Woodward-Fieser Rules for Calculating λmax in Conjugated Dienes and Trienes PDF 2025

Woodward-Fieser Rules for Calculating A, in Conjugated Dienes and Trienes Woodward (1941) formulated a set of empirical rules for calculating or predicting A, in conjugated acyclic and six-membered ring dienes. These rules, modified by Fieser and Scott on the basis of wide experience with dienes and...

Woodward-Fieser Rules for Calculating A, in Conjugated Dienes and Trienes Woodward (1941) formulated a set of empirical rules for calculating or predicting A, in conjugated acyclic and six-membered ring dienes. These rules, modified by Fieser and Scott on the basis of wide experience with dienes and trienes, are called Woodward-Fieser rules and are summarized in Table. Woodward-Fieser rules for calculating A,,, in conjugated dienes and trienes Base value for acyclic or heteroannular diene 214 nm Base value for homoannular diene 253 nm Increment for each: Alkyl substituent or ring residue 5 nm Exocyclic conjugated double bond 5 nm Double bond extending conjugation 30 nm —OR (alkoxy) 6 nm —Cl, —Br 6 nm —OCOR (acyloxy) 0 nm —SR (alkylthio) 30 nm —NR; (dialkylamino) 60 nm In the same double bond is exocyclic to two rings simultaneously 10 nm Solvent correction 0 nm Calculated* A, of the compound Total nm *For m — m* transition (K-band). Woodward-Fieser rules for calculating A,,,, in conjugated dienes and trienes Base value for acyclic or heteroannular diene 214 nm Base value for homoannular diene 253 nm Increment for each: Alkyl substituent or ring residue 5 nm Exocyclic conjugated double bond S nm Double bond extending conjugation 30 nm —OR (alkoxy) 6 nm —Cl, —Br 6 nm —OCOR (acyloxy) 0 nm —SR (alkylthio) 30 nm —NR; (dialkylamino) 60 nm In the same double bond is exocyclic to two rings simultaneously 10 nm Solvent correction 0 nm Calculated* A, of the compound Total nm *For m — m* transition (K-band). (i) Homoannular Dienes In homoannular dienes, conjugated double bonds are present in the same ring and having s-cis (cisoid) configuration (s = single bond joining the two doubly bonded carbon atoms): O M 0 QO The s-cis configuration causes strain which raises the ground state energy level of the molecule leaving the high energy excited state relatively unchanged. Thus, the transition energy is lowered resulting in the shift of absorption position to a longer wavelength. Acyclic dienes exist mostly in the strainless s-trans (transoid) conformation with relatively lower ground state energy level. Thus, their absorptions appear at shorter wavelengths. For example, 1,3-cyclohexadiene  (I) shows A,y 256 nm, whereas 1,3-butadiene shows A, 217 nm. Also, due to the shorter distance between the two ends of the chromophore, s-cis dienes give lower €max (~10,000) than that of the s-trans dienes (~20,000). NS s-trans =/ \ s-cis 1,3-butadiene 1,3-butadiene ~97.5% ~25% Amax 217 nm (ii) Heteroannular Dienes In heteroannular dienes, conjugated double bonds are not present in the same ring and these have s-trans (transoid) configurations: o O GO (ili) Exocyclic Conjugated Double Bonds The carbon-carbon double bonds projecting outside a ring are called exocyclic double bonds. For example Exocyclic to ring A; Exocyclic to two Endocyclic Endocyclic to ring B rings A and C (Two exocyclic (One exocyclic |é double bond) double bonds) ] Exocyclic (Equivalent to two exocyclic double bonds) Note that the same double bond may be exocyclic to one ring, while endocyclic to the other and sometimes the same double bond may be exocyclic to two rings simultaneously. ' (iv) Alkyl Substituents and Ring Residues Only the alkyl substituents and ring residues attached to the carbon atoms constituting the conjugated system of the compound are taken into account. Following examples indicate such carbon atoms by numbers and the alkyl substituents and ring residues by dotted lines: i 4 3 2 1 2 3 2 1 CH3—CH;+CH=CH—CH = CH-CH; 4 CH—CH=CH, Two alkyl substituents Two ring residues  ) 3 e CH,—CH; Three ring residues Three ring residues and Three ring residues and one alkyl substituent two alkyl substituents In compounds containing both homoanular and heteroannular diene systems, the calculations are based on the longer wavelength (253 nm), i.e. the homoannular diene system.  