Calculate
for the
the lambda max
compound.
Answers
Answer:
) 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