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Robert B. Woodward [1917-1979] ,Nobel Prize Winner in Chemistry 1965 "for his outstanding achievements in the art of organic synthesis”[Synthesis of Vitamin - B 12 ] 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) Woodward and Fieser Rule Several sets of empirical derived rules to estimate the absorption maxima (λ max ) in UV spectrum. It depends upon contribution upon substitution like Polyenes, conjugated carbonyl [ketone & aldehyde]. Louis Frederick Fieser (1899 -1977) Synthesis ofVitamin - K

Unit-II_woodword Fiers Rule

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Page 1: Unit-II_woodword Fiers Rule

Robert B. Woodward [1917-1979] ,Nobel Prize Winner in Chemistry 1965 "for his outstanding achievements in the art of organic synthesis”[Synthesis of Vitamin - B12]

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)

Woodward and Fieser Rule Several sets of empirical derived rules to estimate the absorption maxima (λmax) in UV spectrum.

It depends upon contribution upon substitution like Polyenes, conjugated carbonyl [ketone & aldehyde].

Louis Frederick Fieser (1899 -1977) Synthesis ofVitamin - K

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Woodward-Fieser Rules for Dienes

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Exo-cyclic double bond that lies outside a given ring.

It may lie within one ring even though it is outside anther ring.

Often an exocyclic double bond will be found at a junction point on rings.

Three exocyclic double bond

Exo-cyclic double bond

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Woodward-Fieser Rules – Cyclic DienesBe careful with your assignments – three common errors:

This compound has three exocyclic double bonds; the indicated bond is exocyclic to two rings

This is not a heteroannular diene; you would use the base value for an acyclic diene

This is not a homooannular diene; you would use the base value for an acyclic diene

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For example:

Isoprene - acyclic butadiene = 217 nm one alkyl subs. = + 5 nm calculated 222 nm

Experimental value 220 nm

Isoprene - acyclic butadiene = 217 nm three alkyl subs. (3X5) = +15 nm

calculated 232 nm Experimental value 234 nm

Worked Problem

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Where ,R = H

Transoid : =214 nm Alkyl group (or)Ring residue : (3X5) = 15 nmExocyclic double bond = 5 nm ---------------Calculated: 234 nmObserved: 235 nm

Find : R = -OCH2CH3, OCOCH3

R=-OCH2CH3 , λmax = 240nm,

R=-OCOCH3 , λmax = 234nm.

Worked Problem

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Cisoid : =253 nm Alkyl group (or)Ring residue : (2X5) = 10 nm ---------------Calculated: 263 nmObserved: 256 nm

Cisoid : =253 nm

Alkyl group (or)

Ring residue : (3X5) = 15 nm

---------------

Calculated: 268 nm

Cisoid : =253 nm Alkyl group (or)Ring residue : (4X5) = 20 nmExocyclic double = 5 nmbond ---------------Calculated: 278 nmObserved: 275 nm

Worked Problem

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Cisoid : =253 nm Alkyl group (or)Ring residue : (4X5) = 20 nmExocyclic double = 5 nmbond ---------------

Calculated: 278 nmObserved: 275 nm

Worked Problem

Acyclic : =217 nm

Alkyl group (or)

Ring residue : (5X5)= 25 nm

Exocyclic : (2x5) = 10nm

Double bond ---------------

Calculated: 252 nm

Palustric acid Neoabietic acid

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Transoid : =214 nm Alkyl group (or)Ring residue : (3X5) = 15 nmExocyclic : (1x5) = 05nmDouble bond ---------------Calculated: 234 nm

Transoid : =214 nm Alkyl group (or)Ring residue : (4X5) = 20 nmExocyclic : (2x5) = 10nmDouble bond ---------------Calculated: 244 nm

Cisoid : =253 nm Alkyl group (or)Ring residue : (4X5) = 20 nmExocyclic : (2X5) = 10 nmbond ---------------Calculated: 283 nm

Worked Problem

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Home Work Problems

a) Assign the λmax (275, 239 and 235) of the following steroids ?

(i) (ii) (iii)

b) Assign the λmax (235, 268 and 241) of the following steroids ?

