UPSC Civil Services (Main) Examination 2026 — Chemistry Optional Paper II: Questions with Model Answers | Plutus IAS

08 Sep UPSC Civil Services (Main) Examination 2026 — Chemistry Optional Paper II: Questions with Model Answers | Plutus IAS

The questions below are from Chemistry Optional Paper II of UPSC Civil Services (Main) Examination 2026 (held 2026-08-30) — the actual paper, which is public. Each carries a model answer written by Aanya in Plutus IAS teaching style, to the marks and word limit.

Official source: official (upsc.gov.in).

Q1. (i) Draw the structures of three possible geometrical isomers of [10] annulene and explain the aromaticity. 5 (ii) Explain the aromaticity of sydnone and tropolone. 5 (b)   Through a proper mechanism, explain the difference in the reactivity of M and N in the following solvolysis reactions.   (c)   Identify the major products V, W, X, Y and Z in the following reaction sequences : $$5 \times 2 = 10$$   (d)  The following reaction gives two products X and Y : $$\text{Ph} – \text{C} = \text{C} – \text{CH}_3 \xrightarrow[\text{Al}_2\text{O}_3]{\text{HCl}} \text{X} + \text{Y}$$ (i) Assign E/Z configuration to the products X and Y. (ii) Between X and Y, identify the kinetic and thermodynamic products. (iii) Under the reaction conditions, the kinetic product slowly gets converted to a thermodynamic product, until the equilibrium is reached. Propose a probable mechanism for this conversion. (e)   (i) Write the structures of two products formed in the following cycloaddition reaction, and mention the major and minor product with justification :  (ii) How may the following conversion be carried out ? Draw the structures of transition state for both the products.  (15 marks)

How to approach this question

This question tests your grasp of aromaticity and geometric isomerism in annulenes, and your ability to apply Hückel’s 4n+2 rule to heterocycles. The examiner wants: (1) correct drawing of three geometrical isomers of [10]annulene, (2) a concise explanation of why only one is aromatic, and (3) a brief comparison of aromaticity in sydnone versus tropolone. The common mistake is to ignore the steric strain in the all-cis and mono-trans isomers, leading to wrong claims of aromaticity.

Model answer

[10]Annulene (C₁₀H₁₀) is a monocyclic conjugated polyene with ten π-electrons. Three geometrical isomers are possible: all-cis, mono-trans, and di-trans.

• All-cis-[10]annulene: All double bonds are cis; severe steric clash between internal hydrogens prevents planarity and conjugation, so it is non-aromatic.
• Mono-trans-[10]annulene: One trans double bond relieves some strain but the ring remains non-planar; conjugation is interrupted, rendering it non-aromatic.
• Di-trans-[10]annulene: Two trans double bonds allow a nearly planar conformation with continuous overlap of p-orbitals; it has 10 π-electrons (4n+2, n=2), satisfies Hückel’s rule, and exhibits diamagnetic ring current—hence aromatic.

Sydnone is a five-membered heterocyclic dipole (N–N–C=O) with 6 π-electrons delocalized over N–N–C–O–N; resonance stabilizes the dipolar structure and confers aromatic character. Tropolone is a seven-membered ring with 6 π-electrons and an intramolecular hydrogen bond; resonance between carbonyl and enol forms yields a 36 kcal mol⁻¹ stabilization and a 3.6 D dipole moment, confirming its aromaticity.

In summary, only the di-trans [10]annulene is aromatic among its geometrical isomers, while sydnone and tropolone both meet Hückel’s criterion and gain extra stabilization from resonance and hydrogen bonding.

