Woodward–Fieser rules: calculating λmax of enones and dienes
By Vimal Swarnkar (VKC) · UPSC Chemistry · Civil Services, Combined Geo-Scientist and IFoS
Every year there is a question that gives you a few structures and asks for λmax. It looks like a table to memorise, but it is only a count: one base value and a few additions. In this post I will show you how often UPSC asks it, which book to use, how I plan the answer, and how the answer should look in your booklet.
Using Woodward-Fieser rules, calculate λmax for the following compounds A, B, C and D:
λmax = base value + increments. For an enone the base is 215 nm (open chain or six-membered ring), 202 nm (five-membered ring) or 207 nm (aldehyde). Add 10, 12, 18 nm for each alkyl group or ring residue on the α, β and γ-or-higher carbon, 30 nm for each extra C=C in conjugation and 5 nm for each exocyclic C=C. This gives A = 237 nm, B = 259 nm, C = 237 nm and D = 280 nm.
Full topic notes: UV spectroscopy and Woodward–Fieser rules · two-page short notes
A word from me
I have taught UV spectroscopy to many batches, and the first thing I see is students trying to memorise the whole table of increments. Please do not. The table is small, and the marks are not in the table. They are in the counting: which carbon is α, which is β, which ring residues belong to the chromophore, and which double bond is exocyclic. One wrongly counted ring residue changes the answer by 10, 12 or 18 nm, and then the whole answer is wrong, even if you remembered every number.
So my advice is simple. Keep one small rule card. For every structure, first mark α, β, γ and δ on the chromophore, then add. Do the 13 questions from the last five years once, by hand, and you will find that the same few structures keep coming back. If something does not work out, write to me on Telegram or at [email protected]. I would rather you ask than lose these easy marks.
Asked 13 times in the last five years
From 2021 to 2025, the Woodward–Fieser rules were used in 11 of 15 papers across Civil Services (Paper II), Combined Geo-Scientist (Paper III) and IFoS (Paper II): Civil Services in 4 of 5 years (2022 to 2025), IFoS in 4 of 5 years (2022 to 2025) and Geo-Scientist in 3 of 5 years (2022, 2023 and 2025). That is 13 questions in five years. It came 10 times directly, as “calculate λmax” for given structures, for 5 to 15 marks; the largest was Civil Services 2024 (15 marks). The other 3 times it came indirectly, as “which compound has the higher λmax” (Civil Services 2024) or “assign these structures to 249 nm and 287 nm” (Civil Services 2023 and Geo-Scientist 2025).
Syllabus and books
In Civil Services it is Paper II, Unit 7, Spectroscopy, part (iii) Electronic. IFoS asks it in Paper II and the Geo-Scientist exam in Mains Paper III. The Civil Services syllabus reads:
“Electronic: singlet and triplet states; n→π* and π→π* transitions; application to conjugated double bonds and conjugated carbonyls, Woodward-Fieser rules; charge transfer spectra.”
Two phrases in it decide this question: conjugated double bonds (dienes) and conjugated carbonyls (enones).
You do not need all of these books. Any one of them is enough for this question. I have listed three so that you can use whichever you already have.
- Book 1: Jagdamba Singh, Undergraduate Organic Chemistry, Vol. II, Chapter 1 (Spectroscopy), Section 1.17 “Woodward-Fieser rules for conjugated dienes and polyenes” and Section 1.18 “Woodward-Fieser rules for α,β-unsaturated carbonyl compounds”, with 16 worked examples. Section 1.19 shows where the rules fail. If you have this book, this is the one I would use.
- Book 2: Pavia, Lampman, Kriz and Vyvyan, Introduction to Spectroscopy, the chapter on ultraviolet spectroscopy, with the same tables of base values and increments.
- Book 3: Y. R. Sharma, Elementary Organic Spectroscopy, the UV chapter. Good if you want more practice problems.
How to build the answer
Read the tree from left to right. Every leaf is a point to write, in that order. The highlighted words are the ones the examiner looks for.
Given information
- Four compounds A, B, C, D
- Chromophore C=C–C=O (enone); D has one more C=C
Assumptions
- π → π* (K band), intense
- Empirical, ethanol, ± 5 nm
- Planar, unstrained chromophore
- Only groups on the chromophore count
Key law
- λmax = base value + Σ increments
Calculation
- Base from the ring that holds C=O: 215 / 202 / 207
- Mark α, β, γ, δ from C=O
- Ring residue: α 10, β 12, γ+ 18
- Extended conjugation +30
- Exocyclic C=C +5
Values
- Table for A to D
- 237, 259, 237, 280 nm
Check
- Recount every ring residue
Sketch
- Chromophore with α, β, γ, δ marked
Reading the result
- D longest: extended conjugation
- B above A: extra β residue and exocyclic C=C
- A = C by coincidence
Principle
- More conjugation, smaller HOMO–LUMO gap, longer λ
Application
- Δ4 and Δ1 steroidal enones told apart: 244 and 227 nm
Final answer
- A, B, C, D, double underlined
Diagrams that prove it
What gains marks
- The rule card first: base values and increments, in a small table, before any structure.