Now, for the calculation of the A_,_values for & — n* transition we should use the equation: Base value + Substituent contributions + Solvent corrections (when necessary). A few illustrations of calculations of 4 values for © — n* transitions (solvent ethanol) of different diene systems, based on Woodward-Fieser's empirical rules, are given below. H H CH;, H N\ 7/ H\ /C'=C\ CH3\ C-*C\ C=C H C=C H \H CH, / H o Transoid: 214 nm Transoid: 214 nm Observed: 217 nm Alkyl groups: 3 X 5= 15 229 nm Observed: 228 nm The calculated and observed values of Amax usually match within 5 nm as shown in the following examples illustrating the applications of Woodward- Fieser rules Example 1. Using Woodward-Fieser rules, calculate A, for 2,3-dimethyl-1,3- butadiene. H2C=CI — (I:=CH2 CH3; CHj; It is an acyclic diene with two alkyl substituents. Thus, A,y of this compound is Base value 214 nm Two alkyl substituents (2 X 5) 10 nm Calculated A, 224 nm Observed A,y 226 nm Example 2. Calculate the wavelength of the maximum UV absorption for Myrcene Since, it is an acyclic diene with one alkyl substituent, thus Base value 214 nm One alkyl substituent 5 nm Calculated A, 219 nm Observed Apax 224 nm 10  Example 3. Predict the value of A, for B-phellandrene This is a heteroannular diene (conjugated double bonds are not in the same ring) with two ring residues and one exocyclic double bond, hence Base value 214 nm Two alkyl substituents (2 X 5) 10 nm One exocyclic double bond 5 nm Predicted Amax 229 nm Observed Ay 232 nm Example 4. Applying Woodward-Fieser rules, calculate the value of absorption maximum for the ethanolic solution of This is a homoannular diene with three ring residues and one exocyclic double bond, thus Base value 253 nm Three ring residues (3 X 5) 15 nm One exocyclic double bond 5 nm Calculated A,y 273 nm Observed A,y 275 nm (€, 10,000) 12 Example 5. Calculate A, for the ethanolic solution of '\,‘\ This is a heteroannular diene with three ring residues and one exocyclic double bond, thus Base value 214 nm Three ring residues (3 X 5) 15 nm One exocyclic double bond 5 nm Calculated A,y 234 nm Observed Ay 235 nm (€pax 19,000) 13 Example 6. Hydrogenation of one mole of the triene A with one mole of H, gives isomeric dienes having molecular formula C,oH;4. Show how UV spectroscopy and the expected Ap,, values could distinguish these isomers: 0 The hydrogenation of one mole of the triene A with one mole of H, may give three isomeric dienes B, C and D with molecular formula C,oH,4: 14  g 00 Diene B is a heteroannular conjugated diene with three ring residues and one exocyclic double bond, thus its expected Ay, = 214 + 3 X 5 + 5 = 234 nm. Diene C being a homoannular conjugated diene with three ring residues and one exocyclic double bond, thus its expected A, =253 +3 X5+ 5 =273 nm. Diene D is an unconjugated diene, hence A,,,x < 200 nm. Thus, by comparing the expected A, values of the isomeric dienes B, C and D with their observed values of A,,, we can distinguish these isomers. 15 Example 7. An organic compound can have one of the following structures: L (a) (b) (©) The A, of the compound is 236 nm. Which is the most likely structure of the compound? Explain your choice. Let us calculate the A, for each of these structures: Structure (a) is a homoannular diene with two ring residues and one alkyl substituent, hence its calculated A, = 253 + 3 X 5 = 268 nm. Structure (b) represents a heteroannular diene with four ring residues, hence its calculated A, = 214 + 4 X 5 = 234 nm. Structure (c) shows an acyclic conjugated diene with one alkyl substituent, hence its calculated A =214 + 5 = 219 nm. Since the given Ap,y of the compound is 236 nm, its most likely structure is (b) because the calculated A, (234 nm) for this structure is very close to the given value (236 nm). 