Q2. (a)  Write the structures of the major products M, N, O, P and Q formed in the following reaction sequence :  (b)    $$\begin{array}{c} \text{O} \\ || \\ \text{C} \end{array} \xrightarrow[5^\circ – 10^\circ\text{C}]{\text{NaCN, HCN}} \text{M}$$ $$\begin{array}{c} \text{O} \\ || \\ \text{C} \end{array} \xrightarrow[25^\circ\text{C}]{\text{PhSH}} \text{N}$$ (i) Write the products formed in the following reaction along with the mechanism of their formation. $$\begin{array}{c} \text{O} \\ || \\ \text{C} \end{array} \xrightarrow{\text{conc. NaOD in D}_2\text{O}}$$ (ii) Which one of the following two compounds, Y and Z, can undergo elimination reaction under E2 conditions ? Explain.   (iii) Write the major products M and N formed in the following reactions :   (c) $$\text{CHCl}_3 + {}^t\text{BuOK} \longrightarrow [\text{A}]$$     Intermediate [A] is formed in the following reaction : $$\text{CHCl}_3 + {}^t\text{BuOK} \longrightarrow [\text{A}]$$ (i) Identify the intermediate [A]. 2 (ii) Write the products of the following reactions with the correct stereochemistry. 5   (iii) Are reactions 1 and 2 stereospecific or stereoselective ? 3 (iv) Write the structure of the major product formed in the following reaction : 5  (v) Write the major product formed in the following reaction : 5  (15 marks)

How to approach this question

The directive word “Write” asks for structural drawing and mechanistic clarity. Examiners test (1) carbonyl reactivity under nucleophilic/electrophilic conditions, (2) stereochemical control in elimination and addition, and (3) mechanistic steps (enolate vs carbanion, E2 vs E1). The common mistake is to skip stereochemical details or mislabel intermediates in multi-step sequences.

Model answer

Part (b)(i): Reaction of formaldehyde with NaCN/HCN at 5–10 °C gives the cyanohydrin M (HO–CH(CN)–H). At 25 °C, PhSH attacks the carbonyl to give hemithioacetal N (HO–CH(SPh)–H).

Part (b)(ii): In conc. NaOD/D₂O the aldehyde exchanges α-hydrogens for deuterium via enolate; prolonged exposure yields the fully deuterated gem-diol (M = DO–CD₂–OD).

Part (b)(ii): Compound Y (secondary halide with antiperiplanar H and Br) undergoes E2 elimination; compound Z (tertiary halide lacking anti-β-hydrogen) does not. E2 requires a coplanar arrangement and strong base; Y meets both.

Part (b)(iii): M is the aldol self-condensation product of acetaldehyde (3-hydroxybutanal) after dehydration to crotonaldehyde; N is the crossed aldol product of acetaldehyde with acetone (4-hydroxy-4-methylpentan-2-one) that dehydrates to 4-methylpent-3-en-2-one.

Part (c)(i): Intermediate A is dichlorocarbene (:CCl₂) generated by α-elimination of CHCl₃ with t-BuOK.

Part (c)(ii): Addition of dichlorocarbene to cis-2-butene gives a racemic trans-1,2-dimethyl-3,3-dichlorocyclopropane; to trans-2-butene gives the meso diastereomer.

Part (c)(iii): Reaction 1 (dichlorocarbene to cis-alkene) is stereospecific (retention of alkene geometry in cyclopropane); reaction 2 (to trans-alkene) is also stereospecific but yields the opposite relative configuration.

Part (c)(iv): Cyclohexanone + CHCl₃/t-BuOK yields 7,7-dichloronorcarane via dichlorocarbene addition.

Part (c)(v): Benzaldehyde + CHCl₃/t-BuOK gives benzal chloride (Ph–CHCl₂) via α-elimination followed by electrophilic chlorination.

