1. Solving Unknown Structures Using NMR
Organic Structure Analysis, Crews, Rodriguez and Jaspars

2. Six Simple Steps for Successful Structure Solution
Get molecular formula. Use combustion analysis, mass spectrum and/or 13C NMR spectrum. Calculate double bond equivalents.
Determine functional groups from IR, 1H and 13C NMR
Compare 1H integrals to number of H’s in the MF.
Determine coupling constants (J’s) for all multiplets.
Use information from 3. and 4. to construct spin systems (substructures)
Assemble substructures in all possible ways, taking account of dbe and functional groups. Make sure the integrals and coupling patterns agree with the proposed structure. 
SOLUTION
Organic Structure Analysis, Crews, Rodriguez and Jaspars

3. USING 1H NMR DATA UNKNOWN B
A compound shows an M+. in the EIMS at 154 m/z
Further fragments are at 121, 93, 71, 55 and 39 m/z
The IR shows bands at 3400 cm-1 (broad) & 1450 cm-1
Use the 1H and 13C data to determine the structure of the
compound
Organic Structure Analysis, Crews, Rodriguez and Jaspars

4. 13C NMR DATA UNKNOWN B t t t q q d d q s s A B C D E F GHI J
Organic Structure Analysis, Crews, Rodriguez and Jaspars

5. MOLECULAR FORMULA DETERMINATION
UNKNOWN B
s d t q B E A C D F I G H J
(C)2 + (CH)2 + (CH2)3 + (CH3)3 = C10H17
= 137 Da
The M+. appears at 154 m/z, so there is
a mass difference of 17 Da (= OH)
Therefore molecular formula = C10H18O (2 dbe)
Organic Structure Analysis, Crews, Rodriguez and Jaspars

6. 1H NMR DATA UNKNOWN B Integrals: ONLY 17 H! 5H 3H 3H 3H 2H H
Organic Structure Analysis, Crews, Rodriguez and Jaspars

7. SUBSTRUCTURES UNKNOWN B MF = C10H18O ! dC 125 d, 132 s dH 5.05 t &
Me groups at dH 1.5
dC 112 t, 146 d
dH 5.00 dd, 5.15 dd, 5.85 dd
dC 73 s
dH 1.05 s
MF = C10H18O !
Organic Structure Analysis, Crews, Rodriguez and Jaspars

8. WORKING STRUCTURES UNKNOWN B A – Ha should be ddt B – Ha should be ddt
C – Ha should be dd
Organic Structure Analysis, Crews, Rodriguez and Jaspars

9. MASS SPECTRAL FRAGMENTATION
UNKNOWN B
Fragments at:
121, 93, 71, 55 and 39 m/z
Organic Structure Analysis, Crews, Rodriguez and Jaspars

10. USING MASS SPECTRAL DATA
UNKNOWN G
A compound shows an M+. in the EIMS at 128 m/z
Further fragments are at 99, 83, 72 and 57 m/z
The IR shows bands at 1680 cm-1 (strong) &
bands at cm-1
Use the 1H and 13C NMR and MS data to determine
the structure of the compound
Organic Structure Analysis, Crews, Rodriguez and Jaspars

11. 13C NMR DATA UNKNOWN G t t q d d d t A B C D E FG
Organic Structure Analysis, Crews, Rodriguez and Jaspars

12. MOLECULAR FORMULA DETERMINATION
UNKNOWN G
s d t q A B C D E F G
(C)0 + (CH)3 + (CH2)3 + (CH3)1 = C7H12
= 96 Da
The M+. appears at 128 m/z, so there is
a mass difference of 32 Da (= O2)
Therefore molecular formula = C7H12O2 (2 dbe)
Organic Structure Analysis, Crews, Rodriguez and Jaspars

13. 1H NMR DATA UNKNOWN G Integrals: 12 H Total 3H 2H H H H H H H H
Organic Structure Analysis, Crews, Rodriguez and Jaspars

14. SUBSTRUCTURES UNKNOWN G MF = C7H12O2 !
dC 140 d, 101 d
dH 6.15 d, 4.70 m
dC 96 d
dH 4.9 t
dC 64 t
4 Oxygens in substructures but only 2 in MF
MF = C7H12O2 !
Organic Structure Analysis, Crews, Rodriguez and Jaspars

15. WORKING STRUCTURES UNKNOWN G 13C Shift additivity data
MS Fragmentation
Retro Diels-Alder
Fragments are at 99, 83, 72 and 57 m/z
Organic Structure Analysis, Crews, Rodriguez and Jaspars

16. USING COSY DATA UNKNOWN H
A compound shows an [M + H]+ in the FAB MS at 132 m/z
MW = 131 (Odd) therefore odd number of nitrogens
A further fragment is at 86 m/z
The IR shows bands at 3400cm-1 (broad) & 1640 cm-1 (broad)
Use the NMR data to determine the structure of the compound
Organic Structure Analysis, Crews, Rodriguez and Jaspars

17. 13C NMR DATA UNKNOWN H d t t d s A B C D E
Organic Structure Analysis, Crews, Rodriguez and Jaspars

18. MOLECULAR FORMULA DETERMINATION
UNKNOWN H
s d t q A B C D E
(C)1 + (CH)2 + (CH2)2 + (CH3)0 = C5H6
= 66 Da
The MW is 131, so there is
a mass difference of 65 Da (= NO3H3)
Therefore molecular formula = C6H9NO3 (2 dbe)
Organic Structure Analysis, Crews, Rodriguez and Jaspars

