Calculation of λ using woodward fieser rules
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Transcript of Calculation of λ using woodward fieser rules
Assignment #2
Calculation of λmax using Woodward-Fieser rules.
By
Sayyad Ali
To
Prof. Dr. Jamshed Iqbal
CIIT Abbottabad.
Introduction to Woodward-Fieser Rules
In the middle of the last century, R. B. Woodward studied the UV spectra of conjugated dienes
and developed a set of rules for predicting the wavelength of maximum UV absorption based on
the structure of the diene. Later L. M. Fieser extended the rules to conjugated aldehydes and
ketones.
In order to apply the rules to specific structures, we need to learn how certain structures are
described, so we may apply numeric wavelength values to the structure.
Conjugated Dienes
We know that conjugated dienes must lie in an s-cis conformation in order to undergo a Diels-
Alder reaction. Thus, two fundamental conformations of conjugated dienes are the s-cis
conformation and the s-trans conformation. Figure 1 shows that an s-trans conjugated diene is
one in which the two double bonds lie on opposite sides of the single bond that joins them.
Whereas, a double bond in the s-cis conformation is one in which both double bonds lie on the
same side of the single bond that joins them.
1,3-butadiene
s =single bond that connects the two double bonds (dashed, straight line below)
s-trans
conformation
(double bonds on
opposite sides
of single bond)
s-cis
conformation
(double bonds on
same side of
single bond)
Figure 1. Conformations of 1,3-butadiene.
The first-step in predicting the wavelength of maximum UV absorption for a conjugated diene is
to determine whether it lies in an s-trans or s-cis conformation. If it lies in the s-trans
conformation, its base wavelength is 217 nm. If it lies in the s-cis conformation, its base
wavelength is 253 nm.
Endocyclic vs Exocyclic Double Bonds
An interesting feature of double bonds is that they may be part of a ring system, in which case
they are called endocyclic double bonds because their bond lies “within” the ring. Double
bonds may also project from a ring, in which case they are called exocyclic double bonds
because their bond lies “outside” the ring. If a compound is bicyclic, a double bond might be
endocylic with respect to one ring and exocyclic with respect to the other ring.
endocyclic exocyclic
A B
endo to Aexo to B
Figure 2. Types of double bonds.
Figure 2 shows that a double bond is endocyclic if its bond is part of the ring. The double bond
is exocyclic if its bond projects from the ring. If a double bond is exocyclic to a ring, it adds 5
nm to the base wavelength of a conjugated diene.
Extended Conjugated Double Bonds
Two double bonds separated by a single bond are conjugated. If a third double bond is separated
from one of the original pair of double bonds by a single bond, the three double bonds represent
an extended conjugated system.
s-cis
max) = 217 nm
s-cis extended
max) = 247 nm
s-trans
max) = 253 nm
s-trans extended
max) = 283 nm
Figure 3. Effect on λmax of extending the conjugation.
Each double bond that extends the conjugation adds 30 nm to the wavelength of maximum
absorption.
Effect of Alkyl Groups
Any alkyl group bonded to a carbon atom of the conjugated system (i.e., a carbon sharing a
conjugated bond) adds 5 nm to the wavelength of maximum absorption.
s-cis
+ 2 alkyl groups
max) = 227 nm
s-cis extended
+ 3 alkyl groups
max) = 262 nm
s-trans
+ 2 alkyl groups
max) = 263 nm
s-trans extended
+ 3 alkyl groups
max) = 298 nm
Figure 4. Conjugated systems containing alkyl groups.
The wavelength values shown in figures 3 and 4 are predicted values. The actual values vary
slightly from the predicted values and must be determined by experimentation.
Conjugated Aldehydes and Ketones
A conjugated aldehyde or ketone arises when a double bond is separated by a single bond from
the carbonyl group of an aldehyde or ketone.
O
H
O
conjugated aldehye
(max) = 210
conjugated ketone
(max) = 215
Figure 5. Conjugated aldehyde and ketone.
The base wavelength for a conjugated aldehyde is 210 nm and for a conjugated ketone is 215
nm. The compounds shown in Figure 5 are also called -unsaturated because their carbon-
carbon double bond lies between the alpha and beta carbon atoms. The Greek lettering system
starts with the carbon atom bonded to the carbonyl carbon atom of the aldehyde or ketone. An
alkyl groups bonded to an -carbon adds 10 nm to λmax and an alkyl group bonded to a -
carbon adds 12 nm to the base value. The effects of an exocyclic double bond and extended
conjugation are the same for aldehydes and ketones as for dienes. An exocyclic double bond
adds 5 nm and an extended double bond adds 30 nm to the base values.
A wavelength predictor has been created in Excel for conjugated dienes and for conjugated
aldehydes and ketones. A diene is characterized as lying in an s-cis or s-trans conformation and
by its number of alkyl groups bonded to the conjugated system, exo double bonds, and extended
double bonds. The analyst enters these values into the predictor and the predictor calculates
λmax. The predicted value of λmax is the same as one obtains by calculating the value as
described above. A separate sheet in the Excel workbook handles aldehydes and ketones in a
similar fashion. The structure is identified as an aldehyde or ketone, and its number of α and β
alkyl groups, exo double bonds and extended double bonds are entered into the appropriate
boxes. The predictor calculates λmax.
Use the wavelength predictor to answer the problems in the two exercises found in the folder
with this document.
The λmax of the p p* transition for compounds with < 4 conjugated double bonds can be
calculated using Woodward-Fieser rules.
Start with a base number:
To the base add:
i. 30 for each extra conjugated double bond
ii. 5 each time a conjugated double bond is an exocyclic double bond
iii. 36 for each conjugated double bond frozen s-cis
iv. 5 for each alkyl group or halogen bonded to conjugated system of polyene
v. 10 for an α-substituent of a conjugated aldehyde or ketone
vi. 12 for a β-substituent of a conjugated aldehyde or ketone
Example # 1. Calculate expected λmax for following
Base = 217
3 alkyl substituents @ 5 each = 15
Example # 2. Calculate expected λmax for following
To the base add:
i. 30 for each extra conjugated double bond
ii. 5 each time a conjugated double bond is an exocyclic double bond
iii. 36 for each conjugated double bond frozen s-cis
iv. 5 for each alkyl group or halogen bonded to conjugated system of polyene
v. 10 for an α-substituent of a conjugated aldehyde or ketone
vi. 12 for a β-substituent of a conjugated aldehyde or ketone
Calculated = 232
Base = 215 Base = 217
α substituent = 10 Additional = 30
β substituent = 12 Homoannular or s-cis = 36
Exocyclic = 5 2 alkyl substituents = 10
Example # 3
(i) and similarly (ii).
Calculated = 242 Calculated = 293
Observed = 241 Observed = 293
(iii) (iv)
The above last compounds were calculated for its λmax according to the following rules.