1. Interpreting 1 H nmr spectra L.O.:  Intrepet 1 H nmr spectra using the n+1 rule

2. CHEMICAL SHIFT Approximate chemical shifts The actual values depend on the environment 13 12 11 10 9 8 7 6 5 4 3 2 1 0  ‘deshielding’ -COOH -CHO -C=CH- ROH - C - X H - C - H TMS ‘shielding’ Low  ‘deshielding’ High  Bonding to electronegative atoms (O, N)

3. LOW RESOLUTION LOW RESOLUTION SPECTRUM OF 1-BROMOPROPANE low resolution NMR gives 1 peak for each environmentally different group of protons Strengths of the absorption are proportional to number of equivalent 1 H atoms. It is measured by the are under each peak. Integration.

4. Look at low resolution 1 H NMR of EtOH

5. The simplified NMR spectrum of EtOH shows three single peaks. A detailed, high-resolution spectrum of EtOH shows that some peaks are split into a number of subsidiary peaks. This splitting is caused by spin-spin coupling between protons on neighbouring atoms.

6. High resolution 1 H NMR of EtOH.

7. The ‘n +1’ rule The n.m.r. absorption of a proton which has n equivalent neighbouring protons will be split into n + 1 peaks.

9. O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity MULTIPLICITY (Spin-spin splitting)

10. O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1 MULTIPLICITY (Spin-spin splitting)

11. O adjacent H’s There is no effect 1 adjacent H can be aligned either with  or against  the field there are only two equally probable possibilities the signal is split into 2 peaks of equal intensity 2 adjacent H’s more possible combinations get 3 peaks in the ratio 1 : 2 : 1 3 adjacent H’s even more possible combinations get 4 peaks in the ratio 1 : 3 : 3 : 1 MULTIPLICITY (Spin-spin splitting)

12. Number of peaks = number of chemically different H’s on adjacent atoms + 1 1 neighbouring H2 peaks“doublet”1:1 2 neighbouring H’s 3 peaks“triplet”1:2:1 3 neighbouring H’s 4 peaks“quartet”1:3:3:1 4 neighbouring H’s 5 peaks“quintet”1:4:6:4:1 Signals for the H in an O-H bond are unaffected by hydrogens on adjacent atoms - get a singlet MULTIPLICITY (Spin-spin splitting)

13. the area under a signal is proportional to the number of hydrogen atoms present an integration device scans the area under the peaks lines on the spectrum show the relative abundance of each hydrogen type By measuring the distances between the integration lines one can work out the simple ratio between the various types of hydrogen. before integration after integration INTEGRATION NOTICE THAT THE O-H SIGNAL IS ONLY A SINGLET

14. INTEGRATION HOW TO WORK OUT THE SIMPLE RATIOS Measure how much each integration line rises as it goes of a set of signals Compare the relative values and work out the simple ratio between them In the above spectrum the rises are in the ratio... 1:2:3 IMPORTANT: It doesn’t provide the actual number of H’s in each environment, just the ratio Measure the distance between the top and bottom lines. Compare the heights from each signal and make them into a simple ratio.

15. When is a hydrogen chemically different? NMR SPECTROSCOPY TWO SIGNALS Quartet and triplet :- ratio of peak areas = 3 : 2 Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments. The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical. BUTANE 12 34

16. When is a hydrogen chemically different? NMR SPECTROSCOPY TWO SIGNALS Quartet and triplet :- ratio of peak areas = 3 : 2 Carbons 1 & 4 are the similar and so are carbons 2 & 3 so there are only two different chemical environments. The signal for H’s on carbon 2 is a quartet - you ignore the two neighbours on carbon 3 because they are chemically identical. TWO SIGNALS both singlets :- ratio of peak areas = 2 : 1 Hydrogens on OH groups only give singlets. The signal for H’s on each carbon are not split, because - H’s on the neighbouring carbon are chemically identical... and - H’s on adjacent OH groups do not couple. BUTANE ETHANE-1,2-DIOL 12 34

17. An nmr spectrum provides several types of information :- number of signal groups tells you the number of different proton environments chemical shiftthe general environment of the protons peak area (integration)the number of protons in each environment multiplicityhow many protons are on adjacent atoms In many cases this information is sufficient to deduce the structure of an organic molecule but other forms of spectroscopy are used in conjunction with nmr. NMR SPECTROSCOPY - SUMMARY

18. NMR spectra of –OH and –NH protons o They are usually broad o The is usually no splitting pattern.

19. D 2 O Shake CH 3 CH 2 OH + D 2 O → CH 3 CH 2 OD + HOD

21. 1 H NMR TASK 4 For each of the following compounds, draw the molecule predict the number of signals predict the relative intensity of each signalpredict the approximate chemical shift (of each signal a)propanoic acid b)propanal c)2-chloropropane d)2-methylbutane e)methylpropene f)methyl propanoate

24. Task 5-9

25. Interpreting 1 H nmr spectra L.O.:  Intrepet 1 H nmr spectra using the n+1 rule

26. WHAT IS IT! C 2 H 5 Br 2 3

27. WHAT IS IT! C 2 H 3 Br 3 1 2

28. WHAT IS IT! C 2 H 4 Br 2 1 3

29. WHAT IS IT! C2H4O2C2H4O2 1 1 3

30. C4H8O2C4H8O2 2 3 3

31. C3H6OC3H6O

32. C3H6OC3H6O 1 3 2

33. C4H8OC4H8O 2 3 3

34. C 8 H 16 O 2

35. WHAT IS IT! C 11 H 16

36. WHAT IS IT! C 8 H 10

37. WHAT IS IT! C 8 H 10 2 3

38. WHAT IS IT! C 9 H 12

39. WHAT IS IT! C 4 H 8 Br 2