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Problem O, C 5 H 10 O 2

p. 123. Problem O, C 5 H 10 O 2. DBE = {[(2x5)+2]-10}/2 = 1. 2 ‘O’ and IR peak at 1740 suggests:. ester = 1DBE. C 5 H 10 O 2. Integration ratios: 2 : 3 : 2 : 3 =10. d 4 - what can shift a CH 2 down to 4?. only –O-. p. 123.

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Problem O, C 5 H 10 O 2

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  1. p. 123 Problem O, C5H10O2 DBE = {[(2x5)+2]-10}/2 = 1 2 ‘O’ and IR peak at 1740 suggests: ester = 1DBE

  2. C5H10O2 Integration ratios: 2 : 3 : 2 : 3 =10 d 4 - what can shift a CH2 down to 4? only –O- p. 123

  3. 2 : 3 : 2 : 3 C5H10O2 but this is a triplet so it is next to another CH2 but this is a sextet p. 123

  4. 2 : 3 : 2 : 3 C5H10O2 but this is a sextet so it is next to 6-1 = 5 H’s only one peak left a – CH3 !! p. 123

  5. 2 : 3 : 2 : 3 C5H10O2 could have started here always check to see shifts are reasonable 1 1.3+0.5 3.5+0.5 2.3 p. 123

  6. p. 124 Problem P C6H12O DBE = {[(2x6)+2]-12}/2 = 1 1 ‘O’ and IR peak at 1720 suggests: >C=O NO 2750 peak, so probably ketone what sort of carbonyl? how do we verify this – look in 1H NMR for –CHO at d 10

  7. p. 124 Problem P C6H12O no d 10, so must be a ketone. start here Integrations: 2:3:2:2:3 = 12 d 1 = -CH3 – a triplet so next to a –CH2- , so we have a –CH2CH3 we now have to decide is this the triplet , quintet or the sextet, which are all –CH2- ????

  8. p. 124 Problem P C6H12O Integrations: 2:3: 2:2:3 = 12 ?-CH2CH3 try different approach – only functional group is ketone -CHx-CO-CHY- H’s next to a ketone are in 2-3 region so we have a –CH2- and a –CH3 so we now know we have CH3-CO-CH2-

  9. p. 124 Problem P C6H12O Integrations: 2:3:2:2:3 = 12 ?-CH2CH3 so we now know we have CH3-CO-CH2- triplet, so next to –CH2- but don’t forget we have already found a –CH2-CH3 CH3-CO-CH2-CH2- so now we have found all six carbons!!

  10. p. 124 Problem P C6H12O Integrations: 2:3:2:2:3 = 12 -CH2-CH3 so they must be joined! CH3-CO-CH2-CH2- CH3-CO-CH2-CH2-CH2-CH3now check splittings, and d s t 5 6 t <= expect d 2.3 2.5 1.3 0.9 <= expect between 2 and 1.3

  11. p. 125 Problem Q C5H10O2 DBE = {[(2x5)+2]-10}/2 = 1 2 ‘O’ and IR peak at 1740 suggests: ester = 1DBE

  12. p. 125 Problem Q C5H10O2 Integrations: 3 : 2 : 2 : 3 = 10H ester most downfield peak will be on –O- so CH2 next to a carbonyl is 2-3, so

  13. p. 125 Problem Q C5H10O2 Integrations: 3 : 2 : 2 : 3 = 10H <= triplet, so next to –CH2- <= sextet, so next to 5H 2 + 3

  14. p. 125 Problem Q C5H10O2 Integrations: 3 : 2 : 2 : 3 = 10H check data

  15. p. 126 Problem R C5H12O DBE = {[(2x5)+2]-12}/2 = 0 1 ‘O’ but no –OH or C=O so ether – there is an 1100 C-O str

  16. p. 126 Problem R C5H12O Integrations2:3: 2:2:3 = 12H ether = C-O-C so most downfield H’s next to O CH3-O = singlet CH2 is triplet so next to CH2 most likely this is the next peak so CH3OCH2CH2- but to be sure, we can now go to the end

  17. p. 126 Problem R C5H12O Integrations2:3: 2:2:3= 12H so CH3OCH2CH2- CH3 triplet so next to -CH2- = -CH2CH3 but we have now found all 5 carbon atoms so CH3OCH2CH2-CH2CH3 s t 5 6 t <=check !!!

  18. p. 127 Problem S C8H9BrO DBE = {[(2x8)+2]-(9+1)}/2 = 4 > 6C, d 7 so benzene para (maybe) functional group = -OH

  19. p. 127 Problem S C8H9BrO Integrations: 2:2 : 1 : 1 : 3 = 9 -OH, -Br, benzene (para, maybe) confirmed para-benzene = C6H4 so C8H8-C6H4 left to find = C2H4 dd OH do not couple

  20. p. 127 Problem S C8H9BrO Integrations: 2:2 : 1 : 1 : 3 = 9 para-benzene = C6H4 so C8H8-C6H4 left to find = C2H4 -OH, -Br, we have a CH (q next to 3) and a CH3 (d next to 1) so >CHCH3

  21. Problem S C8H9BrO Integrations: 2:2 : 1 : 1 : 3 = 9 -CHCH3 -OH,-Br,

  22. p. 127 Problem S C8H9BrO normal d ~7.3 shielded d 6.8 as well, Ar-OH d 5-10 and Ar-OH has IR 1220 not at around 1100 correct

  23. p. 128 Problem T C11H12 >6C, d 7 so benzene DBE = {[(2x11)+2]-(12)}/2 = 6 two bands, so MONO 2200 = ? = 6 DBE

  24. p. 128 Problem T C11H12 Integrations5: 2 : 2 : 3 = 12 these are all coupled so not singlet

  25. p. 128 Problem T C11H12 Integrations 5: 2 : 2 : 3 = 12 these are all coupled so not singlet

  26. p. 128 Problem T C11H12 Integrations5: 2 : 2 : 3 = 12 start where we know! CH3 is a triplet, so is next to a CH2, which is a sextet so must have 5 neighbors, 3 of which are CH3, so other 2 must be –CH2

  27. p. 128 Problem T C11H12 Integrations 5: 2 : 2 : 3 = 12 CH2---CH2---CH3 t 6 t

  28. p. 129 Problem U C8H10ClN ? DBE = {[(2x8)+2+1]-(10+1)}/2 = 4 > 6 C, d 7 so benzene so use NMR Functional group -NH2

  29. p. 129 Problem U C8H10ClN Note the ‘leaning’ shape Integrations: 2:2 2:2 2 = 10 -Cl, -NH2 para-benzene = C6H4 should be singlet, no coupling we still need to find C8H10ClN – C6H6ClN = C2H4 -CH2-CH2-

  30. p. 129 Problem U C8H10ClN Integrations: 2:2 2:2 2 = 10 -Cl, -NH2 -CH2-CH2- always then join bifunctional or higher groups together

  31. p. 129 Problem U C8H10ClN Integrations: 2:2 2:2 2 = 10 -Cl, -NH2 now, 2 choices

  32. p. 129 Problem U C8H10ClN Integrations: 2:2 2:2 2 = 10 at least d 3.5 d 7-7.3 shielded to 6.8 correct

  33. p. 130 Problem V C9H9N ? DBE = {[(2x9)+2+1]-(9)}/2 = 6 > 6C, d 7 so benzene maybe mono? Functional Group? -CN Nitrile and benzene account for all DBE

  34. p. 130 Problem V C9H9N =9H Integrations: 5 : 1 : 3 = 9 + mono-subs-benzene C9H9N – (C6H5 + CN) = C2H4 so >CHCH3 q d

  35. p. 130 Problem V C9H9N =9H Integrations: 5 : 1 : 3 = 9 >CHCH3

  36. p. 131 Problem W C10H12O >6 C, d 7 so benzene DBE = {[(2x10)+2]-12)}/2 = 5 probably mono-subs Functional Group ? ketone – confirm in NMR with no d 10 What sort of ketone? >1700, satd, so? 8

  37. p. 131 Problem W C10H12O Integrations: 5 : 2 : 2 : 3 = 12 X satd ketone, not next to benzene ring CH3 triplet so next to CH2 –which?

  38. p. 131 Problem W C10H12O Integrations: 5 : 2 : 2 : 3 = 12 -CH2CH3 next to anything else? C6C3CC= C11 !!! d = 2-3, so on carbonyl

  39. p. 131 Problem W C10H12O Integrations: 5 : 2 : 2 : 3 = 12 d? 2.5 1

  40. p. 132 Problem X C9H10O ? > 6 C, d 7 so benzene DBE = {[(2x9)+2]-10)}/2 = 5 Functional group ? >C=O 2750, so –CHO confirm in NMR

  41. p. 132 Problem X C9H10O 1 : 2 : 2 : 2 : 3 = 10 confirms para-benzene ring d 10 confirms CHO C6H4 Have found C7H5O, so C2H5 left to find -CH2-CH3 qt

  42. p. 132 Problem X C9H10O 1 : 2 : 2 : 2 : 3 = 10 -CH2-CH3 OHC- d 7.8

  43. Problem Y C10H12O2 p. 133 > 6 C, d 7 so benzene DBE = {[(2x10)+2]-12)}/2 = 5 mono subst Functional group = ester can we tell what type of ester??? 1740, so saturated both sides!!!

  44. p. 133 Problem Y C10H12O2 Integrations 5 : 3 : 2:2 = 12H } C7H5O2 so C3H7 left to find satd CH3 CH2 CH2 coupled d 3.5 so on O

  45. p. 133 Problem Y C10H12O2 Integrations 5 : 3 : 2:2 = 12H

  46. p. 134 Problem Z C11H14O2 > 6 C, d 7 so benzene DBE = {[(2x11)+2]-14)}/2 = 5 ?-subst Functional group = acid type of acid??? <1700, so conjugated (to what) benzene ring

  47. p. 134 Problem Z C11H14O2 Integrations 2:2 : 9 = 13 -COOH is at d 10-16 not shown in above spectrum Benzene is para conjugated C4H9 C7H5O2

  48. p. 134 Problem Z C11H14O2 Integrations 2:2 : 9 = 13 (CH3)3 identical C4H9 -C(CH3)3 9H ~ d 1

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