1. “Structure Elucidation”-Comprehensive Spectral Interpretation
Lab 2
“Structure Elucidation”-Comprehensive Spectral Interpretation

2. Objectives
Suggest empirical formulae from given elemental analysis data and molecular weight of molecules.
Identify unknown compounds by combined spectral interpretation of NMR, IR, and MS spectra, and with respect to their empirical formulae.

3. Steps for verifying the structure of an unknown compound
1- CHN (molecular formula)
2- IHD (degree of un saturation)
3- FT-IR (functional groups)
4-1H-NMR (carbon skeleton)
5- MS (M.Wt.)

4. 1- Elemental analysis - CHN : How to Get the Empirical Formula of an Unknown
-Qualitative
-Quantitative
CHNX analysis
Most Common: CHN analysis, is accomplished by combustion analysis. In this technique, a sample is burned in an excess of oxygen—> carbon dioxide, water, and nitric oxide. The weights of these combustion products can be used to calculate the composition of the unknown sample.

5. Molecular weight of an unknown can be measured directly by mass-spectrometry
EXAMPLE: practical elemental analysis data of an unknown are found to be:
C: 42.7% ,H: 4.92% , N: 31.1%, others: 21.3%
the molecular weight of the same unknown was found to be:
now, the number of atoms in the molecule are calculated as:
nC* / = => nC = 8 C-atoms
  nH * / = => nH = 11 H-atoms
nN * / = => nN = 5 N-atoms
empirical formula : C8H11N5 ???  molecular weight :
difference of WHY??
“other” atoms present in the molecule.
So the remaining part could be e.g. 3 O-atoms (3 * = ), this has to be verified by interpretation of the charts.

6. Aromatic Ring+ 1 double bond
2- Calculate IHD
IHD = [(2C + 2) – X + N – H ] / 2
Example: C7H6O
IHD= [(2*7+2)-0+0-6]/2= 5
Aromatic Ring+ 1 double bond

7. 3- How to Interpret an IR Spectrum
Is a carbonyl group present?
cm-1.  strongest in the spectrum and of medium width.
If C=O is present,
ACIDS O-H present? broad band near cm-1
ESTERSC-O present? strong band near cm-1
AMIDESN-H present? medium band near 3500 cm-1
ANHYDRIDES two C=O at 1810 / 1760 cm-1
ALDEHYDE is aldehyde C-H present?
two weak bands near 2850 / 2750 cm-1 (Fermi resonance) KETONES if none of the previous five match -

8. Regions where various common types of bonds absorb (Base values)
It is recommended at first to establish the “broad patterns” given below. Then as a second step a “typical absorption value” can be memorized for each of the functional groups in this pattern.
Fingerprint
Region
C=O
=C-H
str. O=
C-H
str.
acid chlorides
anhydrides
esters
ketones
aldehydes
carboxylic acids
amides
C-H
oop bend
Aromatic compound
CC
CO
CN
Str.
C C
C N
OH
NH
Bond str.
Olefinic
aromatic
C-H
str.
Aldehyde Fermi resonance
C=C
C=N
OH
CH
bend
aliphatic
C=C aromatic
N-H bend
Frequency (cm-1)

9. Other functional groups
Bond
Base Value
Strength / Shape
Comments
C=O
1715 (typical)
s, "finger"
exact position depends on the environment of carbonyl
O—H alcohol
s, brd
broad due to H bonding
O—H acid
s, very brd
N—H m
can tell primary from secondary
C-H
just to the right of 3000 s
Aliphatic, sp3
just to the left of 3000
Olefinic or aromatic, sp2
3300
Acetylenic, sp
w, Fork like
“Fermi resonance” of aldehydic group
C—O
also check for OH and C=O
C=C
1650
w alkene  m-s aromatic 
Alkene w due to low polarity  Aromatic usually in pairs
C≡C
2150
w, sharp
characteristic
C≡N
2250
m, sharp

10. 4- How to Interpret an 1H-NMR Spectrum
The number of signals present in the spectrum.
The chemical shift [ppm] of the signals.
The signal intensity = AUC (area under the curve) = integration of each signal.
The multiplicity of each signal.

11. Identify the compound from its molecular formula and its 1H-NMR spectrum.
Propyl benzene

12. Remember: protons of OH, NH and COOH are often “exchangeable”, this means they can easily be replaced by 2H (deuterium), which will NOT give a signal under common NMR settings.

13. 5- How to Interpret a Mass Spectrum
The signal with the greatest m/z value (on the x-axis) usually is the molecular ion peak. If electron impact ionization was used in the mass detector, the m/z value is at the same time the molecular weight of the analyte.
The signal with the biggest intensity in the spectrum is the base peak. It reflects the most stable fragment of the analyte.
Check for eye-catching isotope peaks of Chlorine (n and n+2, relative intensity 75% and 25%) and Bromine (n and n+2, relative intensity 50% and 50%)
Generally, all the peaks in a mass spectrum should be assigned to fragments of the analyte. These fragments can arise from direct fragmentation, or from re-arrangement including eliminations from the molecular ion or from other fragments.

14. Mass spectrum Base peak (highest relative abundance) Fragments
(daughter peaks)
Molecular ion peak (parent peak)