Spectroscopy Structural Analysis (Application of Woodward-Fieser Rules) U.A. Deokate Copyright © by U. A. Deokate, all rights reserved.

Woodward-Fieser Rules It is used for finding absorption maxima position (λMax) This differs from observed values by 5-6%. Effect of substituent groups can be reliably quantified by use Woodward –Fieser Rule Separate values for conjugated dienes and trines and α-β-unsaturated ketnones are available

Woodward-Fieser Rules Woodward and the Fiesers performed extensive studies of terpene and steroidal alkenes and noted similar substituents and structural features would predictably lead to an empirical prediction of the wavelength for the lowest energy p p* electronic transition This work was distilled by Scott in 1964 into an extensive treatise on the Woodward-Fieser rules in combination with comprehensive tables and examples – (A.I. Scott, Interpretation of the Ultraviolet Spectra of Natural Products, Pergamon, NY, 1964) A more modern interpretation was compiled by Rao in 1975 – (C.N.R. Rao, Ultraviolet and Visible Spectroscopy, 3rd Ed., Butterworths, London, 1975)

Terminology Alkenes are named by dropping the -ane ending of the parent and adding -ene. Conjugated :organic compound; containing two or more double bonds each separated from the other by a single bond Dienes: One of a class of organic compounds containing two ethylenic linkages (carbon-to-carbon double bonds) in the molecules. Also known as alkadiene; diolefin.

Terminology Endocyclic double bonds have both carbons in the ring and exocyclic double bonds have only one carbon as part of the ring. Cyclopentene is an example of an endocyclic double bond. Methylenecylopentane is an example of an exocyclic double bond.

Woodward Fieser Rule for Dienes & Polyenes

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Woodward Fieser rule for undsaturated carbonyl compound

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EMPIRICAL RULES FOR BENZOYL DERIVATIVES

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Fieser Rules If more than four double bonds are present the Woodward rule is not applicable The plant pigments contains double bonds up to 11 for this Fieser Rules is used. λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo n=no of conjugated double bonds M= no of alkyl or alkyl like Substituents Rendo= No of endocyclic double bonds Rexo = No of exocyclic double bonds

λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo Fisher Problem-1 λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo = 114 + 5 (8) + 11(48-1.7x 11) –0 -0 = 476 Fisher formula

λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo Fisher Problem-2 λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo = 114 + 5 (8) + 11(48-1.7x 11) –0 -0 = 476 Fisher formula

λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo Fisher Problem-3 λMax= 114 + 5(M) +n(48-1.7n)- 16.5Rendo-10 Rexo = 114 + 5 (8) + 18(48-1.7x 18) –0 -0 = Fisher formula