Chemistry 2025 Paper II 50 marks Explain

Paper II — Q7

(a) (i) Acetone shows a weak absorption at 280 nm and a strong absorption at 190 nm in the UV spectrum. Account for the…

(a)
(i)

Acetone shows a weak absorption at 280 nm and a strong absorption at 190 nm in the UV spectrum. Account for the observation. 5 marks

(ii)

Using Woodward-Fieser rules, calculate λ_max for the following compounds: A, B, C, D 12 marks

(b)
(i)

Rank the following dienes in order of increasing reactivity in a Diels-Alder reaction (1 = least reactive, 4 = most reactive). Briefly explain your answer: A, B, C, D 10 marks

(ii)

Write the structure of the product in the following reaction: OMe + CN 1) Heat 2) H⁺, H₂O ? 5 marks

(c)
(i)

Write the structure of the product formed in the following sigmatropic rearrangement and categorize it with suitable explanation. Explain the thermal feasibility of this rearrangement by drawing orbital diagram: H H Bu 80 °C CCl₄ → ? 10 marks

(ii)

(1) Identify the mode of ring closure for each of the following electrocyclic reactions: (A), (B) (2) Are the indicated hydrogens cis or trans? 10 marks

हिंदी में प्रश्न पढ़ें

(क) (i) UV स्पेक्ट्रम में ऐसीटोन 280 nm पर दुर्बल अवशोषण और 190 nm पर प्रबल अवशोषण दर्शाता है। इस अवलोकन को समझाइए। (5 अंक)

(ii)

वुडवर्ड-फीजर नियमों का उपयोग करके निम्नलिखित यौगिकों के λmax की गणना कीजिए: A, B, C, D (12 अंक)

(ख) (i) डील्स-एल्डर अभिक्रिया में निम्नलिखित डाइईनों को उनकी अभिक्रियाशीलता के बढ़ते हुए क्रम में क्रमबद्ध कीजिए (1 = निम्न अभिक्रियाशील, 4 = अति अभिक्रियाशील)। अपना उत्तर संक्षेप में स्पष्ट कीजिए: A, B, C, D (10 अंक)

(ii)

निम्नलिखित अभिक्रिया में उत्पाद की संरचना लिखिए: OMe + CN 1) ऊष्मा 2) H⁺, H₂O ? (5 अंक)

(ग) (i) निम्नलिखित सिमानुवर्ती पुनर्विन्यास में बनने वाले उत्पाद की संरचना लिखिए तथा उपयुक्त व्याख्या करते हुए इसे वर्गीकृत कीजिए। कक्षीय आरेख बनाकर इस पुनर्विन्यास की ऊष्मीय व्यवहार्यता की व्याख्या कीजिए: H H Bu 80 °C CCl₄ ? (10 अंक)

(ii)

(1) निम्नलिखित इलेक्ट्रोसाइक्लिक अभिक्रियाओं में प्रत्येक के लिए वलय संवरण विधा की पहचान कीजिए: (A), (B) (2) दर्शाए गए हाइड्रोजन समपक्ष (सिस) हैं या विपक्ष (ट्रांस)? (10 अंक)

Q7 of the 2025 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2025 Chemistry paper

The figure this question refers to, in words

The question paper is a scan and the diagram did not survive as text. This is the figure as read from the original page — every component, value and label — so the question can be worked from the text below.

(ii) Four chemical structures labelled A, B, C, D for Woodward-Fieser calculation of lambda max: A is 3-methylbut-3-en-2-one (CH3-C(=O)-C(CH3)=CH2); B is 2-(cyclohex-1-en-1-yl)cyclohexan-1-one; C is 2-methylcyclopent-2-en-1-one; D is 3,4-dihydronaphthalen-1(2H)-one (a bicyclic enone with a benzene ring fused to a cyclohexenone).

(ii) Two electrocyclic ring closure reactions. Reaction (A): A cyclooctatriene ring (8-membered ring with 3 double bonds) is converted to a bicyclic product. The product is a fused bicyclic system: a cyclobutene ring fused to a cyclohexadiene ring. The bridgehead hydrogens are shown with dashed bonds, indicating they are on opposite faces (trans). Reaction (B): A bicyclic starting material, bicyclo[4.2.0]octa-2,4-diene (a cyclobutene fused to a cyclohexadiene), is converted to a cyclooctatriene ring. The starting material has hydrogens on the bridgehead carbons shown with dashed bonds (trans).