- α, β, γ, δ marked on every structure. The examiner can then see your count without hunting for it.
- The base value with its reason: “six-membered ring ketone, 215” or “C=O is outside the ring, so open-chain base”.
- One increment per line, with the group named: “β ring residue × 2 = 24”.
- The exocyclic double bonds named by ring, so that “exocyclic to both rings” earns +5 twice.
- A table of A to D with the totals, and a result with ± 5 nm.
- The words π → π* transition, extended conjugation and one application.
Match the length to the marks
I always tell my students to let the marks decide the length. For 5 marks, say IFoS 2025 (b)(i), I would write the rule card in one line, the count for each structure and the two totals, about half a page. For 10 marks, as in Civil Services 2025, write the numbered assumptions, the labelled structures, the table and the conclusion, the two pages below. For 15 marks, as in Civil Services 2024 with three dienones, add the homoannular diene component and a line on why the rules work.
How it looks in the booklet
This is the full 10-mark answer, in the order of the tree above. Page 1 sets up the rules and does A and B, and page 2 does C and D, then explains the result.
Q.)Using Woodward-Fieser rules, calculate λmax for the following compounds A, B, C and D. [CSE 2025], 10M
Ans.Given: four compounds, each with a C=C conjugated to C=O (an enone); D has two C=C (a dienone). Positions are marked α, β, γ, δ from C=O.
For a conjugated enone, assume that
- i) the band is the intense π → π* transition (K band, ε about 10,000 to 20,000) of the O=C–C=C chromophore.
- ii) the rules are empirical: λmax = base value + increments, for ethanol solution, good to ± 5 nm.
- iii) the chromophore is planar and unstrained.
- iv) only alkyl groups and ring residues on the chromophore carbons are counted.
Woodward–Fieser rule for enones:
| Item | nm |
|---|---|
| Base: open-chain or six-ring ketone | 215 |
| Base: five-ring ketone | 202 |
| Base: α,β-unsaturated aldehyde | 207 |
| Each C=C extending conjugation | + 30 |
| Homoannular diene component | + 39 |
| Exocyclic C=C (each) | + 5 |
| Alkyl / ring residue on α, β, γ and higher | + 10, + 12, + 18 |
All other symbols have their usual meaning. (For dienes the base values are 214 nm and 253 nm; see the other exams below.)
A, CH3–CH=C(CH3)–CO–CH3: open-chain ketone, base 215. One CH3 on Cα (+10) and one on Cβ (+12).
B, 2-cyclohexylidenecyclohexanone: six-ring ketone, base 215. Cα has one ring residue (+10). Cβ has two ring residues, 2 × 12 = 24. The C=C is exocyclic to both rings, 2 × 5 = 10.
C, 1-acetylcyclopentene: the C=O is outside the ring, so the open-chain base 215 applies, not 202. Cα and Cβ each carry one ring residue (+10, +12). The C=C is inside the ring, so there is no exocyclic term.
D: six-ring ketone, base 215. One extra C=C in conjugation (+30). Cβ has one ring residue (+12), Cδ has one (+18). The α=β bond is exocyclic to the second ring (+5). The CH3 is on a saturated carbon, so it is ignored. The two C=C are not in one ring, so no +39.
| Increment | A | B | C | D |
|---|---|---|---|---|
| Base | 215 | 215 | 215 | 215 |
| Extra C=C | – | – | – | 30 |
| α residue | 10 | 10 | 10 | – |
| β residue | 12 | 24 | 12 | 12 |
| δ residue | – | – | – | 18 |
| Exocyclic C=C | – | 10 | – | 5 |
| λmax / nm | 237 | 259 | 237 | 280 |
Check: the additions above 215 are 22, 44, 22 and 65 nm, and each column adds up to the total shown ✓.
Reading the result:
- i) D is the longest, 280 nm: the extra C=C (+30) and the δ residue (+18) lengthen the conjugation.
- ii) B is above A: the same base, but a second β residue (+12) and two exocyclic C=C (+10).
- iii) A and C are equal at 237 nm although one is open-chain and one is cyclic: each has one α and one β substituent.
Principle: a longer conjugated chain raises the HOMO and lowers the LUMO. The π → π* gap shrinks, and λ = hc/ΔE moves to the red. Alkyl groups on the chromophore add a little more. The weak n → π* band near 310 to 330 nm is not calculated.
Application: the rules fix the position of a C=C in a steroid or terpenoid. A Δ4-3-keto steroid is 215 + 2 × 12 + 5 = 244 nm, and its Δ1 isomer is 215 + 12 = 227 nm, so one spectrum tells them apart.
