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VKC Notes · Spectroscopy · Civil Services P-II · IFoS P-II · Geo-Scientist P-III

13.1 UV Spectroscopy and the Woodward–Fieser Rules

★★★ Asked in every paper of all three exams, 2021–2025: 25 parts, 172 marks (Civil Services 9 parts / 75 m · IFoS 8 / 50 m · Geo-Scientist 8 / 47 m). A 10-mark “calculate λmax” comes almost every year.

Syllabus lines. Civil Services and IFoS: “Ultraviolet spectroscopy: types of electronic transitions, Woodward–Fieser rules for conjugated dienes and enones” (as part of the spectroscopy unit). Geo-Scientist: UV–visible spectroscopy under spectroscopic methods of analysis. IR, NMR and mass spectra have their own notes.

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CSE 2024 Q7(a) = asked there; tap to open the solved answerSN = line kept in the short notes

13.1.1 Electronic transitions and absorption bands ★★★

σ*π*nπσσ→σ*<150 nmn→σ*150–250π→π*strongn→π*weak, longest λE
Fig. 13.1.1 The four transitions.
Draw it in 60 s: five lines σ, π, n, π*, σ* from bottom; four upward arrows; the shortest arrow (n→π*) is the longest wavelength.
  • UV–visible light (200–800 nm; vacuum UV below 200) promotes an electron from a filled to an empty MO. Energy needed: σ→σ* > n→σ* > π→π* > n→π*, so n→π* lies at the longest wavelength.SN
  • Allowed π→π* bands are strong (ε 10³–10⁵): the K band of conjugated systems. Forbidden n→π* bands are weak (ε 10–100): the R band of C=O; the n orbital lies in the molecular plane, perpendicular to π*, so overlap is poor.SN
  • Benzene: E bands 184, 204 nm and the B band at 254 nm (ε about 200, fine structure).
  • Terms: chromophore (the group that absorbs: C=C, C=O, N=O), auxochrome (a lone-pair group that shifts it: OH, NH₂, OR, Cl), bathochromic (red, longer λ), hypsochromic (blue), hyperchromic (ε up), hypochromic (ε down).SN
(13.1.1)

Solved PYQ CSE 2021 Q8(b)(ii) 5m CSE 2025 Q7(a)(i) 5m

Acetone absorbs weakly at 280 nm and strongly at 190 nm. Assign.

  1. 280 nm, ε ≈ 15: n→π* of C=O. The lone pair lies above π, so this needs the least energy; n and π* are orthogonal, so the band is forbidden and weak.
  2. 190 nm, ε ≈ 1000: π→π*, allowed, needs more energy. (Some texts call it n→σ*; either way it is the strong short-wavelength band.)
  • Methyl halides show n→σ*: CH₃Cl 173, CH₃Br 204, CH₃I 258 nm; the lone pair rises and σ* falls from Cl to I. Camphor: c = A/εl = 2.52/(14 × 10) = 0.018 M Geo 2024 Q9(c) 5m.
  • Alkene vs acetone intensity: the allowed π→π* of an alkene is far stronger than acetone's n→π* IFoS 2023 Q6(e). Pentan-3-one (π→π* below 200, n→π* ≈ 280, ε ≈ 20) vs pent-3-en-2-one (π→π* ≈ 224 nm, ε ≈ 10⁴; n→π* ≈ 315, weak) IFoS 2024 Q7(b)(i) 10m.

