Woodwards fischer rules to calculate lambda max in UV spectroscopy?

Answer 1

How to calculate lambda max for Dienes and trienes (etc) in the ring system .

Woodward and Fieser's rules for estimating the position of the maximum absorbance wavelength in ultraviolet spectroscopy:

1. If the parent is heteroannular diene(which is decided by checking the presence of atlist two conjugated double bond in the ring), start with 214. For homoannular is the system containing twoconjugated double bond in six member ring, begin with 253.

2. Add 30 for each double bond which extends the conjugation and check its potion with the double bond of system if it in cis position then add 39 nm,for trans addition is nil

3. Add 5 for each carbon substituent on the double bonds

4. Add 5 for exocyclic double bonds.

5. The numbe you ow have is the calulated maximum absorbance wavelength (lambda max) in nanometers (nm)

NOTE: The terms here are not quite as self explanatory as you might think, and change where carbonyl conjugaion etc come into it (parent conjugated carbonyl system: Start with 215, alpha organic substituent: add 10, beta: add 12, gamma/delta: add 18, exocyclic DB: add 5, presence of a homodiene system: add 39, C1 substituent does not seem to add anything) etc. So look them up; or better yet, find a textbook. To test yourself take a look at cholesta-2,4,6-triene; you should get 303nm (the 4-5 DB is exocyclic)

Answer 2

I) CONJUGATED DIENE CORRELATIONS:

a) Homoannular Diene:-Cyclic diene having conjugated double bonds in same ring.

b) Heteroannular Diene:-Cyclic diene having conjugated double bonds in different rings.

c)Endocyclic double bond:-Double bond present in a ring.

d)Exocyclic double bond: -Double bond in which one of the doubly bonded atoms is a part of a ring system.

Here Ring A has one exocyclic and endocyclic double bond. Ring B has only one endocyclic double bond.

PARENT VALUES AND INCREMENTS FOR DIFFERENT SUBSTITUENTS/GROUPS:

I) CONJUGATED DIENE CORRELATIONS:

i) Base value for homoannular diene = 253 nm

ii) Base value for heteroannular diene = 214 nm

iii) Alkyl substituent or Ring residue attached to the parent diene = 5 nm

iv) Double bond extending conjugation = 30 nm

v) Exocyclic double bonds = 5 nm

vi) Polar groups: a) OAc = 0 nm

b) OAlkyl = 6 nm

c) Cl, Br = 5 nm

Eg:

Base value = 214 nm

Ring residue = 3 x 5 = 15 nm

Exocyclic double bond = 1 x 5 = 5 nm

λmax= 214+15+5=234 nm

II) α, β UNSATURATED CARBONYL COMPOUNDS OR KETONES:

1. Base value: a) Acyclic α, β unsaturated ketones = 214 nm

b) 6 membered cyclic α, β unsaturated ketones = 215 nm

c) 5 membered cyclic α, β unsaturated ketones = 202 nm

d) α, β unsaturated aldehydes = 210 nm

e) α, β unsaturated carboxylic acids & esters = 195 nm

2. Alkyl substituent or Ring residue in α position = 10 nm

3. Alkyl substituent or Ring residue in β position = 12 nm

4. Alkyl substituent or Ring residue in γ and higher positions = 18 nm

5. Double bond extending conjugation = 30 nm

6. Exocyclic double bonds = 5 nm

7. Homodiene compound = 39 nm

8. Polar groups: a) –OH in α position = 35 nm

–OH in β position = 30 nm

–OH in δ position = 50 nm

b) –OAc in α, β, γ, δ positions = 6 nm

c) –OMe in α position = 35 nm

–OMe in β position = 30 nm

–OMe in γ position = 17 nm

–OMe in δ position = 31 nm

d) –Cl in α position = 15 nm

–Cl in β position = 12 nm

e) –Br in α position = 25 nm

–Br in β position = 30 nm

f) –NR2in β position = 95 nm

Eg:

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

III) AROMATIC COMPOUNDS:

1)Base value: for a) ArCOR = 246 nm

b) ArCHO = 250 nm

c) ArCO2H = 230 nm

d) ArCO2R = 230 nm

2) Alkyl group or ring residue in ortho and meta position = 3 nm

3) Alkyl group or ring residue in para position =10 nm

4)Polar groups: a) –OH, –OCH3, –OAlkyl in o, m position = 7 nm

b) –OH, –OCH3, –OAlkyl p position = 25 nm

c) –O (oxonium) in o position = 11 nm

d) –O (oxonium) in m position = 20 nm

e) –O (oxonium) in p position = 78 nm

f) –Cl in o, m position = 0 nm

g) –Cl in p position = 10 nm

h) –Br in o, m position = 2 nm

i) –Br in p position = 15 nm

j) –NH2in o, m position = 13 nm

k) –NH2in p position = 58 nm

l) –NHCOCH3in o, m position = 20 nm

m) –NHCOCH3in p position = 45 nm

n) –NHCH3in p position = 73 nm

o) –N(CH3)2in o, m position = 20 nm

p) –N(CH3)2in p position = 85 nm

Eg:

Base value =246 nm

Ring residue in o- position = 1 x 3 = 3 nm

Polar group -OCH3in p- position = 25 nm

λmax= 274 nm