Question

Calculate
for the
the lambda max
compound.​

Answer 1

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