13.1.2 Solvent and pH effects ★★ Geo-Scientist favourite

π*nπhexaneπ*nπethanol / waterH-bonding lowers n most: n→π* blue-shifts; π→π* red-shifts
Fig. 13.1.2 Why polar solvents move the two bands in opposite directions.
Draw it in 60 s: three lines π, n, π* for hexane; next to it the same with n pushed well down and π* a little down → write blue for n→π*, red for π→π*.
  • A protic solvent hydrogen-bonds to the lone pair and lowers n strongly; the excited state, with one n electron gone, is stabilised much less: n→π* shifts to shorter λ (blue).SN
  • π→π*: the excited state is more polar and is stabilised more by a polar solvent: red shift.SN
  • Acetone n→π*: 279 nm in hexane, 264.5 nm in water; the 23.5 kJ mol⁻¹ difference is about one hydrogen bond Geo 2023 Q6(i) 4m. Acrolein: π→π* ≈ 210 nm (ε ≈ 11 500) moves red, n→π* ≈ 315 nm (ε ≈ 14) moves blue from hexane to ethanol Geo 2022 Q10(b) 5m.
  • Aniline in acid: 230 and 280 nm → 203 and 254 nm, close to benzene. Protonation ties up the N lone pair, so the auxochrome is lost: blue and hypochromic shift. Phenol does the reverse in base (phenoxide, red shift) Geo 2022 Q6(f) 4m.SN
aniline: 230, 280 nmN lone pair conjugatedH+ anilinium: 203, 254 nmlone pair tied up: like benzene
Fig. 13.1.3 Aniline and the anilinium ion.
Draw it in 60 s: draw PhNH2 with a lone pair on N; arrow with H+; PhNH3+ with no lone pair; write 230, 280 → 203, 254 nm under them.

13.1.3 Conjugation lowers the gap ★★

ππ*ethene171 nmψ2ψ3butadiene217 nmψ3ψ4hexatriene258 nmeach extra C=C: HOMO up, LUMO down → longer λ
Fig. 13.1.4 Ethene, butadiene, hexatriene.
Draw it in 60 s: two lines for each molecule, closer each time; arrow between them; write 171, 217, 258 nm under them.
  • Each added C=C raises the HOMO and lowers the LUMO; the gap falls and λ rises (λ = hc/ΔE). Hückel: butadiene gap 1.24|β|, hexatriene 0.89|β| CSE 2022 Q8(b)(ii) 5m.SN
  • Roughly +30 to +40 nm per extra conjugated C=C, with ε rising. About eight conjugated C=C bring the band into the visible: β-carotene (11 C=C) absorbs at 452 nm and looks orange.

13.1.4 Woodward–Fieser rules for dienes and polyenes ★★★

Itemnm
Base: acyclic (s-trans) or heteroannular diene214
Base: homoannular diene (both C=C in one ring, s-cis)253
Each C=C extending conjugation+30
Each alkyl group or ring residue on the conjugated system+5
Each exocyclic C=C+5
OR +6 · SR +30 · Cl, Br +5 · NR₂ +60 · OAc 0
  1. Find the conjugated system; if a homoannular and a heteroannular diene are both present, take the homoannular base (253).SN
  2. Count every substituent and every ring residue (a ring bond leaving a conjugated carbon that is not part of the chosen path).SN
  3. Exocyclic C=C: one carbon in a ring, the other outside it; one C=C can be exocyclic to two rings and then counts twice.SN
  4. No solvent correction for dienes. Agreement is usually within ±5 nm.
rrrexoheteroannular: base 214+ 3 ring residues (15) + 1 exocyclic C=C (5)= 234 nm rrr = ring residue on a diene carbon (+5 each); exo = C=C exocyclic to a ring (+5)homoannular: base 253+ 2 ring residues (10), no exocyclic C=C= 263 nm
Fig. 13.1.5 Counting a heteroannular and a homoannular diene. r marks each ring residue on a diene carbon; exo marks the exocyclic C=C.
Draw it in 60 s: draw the two decalins; put the C=C in; circle each ring bond leaving a diene carbon (r) and tick the C=C that sits outside a ring; add up under each.

Solved PYQ CSE 2022 Q8(b)(i) 10m

λmax of two steroidal polyenes.

  1. (1) Homoannular diene in ring B (C5=C6, C7=C8): 253; two extending C=C (C3=C4, C9=C11) +60; five ring residues +25; three exocyclic C=C +15: 353 nm.
  2. (2) Heteroannular diene: 214; one extending C=C +30; ring residues and alkyls 4 × 5 = +20; two exocyclic C=C +10: 274 nm.

Write the table: base, then each line with its count, then the total. The examiner gives marks line by line.SN

Also asked: IFoS 2022 Q7(a) 10m triene 253 + 30 + 25 + 15 = 323 nm · IFoS 2024 Q7(b)(ii) 5m two steroidal tetraenes · IFoS 2025 Q7(b)(i) 5m two dienes · CSE 2025 Q7(a)(ii) 10m four compounds.

