1. Y. B. Chavan College of Pharmacy
Problems solving skill in Spectroscopy
Dr. K. G. Baheti
Y. B. Chavan College of Pharmacy
Aurangabad

2. Data of spectroscopic problems:
Example
Look at Molecular weight, Molecular formula and % of elements
Look at UV data
Look at the IR absorption bands >1500 cm-1 (Functional Groups)
Look at the NMR to determine (No. & types of proton(s))
Draw some possible structures and see if they "work" with the IR, NMR and unsaturation.
Molecular Formula
% C, %H, %O or %N UV IR
NMR
Mass

3. 1. Calculate the total number of rings plus double bonds(Double bond equivalent)DBE): For the molecular formula:  CxHyNzOn
Rings + double bonds = x - (1/2)y + (1/2)z + 1
Example:
Calculate the double bond eq. for C6H6
Ring + Double bond Equivalent: 6-1/2*3+0+1 = 4
This formula consist of four double bond or one ring +3 double bond
Calculate the double bond eq. for C4H8O3
Ring + Double bond Equivalent: 4-1/2*8+0+1 = 1
This formula contain one ring or one double bond.

4. Calculate the double bond eq. for C3H8O2
Ring + Double bond Equivalent: 3-1/2*8+0+1 = 0
This formula contain one ring or one double bond
Calculate the double bond eq. for C12H12N2
Ring + Double bond Equivalent: 12-1/2*12+1/2*2+1 = 8
This formula contain two rings + six double bonds OR
One ring + seven double bonds.

5. If the molecular formula contain halogens, then while calculating the DBE each halogen may be considered as a hydrogen and it should be added to hydrogen.
Example: C10H7Br
Ring + Double bond Equivalent: 10-1/2*8+0+1 = 7
This formula contain two rings + Five double bonds OR
One ring + six double bonds

6. Determination of C, H and other elements:
Examples:
Given: Mol weight= 72
C 66.66%, H11.1% and O 22.3%.
Ans:
Total % of element =
= 100%
Formula
No. of carbon = 72* 66.66/ 100 * 12 = 3.99 ~ 4
No. of Hydrogen = 72* 11.1/ 100 * 1 = 7.99 ~ 8
No. of Oxygen = 72* 22.3/ 100 * 16 = ~ 1

7. Structural formulas of hundreds of Organic compounds (drugs)have been reported, based upon sound experimental evidence from spectroscopic experiments.
Thus spectroscopy helps in structural determination.
Structure of atoms and molecules comes from studying their interaction with light (electromagnetic radiation).
Different regions of the electromagnetic spectrum provide different kinds of information as a result of interactions.

8. NMR Data interpretation
H ppm
Multi-plicity
No of H
Interpretation
Probable structure
5.60 d 1H
6.60 t m 5H
9.27
0.99 6H
1.73
nonet
3.35 2H
4.21
Broad (s)
C9H8O
C4H10O

9. C4H7O2Cl C4H9ON H ppm Multi-plicity No of H Interpretation
Probable structure
0.92 t 3H
1.95
quintet 2H
2.9 1H
12.7 S,
Broad
0.95
1.60
sextet
2.1
6.6, 7.2 S
C4H7O2Cl
C4H9ON

10. Sample molecules are ionized by ionization method.
Mass Spectrometry:
Sample molecules are ionized by ionization method.
The m/z ratio of these ions is measured accurately by electrostatic acceleration and magnetic field perturbation, providing a precise molecular weight.
Ion fragmentation patterns may be related to the structure of the molecular ion.
m/Z
Relative abundance M+
Base Peak
Mass Spectrum

11. Look at Mass Data/ Spectra
Molecular ion peak – Molecular weight/ even/odd-nitrogen rule
Isotopic peak – Isotopes of atoms present in comp.
Fragmentation pattern - Diff. types of fragments of comp.
3-Phenyl-2-propenal C9H8O MW =

