Woodward Feiser Rules -Calculation of absorbtion maximum for conjugated dienes and trienes - for unsaturated carbonyl compounds - benzene and its derivatives

1. Dr. A. Syed Mohamed
Senior Assistant Professor and the Head
Research Dept. of Chemistry
Sadakathullah Appa College, Tirunelveli,
Tamilnadu, India
E-mail : asm2032@gmail.com
Woodward –Feiser Rules

2. Woodward-Fieser Rules
I. For calculating lmax for conjugated dienes and trienes
Compounds lmax (nm)
a) Base value (or) parent value:
i) Acyclic or Heteroannular dienes 214
ii) Homoannular diene 253
b) Increments:
i) Each alkyl substituent or ring residue 5
ii) Exocyclic double bond 5
iii) Double bond extending conjugation (DBEC) 30
iv) Auxochromes
- OR 6
-SR 30
- Cl, -Br 5
- NR2 60
-OCOCH3 0

3. CH2
CH2
CH3 CH3
Acyclic diene, Base value = 214
One alkyl-substituent = 5
Calculated lmax = 219 nm
CH2
CH2
CH3 CH3
Myrcene
Calculated lmax = -----nm
CH3 CH3
CH3
Base value Homoannular diene = 253
Three alkyl-substituents = 15
Calculated lmax = 268 nm
a- Phellandrene

4. CH3 CH3
CH3
a- Terpinene
Calculate lmax = -----nm
Base value Heteroannular diene = 214
Four ring residues = 20
Calculated lmax = 234 nm
Base value Heteroannular diene = 214
Three ring residues = 15
Exocyclic double bond = 5
Calculated lmax = 234 nm
Calculate lmax = -----nm

5. Base value, Hetero annular diene = 214
One double bond extending conjugation (DBEC) = 30
Four ring residues = 20
Two exocyclic double bonds = 10
Calculated lmax = 274 nm
Base value, Homoannular diene = 253
Two double bond extending conjugation (DBEC) = 60
Five ring residues = 25
Three exocyclic double bonds = 15
Calculated lmax = 353 nm

6. OH
CH3
CH3
A B
Base value, Homoannular diene (Cisoid in Ring B) = 253
Two double bond extending conjugation (DBEC) = 60
Five ring residues = 25
One exocyclic double bonds = 5
Calculated lmax = 343 nm

7. Compounds lmax (nm)
Base value
Acyclic (or) six membered cyclic
a, b - unsaturated ketone
= 215
a, b - unsaturated aldehyde = 210
Increments for
i. Homoannular diene = 39
ii. Double Bond Extending Conjugation (DBEC) = 30
iii. Alkyl substituent or ring residue
a +10
b + 12
g or higher + 18
II. For calculating lmax for a, b – unsaturated carbonyl compounds

8. Compounds lmax (nm)
iv. Exocyclic double bond, + 5
v. Auxochromes Position (nm)
a b g d
-OH 35 30 50 50
-OR 35 30 17 31
-SR - 85 - -
-OCOCH3 6 6 6 6
- Cl 15 12 12 12
- Br 25 30 25 25
-NR2 - 95 - -

9. Solvent correction
Water +8
Methanol 0
Chloroform -1
Dioxan -5
Diethyl ether -7
Hexane -11
Cyclohexane -11

10. Base value = 215
2 b substutuents (2 x 12) = 24
Calculated lmax = 239 nm
CH3
O
CH3
CH3
a
b
Base value = 215
a ring residue = 10
b ring residue = 12
Calculated lmax = 237 nm

11. Base value = 215
b ring residue = 12
d ring residue = 18
DBEC = 30
Exocyclic double bond = 5
Calculated lmax = 280 nm
Base value = 210
a substituent = 10
2 b substituents = 24
Calculated lmax = 244 nm

12. Base value = 215
a ring residue = 10
d ring residue = 18
DBEC = 30
Homoannular diene = 39
Exocyclic double bond = 5
Calculated lmax = 317 nm
Base value = 215
a ring residue = 10
b ring residue = 12
Calculated lmax = 237 nm

13. Base value = 215
b ring residue = 12
d3 ring residue = 18
2 DBEC = 60
Homoannular diene = 39
Exocyclic double bond = 5
Calculated lmax = 349 nm
Base value = 215
OH substituent in a position = 35
2 b substituents = 24
Calculated lmax = 274 nm

14. Base value = 215
b ring residue = 12
Higher d ring residue = 54
2 DBEC = 60
2 Exocyclic double bonds = 10
Calculated lmax = 351nm

15. Rules for calculating absorption maximum of benzene
and its derivatives
Scott derived a set
of rules for
calculating the
absorption
maximum (p-p*)
in substituted
benzene
derivatives such as
C6H5COR,
C6H5CHO,
C6H5COOH and
C6H5COOR.
C6H5 - CO - G lmax
(nm)
Parent chromophore
G=alkyl or ring residue (C6H5COR) 246
G = H (C6H5CHO) 250
G = OH or OR (C6H5COOH or
C6H5COOR)
230
Increments for each substituent
-alkyl or ring residue o, m 3
p 10
-OH, -OMe, -O alkyl o,m 7
p 25
-Cl o,m 0
p 10

16. Base value (C6H5COR) = 246
One o-ring residue(1x3) = 3
One m-OCH3 (1x7) = 7
One p-OCH3 (1x25) = 25
Calculated lmax = 281nm
Base value (C6H5COR) = 246
One o-ring residue(1x3) = 3
One o-OH (1x25) = 7
Calculated lmax = 256nm

17. O
O
CH3
Base value (C6H5COR) = 246
One o-ring residue(1x3) = 3
One p-OMe (1x25) = 25
Calculated lmax = 274 nm
O
O
O
CH3
O
CH3
Cl
O O
CH3
Griseofulvin
Base value (C6H5COR) = 246
One o-OMe (2x7) = 14
One p-OMe (1x25) = 25
Calculated lmax = 285 nm