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Vibrational Spectroscopy - IR

Vibrational Spectroscopy - IR. ‘Near IR’. Most useful region for Functional Groups, especially 1000-3500 cm -1. p. 15 Course Manual. p. 16. Most organic compounds have C-H bonds: not as useful for structural assignment. C-H stretch ~ 3000 cm -1. C-H bend ~ 1500-1300 cm -1.

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Vibrational Spectroscopy - IR

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  1. Vibrational Spectroscopy - IR ‘Near IR’ Most useful region for Functional Groups, especially 1000-3500 cm-1 p. 15 Course Manual

  2. p. 16 Most organic compounds have C-H bonds: not as useful for structural assignment C-H stretch ~ 3000 cm-1 C-H bend ~ 1500-1300 cm-1 ‘salt’ cut-off scale

  3. p. 15 Functional groups involve other vibrations e.g. ALCOHOLS all have –O-H stretches Alcohols also have –C-O stretches ~1200-1000 cm-1 -O-H stretches are always strong and almost always broad, ~3500-3300cm-1

  4. p. 16 ~1100 cm-1 ~3400cm-1

  5. p. 18 Alcohols, amines, acids, amides alkynes, nitriles alkynes carbonyl compounds aldehydes, ketones, acids, esters, amides,.... Functional group regions

  6. p. 18 BAND STRENGTH Depends upon the bond dipole: large dipole = strong band C---O large vs. C---C small d+ d-

  7. p. 19 BAND POSITION 1: k, force constant depends on strength of bond, C=C > C-C 2: masses, easier to stretch bonds of heavy elements, C-H > C-O From a ball and spring model…we will see this again later so don’t worry too much about the details now

  8. p. 19 ALKANES 3000-2800cm-1 saturated-C-H stretch, <3000cm-1

  9. p.19 ALKENES UNSATURATED =C-H stretch, >3000 cm-1 not generally useful bends ~3050cm-1 in alkenes and aromatics!

  10. p.19 ALKYNES 2150cm-1, WEAK but diagnostic 3300cm-1, SHARP Compare: alcohol is normally broad

  11. p. 20 C C C C C C STRETCHES: 2150 cm-1 1650 cm-1 1250 cm-1 C---H C---H C---H 3300cm-1 3050cm-1 ~2900cm-1 LEARN THESE VALUES!

  12. p. 21 BENDS: 1,1

  13. p. 21

  14. p. 21 1,1

  15. p. 21 TRANS ~970 CIS ~700

  16. Similarly for benzene derivatives p. 22 2 3 1

  17. p. 22 MONO TWO META THREE

  18. p. 22 ORTHO <800 PARA >800

  19. p. 23/24 CARBONYL C=O REGION:1650-1800 cm-1 ANHYDRIDES R-CO-O-CO-R 3 O’S 1810, 1760 cm-1

  20. p. 24 ACID CHLORIDES R-CO-Cl 1800 cm-1

  21. p. 24 ESTERS R-CO-O-R’ 1735 cm-1

  22. p. 25 ALDEHYDES R-CO-H (RCHO) & KETONES R-CO-R’ Both 1 “O” H-C=O 2750 H-C=O 1725 C=O 1715

  23. p. 25 ACIDS R-CO-OH But also a huge –O-H 3400-2400 1710 cm-1 Note: esters also have 2 “O’s”, are at 1735cm-1, but no –O-H

  24. p. 24/25 ESTER ACID

  25. AMIDES have both a C=O str and an N-H str(unless 3°; not as broad as an acid) C=O str <1700, ~1690cm-1 p. 26 1o amide RCONH2 two NH peaks 2o amide RCONHR’ one NH peak 3o amide RCONR’R” no NH peak

  26. p. 27 CONJUGATION: alternating single and double bonds Conjugation of a carbonyl group LOWERS frequency by ~25-30 cm-1 oxygen is electronegative removes electrons thru bonds this minor resonance structure has a C—O single bond so it should be easier to stretch The resonance hybrid has a lower frequency

  27. p. 27 >1700 <1700

  28. p. 28 ACIDS >1700 saturated <1700 conjugated

  29. p. 28 ESTERS

  30. p. 29 Strain Generally, as you strain bonds, it becomes harder to stretch them.

