 2 -, -CH 3 and –CH 3, subtract these from our original formula: C 4 H 8 O - CH 2 = C 3 H 6 O C 3 H 6 O – CH 3 = C 2 H 3 O C 2 H 3 O – CH 3 = CO We needed an HDI of 1 and there is no evidence for 1 H-C=C on the 1 H NMR, our missing HDI, and EWG is a C=O!

11. 11 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 5: Analyze the spin-spin coupling multiplets to elucidate the carbon chains of the molecule Hints: Singlets indicate you have protons on carbons that have no chemically non-equivalent protons on any adjoining atom Multiplets mean you have chemically non-equivalent protons on adjoining carbons (or atoms), use the n+1 rule in reverse to find out how many Spin-spin coupling or splitting is MUTUAL, if you observe a multiplet there must be another multiplet it is related to (split by)

12. 12 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis For our example, this is trivial; we have concluded the C=O is the electron withdrawing group The –CH 3 singlet at  2.2 obviously has no 1 Hs on adjoining carbons, as it is next to the carbonyl The  2.5 –CH 2 - quartet is next to a –CH 3 (n+1 = 4, so n = 3, it is next to a –CH 3 ) The  1.2 –CH 3 is a triplet, so it is next to a –CH 2

13. 13 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis – Introductory 1 H NMR Step 6: Construct the molecule and double-check consistency Does the HDI match? Have you accounted for all atoms in the formula? From your constructed molecule, pretend you are trying to verify if that spectrum matches, and quickly re-do the problem

14. 14 Spectral Analysis – 1 H NMRNMR Spectroscopy NMR Spectral Analysis We concluded we have: or 2-butanone C 4 H 8 O, HDI = 1 3 proton resonances 2 mutually coupled (split)

15. 15 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 2: C 9 H 9 BrO HDI =

16. 16 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 2: C 9 H 9 BrO HDI = 9 – ½ (10) + 0 + 1 = 5

17. 17 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 2: C 9 H 9 BrO HDI = 9 – ½ (10) + 0 + 1 = 5

18. 18 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 2: C 9 H 9 BrO HDI = 9 – ½ (10) + 0 + 1 = 5

19. 19 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 3: C 4 H 10 O HDI =

20. 20 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 3: C 4 H 10 O HDI = 4 – ½ (10) + 0 + 1 = 0

21. 21 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 3: C 9 H 9 BrO HDI = 0

22. 22 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 3: C 4 H 10 O HDI = 0

23. 23 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 4: C 5 H 10 O 2 HDI =

24. 24 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 4: C 5 H 10 O 2 HDI = 5 – ½ (10) + 0 + 1 = 1

25. 25 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 4: C 5 H 10 O 2 HDI = 1

26. 26 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 4: C 5 H 10 O 2 HDI = 1

27. 27 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 5: C 10 H 12 O 2 HDI =

28. 28 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 5: C 10 H 12 O 2 HDI = 10 – ½ (12) + 0 + 1 = 5

29. 29 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 5: C 10 H 12 O 2 HDI = 5

30. 30 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 5: C 10 H 12 O 2 HDI = 5

31. 31 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 6: C 6 H 4 ClNO 2 HDI =

32. 32 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 6: C 6 H 4 ClNO 2 HDI = 6 – ½ (5) + ½ (1) + 1 = 5

33. 33 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 6: C 6 H 4 ClNO 2 HDI = 5

34. 34 Spectral Analysis – 1 H NMRNMR Spectroscopy Example 6: C 6 H 4 ClNO 2 HDI = 5