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APPLICATION OF WOODWARD
FIESER RULES IN STRUCTURAL
ELUCIDATION OF ORGANIC
COMPOUNDS
By
CH.LAKSHMI KALYANI
Y14MPH432
M.PHAR...
OBJECTIVES:-
• To familiraze the Woodward fieser rules
•To calculate the maximum wavelength of
organic compound.
INTRODUCTION:-
In 1945 Robert burn woodward gave certain
rules for correlating maximum wavelength
with molecular structure...
ADVANTGE:-
It is used to calculate the position and
maximum wave length for a given structure.
By relating the position an...
TERMINOLOGY:-
 Conjugated diene:-organic compound
contaning two or more double bonds each
separated from other by a singl...
WOODWARD FIESER RULE
 It mean each type of diene or triene system
is having a certain fixed values at which
abosrption ta...
BASIC INTRODUCTION:-
 HOMOANNULAR:- cyclic diene having
conjugated double bonds in same ring.
 HETEROANNULAR:-cyclic die...
BASIC INFORMATION REGARDING
CALCULATION OF MAXIMUM WAVE LENGTH OF
A COMPOUND
ENDOCYCLIC DOUBLE BOND:-double bond
present i...
PARENT VALUES AND INCREMENTS FOR
DIFFERENT SUBSTITUENTS/GROUPS FOR
CALULATING MAXIMUM WAVELENGTH
Conjugated diene correlat...
Hetero atoms:-
 -OR = +6nm
 - Cl =+5nm
 -Br = +5nm
 Example:- Base value = 214
nm
Ring residue = 3 x 5 = 15 nm
Exocycl...
EXAMPLE
Acyclic conjugated diene= 217 nm
2 alkyl substituents (2x5) = 10
nm
__________
λmax = 227 nm
APPLICATION OF WOODWARD
FIESER RULE FOR ALPHA,BETA
UNSATURATED
COMPOUNDS(ENONES)
 In this alpha, beta unsaturated compoun...
BASE VALUES FOR DIFFERENT
FUNCTIONAL GROUPS:-
KETONES
 If it is a acyclic compound =215nm
 If 6 membered ring system =21...
Values for substituents or
groups
 Double bond extended conjugation
=30nm
 Exocyclic double bonds = 5nm
 Homodiene comp...
Group
Alkyl R 10 nm 12 nm 18 nm 18 nm
Alkoxy OR 35 nm 30 nm 17 nm 31 nm
Hydroxyl OH 35 nm 30 nm 30 nm 50 nm
Chlorine -Cl 1...
example
Base value = 214 nm
β- Substituents = 1 x 12 = 12 nm
δ- Substituents = 1 x 18 = 18 nm
Double bond extending
Conjug...
example
Parent conjugated enone
in acyclic compound = 215 nm
2 alkyl residue at position = 24 nm
λmax = 239 nm
example
Parent conjugated enone in six
membered ring = 215 nm
Substitution of alkyl groups at
position = 10 nm
Substitutio...
APPLICATION OF WOODWARDFIESER
RULE FOR AROMATIC COMPOUNDS
1.BASE VALUES :-
 Ar COR=246nm
 Ar CHO=250nm
 Ar CO2H=230nm
...
Values for substituents or
groups
Groups Ortho position nm Meta position nm Para position nm
-OH 7 7 25
-OCH3 7 7 25
-O 11...
example
base value = 246 nm
Hydroxy group at meta position = 07 nm
Hydroxy group at para position = 25 nm
λmax = 278 nm
example
Base value =246
nm
Ring residue = 03 nm
OCH3 group at meta position = 07 nm
λmax = 256 nm
Fieser- kuhn rule
In the number of conjugated double bonds is more
than 4, the woodward and fieser rules may not be
applic...
example
No. of alkyl substituent's = 10
No. of rings with endocyclic double bond=2
λmax= 114+5x10+11(48-1.7x11)-(16.5x2)-(...
example
no. of alkyl substituents =8
No. of conjugated double bonds =11
λmax =114+5(8)+11[(48-1.7(11)]-0-0=476 nm
POINTS TO REMEMBER:-
 Incase for which both types of diene systems
are present then the one with the longer
wavelength is...
