APPLICATION OF WOODWARD FIESER RULES

IN STRUCTURAL ELUCIDATION OF ORGANIC

COMPOUNDS

ByCH.LAKSHMI KALYANIY14MPH432M.PHARMACY(PHARMACEUTICAL ANYALSIS)CHALAPATHI INSTITUTE OF PHARMACEUTICAL SCIENCES

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.

In 1959 luis fedrick fieser modified this rules with more experimental data and modified rules known as woodward fieser rules

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

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)

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

BASIC INTRODUCTION:-HOMOANNULAR:- cyclic diene having

conjugated double bonds in same ring.

HETEROANNULAR:-cyclic diene having conjugated double bonds in different rings

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

PARENT VALUES AND INCREMENTS FOR DIFFERENT SUBSTITUENTS/GROUPS FOR CALULATING MAXIMUM WAVELENGTH

Conjugated diene correlation:-

a. Base value for homoannular diene=253nmb. Base value for heteroannular diene=214nmc. Alkylsubstituent to ring residue attached to

the parent diene =5nmd. Double bond extending conjugation=30nme. Exocyclic double bonds=5nm

Hetero atoms:--OR = +6nm- Cl =+5nm-Br = +5nmExample:-     Base value = 214

nm

      Ring residue = 3 x 5 = 15 nm

         

Exocyclic double bond = 1x 5 = 5 nm

      λmax = 214 + 15 + 5 = 234

nm

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

compounds the compound may be a

aldehydes or ketones.

It may be acyclic or 6 membered or 5

membered ring systems

BASE VALUES FOR DIFFERENT FUNCTIONAL GROUPS:-KETONESIf it is a acyclic compound =215nmIf 6 membered ring system =215 nmIf 5 membered ring system =202 nm

ALDEHYDESIf it is a acyclic compound =210 nmIf 6 membered ring system =215 nmIf 5 membered ring system =202nm

CARBOXYLICACID AND ESTERIf compound is carboxylicacid or ester

=197nm

Values for substituents or groups

Double bond extended conjugation

=30nm

Exocyclic double bonds = 5nm

Homodiene compound=39nm

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

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

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 nmSubstitution of alkyl groups at

position = 10 nm

Substitution of alkyl residue at position = 12

nm

λmax =

237 nm

APPLICATION OF WOODWARDFIESER RULE FOR AROMATIC COMPOUNDS1.BASE VALUES :-Ar COR=246nmAr CHO=250nmAr CO2H=230nm

Ar CO2R=230nmAlkyl groups or ring residues in ortho and

meta positions=3nmAlkyl groups or ring residue in Para

position=10nm

Values for substituents or groupsGroups Ortho position

nmMeta 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

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 nmRing residue = 03 nmOCH3 group at meta position = 07

nmλmax = 256 nm

Fieser- kuhn ruleIn 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 substitutentsN = no. of conjugated double bonds Rendo= no. of rings with endocyclic bondsRexo= no. of rings with exocyclic bonds

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

example

no. of alkyl substituents =8No. of conjugated double bonds =11λmax =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 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.

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)