Interpretation of NMR-Spectra
Strategy and Conclusions
Mass Spectrum : M= 150 g/mol
13
C{1H} :
Number the 13C signals sequentially, starting with No. 1 for the most
high field signal.
10 carbons
HSQC :
Assign the 1H signals to the corresponding 13C signals.
1
Copy the numeration of the 1H signals in the HSQC to the 1H spectra.
Addressing of the integration and the chemical shift, we get the
following list:
H:
1
2
3
4
5
6
7
8
9
10
CH3
CH3
CH2
CH
CH2
CH2
C
CH
C
C
╤
- range
(C=O)
Mass Spectrum : Counting all carbons and protons gives a sum formula of C10H14 with a
molecular mass of 134. The difference to the measured mass is 16, an
–O–.
Calculate the double-bond equivalents from the sum formula C10H14O.
DBE 
(2 *10  2)  14
2
=4
-> 4 double bonds / rings
DQF-COSY :
Label the 1H signals on the F1- and F2-axis.
Put several structure fragments together.
H(3a,b)
H(5a,b)
H(1)
H(2)
H(8)
H(3b,a), H(4)
H(5b,a), H(4)
H(3a), H(3b), H(8)
H(6a/b)
H(3a), H(3b)
Label the 1H signals on the F2-axis and the 13C signals on the F1-axis.
We recommend to make a list of the C→H and H→C correlations.
HMBC :
13
1
1
1
2
3
4
5
6
7
8
9
10
8
6
5a
2
3a?
2
1
1
2
2
1
2
3a
3b
4
5a
5b
6
8
C
H
H
9
5b
3a
4
4
3a
3a
5a
5a
10
4 8
5a 5b 6
3b 5a
3b
5b 4
5b 4 8
8
C(10)
C(9)
C(9)
H(6, ╤)
H(6)
H(6)
H(4)
C(9, ╤), C(2), C(4)
H(4)
H(8)
H(8)
C(8)
C(3)
C(5)
13
C(7)
H(3a), H(3b), H(1), H(5a)
C(10,C=O)
H(8), H(5a), H(5b)
C
7
4
4
7
2
3
3
2
1
8
6
5?
8
3
4
4
4
3
9
7
10
8
5
7
9
9
4
6 9 10
9 10
10
10
Carvon