13.1.5 Woodward–Fieser rules for α,β-unsaturated carbonyls ★★★ most asked

CδCγCβCαCORbase: acyclic or 6-ring ketone 215 · 5-ring ketone 202 · aldehyde 207alkyl or ring residue: α +10, β +12, γ and δ +18extra C=C +30 · exocyclic C=C +5 · homoannular diene +39
Fig. 13.1.6 Positions are counted from the carbonyl.
Draw it in 60 s: write C=C–C=C–C=O; label α, β, γ, δ from the C=O side; write the three lines of numbers below.
Itemαβγδ and higher
Alkyl group or ring residue10121818
OH3530–50
OMe35301731
OAc6666
Cl / Br15 / 2512 / 30
SR / NR₂85 / 95
  • Base: acyclic or six-ring enone 215, five-ring enone 202, α,β-unsaturated aldehyde 207 (some books 210), acid or ester 195.SN
  • +30 for each C=C extending conjugation, +5 for each exocyclic C=C, +39 for a homoannular diene component.SN
  • Values are for ethanol. Solvent correction: water −8, chloroform +1, dioxane +5, ether +7, hexane +11.SN

Solved PYQ CSE 2023 Q8(a)(ii) 5m Geo 2025 Q10(b)(i) 5m

A ketone C₇H₁₂O absorbs at 249 nm. Structure?

  1. DBE = 2: one C=O and one C=C, no ring. 249 nm means a conjugated enone.
  2. 215 + α-alkyl 10 + two β-alkyl 24 = 249 nm: three alkyl groups on the C=C. With seven carbons all three are methyl.
  3. 3,4-Dimethylpent-3-en-2-one, CH₃CO–C(CH₃)=C(CH₃)₂.
α-CH3 +10β-CH3 +12β-CH3 +123,4-dimethylpent-3-en-2-oneacyclic enone base215α-alkyl (1)+10β-alkyl (2 × 12)+24λmax (calc.)249 nmfound 249 nm: C7H12O, DBE 2
Fig. 13.1.7 The answer drawn, with each counted group marked.
Draw it in 60 s: draw the enone; label α and β from the C=O; write +10 by the α-CH3 and +12 by each β-CH3; total in a small table.

Solved PYQ Geo 2021 Q6(c) 4m

Three linear dienones: order of λmax.

  1. I: 215 + 30 (extra C=C) + γ ring residue 18 + δ ring residue 18 = 281 nm.
  2. II: 215 + 30 + α-CH₃ 10 + γ 18 + δ 2 × 18 = 309 nm.
  3. III: 215 + 30 + β-CH₃ 12 + γ 18 + δ 2 × 18 = 311 nm. I < II < III.
  • Aldehyde: OHC–C(Me)=C(OMe)Me = 207 + α-Me 10 + β-Me 12 + β-OMe 30 = 259 nm. Check that a printed name is a real structure before calculating IFoS 2023 Q5(d) 5m.
  • Telling isomers apart: calculate both and match; a twisted (non-planar) dienone also has a lower ε CSE 2022 Q8(c)(ii) 10m.

Also asked: CSE 2024 Q7(a) 15m three dienones · Geo 2023 Q9(a) 10m a dienone, a tetraene and a cyclopentenone · Geo 2022 Q8(a) 10m three unsaturated ketones · IFoS 2025 Q7(b)(ii) 5m two carbonyl compounds.

13.1.6 Aromatic carbonyl compounds (Scott's rules) ★

  • Base for ArCOR 246, ArCHO 250, ArCOOH or ArCOOR 230 nm; add for ring substituents (o / m / p): alkyl 3 / 3 / 10, OH or OMe 7 / 7 / 25, Cl 0 / 0 / 10, Br 2 / 2 / 15, NH₂ 13 / 13 / 58, NMe₂ 20 / 20 / 85.SN
  • A para donor and acceptor give an intense charge-transfer band; the meta isomer cannot conjugate through the ring CSE 2024 Q5(d) 10m.
  • An ortho group that twists the C=O out of the ring plane cuts the conjugation (steric inhibition of resonance): lower λ and ε.

