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

5. Sample Problem #1 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 2 ( 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 #1 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 3 ( cm-1) C9H12 DBE = 4
Alkyne CC:
Nitrile CN:
Both absent – no peaks

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

8. Sample Problem #1 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 5 ( 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 #1 Step 4: C-H Skeleton from 1H-NMR
1H-NMR: 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
Shift Splitting Integral # of H Implications
ppm multiplet
2.57 ppm triplet 2
1.64 ppm sextet 2
0.94 ppm triplet 3

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

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

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

13. Sample Problem #1 Step 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 #1 Step 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 2 Lecture Supplement page 166+ +

16. Implications, Blind Men, and Elephants
An NMR Fable + + +
Ph- +
-CH2CH2- +
CH3CH2CH2- +
-CH3

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

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

22. Sample Problem #2 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 5 ( 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 #2 Step 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
Shift Splitting Integral # of H Implications
4.22 ppm triplet
3.59 ppm triplet
3.39 ppm singlet
2.09 ppm singlet

24. Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR
Step 4b: Divide hydrogens according to integrals
Shift Splitting Integral # of H Implications
4.22 ppm triplet
3.59 ppm triplet
3.39 ppm singlet
2.09 ppm singlet
2 H
3 H
Totals
 10 H

25. Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR
Step 4c: Combine splitting with number of hydrogens to get implications
Shift Splitting Integral # of H Implications
2 H
3 H
CH2 in CH2CH2
CH2 in CHCH2CH
2 x CH in CHCH2
2 x CH in CHCHCH
4.22 ppm triplet
3.59 ppm triplet
3.39 ppm singlet
2.09 ppm singlet
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
 10 H

26. Sample Problem #2 Step 4: C-H Skeleton from 1H-NMR
Step 4d: Select most likely implications, then total the atoms
Most likely implication = least number of atoms
Shift Splitting Integral # of H Implications
2 H
3 H
CH2 in CH2CH2
CH2 in CHCH2CH
2 x CH in CHCH2
2 x CH in CHCHCH
4.22 ppm triplet
3.59 ppm triplet
3.39 ppm singlet
2.09 ppm singlet
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
 10 H
CH2 + CH2 + CH3 + CH3 = C4H10

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

29. Sample Problem #2 Step 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 = ppm
Typical O=CCH2 chemical shift = ppm
More consistent with middle structure
Step 7: Check your work Left as a student exercise

30. Sample Problem #3 Steps 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 #3 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 1 ( 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 #3 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 2 ( 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 #3 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 3 ( cm-1) C4H9NO DBE = 1
CC:
CN:
Absent - no peaks; not enough DBE

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

35. Sample Problem #3 Step 3: Functional Groups from IR
100
Transmittance (%)
4000
3000
2000
1500
1000
Stretching frequency (cm-1)
Zone 5 ( 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 #3 Step 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
Shift Splitting Integral # of H Implications
3.67 ppm triplet
2.86 ppm triplet
2.59 ppm singlet

37. Sample Problem #3 Step 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
Shift Splitting Integral # of H Implications
3.67 ppm triplet
2.86 ppm triplet
2.59 ppm singlet
4 H
1 H
Total = 9.0
 H

38. Sample Problem #3 Step 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
Shift Splitting Integral # of H Implications
2 x CH2 in CH2CH2
2 x CH2 in CHCH2CH
4 x CH in CHCH2
4 x CH in CHCHCH
3.67 ppm triplet
2.86 ppm triplet
2.59 ppm singlet
4 H
1 H
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
 H

39. Sample Problem #3 Step 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
Shift Splitting Integral # of H Implications
2 x CH2 in CH2CH2
2 x CH2 in CHCH2CH
4 x CH in CHCH2
4 x CH in CHCHCH
3.67 ppm triplet
2.86 ppm triplet
2.59 ppm singlet
4 H
1 H
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
 H
(2 x CH2) + (2 x CH2) + NH = C4H9N

40. Sample Problem #3 Step 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 #3 Step 6: Assembly
Pieces
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: