The HL content for spectroscopy. IB Chemistry.

1. 21.1- ‘Spectroscopy II’
(1 week)
High-resolution 1H-NMR AND X ray crystallography.

2. Topic outline….
• Lesson 1- Review of IR, MS and H1-NMR (including I.H.D. concept)
• Lesson 2- Answering exam questions using these 3 techniques.
• Lesson 3- High resolution H1-NMR (‘proton’) spectroscopy.
• Lesson 4- Interpreting high resolution NMR spectra.
• Lesson 5- A brief look at X-ray crystallography & some review Qs.
• Lesson 6- Test (T20& 21)!

3. Lesson 1- Review of IR, MS and H1-NMR
(including I.H.D. concept)
Level 4: Recall the key points of each of
the 3 techniques.
Level 7: Explain the theory behind H1-NMR
spectroscopy.
Level 5/6: determine the IHD for a series of
organic compounds.

4. Starter….
In SL last year we covered the following:
1. Index of hydrogen deficiency (I.H.D.)
2. Infra red spectroscopy (I.R.)
3. Mass spectroscopy (M.S.)
4. H1- NMR spectroscopy (H1-NMR)

5. 1. Index of Hydrogen deficiency (I.H.D.)
a.k.a- degree of unsaturation.
A measure of how many H2 molecules would be needed to convert an
organic compound to the corresponding saturated, non-cyclic
molecule.
e.g. cyclobutane (IHD = 1):
+ H2
C4H8
C4H10

6. The magic formula!
(2x + 2)-y = IHD
2
Where x= number of carbons
y= number of hydrogens
Try it out for cyclobutane, C4H8…..

7. Summary of the rules for determining I.H.D.
(leave on board and then use white boards to practice set of questions which follows…)
1. Draw the full structural formula of the compound.
2. Ignore O and S atoms (in the formula) if present.
3. Count halogens as hydrogen atom place holders (in the formula).
4. Consider Nitrogen atoms as carbon atoms and add one H atom for
each N atom that you have (to get valency of 4 that a carbon would
have).

8. Now your turn…☺

9. 2. Infra red spectroscopy (I.R.)
*Key- Energy from IR is only transferred to a bond if the bond has a
dipole that changes as it vibrates. The frequency at which these bonds
vibrate is characteristic of each functional group.
Non-polar molecules like CO2 and SO2 often come up in MCQs testing
this concept.
E.g. Q. CO2 has a mode of vibration which is IR inactive and one which
is IR active. Draw each.

11. 3. Mass spectroscopy (M.S.)
*Key: Can be used to identify the overall mass of an organic/inorganic
molecule (i.e. the molecular ion mass M+)
For organic molecules we go a step further and fragment it into as
many possible pieces and identify which fragment is responsible for
which mass peak on the spectrum.
* Remember, if dealing with halogenoalkanes, then there will be two small peaks for the two isotopes of Br or
Cl and also two peaks ‘upstream’ for the molecular ion.

13. Lesson 2- Answering exam questions using all
3 techniques.
.
Level 4: Interpret IR,MS and H1-NMR
spectra.
Level 7: Complete all of the practice
questions which you have been set.
Level 5/6: Deduce the correct structure of an
organic compound using all 3 techniques.

14. 4. H1- NMR spectroscopy (H1-NMR)
(low resolution)
*Key: The nuclei of H1 atoms have one unpaired
proton (no neutron) which can behave like a
tiny magnet.
These nuclei have two possible spin states when
subjected to an external magnetic field (high
and low).
The radio wave frequency required to make
them flip from one energy state to the other is
called the ‘resonance energy.’
Different local environments affect this signal
and so you get different shift patterns on an
NMR spectrum.

15. CH3 CH CH2 CH3
Br
CH3 CH CH2 CH2 CH3
Br
CH3 CH2 CH3 *2 different H environments: ratio 6:2 (3:1)
CH3 C
CH3
CH2
OH
CH3
4 different H environments: ratio 3:1:2:3
5 different H environments: ratio 3:1:2:2:3
*4 different H environments: ratio 6:1:2:3
Identifying the number of different H environments.

16. Interpreting a H1 NMR spectrum.
The position of the NMR signal
relative to tetra methyl silane
(T.M.S.) is called the chemical
shift of the proton and is
measured in parts per million
(ppm).
See databook section 27 for some
common examples in organic
compounds.
*TMS is used because:
• All the protons are in the same
environment so it gives a strong,
single peak.
• Unreactive.
• Volatile and easy to remove at the
end.

17. Starter (See worked example from Pearson)
• Now try the rest of the questions on your handout☺

18. Lesson 3- High resolution H1-NMR
(‘proton’) spectroscopy.
Level 4: State what ‘spin-spin coupling’ is.
Level 7: Explain why H atoms bonded to O or N
atoms do not spin-spin couple with adjacent H
atoms.
Level 5/6: Apply the ‘n+1’ rule to determine the
splitting pattern for the signal of a H atom.

19. Interpreting 1H NMR spectra activity

20. Integration and the number of hydrogens
The height of the peaks in an NMR spectrum does not give us any useful information.
The spectrum can be integrated to find this information.
However, the area under the
peaks on a 1H NMR spectrum
is proportional to the number
of hydrogen atoms causing
the signal. The ratio of the
areas under the peaks tells
you the ratio of 1H atoms in
each environment.
2
1
3

22. A more in depth look at the ‘n+1 rule’….
Since there are 2 possible spin states (up and down) for any one H1 nucleus, an alternative formula to use is:
2n= total number of possible combinations of nuclei (where n= number of neighbouring H atoms)
Eg, consider a single H atom next to the 3 Hs of –CH3. The formula tells us that we should have 23= 8 possible
combinations of nuclear spin:
1 , 3 , 3 , 1 = a quartet!

23. For the mathematicians amongst us☺
Pascal’s triangle

24. In summary…
1. The number of peaks will indicate the number of different H environments.
2. The chemicalshift values will indicate the presence of particular functional groups.
3. High resolution H1-NMR spectroscopyallows us to interpret the influence of neighbouring H atoms on the signal of
another (because of the ‘spin-spin coupling effect’).
4. The n+1 rule allows us to determine the splitting pattern of the signal of a H atom relative to what neighbours it.
5. Protons bonded to the same atom do not interact with each other as they are equivalent and behave as a group.
6. Protons on non-adjacent carbon atoms do not generally interact with one another.
7. Any H atoms bonded to O or N (i.e. not carbon) will not affect adjacent Hs because they are occupied in H-bonding
to other molecules ! This causes deshielding of these H atoms and so they are far more affected by the external
magnetic field and would give a signal well up the x-axis.

25. Check for learning… (Pearson pg 571)

26. Lesson 4- Interpreting high resolution NMR
spectra.
Level 4: List the possible organic families
from a given high resolution NMR spec.
Level 7: Deduce the structure of a
compound from its H1-NMR spectrum.
Level 5/6: Identify the different H environments
in a compound from its NMR splitting pattern.

27. SPIN-SPIN COUPLING
signal singlet doublet triplet quartet
appearance
number of
lines
1 2 3 4
number of
neighbouring
inequivalent
H atoms
0 1 2 3
relative size 1:1 1:2:1 1:3:3:1

28. butanone CH3 CH2 C CH3
O
8 7 6 5 4 3 2 1 0
 chemical shift
shift () assignment
relative
intensity
coupling coupled to
1.0 CH3CH2 3 triplet CH2
2.0 CH3CO 3 singlet
2.4 CH2 2 quartet CH3

29. butane
shift () Assignment
relative
intensity
coupling coupled to
1.3 CH2 2 quartet CH3
0.8 CH3 3 triplet CH2
CH3 CH2 CH2 CH3
8 7 6 5 4 3 2 1 0
 chemical shift

30. SUMMARY
Number of signals
Position of signals
Relative intensities
Splitting
how many different sets of
equivalent H atoms there are
information about chemical
environment of H atom
gives ratio of H atoms for peaks
how many H atoms on adjacent C
atoms

31. For each of the following compounds, predict the
number of signals, the relative intensity of the signals,
and the multiplicity of each signal- white boards☺
a) *2-methylprop-1-ene
b) Prop-1-ene
c) 2-chloropropane
d) propanone
f) ethyl propanoate
g) 1,2-dibromopropane
h) 5,5-dimethylethyl
propanoate
i) *but-2-ene

32. 2 signals: ratio 6 : 2 (3 :1)
s s
C CH3CH3
O
CH CH3CH3
Cl
CH2 C CH3
CH3
CH2 CH CH3
3 signals: ratio 2 : 1 : 3
d m d
2 signals: ratio 6 : 1
d m
1 signal
CH3 CH2 C
O
O CH2 CH3
4 signals: ratio 3 : 2 : 2 : 3
t q q t
Where ‘m’ = multiple which is going to be greater than a quartet (not on IB course)

33. 3 signals: ratio 2 : 1 : 3
d m d
2 signals: ratio 6 : 2 (3 :1)
d q
CH CH3CH2
BrBr
CH3 CH2 C
O
O C CH3
CH3
CH3
3 signals: ratio 3 : 2 : 9
t q s
CH CHCH3 CH3

34. Putting it all together-
Q11 Pearson HL pg578

35. The answer!
C4H8O2- (check that the question says that this is the molecular formula!)
Ester or Carboxylic acid.
Can eliminate C.acid straight away using section 27 of d.book as there is no signal around 9-13.0ppm for a carboxylic acid OH
hydrogen.
Just need to look at the splitting patterns to determine the correct ester structure.
The quartet is for –CH2
The triplet is for - CH3
The singlet is for – CH3 attached to the C=O (no splitting effect).
Ethyl methanote

36. Lesson 5- X-ray crystallography & some review Qs.
Level 4: Recall that x-ray crystallography can be used to
determine the bond lengths and angles in a molecule.
Level 7: Interpret an electron density map
which was generated by X-ray crystallog.
Level 5/6: Describe the technique of x-ray
crystallography.

37. Starter….
See question 12 from Pearson- good question which brings together all
3 analytical techniques to identify the best structure of an unknown
compound. (print for students)- This is as tough as they get!
There are other questions there for you to practice for homework
revision…. (from inthinking)

38. X-ray crystallography….
Gurate wee video to explain ☺
https://www.youtube.com/watch?v=php6R7qffn0

39. Interpreting x- ray crystallography images

40. Electron density maps (from x-ray crystallography)
for simple organic molecules….
Benzene Anthracene

41. Check for learning…

42. The answers!