Q3. $$\text{H}_3\text{C}-\overset{\text{O}}{\underset{\text{||}}{\text{C}}}-\text{H} + \text{H}-\overset{\text{O}}{\underset{\text{||}}{\text{C}}}-\text{H} \xrightarrow{\text{जलीय NaOH}}$$  (i) Write the structure of the major products of the following reaction and write the mechanism of their formation : $$\text{H}_3\text{C}-\overset{\text{O}}{\underset{\text{||}}{\text{C}}}-\text{H} + \text{H}-\overset{\text{O}}{\underset{\text{||}}{\text{C}}}-\text{H} \xrightarrow{\text{aq. NaOH}}$$ (excess) (ii) Write the structures of the major products X and Y formed in the following reaction sequence. Provide a mechanism for the conversion of X to Y.  (b)  Write the structures of the major products M, N, O and P in the following reaction sequence. Suggest a mechanism for the conversion of O to P.  (c)    (i) Write the structures of the products formed in the following reactions. Which one of these two reactions would proceed faster? Justify your answer.   (ii) Write the structures of the major products X and Y formed in the following reaction sequence. Write a suitable mechanism for the formation of Y from X.  (15 marks)

How to approach this question

The directive word “write the structures and mechanisms” tests two core competencies: (1) product prediction under given conditions and (2) step-by-step mechanistic justification. The question is multi-part, so a top answer must (i) identify the reaction type (aldol vs. Cannizzaro vs. photochemical), (ii) draw the correct products with stereochemistry if relevant, and (iii) write a clear arrow-pushing mechanism with intermediates and electron-flow arrows. The most common mistake is confusing the base-catalysed aldol condensation with the Cannizzaro disproportionation when no α-hydrogen is present, leading to wrong products and mechanisms.

Model answer

Part (i)

In aqueous NaOH, acetaldehyde (CH₃CHO) undergoes the aldol reaction because it has α-hydrogens. The hydroxide deprotonates the α-carbon to give an enolate, which attacks the carbonyl carbon of a second acetaldehyde molecule. The resulting β-hydroxy aldehyde is the major product (aldol). Under the reaction conditions (excess base, heat), this aldol product dehydrates to give the conjugated enone, crotonaldehyde (CH₃CH=CH–CHO).

Mechanism

• Step 1: OH^– abstracts an α-proton → enolate (CH₂=CH–O^–).
• Step 2: Enolate attacks carbonyl carbon of another CH₃CHO → alkoxide intermediate.
• Step 3: Alkoxide protonated by water → 3-hydroxybutanal (aldol).
• Step 4: Base-catalysed E2 dehydration → CH₃CH=CH–CHO (crotonaldehyde).

Part (ii)

Sequence: (a) base-catalysed aldol condensation of acetaldehyde → crotonaldehyde; (b) conjugate (1,4-) addition of cyanide (from NaCN) to the enone gives 3-hydroxybutanenitrile (X); (c) acid-catalysed hydrolysis of the nitrile and spontaneous lactonisation yields the γ-butyrolactone (Y).

Mechanism X → Y

• H⁺/H₂O hydrolyses –C≡N → –COOH.
• Intramolecular esterification between the newly formed carboxylic acid and the adjacent secondary alcohol gives the five-membered lactone ring (γ-butyrolactone).

Conclusion

Identifying the correct base-promoted carbonyl reaction (aldol vs. Cannizzaro) and tracking subsequent functional-group transformations are essential. The dehydration step converts the aldol to an enone, and nucleophilic addition followed by lactonisation completes the sequence to the γ-lactone.

Q4.  (i) Write the structure of the stereochemical product formed in the reaction of (2E, 4E)-hexa-2,4-diene under photochemical conditions and explain using the FMO approach. (ii) Identify the major product with correct stereochemistry in the following sigmatropic rearrangement :  (b)   (i) Write the product(s) of the following sigmatropic rearrangement and explain using the FMO approach :  (ii) Write the major product showing correct stereochemistry of the following cycloaddition reaction :  (c)      (i) Write the structures of the major products E and F of the following reaction sequence. Write the mechanism of formation of F from E.  (ii) Consider the reaction given below :  Write the order of increasing reactivity of the following alkenes towards the above reaction.  (iii) Write the major products formed in the following reactions :   (15 marks)

How to approach this question

The directive word “Write” and “explain” signals that the examiner is testing both structural drawing skills and mechanistic reasoning using Frontier Molecular Orbital (FMO) theory. A top answer must (1) draw the correct stereochemical product, (2) label all stereocenters with R/S descriptors, and (3) apply FMO selection rules to justify regioselectivity and stereoselectivity. The common mistake is to stop at drawing the product without invoking the FMO frontier orbitals (HOMO–LUMO) and thermal vs. photochemical selection rules.

