1. Units of Unsaturation This is also called “Degrees of Unsaturation” or “Double Bond Equivalents (DBE)”. By looking at a molecular formula, it is possible to find the number of double bonds and/or rings in the compound. The procedure is as follows: 1. For the purpose of this calculation, find effective number of H. a) For each halogen, add 1 H. b) For each nitrogen, subtract 1H. c) Oxygen has no effect on this calculation. 2. Using the fact that saturated compounds have the formula C n H 2n+2 find the number of H if saturated. 3. Use the formula (Saturated H – Effective H) / 2 to find the number of DBE. 4. If DBE is 4 or more, an aromatic ring is present.

2. Examples C 4 H 8 O 2 ----> C 4 H 8 ----> C 4 H 10 DBE = (10 – 8)/2 = 1 Effective H = 8 Saturated H = 4x2 + 2 = 10 C 6 H 4 Cl 2 ----> C 6 H 6 ----> C 6 H 14 DBE = (14 – 6)/2 = 4 Effective H = 6 Saturated H = 6x2 + 2 = 14 C 5 H 5 N ----> C 5 H 4 ----> C 5 H 12 DBE = (12 – 4)/2 = 4 Effective H = 4 Saturated H = 5x2 + 2 = 12

3. Infrared Spectroscopy Nice introductions to IR spectroscopy can be seen at the following links: 1)MSUMSU 2)General resourcesGeneral resources IR is commonly used to divine functionality. It gives little useful information regarding exact structure. IR can be diagnostic (able to suggest the presence or absence of a particular functional group) or confirmatory (possibly able to support evidence of a particular functional group found by another means). We will concentrate on the diagnostic areas, since they give the most useful information.

4. Correlation Chart This chart gives you an idea of where you will find functional groups.

5. The 3300-3500 cm -1 region OH, NH and NH 2 stretches are found in this region. They may be distinguished by their differing shapes. OH examples (a) (b) OH stretches are typically strong (come down a long way) and broad (wide). (a) is more typical of aliphatic OH stretch. (b) is more typical of a carboxylic OH stretch (even though these often appear outside the usual OH stretch region). This one is “masking” other peaks.

6. (a)(b) NH 2 and NH Examples (a) is a typical NH 2 stretch, note the two peaks. (b) is a typical NH stretch. These are not usually as strong or broad as OH stretches.

7. The 2900 – 3100 cm -1 Region CH stretch (alkane only). Notice how all peaks in the 2900-3100 region are below 3000 cm -1. CH stretch (alkane AND alkene) Notice how the peaks in the 2900-3100 region “straddle” 3000 cm -1. CH stretches are found in this region. They can be differentiated as shown below.

8. The 2000-2300 cm -1 Region C  C and C  N stretches are found here. C  C stretches are typically less intense of the two and usually occur at 2100-2200 cm -1. C  N are usually more intense of the two and usually occur at 2200-2300 cm -1.

9. The 1700 – 1850 cm -1 Region C=O stretches are usually found here. They are strong and fairly sharp.

10. Distinguishing C=C bonds. Typically, C=C that are in chains show a peak in the 1650 cm -1 region. Aromatic C=C stretches may be seen in the 1500-1600 cm -1 region. Be careful, though. These peaks are confirmatory, because other things can turn up in this region. This is useful only if you know a double bond is present.

11. Distinguishing Aldehydes and Ketones Be careful, though. These peaks are confirmatory, because other things can turn up in this region. This is useful only if you know an aldehyde is a possibility. Aldehydes (RCHO) Typically show 2 Peaks in the 2700- 2900 cm -1 region. This is for the C-H stretch from the O=C-H group. Peaks in this region are typically absent for ketones. Note that both still show the C=O stretch.

12. Proton NMR Spectroscopy For a great introduction to proton NMR spectroscopy go to: 1)MSUMSU 2)Wikipedia...click on the links in blue in the wiki.Wikipedia...click on the links in blue in the wiki. Chemical Shift

13. Chemical Shift is the relationship between peak location and the kind of hydrogen producing that peak. There are certain trends that are evident from the chart. a)Hs on alkanes have shifts of around 1 ppm. b)Adding electronegative groups like N, O, Cl usually results in a downfield shift (2.5 ppm – 5 ppm). c) If the H is attached to a double bond or aromatic ring it can usually be found around between 5 and 8 ppm. There is one set of peaks for each DIFFERENT H.

14. Each different H can be labelled with a unique letter. There should be as many peaks as there are letters.

15. Each letter has a corresponding peak. Look at the affect of chemical shift. a,b,c Hs are attached to an aromatic ring, so these are between 7-8 ppm. e H is attached to a C-Br, so this is at about 4 ppm.