13.1.7 Shape, strain and intensity ★★

  • s-cis (homoannular) dienes absorb at longer λ but with lower ε (about 10⁴) than s-trans ones (about 2 × 10⁴).SN
  • Non-planarity (crowding, twisted rings, cross-conjugation) lowers ε and the λ falls short of the calculated value.
  • trans-Dienes and trans-stilbene absorb at slightly longer λ and with higher ε than the cis isomers (less crowded, more planar) CSE 2024 Q5(d).
  • Tautomers have different chromophores: but-3-en-2-one (enone chromophore) and its enol CH₂=C(OH)–CH=CH₂ (diene chromophore with an OH auxochrome) IFoS 2022 Q7(b) 5m.

13.1.8 ★ Add-on beyond the syllabus

Fieser–Kuhn rule for polyenes with more than four conjugated C=C: λmax = 114 + 5M + n(48.0 − 1.7n) − 16.5 Rendo − 10 Rexo, where n = number of conjugated C=C, M = alkyl substituents and ring residues, Rendo and Rexo = rings with endocyclic and exocyclic C=C. For β-carotene (n = 11, M = 10, Rendo = 2): 114 + 50 + 11(48.0 − 18.7) − 33 ≈ 453 nm, observed 452 nm. One line of this in a carotenoid or colour question shows range.

13.1.9 How UPSC asks this

Question typeAskedWhat earns the marksTime
Calculate λmax (dienes, enones, steroids)CSE 2025 Q7(a)(ii) CSE 2024 Q7(a) CSE 2022 Q8(b)(i) IFoS 2025 Q7(b)(i) IFoS 2025 Q7(b)(ii) IFoS 2024 Q7(b)(ii) IFoS 2022 Q7(a) Geo 2023 Q9(a) Geo 2022 Q8(a) Geo 2021 Q6(c) Geo 2025 Q10(b)(i) CSE 2023 Q8(a)(ii)Base value named, each increment on its own line with the count, total in nm3–11 min
Assign transitions, intensitiesCSE 2025 Q7(a)(i) CSE 2021 Q8(b)(ii) IFoS 2024 Q7(b)(i) IFoS 2023 Q6(e) Geo 2022 Q10(b)n→π* weak and long λ, π→π* strong; reason (orthogonal orbitals); ε values4–8 min
Solvent and pH shiftsGeo 2023 Q6(i) Geo 2022 Q10(b) Geo 2022 Q6(f)Level diagram (Fig. 13.1.2); direction of each shift; hydrogen bonding3–4 min
Compare λmax, identify isomersCSE 2024 Q5(d) CSE 2022 Q8(c)(ii) CSE 2022 Q8(b)(ii) IFoS 2022 Q7(b) IFoS 2023 Q5(d) Geo 2024 Q9(c)Calculation for each, then the comparison; planarity and ε4–8 min

Time at 0.72 min per mark for Civil Services and 0.8 min per mark for IFoS and Geo-Scientist.

13.1.10 Traps

✗ Using the diene table (5 per alkyl) for an enone. Enones use α 10, β 12, γ and δ 18.
✗ Missing ring residues and exocyclic C=C in steroids. Go round every conjugated carbon and count each bond that leaves the path.
✗ Saying a polar solvent red-shifts the n→π* band. It is blue-shifted; only π→π* moves red.

13.1.11 Where to read

Any one book is enough.

  1. Book 1: Jagdamba Singh and L.D.S. Yadav, Undergraduate Organic Chemistry, Vol. II, Ch. 1 “Spectroscopy”: types of electronic transitions (p. 10), terminology (p. 14), effect of conjugation (p. 15), solvent effects (p. 17), Woodward–Fieser rules for polyenes (§1.17, p. 21) and for enones (§1.18, p. 28) with worked examples, angular distortion and cross-conjugation (§1.19, p. 30). Best for this exam: every table above is there.
  2. Book 2: Clayden, Greeves and Warren, Organic Chemistry, 2nd ed., Ch. 7 “Delocalization and conjugation”: UV–visible spectra and why conjugation brings colour (pp. 148–149). Best for the reason behind the numbers.

Online: the solved blog post Woodward–Fieser rules: calculating λmax shows a full answer as written in the booklet.

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