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Find the λmax of the following compound

Home work problems [Out of syllabus]

353

249284

323

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Woodward-Fieser Rules for α,β-unsaturated carbonyl a,b-unsaturated corbony compounds or Enol are a conjugated system of an alkene and a ketone.

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Woodward-Fieser Rules for α,β-unsaturated carbonylGroup Increment (nm)

6-membered ring or acyclic enone 215

5-membered ring parent enone

Acyclic dienone

202

245

Double bond extending conjugation 30

Alkyl group or ring residue α, β, γ and higher

10, 12, 18

-OH α, β, γ and higher

35, 30, 18

-OR α, β, γ, δ 35, 30, 17, 31

-OCOR (OAC) α, β, δ 6

-Cl α, β 15, 12

-Br α, β 25, 30

-NR2 β 95

Exocyclic double bond 5

Homocyclic diene component 39

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Acyclic enone : = 215 nm

2 X β Alkyl group (or)

Ring residue : (2X12) = 24 nmExocyclic : (0) = 0nmDouble bond ---------------Calculated: 239 nmObserved: 237 nm

Worked problems for α,β-unsaturated carbonyl compound

Cyclicenone (6-member) : = 215 nm

2 X β Alkyl group (or)

Ring residue : (2X12) = 24 nm

OH sub. at α Position : (1X35) = 35nm

---------------Calculated: 274 nmObserved: 275 nm

Acyclic enone : = 215 nm

1 X β Alkyl group : (1X12) = 12 nm

1Xα Alkyl group : (1X10) = 10nm

---------------Calculated: 237 nmObserved: 236 nm

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Cyclic enone (6-member) : = 215 nm

2 X β Alkyl group (or)

Ring residue : (2X12) = 24 nmExocyclic : (0) = 0nmDouble bond ---------------Calculated: 239 nmObserved: 238 nm

Cyclic enone (6-member) : = 215 nm

2 X β Alkyl group (or)

Ring residue : (2X12) = 24 nmExocyclic : (1X5) = 5nmDouble bond ---------------Calculated: 244 nmObserved: 241 nm

Cyclic enone (6-member) : = 215 nm 2 X β Alkyl group (or)Ring residue : (2X12) = 24 nm1Xα Alkyl group : (1X10) = 10nmExocyclic : (2X5) = 10nmDouble bond ---------------Calculated: 259 nmObserved: 256 nm

Worked problems for α,β-unsaturated carbonyl compound

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Can these two isomers be discerned by UV-spectrum

Cyclic enone (6-member) : = 215 nm 1 X β Alkyl group (or)Ring residue : (1X12) = 12 nm1Xα Alkyl group : (1X10) = 10nmExocyclic : (1X5) = 05nmDouble bond ---------------Calculated: 242 nm

Cyclic enone (6-member) : = 215 nm 1 X β Alkyl group (or)Ring residue : (2X12) = 24 nm0 Xα Alkyl group : (0) = 0nmExocyclic : (1X5) = 05nmDouble bond ---------------Calculated: 244 nm

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Cyclic enone (6-member) : = 215 nm 1 X β Alkyl group (or)Ring residue : (1X12) = 12 nm0X α Alkyl group : (0) = 0 nm0X γ Alkyl group : (0) = 0 nm

1X δ Alkyl group : (1X18) = 18 nm

Exocyclic = bond : (1X5) = 5 nmExtended Double bond : (1X30) = 30 nm ---------------Calculated: 280 nmObserved: 277 nm

Cyclic enone (6-member) : = 215 nm 0 X β Alkyl group (or)Ring residue : (0) = 0 nm0X α Alkyl group : (0) = 0 nm1X γ Alkyl group : (1X18) = 18 nm

1X δ Alkyl group : (1X18) = 18 nm

Exocyclic = bond : (1X5) = 5 nmExtended Double bond : (1X30) = 30 nm ---------------Calculated: 286 nmObserved: 290 nm

Worked problems for α,β-unsaturated carbonyl compound

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Cyclic enone (5-member) : = 202 nm 1 X β Alkyl group (or)Ring residue : (1X12) = 12 nm0X α Alkyl group : (0) = 0 nm1X γ Alkyl group : (1X18) = 18 nm

