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CHEM 212 – NMR Spectroscopy

Spring 2014. CHEM 212 – NMR Spectroscopy. Spectral Analysis – 1 H NMR. NMR Spectroscopy. NMR Spectral Analysis – Introductory 1 H NMR NMR is rarely used in a vacuum to do a “ forensic ” analysis of an unknown NMR (all nuclei) is usually used:

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CHEM 212 – NMR Spectroscopy

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  1. Spring 2014 CHEM 212 – NMR Spectroscopy

  2. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H NMR • NMR is rarely used in a vacuum to do a “forensic” analysis of an unknown • 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 • In this course, you will be given one of three pieces of data with an 1H NMR for consideration: • A molecular formula • An IR spectrum • The first part of a chemical reaction – for example:

  3. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H 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: CxHyNzO (halogens count as Hs) • HDI = x – ½ Y + ½ Z + 1 • Example: C4H8O -- HDI = 4 – ½ (8) + ½ (0) + 1 • HDI = 1 This compound contains either one double bond or one ring

  4. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H NMR • Step 2: Do a quick assessment of the 1H 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, C10H14 • 1H NMR spectrum consists of two singlets!

  5. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H NMR • Example: 1H NMR for C4H8 (which has an HDI of 1)

  6. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H NMR • Step 3: Use the integration along with the molecular formula to make sure you can find all of the 1Hs (this is 1H NMR after all!) • If you do not have a molecular formula, use the integration to attempt to tabulate the number of 1Hs 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, -CH2- ’s, -CH3 ’s and multiples of chemically equivalent ones!

  7. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis • Continue our example: • For C4H8O we need to find 8 protons • The integral for the quartet at d 2.5 measures 30 mm, the singlet at d 2.2 measures 44 mm, the triplet at d 1.1 measures 43 mm • The ratio: • 30:43:42 is roughly 2:3:3 • 2 + 3 + 3 = 8 • We found all 8 protons 43 44 30

  8. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H 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. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis • Continue our example: • For C4H8O we have 3 families of chemically equivalent protons in a ratio of 2:3:3 or –CH2-, -CH3, -CH3 • Both the d 2.2 and 2.5 resonance correspond to protons on carbons next to electron withdrawing groups • The d 1.1 resonance corresponds to protons on carbons bound to other aliphatic carbons

  10. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis • Continue our example: • We found –CH2-, -CH3 and • –CH3, subtract these from our original formula: • C4H8O - CH2 = C3H6O • C3H6O – CH3 = C2H3O • C2H3O – CH3 = CO • We needed an HDI of 1 and there is no evidence for • 1H-C=C on the 1H NMR, our missing HDI, and EWG is a C=O!

  11. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H 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. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis • For our example, this is trivial; we have concluded the C=O is the electron withdrawing group • The –CH3 singlet at d 2.2 obviously has no 1Hs on adjoining carbons, as it is next to the carbonyl • The d 2.5 –CH2- quartet is next to a –CH3 (n+1 = 4, so n = 3, it is next to a –CH3) • The d 1.2 –CH3 is a triplet, so it is next to a –CH2

  13. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis – Introductory 1H 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. Spectral Analysis – 1H NMR NMR Spectroscopy • NMR Spectral Analysis • We concluded we have: • or 2-butanone • C4H8O, HDI = 1 • 3 proton resonances • 2 mutually coupled (split)

  15. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 2: C9H9BrO • HDI =

  16. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 2: C9H9BrO • HDI = 9 – ½ (10) + 0 + 1 = 5

  17. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 2: C9H9BrO • HDI = 9 – ½ (10) + 0 + 1 = 5

  18. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 2: C9H9BrO • HDI = 9 – ½ (10) + 0 + 1 = 5

  19. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 3: C4H10O • HDI =

  20. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 3: C4H10O • HDI = 4 – ½ (10) + 0 + 1 = 0

  21. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 3: C9H9BrO • HDI = 0

  22. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 3: C4H10O • HDI = 0

  23. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 4: C5H10O2 • HDI =

  24. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 4: C5H10O2 • HDI = 5 – ½ (10) + 0 + 1 = 1

  25. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 4: C5H10O2 • HDI = 1

  26. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 4: C5H10O2 • HDI = 1

  27. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 5: C10H12O2 • HDI =

  28. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 5: C10H12O2 • HDI = 10 – ½ (12) + 0 + 1 = 5

  29. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 5: C10H12O2 • HDI = 5

  30. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 5: C10H12O2 • HDI = 5

  31. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 6: C6H4ClNO2 • HDI =

  32. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 6: C6H4ClNO2 • HDI = 6 – ½ (5) + ½ (1) + 1 = 5

  33. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 6: C6H4ClNO2 • HDI = 5

  34. Spectral Analysis – 1H NMR NMR Spectroscopy • Example 6: C6H4ClNO2 • HDI = 5

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