A sigmatropic rearrangement reaction scheme. A bicyclic starting material with a cyclobutene ring fused to a cyclopentene ring. The cyclobutene ring has a vinyl group (-CH=CH2) attached to one bridgehead carbon and a butyl group (-Bu) attached to the other bridgehead carbon. The vinyl group is shown with a hydrogen on the terminal carbon. The reaction arrow is labeled with conditions: 80 degrees C and CCl4. The product is indicated by a question mark.

Model answer

Drafted with AI by UPSC Answer Check against this question's marking rubric, to the expected length. UPSC does not publish answers for Mains — this is one way to score well, not an official key. How we build rubrics and answers.

(a)(i)

Acetone shows two bands because two different electronic transitions of the carbonyl group fall in this region.

  • 280 nm, weak (ε about 15 L mol⁻¹ cm⁻¹): n → π∗. A non-bonding (lone-pair) electron on oxygen is promoted into the π∗ orbital of C=O. The n–π∗ gap is the smallest in the molecule, so the band lies at the longest wavelength. The n orbital lies in the molecular plane and π∗ is perpendicular to it, so overlap is poor and the transition is symmetry-forbidden, hence the low intensity.
  • 190 nm, strong (ε of the order of 10³): π → π∗. An electron of the C=O π bond is promoted into π∗. The bonding π level lies well below the non-bonding level, so more energy (a shorter wavelength) is needed, but the two orbitals overlap well and the transition is allowed, hence the high intensity. (The n → σ∗ transition of acetone lies at still shorter wavelength, near 166 nm.)

Thus the weak long-wavelength band is the forbidden, low-energy n → π∗ transition and the strong short-wavelength band is the allowed, higher-energy π → π∗ transition.

(a)(ii)

Woodward–Fieser rules for α,β-unsaturated ketones (ethanol): base value for an acyclic or six-membered-ring ketone 215 nm, five-membered-ring ketone 202 nm; α-alkyl/ring residue +10, β +12, γ and higher +18; each exocyclic C=C +5; extended conjugation (each extra C=C) +30; homoannular diene component +39.

A — (E)-3-methylpent-3-en-2-one, CH₃–CH=C(CH₃)–CO–CH₃

Base (acyclic enone): 215; α-alkyl (CH₃ on C-3): +10; β-alkyl (CH₃ on C-4): +12; λmax = 237 nm

B — 2-cyclohexylidenecyclohexan-1-one

Base (six-membered ring ketone): 215; α ring residue (C-3 of cyclohexanone): +10; β ring residues (two, of the cyclohexylidene ring): +24; C=C exocyclic to both rings (2 × 5): +10; λmax = 259 nm

C — 1-acetylcyclopentene

The C=O is acyclic (a methyl ketone); only the C=C is in the ring, so the base is 215 nm, not the cyclopentenone value.

Base: 215; α ring residue (C-5): +10; β ring residue (C-3): +12; λmax = 237 nm

D — 4-methyl-4,4a,5,6-tetrahydronaphthalen-2(3H)-one (a heteroannular dienone: O=C2–C1=C8a–C8=C7)

Base (six-membered ring ketone): 215; Extended conjugation (second C=C): +30; β ring residue (C-4a on C-8a): +12; δ ring residue (C-6 on C-7): +18; Exocyclic C=C (C-1=C-8a, exocyclic to the left ring): +5; λmax = 280 nm

(b)(i)

A Diels–Alder reaction needs the diene in the s-cis conformation, so that C-1 and C-4 can bond to the two dienophile carbons at the same time.

Order of increasing reactivity: A (1) < B (2) < C (3) < D (4).

  1. A (least reactive) — the bicyclic hexahydronaphthalene diene has its two C=C bonds on opposite sides of the ring-fusion bond. It is locked s-trans; rotation to s-cis is impossible, so it does not undergo Diels–Alder reaction.
  2. B — buta-1,3-diene drawn in the s-trans conformation (the more stable conformer, which predominates at room temperature). It must first rotate to s-cis, which costs energy.
  3. C — the same buta-1,3-diene drawn in the s-cis conformation, the conformer that reacts directly (B and C interconvert rapidly).
  4. D (most reactive) — cyclohexa-1,3-diene is locked s-cis by the ring. No conformational energy is needed, so it reacts most readily.

