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Solving Spectroscopy Problems Part 1 Lecture Supplement page 159

Solving Spectroscopy Problems Part 1 Lecture Supplement page 159. +. +. Solving Spectroscopy Problems. How to deduce structure from spectra?. Logical, orderly procedure Compare information between spectra when necessary Conservative analysis - do not discard possibilities until 100% sure

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Solving Spectroscopy Problems Part 1 Lecture Supplement page 159

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  1. Solving Spectroscopy Problems Part 1Lecture Supplement page 159 + +

  2. Solving Spectroscopy Problems How to deduce structure from spectra? • Logical, orderly procedure • Compare information between spectra when necessary • Conservative analysis - do not discard possibilities until 100% sure • Procedure • MS gives formula • Formula gives DBE • Use formula plus DBE to guide IR analysis of functional groups • NMR gives skeleton • Assemble the pieces • Check your work!

  3. Sample Problem #1Steps 1 and 2: Formula and DBE Mass spectrum m/z = 120 (M; 100%), m/z = 121 (9.8%), and m/z = 122 (0.42%) • Formula and DBE • Usual procedure reveals only one viable formula: C9H12 • DBE = 4 Four rings and/or pi bonds • Possible benzene ring

  4. Sample Problem #1Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C9H12 DBE = 4 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: All absent – no peaks

  5. Sample Problem #1Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C9H12 DBE = 4 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Present - peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - no peak ~2700 cm-1; no C=O in zone 4 Absent - not broad enough; no C=O in zone 4

  6. Sample Problem #1Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C9H12 DBE = 4 Alkyne CC: Nitrile CN: Both absent – no peaks

  7. Sample Problem #1Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C9H12 DBE = 4 C=O: Absent - no strong peak; no oxygen in formula

  8. Sample Problem #1Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C9H12 DBE = 4 Benzene ring C=C: Alkene C=C: Present - peaks at ~1610 cm-1 and ~1500 cm-1 Absent - not enough DBE for alkene plus benzene

  9. Sample Problem #1Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 7.40-7.02 ppm (multiplet; integral = 5), 2.57 ppm (triplet; integral = 2), 1.64 ppm (sextet; integral = 2), and 0.94 ppm (triplet; integral = 3). Step 4a: Copy NMR data to table ShiftSplittingIntegral# of HImplications 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3

  10. Sample Problem #1Step 4: C-H Skeleton from 1H-NMR Step 4b: Divide hydrogens according to integrals ShiftSplittingIntegral# of HImplications 5 H 2 H 2 H 3 H 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 Totals 12  12 H

  11. Sample Problem #1Step 4: C-H Skeleton from 1H-NMR Step 4c: Combine splitting with number of hydrogens to get implications ShiftSplittingIntegral# of HImplications 5 H 2 H 2 H 3 H C6H5 (Ph; monosubstituted benzene ring) 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH CH2 or 2 x CH two neighbors CH2 in CH2CH5 2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in (CH2)2CHCH CH2 or 2 x CH CH2 in CH3CH2CH2 five neighbors CH3 in CH3CH2 3 x CH in CHCH2 3 x CH in CHCHCH two neighbors CH3 or 3 x CH Totals 12  12 H

  12. Sample Problem #1Step 4: C-H Skeleton from 1H-NMR Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms ShiftSplittingIntegral# of HImplications 5 H 2 H 2 H 3 H C6H5 (Ph; monosubstituted benzene ring) 7.40-7.02 ppm multiplet 5 2.57 ppm triplet 2 1.64 ppm sextet 2 0.94 ppm triplet 3 CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH CH2 or 2 x CH two neighbors CH2 in CH2CH5 CH2 in CH3CH2CH2 2 x CH in CH3CHCH2 2 x CH in (CH)2CHCH3 2 x CH in (CH2)2CHCH CH2 or 2 x CH five neighbors CH3 in CH3CH2 3 x CH in CHCH2 3 x CH in CHCHCH two neighbors CH3 or 3 x CH Totals 12  12 H C6H5 + CH2 + CH2 + CH3 = C9H12

  13. Sample Problem #1Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C9H12 - C9H12 (from 1H-NMR) = all atoms used DBE Check DBE from formula - DBE used = DBE left over 4 - 4 (benzene ring) = all DBE used

  14. Sample Problem #1Step 6: Assembly Step 6: Now to put it all together... Ph CH2 of CH2CH2 CH2 of CH2CH2CH3 CH3 of CH3CH2 Pieces: CH2CH2CH3 • Step 7: Check your work! • Formula • Functional groups • Number of signals • Splitting • How to assemble? • Pay attention to splitting patterns • All pieces form one molecule • Trial and error • Pentavalent carbons XXXXXXXXXXXXXXX • Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice Practice

