1. 1H-NMR spectra interpretation
Examples , Tables and Problems

2. Nuclear Magnetic Resonance
1H-nmr spectrum: Chemical Shift

6. How 1H NMR spectrum for p-xylene will appear?
How many H type are their?
This will inform us the number of signals in the spectra
How many H atoms in each type?
This will inform us the area of each signal
What are the functional groups near each H type?
This will inform us where the signal will appear (chemical shift d)
How many H atoms are attached to the carbon atom s nearby each H type?
This will inform us how the splitting pattern will appear for each signal

7. How 1H NMR spectra for p-xylene will appear?b
a 6H at about 2.2 ppm singlet
b 4H at about 7.0 ppm singlet

8. How 1H NMR spectra for tert-butyl bromide will appear?
a singlet 9H at about 1.8ppm

9. How 1H NMR spectra for ethyl bromide will appear?
a b
CH3CH2-Br
a triplet 3H
b quartet 2H
ethyl bromide

10. Draw the chemical structure for the compound
Number of signals in the spectra  will inform us the number of H type in the molecule
Area of each signal  will inform us the number of H atoms in each type
Chemical shift d of the signal  will inform us about closeness of H type to the functional group
Splitting pattern for each signal  will inform us how many H atoms surrounding each type

11. Draw the chemical structure for the compound C3H7Br
a triplet H
b complex 2H
c triplet H
c b a

12. Draw the chemical structure for the compound C3H7Br
C3H7Br has IHD = 0  no unsaturation (no double bonds nor cycles), so it is chain molecule (straight or branched).
From the given data, there are three signals (a, b and c) so there are three types of H atoms, each type will be at a unique carbon atom.
It is obvious that there is no symmetry in the molecule, the formula has 7H atoms and they are distributed on three separate signals (3+2+2). We have CH3-, -CH2- and -CH2-
c signal has the higher d value (more deshielded atoms), this mean that this H type of atoms are the nearest to the Br atom. We could use tables to check for this, so c signal should be like this -CH2-Br
The splitting pattern shows that a signal is triplet, this mean that it has only 2H atoms surrounding it, so this side of the molecule will be CH3-CH2- because it has only one side to attach with.
Now, the final touch, the like between the tow ends, it is the b hydrogens. This will form CH3-CH2-CH2-Br. This justify the splitting patterns for b and c.
a b c c a
a b c

13. Draw the chemical structure for the compound C3H7Br
1-bromopropane
a b c
CH3CH2CH2-Br
a triplet 3H
b complex 2H
c triplet H

14. Try the same methodology for C3H7Cl spectra below
a doublet 6H
b septet H
b a
Answer: isopropyl chloride

15. Draw the chemical structure for the compound C4H9Br
a triplet H
b doublet 3H
c complex 2H
d sextet H

16. Draw the chemical structure for the compound C8H9Cl

17. Draw the chemical structure for the compound C2H6O
a triplet 3H
b singlet 1H
c quartet 2H

18. Draw the chemical structure for the compound C8H10
a triplet 3H
b quartet 2H
c ~singlet 5H

19. Draw the chemical structure for the compound C6H14O
a triplet H
b complex 4H
c triplet H

20. Draw the chemical structure for the compound C10H14

21. Draw the chemical structure for the compound C5H11Br
a triplet 3H
b singlet 6H
c quartet 2H b & c overlap

22. Draw the chemical structure for the compound C3H8O
a triplet 3H
b complex 2H
c singlet 1H
d triplet 2H

23. Draw the chemical structure for the compound C11H16
a 9H = 3CH3, no neighbors
c 5H = monosubstituted benzene
b 2H, no neighbors 9H 2H 5H

24. Draw the chemical structure for the compound C4H8Br2
a = 6H, two CH3 with no neighbors (CH3)2C—
b = CH2, no neighbors & shifted downfield due to Br 2H

25. C7H8O
c = monosubst. benzene
b = CH2
c = OH 5H 2H 1H