(b)(ii)

The diene is 2-methoxybuta-1,3-diene (an electron-rich diene) and the dienophile is acrylonitrile (electron-poor).

  • Step 1 (heat): Diels–Alder. The +M OMe group at C-2 puts the largest HOMO coefficient on C-1; the largest LUMO coefficient of acrylonitrile is on its CH₂. Joining these gives the "para" (1,4) adduct, 4-methoxycyclohex-3-ene-1-carbonitrile.
  • Step 2 (H⁺, H₂O): the product contains an enol ether, which is protonated at C-3 and hydrolysed to a ketone. The nitrile survives dilute aqueous acid.

Product: 4-oxocyclohexane-1-carbonitrile (4-cyanocyclohexanone). C-1 is not a stereocentre.

(c)(i)

The substrate is a cis-1,2-divinylcyclopropane: one vinyl group carries the n-butyl group at its end (an (E)-hex-1-enyl group), and the two ring hydrogens are cis. The system C=C–C(ring)–C(ring)–C=C is a 1,5-diene whose central σ bond is a strained cyclopropane bond.

Category: a [3,3]-sigmatropic (Cope) rearrangement, the divinylcyclopropane rearrangement. A σ bond flanked by two allyl units migrates from position 1,1′ to 3,3′: the cyclopropane C1–C2 bond breaks, a new σ bond forms between the two vinyl termini, and the π bonds move into the old C1–C=C and C2–C=C positions. The cis geometry allows a boat-like transition state, and relief of cyclopropane ring strain explains the mild conditions (80 °C).

Product: 6-butylcyclohepta-1,4-diene (seven-membered ring; Bu on the carbon formed from the vinyl terminus).

Thermal feasibility (FMO). Treat the transition state as two allyl radicals. Each has ψ₂ as its singly occupied frontier orbital, with a node at the central carbon and opposite phases on the two termini. In the suprafacial–suprafacial geometry the lobes are in phase both at the bond that breaks (C1…C1′) and at the bond that forms (C3…C3′). The reaction is therefore thermally allowed ([π2s + σ2s + π2s]).

Diagram: Divinylcyclopropane [3,3] Cope rearrangement and the allyl psi2 FMO picture

(c)(ii)

Woodward–Hoffmann rules for thermal electrocyclic reactions: 4n π electrons → conrotatory; (4n + 2) π electrons → disrotatory. The stereochemistry follows from the geometry of the terminal double bonds of the polyene.

(1) Mode of ring closure

  • (A) Cycloocta-1,3-diene → bicyclo[4.2.0]oct-7-ene. This is a 4π (n = 1) diene → cyclobutene closure, so under heat it is conrotatory (HOMO ψ₂ has opposite phases at the termini).
  • (B) Cyclonona-1,3,5-triene → bicyclo[4.3.0]nona-2,4-diene. This is a 6π (4n + 2) triene → cyclohexadiene closure, so under heat it is disrotatory (HOMO ψ₃ has the same phase at the termini).

(2) Relative configuration of the indicated hydrogens

  • (A) trans. As drawn, both ring double bonds are cis (Z). The ring carbons attached to the termini (C-8 and C-5) therefore both occupy "inside" positions, and the hydrogens both occupy "outside" positions. Conrotation turns the two termini in the same sense, so like groups end up on opposite faces: the ring-fusion hydrogens are trans.
  • (B) trans. The printed triene has the C-5=C-6 double bond drawn E, with the other two Z (a Z,Z,E-cyclononatriene). One ring-attached carbon at a terminus is therefore "inside" and the other "outside", so the hydrogens are likewise one inside and one outside. Disrotation turns the termini in opposite senses and puts such a pair on opposite faces. The ring-fusion hydrogens are trans, giving trans-bicyclo[4.3.0]nona-2,4-diene.

How this answer will be evaluated

What earns full marks: Precise mechanisms, correct orbital diagrams, and accurate numerical calculations.