  31. p. 29 conj on C=O LOWER n conj on O HIGHER n strain increases

  32. p. 30 ALCOHOLS & ETHERS If it has ONE “O” no C=O at 1700 no OH at 3400 must be an ether C-O Stretch Ar-OH 1220 R3C-OH 1150 R2CH-OH 1100 RCH2-OH 1050 in simple cases

  33. AMINES p. 31 3400 N-H stretch 1o R-NH2 - two peaks Amines often weak, amides strong (and no C=O) 2o R2NH - one peaks 3o R3N NO NH peak

  34. p. 32 NITRILES 2250 cm-1 [medium to strong] Alkynes 2150cm-1 [weak, sometimes absent]

  35. p. 33-34 SOLVING PROBLEMS: WHERE TO START? Acyclic saturated hydrocarbons have the formula CnH2n+2 so if the formula of your compound does not fit, it must have unsaturation (double bonds) or rings present When solving problems WE ALWAYS work out the number of double bond equivalents (DBE’s) FIRST

  36. p. 33 So, if your compound is a hydrocarbon: # DBE’s = {CnH2n+2 - your formula} / 2 n = number of carbon atoms in your formula e.g. if your unknown is C3H4 # DBE’s = {C3H8 – C3H4}/2 = 4/2 = 2 There are then formally two double bonds, two rings, or one of each all fit, but alkyne is most likely! alkynes count as two double bonds

  37. p. 34 Benzene has 3 double bonds + one ring so has 4 DBE’s If you have >6C’s and 4 or more DBE’s, HIGHLY likely to be aromatic

  38. p. 34 OTHER ELEMENTS: • If valency = 1 (e.g. halogens) count each halogen as a hydrogen • If valency = 2 (e.g. oxygen, sulfur) ignore in the count • If valency = 3 (e.g. nitrogen) add one to the CnH2n for each atom present so if one nitrogen use CnH2n+3, if two nitrogens use CnH2n+4 etc

  39. PROBLEM A, p. 35 C4H10O -OH str DBE’s = {C4H2n+2=10 – C4H10}/2 ignore oxygen rule 2 = zero what is main peak? so its an alcohol can we tell what type? PRIMARY 1042 cm-1 so

  40. C C--H PROBLEM B, p. 36 C3H3Br what is main peak? 3300 cm-1 sharp! additional hint 2150 cm-1 2 DBE’s - OK DBE = {C3H2n+2=8 – C3H3+1}/2 = 4/2 = 2

  41. PROBLEM B, p. 36 C3H3Br DBE = {C3H2n+2=8 – C3H3+1}/2 = 4/2 = 2 -C C--H 2 DBE’s - OK Subtract what we have found from what we started with: C3H3Br – C2H = CH2Br so we have CH2 and a Br but cannot join Br to alkyne so must be: H-CC-CH2Br

  42. PROBLEM C, p. 37 C5H8O = 1675 DBE = [(2x5 + 2)-8]/2 = 2 MAIN PEAK = conjugated carbonyl nothing at 2750, so KETONE KETONE or ALDEHYDE?

  43. PROBLEM C, p. 37 C5H8O DBE = 2 so conjugated ketone so what is other DBE? peak at 970 cm-1 suggests? trans-alkene C3H2O here C2H6 left to find

  44. p. 37 unique solution

  45. PROBLEM D, p. 38 C10H14 DBE = {(2 x10 + 2) – 14}/2 = 4 4DBE, >6C suggests? other functional groups? BENZENE ring NO what is substitution pattern? 829 cm-1 = PARA BUT 4 C’s still to find C1 C3 C2 C4

  46. PROBLEM E, p. 39 C7H9N DBE = {(2 x7 + 2 + 1) – 9}/2 = 4 add one for N BENZENE DERIV (6C, 4DBE) What is the N? PRIMARY AMINE R-NH2 so we still need to find one more carbon benzene carbon -NH2 but how joined?

  47. PROBLEM E, p. 39 C7H9N benzene carbon -NH2 but how joined? 750 cm-1 = ? ortho-benzene so only one choice:

  48. PROBLEM F, p. 40 C7H8O DBE = {(2 x7 + 2) – 8}/2 = 4 = BENZENE substitution pattern? 3 peaks = META (6C) What is the oxygen? broad, 3300 = -OH meta-BENZENE + carbon + OH

  49. PROBLEM G, p. 41 C9H10O <1700 Benzene DBE = {(2 x 9 + 2) – 10}/2 = 5 + 1 more DBE aldehyde or ketone? What is the oxygen? ~1700 =carbonyl (C=O) check 2750 none, so KETONE <1700 or >1700 <1700, so conjugated

  50. PROBLEM G, p. 41 C9H10O so benzene + conjugated ketone, but only 5DBE, so where is the ketone? 2 peaks, so mono-sub on the benzene ring: but what substitution? C9H10O – C7H5O = C2H5

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