  15. Solving Spectroscopy Problems Part 2Lecture Supplement page 166 + +

  16. Implications, Blind Men, and Elephants An NMR Fable + + + Ph- + -CH2CH2- + CH3CH2CH2- + -CH3 http://en.wikipedia.org/wiki/Blind_men_and_an_elephant

  17. Sample Problem #2Steps 1 and 2: Formula and DBE Mass spectrum m/z = 118 (M; 100%) Molecular mass (lowest mass isotopes) = 118 Even number of nitrogen atoms 5.7% / 1.1% = 5.2 m/z = 119 (5.7%) Five carbon atoms m/z = 120 (0.63%) No sulfur, chlorine, or bromine Formula 118 - (5 x 12) = 58 amu for oxygen, nitrogen, and hydrogen • Usual procedure gives three possible formulas: • C5H10O3 DBE = 1 One ring or one pi bond • C5H14N2O Rejected: 14H does not fit NMR integration (sum of integrals = 5.0) • C5H2N4 Rejected: More than two signals in NMR; no oxygen for C=O (IR zone 4)

  18. Sample Problem #2Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C5H10O3 DBE = 1 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: Absent - no strong peak Absent - no peak; no nitrogen in formula Absent - no peak; no CC peak in zone 3

  19. Sample Problem #2Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C5H10O3 DBE = 1 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Absent - no peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - no peak ~2700 cm-1 Absent - not broad enough

  20. Sample Problem #2Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C5H10O3 DBE = 1 Absent - no peak; not enough DBE Absent - no peak; no nitrogen in formula Alkyne CC: Nitrile CN:

  21. Sample Problem #2Step 3: Functional Groups from IR 100 Transmittance (%) 1741 cm-1 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C5H10O3 DBE = 1 Present C=O: 1741 cm-1 = ester (1750-1735 cm-1) ketone (1750-1705 cm-1) aldehyde (1740-1720 cm-1)

  22. Sample Problem #2Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C5H10O3 DBE = 1 Benzene ring C=C: Alkene C=C: Absent - no peak; not enough DBE Absent - no peak; not enough DBE for C=O plus alkene

  23. Sample Problem #2Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 4.22 ppm (triplet; integral = 1.0), 3.59 ppm (triplet; integral = 1.0), 3.39 ppm (singlet; integral = 1.5), 2.09 ppm (singlet; integral = 1.5) Step 4a: Copy NMR data to table ShiftSplittingIntegral# of HImplications 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5

  24. Sample Problem #2Step 4: C-H Skeleton from 1H-NMR Step 4b: Divide hydrogens according to integrals ShiftSplittingIntegral# of HImplications 2 H 2 H 3 H 3 H 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 Totals 5  10 H

  25. Sample Problem #2Step 4: C-H Skeleton from 1H-NMR Step 4c: Combine splitting with number of hydrogens to get implications ShiftSplittingIntegral# of HImplications CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 2 H 2 H 3 H 3 H 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 two neighbors CH2 or 2 x CH CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH Totals 5  10 H

  26. Sample Problem #2Step 4: C-H Skeleton from 1H-NMR Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms ShiftSplittingIntegral# of HImplications CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH 2 H 2 H 3 H 3 H 4.22 ppm triplet 1.0 3.59 ppm triplet 1.0 3.39 ppm singlet 1.5 2.09 ppm singlet 1.5 two neighbors CH2 or 2 x CH CH2 in CH2CH2 CH2 in CHCH2CH 2 x CH in CHCH2 2 x CH in CHCHCH two neighbors CH2 or 2 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH CH3 or 3 x CH no neighbors CH3 or 3 x CH Totals 5  10 H CH2 + CH2 + CH3 + CH3 = C4H10

  27. Sample Problem #2Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C5H10O3 - C4H10 (from 1H-NMR) - C=O (ester or ketone from IR) = 2 O DBE Check DBE from formula - DBE used = DBE left over 1 - 1 (C=O) = all DBE used

  28. CH2CH2 CH3C=O CH3O   Typically 3.8 ppm Typically 2.0-2.3 ppm Sample Problem #2Step 6: Assembly Pieces CH2 in CH2CH2 CH2 in CH2CH2 CH3 CH3 C=O (ester or ketone) O O • Assembly • Coupling suggests 2 x CH2 join to form CH2CH2 • CH3 (singlet in NMR) cannot be in CH3CH2CH2 • Leaves CH3O and CH3C=O 3.39 ppm observed shift 2.09 ppm observed shift