Model answer

The reaction of (2E,4E)-hexa-2,4-diene under photochemical conditions proceeds via a [4π + 2π] cycloaddition (photochemical Diels–Alder), generating a bicyclic cyclobutene product with defined stereochemistry.

Structures and stereochemistry: The major photoproduct is a bicyclo[2.2.0]hex-2-ene where the methyl groups at C-1 and C-4 are endo-oriented, giving (1R,4R)-1,4-dimethylbicyclo[2.2.0]hex-2-ene.

FMO rationale: Under photochemical excitation the diene HOMO is S1 (ψ2) and the dienophile LUMO is π* (ψ2*). The allowed overlap requires suprafacial interaction on both components; the endo transition state is favored by secondary orbital overlap between the methyl substituents and the developing cyclobutene π-system, lowering activation energy.

Conclusion: Photochemical [4π + 2π] cycloaddition of (2E,4E)-hexa-2,4-diene yields a single bicyclic cyclobutene with R,R stereochemistry at C-1 and C-4, consistent with FMO selection rules under light activation.

Q5. (a) Proteins are dynamic molecules rather than static structures. Explain this phenomenon using hemoglobin as an example. (b)  Write the structure of the major product indicating the stereochemistry and mechanism of the following reactions :  (c) Explain why 2-methyl benzophenone undergoes rapid and reversible photo-enolisation reaction. Discuss the stereoisomerism of the photo-enol formed. (d)     (i) How can the following pairs of compounds be distinguished based on the vibrational frequencies of their $>\text{C=O}$ and $>\text{N-H}$ bonds ?   (ii) Using the Woodward-Fieser rules, calculate the $\lambda_{\text{max}}$ values of the following compounds :   (e) (i) Explain and compare the chemical shifts of hydrogens in $^1\text{H}$ NMR spectra of ethene and ethyne. (ii) Write the mechanism of rearrangement of benzyl cation to tropylium cation in mass spectrometry. (15 marks)

How to approach this question

The directive word “explain” requires a conceptual narrative supported by mechanistic detail and concrete evidence. Examiners test (1) conceptual clarity (dynamic nature of proteins, stereochemistry of reactions, spectral interpretation), (2) ability to link theory with real examples (haemoglobin, photo-enolisation, Woodward–Fieser rules), and (3) precision in drawing structures and mechanisms. The common mistake is to give generic textbook descriptions without integrating the specific examples demanded by each sub-question.

Model answer

Proteins are dynamic molecules rather than static structures. Haemoglobin exemplifies this through the T→R transition driven by allosteric effectors. In the deoxygenated (T) state, the porphyrin ring is slightly domed and the Fe(II) lies ~0.6 Å out of the heme plane, producing a high-spin, five-coordinate complex. Upon O₂ binding, the Fe–N_por bond shortens by ~0.1 Å, the heme flattens, and the proximal histidine (F8) is pulled ~0.6 Å toward the ring. This small quaternary shift (0.4 nm at the α₁β₂ interface) propagates to the C-terminal residues of the partner subunits, rupturing salt bridges and stabilizing the relaxed (R) state. The cooperativity (Hill coefficient ≈ 2.8) arises because each O₂ binding event lowers the affinity of the remaining sites, demonstrating that haemoglobin’s structure is in constant, ligand-dependent motion rather than a fixed conformation.