19. 1H NMR DATA UNKNOWN H Integrals: D2O so no XH (OH, NH) ONLY 6 H! H H H
b c d d’ e e’
Organic Structure Analysis, Crews, Rodriguez and Jaspars

20. SUBSTRUCTURES UNKNOWN H MF = C5H9NO3 ! dC 176 s dC 70 d
IR band at 3400
dC 70 d
dH 4.6 m
dC 60 d
dH 3.9 m
MF = C5H9NO3 !
Organic Structure Analysis, Crews, Rodriguez and Jaspars

21. 1H – 1H COSY NMR SPECTRUM UNKNOWN H b c d d’ e e’ d-e’ c-e/e’ e-e’
b-e/e’
b-d/d’
d-d’
Organic Structure Analysis, Crews, Rodriguez and Jaspars

22. Also 4-bond correlation d-e’ Diastereotopic pairs d-d’ and e-e’
1H – 1H COSY NMR DATA
UNKNOWN H
c-e/e’
b-e/e’
b-d/d’
Also 4-bond correlation d-e’
Diastereotopic pairs d-d’ and e-e’
Organic Structure Analysis, Crews, Rodriguez and Jaspars

23. SUBTRUCTURES UNKNOWN H MF = C5H9NO3 Working structures:
Organic Structure Analysis, Crews, Rodriguez and Jaspars

24. ASSIGNING NMR DATA TO A KNOWN STRUCTURE
GUAIAZULENE
Expect:
(C)5
(CH)6
(CH2)0
(CH3)4
MF = C15H18
Organic Structure Analysis, Crews, Rodriguez and Jaspars

25. (C)5 + (CH)6 + (CH2)0 + (CH3)4 = C15H18
13C NMR DATA
GUAIAZULENE
(C)5 + (CH)6 + (CH2)0 + (CH3)4 = C15H18 qq LM d J d H d K
ddd
EFG q N q O s B s D s I s A s C
Organic Structure Analysis, Crews, Rodriguez and Jaspars

26. HSQC NMR DATA GUAIAZULENE n o lm k O O N N LM LM K K
Organic Structure Analysis, Crews, Rodriguez and Jaspars

27. HSQC NMR DATA GUAIAZULENE g e f h j J J H H G G F F E E
Organic Structure Analysis, Crews, Rodriguez and Jaspars

28. Label spectrum according to HSQC:
1H NMR DATA
GUAIAZULENE
Label spectrum according to HSQC: 3H l 3H m 3H n 3H o H g H e H f H j H h H k
Organic Structure Analysis, Crews, Rodriguez and Jaspars

29. We will need expansions:
HMBC NMR DATA
GUAIAZULENE
We will need expansions:
Organic Structure Analysis, Crews, Rodriguez and Jaspars

30. HMBC NMR DATA GUAIAZULENE g e f j h I-g I-j G-f F-g C-h D-j C-g D-e
CDE B A
I-g
I-j
G-f
F-g
C-h
D-j
C-g
D-e
B-h
A-f
Organic Structure Analysis, Crews, Rodriguez and Jaspars

31. HMBC NMR DATA GUAIAZULENE g e f j h N-h K-g K-f O N LM K
Organic Structure Analysis, Crews, Rodriguez and Jaspars

32. HMBC NMR DATA GUAIAZULENE n o lm LM-lm K-lm H-n I-o C-n D-o B-lm A-n
Organic Structure Analysis, Crews, Rodriguez and Jaspars

33. HMBC NMR DATA GUAIAZULENE ‘Obvious’ assignments: N/O F/H J/E H/F E/J
Carbon
Proton A
f, n B
h, l/m C
g, h, n D
e, j, o E F g G f H n I
g, j, o J K
f, g, l/m L m M l N h O
‘Obvious’ assignments:
N/O
F/H
J/E
H/F
E/J
O/N M K L
1H-1H COSY data indicates that
e and j are adjacent (J(e-j) = 4 Hz)
as are f and h (J(f-h) = 11 Hz)
Organic Structure Analysis, Crews, Rodriguez and Jaspars

34. HMBC NMR DATA GUAIAZULENE N/O H J/E F E/J G O/N M K L
Carbon
Proton A
f, n B
h, l/m C
g, h, n D
e, j, o E F g G f H n I
g, j, o J K
f, g, l/m L m M l N h O
N/O H
J/E F
E/J G
O/N M K L
Signal for f is a dd
long-range coupling
to remaining proton g
Organic Structure Analysis, Crews, Rodriguez and Jaspars

35. HMBC NMR DATA GUAIAZULENE N A H J/E C F E/J B G O M K L
Carbon
Proton A
f, n B
h, l/m C
g, h, n D
e, j, o E F g G f H n I
g, j, o J K
f, g, l/m L m M l N h O N A H
J/E C F
E/J B G O M K L
HMBC data can’t decide positions of E, J
HMBC data can’t decide positions of D, I
How do we decide?
Organic Structure Analysis, Crews, Rodriguez and Jaspars

36. FINALISING THE ASSIGNMENTS
GUAIAZULENE
NOE N A H J C F E I D B G O M K L
Placing D (134 ppm) here
puts it in a similar environment to C (136 ppm)
This puts I (125 ppm) here
Organic Structure Analysis, Crews, Rodriguez and Jaspars