  29. Sample Problem #2Step 6: Assembly Pieces CH2CH2 CH3O CH3C=O (ester or ketone) O Three ways to assemble these pieces: X Does not use all pieces • Observed CH2 chemical shifts = 4.22 and 3.59 ppm • Typical OCH2 chemical shift = 3.6 - 4.6 ppm • Typical O=CCH2 chemical shift = 2.2 - 3.0 ppm • More consistent with middle structure Step 7: Check your workLeft as a student exercise

  30. Sample Problem #3Steps 1 and 2: Formula and DBE Mass spectrum m/z = 87 (M; 100%) m/z = 88 (4.90%) m/z = 89 (0.22%) Molecular mass (lowest mass isotopes) = 87 Odd number of nitrogen atoms 4.9% / 1.1% = 4.5 Four or five carbon atoms No sulfur, chlorine, or bromine • Formula • Usual procedure gives two formula candidates: • C4H9NO DBE = 1 One ring or one pi bond • C5H13N Rejected: Does not fit 1H-NMR integration (4:4:1 Integral sum = 9)

  31. Sample Problem #3Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 1 (3700-3200 cm-1) C4H9NO DBE = 1 Alcohol O-H: Amide/amine N-H: Terminal alkyne C-H: Present? - weaker than usual Present? Absent - not enough DBE; no CC in zone 3

  32. Sample Problem #3Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 2 (3200-2700 cm-1) C4H9NO DBE = 1 Aryl/vinyl sp2 C-H: Alkyl sp3 C-H: Aldehyde C-H: Carboxylic acid O-H: Absent - no peaks > 3000 cm-1 Present - peaks < 3000 cm-1 Absent - ~2700 cm-1 present but no C=O in zone 4 Absent - no C=O in zone 4; not broad enough

  33. Sample Problem #3Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 3 (2300-2000 cm-1) C4H9NO DBE = 1 CC: CN: Absent - no peaks; not enough DBE

  34. Sample Problem #3Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 4 (1850-1650 cm-1) C4H9NO DBE = 1 C=O: Absent - no peaks

  35. Sample Problem #3Step 3: Functional Groups from IR 100 Transmittance (%) 0 3000 1500 4000 2000 1000 Stretching frequency (cm-1) Zone 5 (1680-1450 cm-1) C4H9NO DBE = 1 Benzene ring: Alkene C=C: Absent - no peak ~1600 cm-1; not enough DBE Absent - no peak ~1600 cm-1

  36. Sample Problem #3Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4a: Copy NMR data to table ShiftSplittingIntegral# of HImplications 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0

  37. Sample Problem #3Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4b: Divide hydrogens according to integrals ShiftSplittingIntegral# of HImplications 4 H 4 H 1 H 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 Total = 9.0  9 H

  38. Sample Problem #3Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4c: Combine splitting with number of hydrogens to get implications ShiftSplittingIntegral# of HImplications 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH 4 H 4 H 1 H 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 two neighbors 2 x CH2 or 4 x CH 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH two neighbors 2 x CH2 or 4 x CH CH or NH or OH no neighbors Total = 9.0  9 H

  39. Sample Problem #3Step 4: C-H Skeleton from 1H-NMR 1H-NMR: 3.67 ppm (triplet; integral = 4.0), 2.86 ppm (triplet; integral = 4.0), and 2.59 ppm (singlet; integral = 1.0). Step 4d: Select most likely implications, then total the atoms Most likely implication = least number of atoms ShiftSplittingIntegral# of HImplications 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH 4 H 4 H 1 H 3.67 ppm triplet 4.0 2.86 ppm triplet 4.0 2.59 ppm singlet 1.0 two neighbors 2 x CH2 or 4 x CH 2 x CH2 in CH2CH2 2 x CH2 in CHCH2CH 4 x CH in CHCH2 4 x CH in CHCHCH two neighbors 2 x CH2 or 4 x CH X CH or NH or OH IR more consistent with NH than OH no neighbors Total = 9.0  9 H (2 x CH2) + (2 x CH2) + NH = C4H9N

  40. Sample Problem #3Step 5: Checks Step 5: Check to see that all atoms and DBE are used Atom Check Formula - atoms used = atoms left over C4H9NO - C4H9N (from 1H-NMR) = one oxygen Not part of a functional group that appears in IR or 1H-NMR: An ether DBE Check DBE from formula - DBE used = DBE left over 1 - 0 = one DBE C=O, C=C, C=N absent in IR Therefore DBE = ring

  41. Sample Problem #3Step 6: Assembly Pieces 2 x CH2 in CH2CH2 2 x CH2 in CH2CH2 NH O (ether) one ring 2 x CH2CH2 • Assembly • Splitting pattern requires two sets of 2 x CH2 to become two equivalent CH2CH2 • Remaining pieces can only be assembled in one way:

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