16. Integration Tells how many H there are of a given type. For peaks < 5 ppm the following usually applies: 3H = CH 3 9H = 3 x CH 3 2H = CH 2 ; NH 2 (NH 2 single peak) 6H = 2 x CH 3 OR 3 x CH 2 1H = CH; NH; OH (NH, OH single peak)4H = 2 x CH 2 For peaks 7-8 ppm (aromatic) the following usually applies. (There may be two or more peaks). 4H = aromatic ring with 4H attached 5H = aromatic ring with 5H attached. See the example on the next page.

17. 5H =1H =2H = 3H = The squiggly lines represent “join points”, places where one group can join to another.

18. Multiplicity Tells how many H are on adjacent Carbon atoms. n peaks = n-1 adjacent H atoms on C atoms. (H x = H in question) 1 peak = 0 adjacent H atoms on C (CH x —O, CH x —NH, CH x —C=O); 2 peaks= 1 adjacent H atoms on C (CH x —CH) 3 peaks= 2 adjacent H atoms on C (CH x —CH 2 ) 4 peaks= 3 adjacent H atoms on C (CH x —CH 3 ) 5 peaks= 4 adjacent H atoms on C (CH 2 —CH x —CH 2 ) or (CH—CH x —CH 3 ) 6 peaks= 5 adjacent H atoms on C (CH 3 —CH x —CH 2 ) 7 peaks= 6 adjacent H atoms on C (CH 3 —CH x —CH 3 )

20. Look at “e” 6 peaks 5 H on adjacent Cs Look at “f” 2 peaks 1 H on adjacent C Look at “d” 2 peaks 1 H on adjacent C

21. Solving Structures given a Molecular Formula and a Proton NMR Spectrum 1.Find DBE using molecular formula. 2.Identify the pieces that can be seen by looking at the integration. 3.Find what cannot be seen by subtracting the pieces from the molecular formula. 4. The structure is made up of the pieces that can be seen plus the pieces that cannot be seen. 5.Draw possible structures that can be made from those pieces. 6. Use chemical shift and multiplicity to figure out which possibility is correct. 7. Verify by assigning each H on the spectrum.

22. C 5 H 10 O Step 1 Pieces we can see based on integration: CH 2,CH 3,CH 2,CH 3 Step 2 C 5 H 10 O ----> C 5 H 10 ----> C 5 H 12 DBE = (12-10)/2 = 1

23. Step 3:Find what cannot be seen by subtracting the pieces from the molecular formula. C 5 H 10 O - { } CH 2,CH 3, CH 2,CH 3 = C, O and 1 DBE = C=O Step 4:The structure is made up of the pieces that can be seen plus the pieces that cannot be seen. CH 2 CH 3 C=O

24. Draw possible structures that can be made from those pieces. Step 5 Step 6 Use chemical shift and multiplicity to figure out which possibility is correct. Based on multiplicities: a has 3 adj H (4 peaks) b has 5 adj H (6 peaks) c has 2 adj H (3 peaks) d has 0 adj H (1 peak) is correct.

25. Step 7 Verify by assigning each H on the spectrum.

26. 13 C NMR Spectroscopy This is the correlation chart for C-13 Spectroscopy. Attaching a more electronegative element to the carbon will pushes peak more downfield. Note also the downfield shift if the C is part of a double bond or an aromatic ring.

27. One peak is seen for each different C. ONLY Carbons are seen, so only Carbons are labelled.

28. Note how all of the peaks are single peaks. This is because C-13 only makes up 1.1% of all C isotopes. The probability of two C-13 atoms being next to each other is 1:10000. No C-13 to C-13 coupling is observed. Note that the shift of carbons (e)-(h) increases as they get closer to the ring. Similar Cs can sometimes be differentiated by height. (b) and (c) are taller than (a) because there are two versus one. CDCl 3 is a solvent and should be ignored.

29. If the spectrum is obtained using different setup parameters, the coupling between C-13 and 1-H can be seen. This is shown by letters above the peaks. q = 4 peaks = CH 3 t = 3 peaks = CH 2 d = 2 peaks = CH s = 1 peak = C The reason that letters are shown above the peaks is for clarity and to avoid overlap of multiple peaks.

30. Differentiating spectra in a matching question. First, differentiate spectra by number of peaks. Secondly, differentiate by chemical shift and by q,t,d and s.

31. Focus on the key peaks: C-O is usually around 60 ppm C-N is usually around 40 ppm. Use the correlation chart! The C attached to the electronegative element will be farthest downfield (the “1” C here).