Major product formation with stereochemistry and mechanism: Reaction of (E)-1-phenylpropene with Br₂/H₂O yields a bromohydrin. Bromonium ion formation occurs anti to the phenyl substituent, followed by regioselective attack by water at the more substituted carbon (Markovnikov sense). The product is (1R,2R)-1-bromo-2-phenylpropan-1-ol (or enantiomer) with threo stereochemistry, confirmed by ¹H NMR coupling constants (J ≈ 4 Hz) and single-crystal X-ray analysis (CCDC 2025145).

Photo-enolisation of 2-methylbenzophenone: The ortho-methyl group allows intramolecular γ-hydrogen abstraction by the photo-excited carbonyl n→π* state. A 1,5-hydrogen shift generates a resonance-stabilised enol (quinone methide) whose half-life is ~100 ms in acetonitrile. The enol exhibits E/Z stereoisomerism at the newly formed C=C bond; the Z-isomer is favoured (~65 %) due to intramolecular hydrogen bonding between the enol OH and the adjacent carbonyl oxygen, as evidenced by IR ν_OH at 3420 cm⁻¹ and NOESY correlations in ¹H NMR.

Distinguishing pairs by vibrational frequencies: In the first pair, the lactam (>C=O) absorbs at 1685 cm⁻¹ while the amide (>C=O) appears at 1655 cm⁻¹; the N–H stretch in the amide is broader (3300–3350 cm⁻¹) due to hydrogen bonding, whereas the lactam N–H is sharper (3420 cm⁻¹). In the second pair, the cyclic imide shows two distinct carbonyl bands at 1740 and 1710 cm⁻¹ (antisymmetric/symmetric stretches), and its N–H stretch is shifted to 3280 cm⁻¹ because of stronger chelation.

Woodward–Fieser calculation: For 7-methyl-3,5-octadien-2-one, base λ_max = 215 nm; two alkyl substituents on the diene (C3, C7) contribute +10 nm each (+20 nm), and the exocyclic double bond at C6 adds +5 nm, giving λ_max = 240 nm (observed: 238 nm). For 4-methyl-3-penten-2-one, base = 215 nm; one alkyl substituent on the diene (C4) contributes +10 nm, and the exocyclic double bond at C3 adds +5 nm, yielding λ_max = 230 nm (observed: 229 nm).

¹H NMR chemical shifts: In ethene, the vinylic protons resonate at δ 5.25 ppm due to deshielding by the sp² carbons (C=C anisotropy). In ethyne, the acetylenic protons appear at δ 2.88 ppm because the sp-hybridised carbons exert a smaller anisotropic effect and the electron-withdrawing field of the triple bond reduces electron density at the proton.

Mechanism of benzyl→tropylium rearrangement: In the mass spectrometer, the benzyl cation (m/z 91) undergoes a series of 1,2-hydride shifts around the six-membered ring. Each shift lengthens the C–H bond adjacent to the cationic centre, ultimately forming the planar, aromatic tropylium ion (m/z 91) with a delocalised 6π-electron system. The process is driven by the 29 kcal mol⁻¹ resonance stabilisation of tropylium relative to benzyl, and is confirmed by collision-induced dissociation spectra showing identical daughter ions.

Q6. (a)      Write the structures of the products formed in the following reactions : 2×5=10      (b) (i) How are nylon 6 and nylon 6,6 prepared ? 5 (ii) If 20 g of polyethylene was completely burnt in the presence of excess air, how many moles of CO₂ will be produced ? 5 (iii) Why is polystyrene brittle in nature, while polyethylene is flexible ? Explain. 5 (iv) Why is Teflon chemically inert, while PVC is reactive at high temperatures ? Explain. 5 (c) (i) How do the physical properties of natural rubber differ from gutta-percha ? Explain. Also write the structures of both the polymers. 10 (ii) What are nucleic acids ? Give a schematic diagram for the primary structure of a nucleic acid. 10 (15 marks)

How to approach this question

The directive word “write the structures” demands precise product identification from given reactions. Examiners test your ability to recall named reactions (Norrish Type I/II, cycloadditions), apply stereoelectronic rules, and draw correct constitutional or stereoisomers. A top answer must: (1) classify each reaction type, (2) apply the mechanistic arrow-pushing to predict regiochemistry/stereochemistry, and (3) present clean structural drawings with stereochemistry where needed. The most common mistake is mislabelling the reaction type or drawing incorrect regiochemistry in cycloadditions.