REFRENCES
 A TEXT BOOK OF ORGANIC
SPECTROSCOPY BY P.S KALSI( PAGE
.NO-40)
 A TEXT BOOK OF UV-VISIBLE AND
INFRARED SPECTR...
Application of woodward fieser rules in structural elucidation
Application of woodward fieser rules in structural elucidation
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Application of woodward fieser rules in structural elucidation

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Calculation of lambda max for the compounds

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Application of woodward fieser rules in structural elucidation

  1. 1. APPLICATION OF WOODWARD FIESER RULES IN STRUCTURAL ELUCIDATION OF ORGANIC COMPOUNDS By CH.LAKSHMI KALYANI Y14MPH432 M.PHARMACY(PHARMACEUTICAL ANYALSIS) CHALAPATHI INSTITUTE OF PHARMACEUTICAL SCIENCES
  2. 2. OBJECTIVES:- • To familiraze the Woodward fieser rules •To calculate the maximum wavelength of organic compound.
  3. 3. INTRODUCTION:- In 1945 Robert burn woodward gave certain rules for correlating maximum wavelength with molecular structure. In 1959 luis fedrick fieser modified this rules with more experimental data and modified rules known as woodward fieser rules
  4. 4. ADVANTGE:- It is used to calculate the position and maximum wave length for a given structure. By relating the position and degree of substitution of chromophore
  5. 5. TERMINOLOGY:-  Conjugated diene:-organic compound contaning two or more double bonds each separated from other by a single bond  Dienes:-it is also known as a alkadiene, diolefin .  It is one class of organic compound contaning two ethyelnic linkages (carbon to carbon double bond)
  6. 6. WOODWARD FIESER RULE  It mean each type of diene or triene system is having a certain fixed values at which abosrption takes place; this constitutes the base value or parent value.  The contribution made by various alkyl substituents or ring residue, double bond extending conjugation and polar groups such as -Cl,-Br etc … are added to the base value to obtain for a particular compound
  7. 7. BASIC INTRODUCTION:-  HOMOANNULAR:- cyclic diene having conjugated double bonds in same ring.  HETEROANNULAR:-cyclic diene having conjugated double bonds in different rings
  8. 8. BASIC INFORMATION REGARDING CALCULATION OF MAXIMUM WAVE LENGTH OF A COMPOUND ENDOCYCLIC DOUBLE BOND:-double bond present in a ring. EXOCYCLIC DOUBLE BOND:-double bond in which one of the doubly bonded atoms is a part of ring system. Here ring B has one exocyclic and endocyclic double bond .ring A has only one endocyclic double bond
  9. 9. PARENT VALUES AND INCREMENTS FOR DIFFERENT SUBSTITUENTS/GROUPS FOR CALULATING MAXIMUM WAVELENGTH Conjugated diene correlation:- a. Base value for homoannular diene=253nm b. Base value for heteroannular diene=214nm c. Alkylsubstituent to ring residue attached to the parent diene =5nm d. Double bond extending conjugation=30nm e. Exocyclic double bonds=5nm
  10. 10. Hetero atoms:-  -OR = +6nm  - Cl =+5nm  -Br = +5nm  Example:- Base value = 214 nm Ring residue = 3 x 5 = 15 nm Exocyclic double bond = 1x 5 = 5 nm λmax = 214 + 15 + 5 = 234 nm
  11. 11. EXAMPLE Acyclic conjugated diene= 217 nm 2 alkyl substituents (2x5) = 10 nm __________ λmax = 227 nm