16 Example 8. Applying Woodward-Fieser rules, calculate the value of absorption maximum for Itcontains both homoannular and heteroannular diene systems but the calculation of its Apax Will be based on the homoannular diene system. There are six ring residues attached to the carbon atoms of the entire conjugated system, one double bond extending conjugation, two exocyclic double bonds and one double bond exocyclic to tworings simultaneously. Thus, A, of this compound is calculated as: Base value 253 nm Six ring residues (6 x 5) 30 nm One double bond extending conjugation 30 nm Two exocyclic double bonds (2 X 5) 10 nm One double bond exocyclic to two rings simultaneously 10 nm Calculated Ap,, 333 nm  CH\ CH; CH; CH;4 H,C COOH CH;COO Cisoid: 253 nm Cisoid: 253 nm Alkyl substituent: 5 Ring residues: 5 X 5= 25 Ring residues: 3 X 5= 15 Double-bond-extending conjugation: 2 X 30 = 60 Exocyclic double bond: __ 5 Exocyclic double bond: 3 X 5 = 15 278 nm CH;COO0—: 0 Observed: 275 nm =58 nm Observed: 355 nm 18 Example 4 Ring residue Base value =215 nm Three ring residues = 3x 5 nm eo --Ring residue One exocyclic double bond = 5 nm Ring residue Calculated value 4 =235 nm Heteroannular Observed value = 235 nm diene system Example 5 B Base value =253 nm Ring residue Two extended double bonds =2 x 30 nm fing residu Five ring residues = 5 X 5 nm @ Ring residue One exocyclic double bond = 5 nm Seaniiad L il Calculated value 2, =343 nm conjugation diene system Observed value = 345 nm 19  Ring residue Base value =253 nm Ring residue One extended double bond =1 x 30 nm Four ring residues =4 x 5 nm Two exocyclic double bonds =2 x 5 nm Ring residue-- Iilng residue Calculated value 4 =313 nm Extended~~ Homoannular conjugation diene system Observed value =312 nm Example 7 Ring residue Ring residue Base value =215 nm Ring residue One extended double bond = 1 x 30 nm Five ring residues = 5 x 5 nm Ring residue Three exocyclic double bonds =3 x 5 nm Ring residue Heteroannular Calculated value 4, =285 nm Extended diene system Observed value = 287 nm conjugation Example 8 Ring residue Ring residue. Homoannular Base value = 253 nm diene system Five ring residues = 5 x 5 nm _ Ring residue Three exocyclic double bonds = 3 x 5 nm Calculated value 2, =293 nm When both heteroannular and homoannular diene Observed value =295 nm systems are present, then A value based on 20 homoannular diene is considered to be the base value. Example 9 i i Extended Ring residue conjugation Base value = 253 nm Ring residue One extended double bonds = 30 nm Five ring residues = 5 x 5 nm Three exocyclic double bond =3 x 5 nm Ring residue Calculated value 4, =323 nm o H nular Ring residue 3 omoanmLTa) Observed value = 325 nm diene system Example 10 Extended conjugation ) ) Base value = 253 nm Sogradin S resion Two extended double bonds =2 x 30 nm Five ring residues = 5 X 5 nm P> Three exocyclic double bond = 3 x 5 nm Ring residue - Ring residue Calculated value 4 =353 nm Homoannular Observed value = 355 nm Extended diene system conjugation 21 e = It is an acyclic diene with two alkyl substituents and one double bond which extends conjugation. Base value = 215 nm Increments for: Two alkyl substituents (2 x5 nm) = 10 nm One conjugated C=C = 30 nm s Calculated 4, = 255 nm oul ’ \r\_/ 7 exocyclic double bonds It is a heteroannular diene with two exocyclic double bond, four ring residues and one double bond that extends conjugation. Base value = 215 nm "Increments for: Four ring residues (4 x 5 nm) = 20 nm Two exocyclic double bonds (2 x 5 nm) = 10 nm One conjugated C=C = 30 nm Calculated A, = 275 nm  €Xy It is a homoannular diene with four ring residues and