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13 C NMR SPECTRA

p. 149. 13 C NMR SPECTRA. 12 C: I = 0 13 C: I = ½ 99% 1.1% not active + low sensitivity relative to 1 H Therefore many scans (FT methods). 13 C H 3 13 C H 2 Br 1% x 1% = .01% chance of being in same molecule. SO WE DO NOT SEE COUPLING BETWEEN ADJACENT CARBONS.

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13 C NMR SPECTRA

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  1. p. 149 13C NMR SPECTRA 12C: I = 013C: I = ½ 99%1.1% not active+ low sensitivity relative to 1H Therefore many scans (FT methods) 13CH313CH2Br 1% x 1% = .01% chance of being in same molecule SO WE DO NOT SEE COUPLING BETWEEN ADJACENT CARBONS BUT DO SEE COUPLING TO HYDROGENS 13CH3 is a 1Jso is large, 125-250 Hz, see a quartet however because 13C is only 1% does not affect 1H spec

  2. p. 150 see coupling suppress coupling

  3. p. 150 d(C) ~ 15 x d(H) d scale now 200 ppm

  4. p. 150 We usually run proton decoupled spectra: NOE, nuclear Overhauser enhancement effect transfers magnetization from H to directly bonded C greater signal strength for C chemically different carbons usually have different d so # PEAKS = # of CHEMICALLY DIFFERENT CARBONS -CH3 >CH2 -CH >C< } only 4 types q t d s However if we want to know this coupling information we need to run a separate experiment

  5. p. 150 We usually run DEPT-135 experiments Complicated pulse sequence but the NET EFFECT is simple and useful: peaks appearance governed by multiplicity } only 4 types -CH3 >CH2 -CH >C< q t d s +ve -ve +ve gone This is much faster than simply turning off the decoupler because the NOE effect still applies to DEPT

  6. A typical DEPT-135 spectrum: note phasing – CH2 are down and CH or CH3 are up

  7. p. 151 Proton decoupled 13C spectra : see only singlets CH3-CH2-CHBr-CH3 12.1 (q) CH3 26.0 (q) CH3 34.2 (t) CH2 53.1(d) CH

  8. p. 151 Chemical Shifts CH3 >C=O -COO C=C C-O

  9. p. 152 ALKANES 10-50 ppm CH3---CH2----CH2----CH2---- 1423 ~30 CH3CH2CH2CH2CH2CH2CH3 Integrations are NOT reliable in C NMR relaxation times are longer and more variable (if we wanted them to be reliable the experiment would take a very long time to do!!) Peak height is roughly correlated with number of attached H NOT so much with number of C

  10. p. 152 ALKENES 100-160 ppm CH2=CH-- ~115~140 more substituted, more downfield 14 23 30

  11. p. 152 ALKYNES 65-95 ppm shielded 8568 sd 18, shielded from 30 4oC, so weak

  12. p. 153 AROMATICS X most X deshield this ipso C by ~10 ppm so at ~ 140 ppm 130 d128.5 D: donor groups (OR, NR) deshield this carbon to ~ 155-165 and shield the ortho carbons to ~ 115 pm Aromatic shifts to remember: 160140130115

  13. Full data is in table, yellow pages, Appendix p. A8 Shifts are additive

  14. OCH3 31.4 -14.4 1.0 -7.7 C = 128.5 + 31.4 + 1.0 = 160.9 p.153

  15. OCH3 31.4 -14.4 1.0 -7.7 C = 128.5 + 31.4 + 1.0 = 160.9 C = 128.5 -14.4 -14.4 = 99.7 p.153

  16. p.153 OCH3 31.4 -14.4 1.0 -7.7 C = 128.5 + 31.4 + 1.0 = 160.9 C = 128.5 -14.4 -14.4 = 99.7 C = 128.5 -14.4 -7.7 = 106.4 C = 128.5 +1 +1 = 130.5

  17. p. 154 Patterns:

  18. p. 155 CARBONYL COMPOUNDS -CHO d d 190-210 >C=O s d 200-220 ACID DERIVS -COOH, -COOR, -CON<, -COO- all are singlets, and all at d 160-180 211 27

  19. p. 156 Problem AA C5H11Cl DBE = {(2x5 + 2)-(11 + 1) = 0 only 4 lines in carbon spectrum, so two C’s identical 43t 42t 26d 22q -CH3 -CH2- -CH2- >CH- Now add these up = C4H8 – C5H11 = CH3 missing

  20. p. 156 Problem AA C5H11Cl -Cl -CH3 -CH3 -CH2- -CH2- >CH- The 26d is not down field enough to hold -Cl so –Cl must be on a –CH2- which now becomes –CH2Cl

  21. p. 156 Problem AA C5H11Cl so –Cl must be on a –CH2- which now becomes –CH2Cl -CH3 -CH3 -CH2Cl join bi or higher groups -CH2- >CH- -CH2-CH<

  22. p. 156 Problem AA C5H11Cl -CH3 -CH3 -CH2Cl -CH2-CH< now add rest, BUT –CH3’s identical ClCH2-CH2-CH(CH3)2 Are you happy with the chemical shifts?

  23. p. 157 Problem AB C4H8O2 171s60t 21q 14q 1 DBE -CH2- -CH3 -CH3 >C< acid (deriv) on -O- Now add up to check C4H8 all H’s found, so deriv is an CH3-CH2OCOCH3 d14 ESTER -CH2OCO-

  24. p. 158 Problem AC C6H8O 3DBE 193d152d 142d 131d 130d19q }C6H8O ALKENE C’s =CH =CH =CH =CH -CH3 -CHO put ½ ‘s together -CH=CH--CH=CH- these must be joined together -CH=CH-CH=CH- CH3-CH=CH-CH=CH-CHO

  25. p. 159 Problem AD C9H12 4DBE = aromatic 137s 127d21q Highly symmetric =C< =CH--CH3 benzene has 6C so C9 total so must be 3 (CH3) identical =9H so must be 12H – 9H = 3H on ring so answer is a trimethyl-benzene which? mesitylene

  26. p. 160 Problem AE C9H10O 5DBE, aromatic + 1 200s137s 133d 129d 128d 32t 8q benzene ring -CH2- -CH3 >C=O how do we tell what type of benzene? 1] pattern, 1s+3d=mono 2] add up rest and subtract C9H10O – C3H5O = C6H5 = mono

  27. p. 160 Problem AE C9H10O Ph- >C=O -CH2- -CH3 -CH2CO- join bifunctionals so is PhCH2COCH3 or CH3CH2COPh

  28. p. 160 Problem AE C9H10O PhCH2COCH3 or CH3CH2COPh X X Also, only the CH2 is far enough downfield to be next to C=O

  29. p. 161 Problem AF C5H8O2 2DBE 166s130d 129t 60t 14q =CH =CH2 -OCH2- -CH3 C5H8O2{ -COO- CH3- -CH=CH2-COOCH2- CH3COOCH2CH=CH2 or CH3CH2OCOCH=CH2

  30. E1, p. 162 C6H10O 2DBE 195d130s 115t42d 15q =C< =CH2 >CH- -CH3 }C5H7O -CHO C6H10O >C=CH2 x2 CH3 so we have (CH3)2 -CHO to go on the joined other units >CH-C=CH2 IR: 2750 1695 950 900 3050 identical! (CH3)2CH-C=CH2 CHO

  31. E2, p. 162C4H9N DBE = [(2x4 +2+1)-9]/2 =1 IR (cm-1): no bands at 3400, 2200 or 1700 cm-1 13C NMR: d 50 t, 45q, 26t NCH3 -CH2- -CH2N Functional group = 30 amine = R3N