So, A = 237 nm, B = 259 nm, C = 237 nm, D = 280 nm (± 5 nm)
The same question in other exams
Civil Services 2022, Paper II, Q8(b)(i), 10 marks: two steroid polyenes. The first is a homoannular diene, so λmax = 253 + 60 + 25 + 15 = 353 nm. The second is heteroannular, so λmax = 214 + 30 + 20 + 10 = 274 nm.
Civil Services 2024, Paper II, Q7(a), 15 marks: three dienones. Use the enone card with the homoannular component (+39) added: 357, 324 and 317 nm.
Civil Services 2023, Paper II, Q8(a)(ii), 5 marks: run the other way. A ketone C7H12O with λmax = 249 nm is 3,4-dimethylpent-3-en-2-one: 215 + 10 + 2 × 12 = 249.
Combined Geo-Scientist 2025, Paper III, Q10(b)(i), 5 marks: two ketones at 249 nm and 287 nm. The 249 nm one is that same ketone, (CH3)2C=C(CH3)COCH3. The 287 nm one is the fused cyclopentenone: 202 + 30 (extra C=C) + 12 (β) + 18 + 18 (γ and δ ring residues) + 5 (exocyclic) = 285 nm. The same fused ketone is compound III in Geo-Scientist 2022, Q8(a).
IFoS 2023, Paper II, Q5(d), 5 marks: two named compounds. Read as 4-methoxypent-3-en-2-one, A = 215 + 30 (β-OMe) + 12 (β-CH3) = 257 nm. 3-Methoxy-2-methylbut-2-enal, B = 207 + 35 (α-OMe) + 2 × 12 = 266 nm.
The diene version has its own two base values: 214 nm for an open-chain or heteroannular diene and 253 nm for a homoannular diene, then +30 for each extra C=C, +5 for each alkyl group or ring residue and +5 for each exocyclic C=C. If a compound has both kinds of diene, use 253. Learn the two cards once and every version of the question becomes the same question.
Questions students ask
How do you calculate λmax by the Woodward–Fieser rules?
Pick the base value from the chromophore: 215 nm for an open-chain or six-ring enone, 202 nm for a five-ring enone, 207 nm for an enal, 214 nm for a heteroannular or open-chain diene and 253 nm for a homoannular diene. Then add 30 nm for each extra conjugated C=C, 5 nm for each exocyclic C=C, and for an enone 10, 12 or 18 nm for each alkyl group or ring residue on the α, β or γ-and-higher carbon. In Civil Services 2025, Paper II, Q7(a)(ii), CH₃CH=C(CH₃)COCH₃ gives 215 + 10 + 12 = 237 nm.
What is the base value for a cyclopentenone, a cyclohexenone and an enal?
Cyclopentenone 202 nm, cyclohexenone and open-chain enone 215 nm, α,β-unsaturated aldehyde 207 nm. The base depends on the ring that holds the C=O. In Civil Services 2025, Q7(a)(ii), compound C, 1-acetylcyclopentene, has its C=C in a five-membered ring but its C=O outside it, so the base is 215 nm, not 202 nm.
What counts as an exocyclic double bond in the Woodward–Fieser rules?
A C=C with one carbon inside a ring and the other outside it. Each such ring adds 5 nm, so a C=C that is exocyclic to two rings adds 10 nm. In 2-cyclohexylidenecyclohexanone (compound B, Civil Services 2025, 10 marks) the C=C is exocyclic to both rings, so 2 × 5 = 10 nm is added.
What is a homoannular diene, and why is its base value 253 nm?
A diene whose two C=C are in the same ring, held s-cis. The s-cis arrangement is strained, which raises the ground-state energy, so the transition needs less energy and the band moves to a longer wavelength: 253 nm against 214 nm for a heteroannular, s-trans diene. If a compound has both kinds, the calculation uses 253 nm. Civil Services 2022, Q8(b)(i), 10 marks, has both in one question: 353 nm and 274 nm.
Why do the Woodward–Fieser rules sometimes fail?
The rules assume a planar, unstrained chromophore. In the worked examples of Jagdamba Singh, a strained ring calculates to 229 nm but is observed at 246 nm, a twisted diene calculates to 234 nm but is observed at 220 nm, and a cross-conjugated dienone calculates to 280 nm but is observed at 298 nm. Twelve of the 16 worked examples agree to within 5 nm.
How many times has UPSC asked the Woodward–Fieser rules?
From 2021 to 2025 it appeared in 11 of 15 papers across Civil Services Paper II, Geo-Scientist Paper III and IFoS Paper II, 13 questions in all. Ten were direct “calculate λmax” questions, for 5 to 15 marks, and three were indirect. The four papers with no Woodward–Fieser question were Civil Services 2021, Geo-Scientist 2021 and 2024, and IFoS 2021.
Read the complete answers
By VKC, your mentor. Integrated M.Sc. Chemistry, IIT Roorkee. CSIR NET Chemical Sciences, All India Rank 4. New posts are announced first on Telegram.