  12. 12. APPLICATION OF WOODWARD FIESER RULE FOR ALPHA,BETA UNSATURATED COMPOUNDS(ENONES)  In this alpha, beta unsaturated compounds the compound may be a aldehydes or ketones.  It may be acyclic or 6 membered or 5 membered ring systems
  13. 13. BASE VALUES FOR DIFFERENT FUNCTIONAL GROUPS:- KETONES  If it is a acyclic compound =215nm  If 6 membered ring system =215 nm  If 5 membered ring system =202 nm ALDEHYDES  If it is a acyclic compound =210 nm  If 6 membered ring system =215 nm  If 5 membered ring system =202nm CARBOXYLICACID AND ESTER  If compound is carboxylicacid or ester =197nm
  14. 14. Values for substituents or groups  Double bond extended conjugation =30nm  Exocyclic double bonds = 5nm  Homodiene compound=39nm
  15. 15. Group Alkyl R 10 nm 12 nm 18 nm 18 nm Alkoxy OR 35 nm 30 nm 17 nm 31 nm Hydroxyl OH 35 nm 30 nm 30 nm 50 nm Chlorine -Cl 15 nm 12 nm 12 nm 12 nm Bromine –Br 25 nm 30 nm 25 nm 25 nm
  16. 16. example Base value = 214 nm β- Substituents = 1 x 12 = 12 nm δ- Substituents = 1 x 18 = 18 nm Double bond extending Conjugation =1 x 30 = 30 nm Exocyclic double bond = 5 nm λmax = 279 nm
  17. 17. example Parent conjugated enone in acyclic compound = 215 nm 2 alkyl residue at position = 24 nm λmax = 239 nm
  18. 18. example Parent conjugated enone in six membered ring = 215 nm Substitution of alkyl groups at position = 10 nm Substitution of alkyl residue at position = 12 nm λmax = 237 nm
  19. 19. APPLICATION OF WOODWARDFIESER RULE FOR AROMATIC COMPOUNDS 1.BASE VALUES :-  Ar COR=246nm  Ar CHO=250nm  Ar CO2H=230nm  Ar CO2R=230nm  Alkyl groups or ring residues in ortho and meta positions=3nm  Alkyl groups or ring residue in Para position=10nm
  20. 20. Values for substituents or groups Groups Ortho position nm Meta position nm Para position nm -OH 7 7 25 -OCH3 7 7 25 -O 11 20 78 -Cl 0 0 10 -Br 2 2 15 -NH2 13 13 58
  21. 21. example base value = 246 nm Hydroxy group at meta position = 07 nm Hydroxy group at para position = 25 nm λmax = 278 nm
  22. 22. example Base value =246 nm Ring residue = 03 nm OCH3 group at meta position = 07 nm λmax = 256 nm
  23. 23. Fieser- kuhn rule In the number of conjugated double bonds is more than 4, the woodward and fieser rules may not be applicable and hence fieser with kuhn has derived an equation for predicting the λmax λmax= 114+5M+n(48.0-1.7n)-16.5 Rendo-10 Rexo M= no. of alkyl substitutents N = no. of conjugated double bonds Rendo= no. of rings with endocyclic bonds Rexo= no. of rings with exocyclic bonds
  24. 24. example No. of alkyl substituent's = 10 No. of rings with endocyclic double bond=2 λmax= 114+5x10+11(48-1.7x11)-(16.5x2)-(10x0)=453.3 nm
  25. 25. example no. of alkyl substituents =8 No. of conjugated double bonds =11 λmax =114+5(8)+11[(48-1.7(11)]-0-0=476 nm
  26. 26. POINTS TO REMEMBER:-  Incase for which both types of diene systems are present then the one with the longer wavelength is designated as a parent system  When ever there is a increasing conjugation leads to increase in wavelength and required less amount of energy  Up to four conjugations woodward fieser rule is applied  > four conjugation we have to use fieser khun rule is applied.
  27. 27. REFRENCES  A TEXT BOOK OF ORGANIC SPECTROSCOPY BY P.S KALSI( PAGE .NO-40)  A TEXT BOOK OF UV-VISIBLE AND INFRARED SPECTROSCOPY BY RAJASHEKARAN( PAGE NO.88-100)

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