  32. E2, p. 162C4H9N DBE=1 R3N 50t 45q 26t } C3H7 -CH2- -CH2N< >N-CH3 C4H9 ? CH2 how many C’s on N ? 3, so must be –CH2N -CH2- DBE=1

  33. E3, p. 163C6H8O 3DBE Find functional groups from IR -OH 3300+ 2100 ?

  34. E3, p. 163C6H8O C6H8O 3DBE -OH 137up 119x86x 75up59down 17up -OCH2 -CH3 }C6H7O =CH- =C< HOCH2- CH3- >C=CH-

  35. p. 163 HOCH2- CH3- >C=CH- so what is next to the =CH ? if t, it is –CH2- or

  36. E4, p. 164C10H15NO2 DBE= 4 so aromatic IR (cm-1): 3400 m, 3300 m IR = ? -NH2 O ? not >C=O

  37. E4, p. 164 C10H15NO2 -NH2 13C NMR: 161s 142s 104d 97d55q 42t 38t benzene C6 -OCH3 -CH2- -CH2- }C9

  38. E4, p. 164 C10H15NO2 -NH2 13C NMR: 161s 142s 104d 97d55q 42t 38t benzene C6 -OCH3 -CH2- -CH2- }C9 coupled 1H NMR: 6.5s (2H) 6.3s (1H) 3.8s (6H) 2.9t (2H) 2.7t (2H) 1.1 br s (2H) exchanges with D2O [-OCH3]2 identical

  39. E4, p. 164 C10H15NO2 -NH2 [-OCH3]2 -CH2CH2- -NH2 to be identical, have to be on ring but that means

  40. E4, p. 164 C10H15NO2 -NH2 = shielded = very shielded 13C: 161s 142s 104d97d 55q 42t 38t 1H: all 3H are shielded, 1H at d 6.3, 2H at d 6.5

  41. E5, p. 165C13H14O DBE = 7 4 of these aromatic 1H NMR: d 7.8d (2H), 7.6d (2H), 2.9t (2H), 2.5s (1H), 1.7 pentet (2H), 1.4 sextet (2H), 0.9t (3H) para-benzene C6H4 IR: 825 3300, 21501680 conjugated ketone

  42. E5, p. 165 C13H14O = 7DBE conj >C=O 13C NMR 199s136s 132d 129d 127s82s 79d 38t 26t 23t 14q -CH2- -CH2- -CH2- -CH3 } C13H14O

  43. E5, p. 165C13H14O DBE = 7 4 of these aromatic 1H NMR: d 7.8d (2H), 7.6d (2H), 2.9t (2H), 2.5s (1H), 1.7 pentet (2H), 1.4 sextet (2H), 0.9t (3H) conj >C=O t 5 6 t -CH2- -CH2- -CH2- -CH3 CH3-CH2- -CH2-CH2- so CH3CH2CH2CH2- t 6 5 t t t 6 (or 5) 5(or6)

  44. E5, p. 165C13H14O DBE = 7 4 of these aromatic 1H NMR: d 7.8d (2H), 7.6d (2H), 2.9t (2H), 2.5s (1H), 1.7 pentet (2H), 1.4 sextet (2H), 0.9t (3H) conj >C=O CH3CH2CH2CH2- end groups!

  45. E5, p. 165C13H14O DBE = 7 4 of these aromatic 1H NMR: d 7.8d (2H), 7.6d (2H), 2.9t (2H), 2.5s (1H), 1.7 pentet (2H), 1.4 sextet (2H), 0.9t (3H) Predicted = 126.6 Predicted = 148.3 In C NMR one aromatic peak is at d 127(s)

  46. DBE=5, so aromatic +1 E6, p. 166 C9H10O4 conj-COOH IR 3400-2400(br) 1680 = 167x 156x 130up 114x 104up 56up -OCH3 -COOH benzene x2 (identical) but which isomer?

  47. E6, p. 166 C9H10O4 shielded 103s d13C d1H

  48. ASSIGNMENT 6

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