1X δ Alkyl group : (1X18) = 18 nm

Exocyclic = bond : (1X5) = 5 nmExtended Double bond : (1X30) = 30 nm ---------------Calculated: 285 nmObserved: 287 nm

Worked problems for α,β-unsaturated carbonyl compound

Cyclic enone (6-member) : = 215 nm 0 X β Alkyl group (or)Ring residue : (0) = 0 nm1X α Alkyl group : (1X10) = 10 nm0X γ Alkyl group : (0) = 0 nm

1X δ Alkyl group : (1X18) = 18 nm

Exocyclic = bond : (1X5) = 5 nmExtended Double bond : (1X30) = 30 nmHomoannular conjugated : (39) = 39 nm diene ---------------Calculated: 317 nmObserved: 319 nm

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Worked problems for cross- conjugated system:

Cyclic enone (6-member) : = 215 nm 2 X β Alkyl group (or)Ring residue : (2X12) = 24 nm0Xα Alkyl group : (0X10) = 0nmExocyclic : (1X5) = 05nmDouble bond ---------------Calculated: 244 nm

Find the λmax max of the following

compounds

Cyclic enone (6-member) : = 215 nm 1 X β Alkyl group (or)Ring residue : (1X12) = 12 nm0Xα Alkyl group : (0X10) = 0 nmExocyclic : (0) = 0 nmDouble bond ---------------Calculated: 237 nm

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Unsaturated Carboxylic acid: = 195 nm 1X β Alkyl group : (1X12) = 12 nm1Xα Alkyl group : (1X10) = 1 0 nmExocyclic = bond : (0) = 0 nm ---------------Calculated: 217 nmObserved: 217 nm

Unsaturated aldehyde: = 208 nm 1X β Alkyl group : (2X12) = 24 nm1Xα Alkyl group : (1X10) = 1 0 nmExocyclic = bond : (0) = 0 nm ---------------Calculated: 242 nmObserved: 242 nm

Unsaturated aldehyde: = 208 nm 1X α Alkyl group : (1X10) = 10 nm1X δ Alkyl group : (1X18) = 18 nm

Exocyclic = bond : (0) = 0 nmExtended Double bond : (1X30) = 30 nmHomoannular conjugated : (39) = 39 nm diene ---------------Calculated: 305 nmObserved: 302 nm

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• Detection of functional group: To dectect the presence and absence of the funtional group (chromophore) like Conjugation, carbonyl (aldehyde, ketone), benzene (Aromatic compounds) and halo compound ( bromo or iodo atom).

• Extent of conjugation: The extent of conjugation in polyenes can be estimated. An increase in the double bonds shifts the absorption to longer wave length region (Bathochromic shift). it is found that the absorption occurs in the visible region. i.e at about 420 nm, if n=8 in the above polyene.

• n=8 and above alkenes appears colored to human eye.

• Distinction in conjugated and non-conjugated compounds:

• The forbidden n=>π* band for the carbonyl group in the compound (2)

will appear at longer wave-length compared to that for the compound (1).

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Application of UV Spectroscopy

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• Elucidation of the structure of Vitamins A and K:

• Due to presence of additional double bond.

• Preference over two tautomeric foam:

• The spectra of the two compound were found to favour 2-pyridone which is an a,b-unsaturated ketone and the equilibrium shifted towards the right in polar solvents.

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Application of UV Spectroscopy

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• Identification of a compound in different solvents:

• some times the structure of the compound changes with the change of the solvents. Eg.Chloral hydrate CCl3.CHO.H2O.

• In Hexane shows absorption at 290nm

• In water no absorption observed due to

• Determination of configuration of Geometrical isomers:

• Generally trans forms are more stable than cis form.

• UV absorbance for trans-resveratrol UV lambda(max) (EtOH) nm (epsilon) 308 (30 000) and cis-resveratrol UV lambda(max) (EtOH) nm (epsilon) 288 (12 600).

• Cis form absorbs at lower wave length and trans forms absorbs at higher wave length.

• http://www.mendeley.com/research/resveratrol-isomeric-molar-absorptivities-stability/ 23

Application of UV Spectroscopy