two exocyclic double bonds. Base value = 253 nm Increments for: Four ring residues (4 x 5nm) = 20 nm Two exocyclic double bonds (2 x 5 nm) = 10 nm r Calculated 4, = 283 nm Wy rz/ P Al It is a heteroannular diene with four ring residues. Base value = 215 nm Increment for: Four ring residues (4 x 5 nm) = 20 nm Calculated 4, = 235 nm 23 AcO Tt is 2 homoannular diene with five ring residues, three exocyclic double bonds, one —OAc substituent and two double bonds which extend conjugation. Base value = 253 nm Increments for: Five ring residues (5 x 5 nm) = 25nm Three exocyclic double bonds (3 x 5 nm)= 15 nm One —OAC substituent = 0 nm Two conjugated C=C (2 x 30 nm) = 60 nm Calculated Z,,,, = 353 nm It is a homoannular diene with five ring residues, one exocyclic double bond and two double bonds which extend conjugation. Base value = 253 nm Increments for: Five ring residues (5 x 5nm) = 25 nm One exocyclic double bond = 5 nm Two conjugated C=C (2 x 30 nm) = 60 nm Calculated 2, = 343 nm 24 Example 9. The following dienes have the experimental A, 243 and 265 nm in ethanol. Giving reasons, correlate the A, values to the structures (a) and (b): (a) (b) The calculated A, for (a) (a homoannular diene with two ring residues and two alkyl substituents) is 253 + 2 X 5 + 2 X 5 = 273 nm. Thus, structure (a) is the diene having experimental A.,,, 265 nm as this value is in fair agreement with the calculated A, for (a). The calculated Ay, for (b) (a heteroannular diene with two ring residues, two alkyl substituents and one exocyclic double bond) is 214 + 2 X5+ 2 X5+ 5= 239 nm. Thus, the structure (b) is the diene having the experimental A, 243 nm because this value is in agreement with the calculated A, for (b). Example 10. Which of the following compounds is expected to have higher value of A, and why? Ol (a) Oroween (b) Compound (a) is an unconjugated diene. Hence, it will have A, < 200 nm. The compound (b) is a heteroannular conjugated diene with two ring residues, one alkyl substituent and one exocyclic double bond. Thus, it is expected to have Amax (214 + 2 X 5 + 5+ 5) = 234 nm, i.e. higher than that of the compound (a). 26 (v) Exceptions to Woodward-Fieser Rules Distortion of the chromophore is the most important factor responsible for deviations from the predicted values of A, for dienes and trienes. Distortion of the chromophore may cause red or blue shifts depending on the nature of the distortion. Thus, the strained diene verbenene (III) has 4,,,, 245.5 nm compared to 232 nm of B-phellandrene (II), whereas the calculated value for both is 229 nm. In such bicyclic compounds, 15 nm is added as the ring strain correction to the calculated values of absorption maxima. The diene (IV) which is expected to have Apa, 234 nm but actually it is 220 nm (€y,, 10,050). This blue shift is due to the distortion of the chromophore resulting in the loss of coplanarity of the double bonds with consequent loss of conjugation. On the other hand, the expected coplanarity of the double bonds in the diene (V) is confirmed by its A, 243 nm (Emax 15,800) although it still does not agree with the calculated value 234 nm. 27  k. (I 1)) %) @ In general, if the strain in the molecule does not considerably affect the coplanarity of the conjugated system, then a bathochromic shift occurs. If the strain forces the chromophore out of coplanarity, then there is loss of conjugation and a hypsochromic shift occurs. The change of ring size of homoannular dienes from six-membered to any other, say five- or seven-membered, leads to deviations from predicted values due to distortion of the chromophore resulting in the loss of conjugation. Thus, calculations for homoannular dienes are valid only for six-membered rings. Woodward-Fieser rules work well only for conjugated systems containing up to four double bonds. These rules do not apply satisfactorily to cross (branched)- conjugated svstems like :)=