Exercise plus Solution Quick overview
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1
H NMR spectrum
recorded at 250.13 MHz
C
3
H
9
N measured in CDCl
3
Get the constitution.
Solution
C
3
H
9
N
- no degree of unsaturation (double
bond equivalents)
- four signal group
- integral ratio: 2 : 2 : 2 : 3
- there are only 4 possible structures
2H 2H2H 3H
2H 2H2H 3H
N CH
3
H
3
C
H
3
C
N CH
3
H
CH
2
H
3
C
NH
2
H
2
C
CH
2
H
3
C
CH NH
2
H
3
C
H
3
C
Isopropylamine
n-Propylamine
N-Ethylmethylamine
Trimethylamine
9H
6H
2H
1H
3H
3H
2H
1H
3H
2H
2H
2H
1
H distribution
expected measured
3H (triplet)
2H (singlet)
2H (sextet ?)
2H (triplet)
In which of the four possible isomers do four
chemically distinguishable protons occur in the
measured integral ratio?
One signal group only.
Only three signal groups.
Four signal groups but
not the measured
integral ratio.
Four signal groups,
integral ratio as
measured.