Exercise plus Solution Quick overview
It is recommended to use this version only for a quick overview of the NMR challenge. All animations of the PowerPoint
version are missing, under certain circumstances quality deficiencies may also occur.
The higher quality PowerPoint files are freely available for download at any time.
2.39
0.81.01.21.41.61.82.02.22.42.62.83.03.23.4
ppm
1
H
185.20
190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20
ppm
13
C
36.92
13
C{
1
H} NMR spectrum
recorded at 62.90{250.13} MHz
1
H NMR spectrum
recorded at 250.13 MHz
C
4
H
6
O
4
measured in D
2
O
Deduce the structure.
Solution
𝒏
𝐃𝐁Ä
=
𝟐𝒏
𝐂
𝒏
𝐇
+ 𝟐
𝟐
C
4
H
6
O
4
First results:
- two double bond equivalents
- four carbon atoms, but only two carbon signals in the NMR spectrum
- six protons, but one singlet only
Probable explanation:
- a symmetrical molecule R R, where R corresponds to the molecular
formula C
2
H
3
O
2
, contains an open binding site and a double bond
equivalent
𝒏
𝐃𝐁𝐄
=
𝟐𝒏
𝐂
𝒏
𝐇
+ 𝟐
𝟐
C
4
H
6
O
4
Let's write down the half-molecule fragment we are looking for
and the little known information on sticky notes.
C : 36.92 ppm C : 185.20 ppmH : 2.39 ppmwanted: -C
2
H
3
O
2
with 1 DBE
Solution