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a)(i) Explain the origin of the two UV absorptions in acetone. 5 marks

    account for— state the phenomenon → the causes in order of weight → conclusion

    Must cover

    • 280 nm weak band = symmetry-forbidden n→π* (ε ≈ 15); n orbital in plane, π* perpendicular
    • 190 nm strong band = allowed π→π* (ε ≈ 10³); well-overlapped orbitals
    • n→σ* of acetone lies at shorter wavelength (~166 nm)
    • Energy order: n–π* gap smallest (longest λ) vs π–π* requiring more energy

    Loses marks

    • Confusing σ→σ* with n→σ*
    • Failing to distinguish transition types

    Earns more

    • Mention forbidden nature of n→π*
    • Reference HOMO-LUMO gap

    Extra mark

    • Molar absorptivity values (ε)
  2. (a)(ii) Determine λmax for compounds A, B, C, D using Woodward-Fieser rules. 10 marks

    calculate— given → formula → substitution → result with units → interpretation

    Must cover

    • Correct base value (215 nm acyclic/six-ring; 202 nm for five-ring ketones)
    • Correct α/β ring-residue and alkyl increments
    • Correct exocyclic C=C and extended conjugation increments
    • Final sums: A 237 nm; B 259 nm; C 237 nm; D 280 nm

    Loses marks

    • Missing exocyclic double bond count
    • Incorrect base value selection

    Earns more

    • Explicit listing of each increment
    • Correct identification of ring size

    Extra mark

    • Comparison of calculated vs experimental
  3. (b)(i) Order dienes A-D by increasing Diels-Alder reactivity with explanation. 10 marks

    enumerate— list the items in order → one line each → no commentary

    Must cover

    • Increasing reactivity: A(1) < B(2) < C(3) < D(4)
    • A: bicyclic diene locked s-trans (cannot reach s-cis) — least reactive
    • B: butadiene s-trans (must rotate); C: butadiene s-cis (reactive conformer)
    • D: cyclohexa-1,3-diene locked s-cis — most reactive

    Loses marks

    • Ranking A (locked s-trans) as reactive, or D (locked s-cis) as unreactive
    • Failing to explain conformational lock

    Earns more

    • Mentioning steric hindrance in A
    • Drawing conformers

    Extra mark

    • Mentioning orbital coefficients
  4. (b)(ii) Draw the product of the Diels-Alder reaction followed by hydrolysis. 5 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Diels–Alder of electron-rich 2-methoxybuta-1,3-diene with acrylonitrile
    • Regiochemistry: the 'para' (1,4) adduct 4-methoxycyclohex-3-ene-1-carbonitrile (largest HOMO coeff. at C-1; largest LUMO coeff. on the dienophile CH₂)
    • Step 2: hydrolysis of the enol ether → 4-oxocyclohexane-1-carbonitrile; the nitrile survives dilute acid

    Loses marks

    • Wrong regiochemistry (meta product)
    • Hydrolysing the nitrile to an acid under the mild aqueous acid work-up

    Earns more

    • Showing intermediate nitrile product
    • Labeling endo/exo selectivity

    Extra mark

    • Mechanism of hydrolysis
  5. (c)(i) Draw product, categorize rearrangement, and explain thermal feasibility. 10 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • Identify as [3,3]-sigmatropic (Cope) divinylcyclopropane rearrangement (cis-1,2-divinylcyclopropane substrate)
    • Draw the correct product: 6-butylcyclohepta-1,4-diene
    • Explain thermal feasibility via orbital symmetry: ψ₂ of the two allyl fragments, suprafacial–suprafacial [π2s+σ2s+π2s], always allowed
    • Draw the HOMO/allyl orbital diagram (boat-like TS; strain relief explains the mild 80 °C)

    Loses marks

    • Wrong product connectivity
    • Missing orbital diagram

    Earns more

    • Mentioning suprafacial-suprafacial mode
    • Conservation of orbital symmetry

    Extra mark

    • Transition state drawing
  6. (c)(ii) Identify ring closure mode and hydrogen stereochemistry for A and B. 10 marks

    map— locate accurately → label → one line on why it matters

    Must cover

    • (A) 4π → conrotatory (thermal); (B) 6π → disrotatory (thermal)
    • Stereochemistry: (A) ring-fusion hydrogens trans (Z,Z-diene, conrotation)
    • (B) ring-fusion hydrogens trans (Z,Z,E-triene, disrotation) — the printed product shows H on opposite faces
    • Orbital symmetry fixes the rotation sense; polyene geometry then fixes the junction stereochemistry

    Loses marks

    • Swapping conrotatory/disrotatory modes
    • Wrong cis/trans assignment

    Earns more

    • Citing Woodward-Hoffmann rules
    • Counting pi electrons (4n vs 4n+2)

    Extra mark

    • Orbital rotation diagrams
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