12. n-Methylbenzylamine C8H11N MW = 121.18

13. 2-Butenoic acid C4H6O2 MW = 86.09
Mass: 323

14. Problems Fragments: C=O, COOH, CH2, CH2 CH3 Structure/Ans:
Mol wt.: 116
UV: 283nm, max 22
IR(Cm-1) (b), 1715(s), 1342(w)
NMR  ppm: 2.15, s, 3H
2.6, t, 2H
2.30, t, 2H
11.2, s, 1H
Fragments:
C=O, COOH, CH2, CH2 CH3
Structure/Ans:
Data
Discussion
Inference
UV: 283nm,  max 22
Low intensity at 283nm
C=O
may present
IR(Cm-1)
(b), 1715(s), 1342(w)
Broad peak of OH, H-bonding
Carbonyl str.
COOH
NMR  ppm:
2.15, s, 3H
2.6, t, 2H
2.30, t, 2H
11.2, s, 1H
Singlet indicate no neighboring H
t, 2 neighboring H, deshilded,-CH2
t-neighboring proton, -CH2
Singlet, OH
-CH3
-CH2-
-OH

15. Problems Mol Formula: C9H10O2 UV: 274nm, max 2050
IR(Cm-1) 3031v, 2941w, 1725s,1608, 1504w, 1060s,
NMR  ppm: 2.35, s, 3H
3.8,s, 3H
,dd,4H
Fragments:
CH3, CH3 attached to COO, COO, C6H4(Ar)
Structure/Ans:
Data
Discussion
Inference
Mol Formula: C9H10O2
CxHyNzOn
DBE= x - (1/2)y + (1/2)z + 1
DBE=5
5 DBE present.
May be one aromatic ring
UV: 274nm,  max 2050
Med intensity,
Aromatic with acid or ester group
IR(Cm-1) 3031v, 2941w, 1725s,1608, 1504w, 1060s
Ar. Str, Alkyl str, str., Ester str, Ar C=C str, C-O str.
Aromatic, CH3, C=O in ester
NMR  ppm:
2.35, s, 3H 3.8,s, 3H
,dd,4H
S indicate no neighboring H
S indicate no neighboring H,Deshilded
Ar. H, Substitution at 1, 4 position
-CH3
-CH3 attached with CO
Ar. ring with 1, 4 sub.

16. Problems Mol Wt: 191 %C 56.50, %H 2.60 %N 7.30
IR(Cm-1) (b), 2225(m), 1715(s), 1605, 1518(S), 1344(s), (s)
NMR  ppm:
11.3, s, 1H
, m, 4H
Fragments:
COOH, NO2, CC, C6H4
Structure/Ans:
Data
Discussion
Inference
Mol Wt: 191
%C 56.50, %H 2.60 %N 7.30
% O = = 33.6
Number of (C 9, H 5, N 1, O 4)
CxHyNzOn
DBE= x - (1/2)y + (1/2)z + 1
DBE=8
8 DBE present
May be one aromatic ring and highly unsaturated
IR(Cm-1) (b), 2225(m), 1715(s), 1605, 1518(S), 1344(s), (s)
-OH str, H-bonded, CN/CC str, C=O of COOH str, C=C Ar str, NO2 Str, O-disubstitution in benzene
COOH, CN/CC, NO2
NMR  ppm:
11.3 s, 1H 7.3, t,1H
7.5, d, 1H
7.7, t, 1H
7.9,d,1H
S indicate no neighboring H, highly dedshided
Aromatic H
-COOH
Benzene ring

17. Problems Mol Wt: 158, UV: 225nm, max 50 hexane
IR(Cm-1) (m), 1828(s), 1757(m), 1457(m)
NMR  ppm:
2.7, septet, 6.4 square
1.2, d, 37.2, square
Fragments:
Two set CH(CH3)2 , CO-O-CO
Structure/Ans:
Data
Discussion
Inference
UV: 225nm,  max 50
Low intensity at 225nm
Does not give much information
IR(Cm-1) (m), 1828(s), 1757(m), 1457(m)
C-H str, 1828(s) and 1757(m) characteristic of anhydride, C-C str,
CH3, CO-O-CO,
NMR  ppm:
2.7, septet, 6.4 sq.
1.2, d, 37.2, square
Septet indicate six neighboring H, one H (two sets)
37.2/6.4=5.816H, d, neighboring 1H
CH(CH3)2 two set
(CH3)2 two set

18. 1.3, t, 3H, CH3 3.4, s, 2H, CH2 4.3, q, 2H, CH2 (attached to
Comp. with mol. formula C5H7NO2 gave following spectra analyze the spectra and
propose the structure.
2260cm-1- CN
1747 cm-1 ester
Str at 1200 confirm
C-O and hence ester
1.3, t, 3H, CH3
3.4, s, 2H, CH2
4.3, q, 2H, CH2 (attached to
electronegative O2,
shifted to downfield)

19. Thank You