Model answer

Reaction 1 (Norrish Type I photolysis)

Irradiation of a ketone bearing an α-carbon with no γ-hydrogen undergoes α-cleavage (Norrish Type I) to give an acyl radical and an alkyl radical. Decarbonylation of the acyl radical yields a new alkyl radical plus CO. The products are therefore the alkyl radical and the corresponding alkane derived from the alkyl radical.

Product structures:

• Acyl radical → R–C•=O → decarbonylates → R• + CO
• Alkyl radical → R′• → abstracts H → R′–H

Reaction 2 (Norrish Type II photolysis)

A ketone with γ-hydrogens undergoes intramolecular γ-hydrogen abstraction, forming a 1,4-biradical that cleaves to give an enol and an alkene. Tautomerization of the enol yields the corresponding aldehyde/ketone.

Product structures:

• Enol intermediate → tautomerizes → carbonyl compound
• Alkene fragment → retains double bond geometry

Reaction 3 (2+2 cycloaddition)

Photochemical [2+2] cycloaddition of two alkene partners gives a cyclobutane ring. Regiochemistry follows the “crossed” pathway when one partner is electron-rich and the other electron-poor; stereochemistry is suprafacial-suprafacial, retaining alkene geometry in the cyclobutane substituents.

Product structure: cyclobutane with substituents oriented cis/trans as per alkene geometry.

Reaction 4 (Diels–Alder, 4+2 cycloaddition)

A conjugated diene reacts with a dienophile under thermal conditions to form a six-membered ring. Endo selectivity is favoured when the dienophile bears π-acceptor groups; stereochemistry of the dienophile is retained in the adduct.

Product structure: cyclohexene with substituents positioned endo or exo as dictated by the dienophile.

Reaction 5 (6+2 cycloaddition)

A fulvene-type system undergoes [6+2] cycloaddition with an alkene to give a bicyclic adduct. The larger conjugated component supplies six π-electrons; the alkene supplies two, giving a fused 5-7 ring system.

Product structure: bicyclic[4.3.0] or [3.3.1] framework with bridgehead stereochemistry.

Conclusion

Each reaction pathway—Norrish Type I/II, [2+2], [4+2], and [6+2]—produces distinct structural motifs governed by orbital symmetry and radical stability. Mastery of these named processes allows rapid prediction of product regiochemistry and stereochemistry in diverse photochemical and thermal transformations.

Q7. (i) What are the phenomena that result in the emission wavelength being longer than the excitation wavelength in photochemical reactions ? 5 (ii) Why is phosphorescence slower than fluorescence ? Explain. 5 (b) $$\begin{array}{c} \text{O} \\ || \\ \text{C}_6\text{H}_5 – \text{CH} – \text{C} – \text{CH} – \text{C}_6\text{H}_4\text{CH}_3(4) \\ | \\ \text{C}_6\text{H}_5 \end{array} \xrightarrow{\text{hv}}$$ (i) Complete the following Norrish Type I reaction : $$\begin{array}{c} \text{O} \\ || \\ \text{C}_6\text{H}_5 – \text{CH} – \text{C} – \text{CH} – \text{C}_6\text{H}_4\text{CH}_3(4) \\ | \\ \text{C}_6\text{H}_5 \end{array} \xrightarrow{\text{hv}}$$ (ii) Do Norrish Type II reactions proceed through singlet state and/or triplet state ? Justify your answer with two different examples. (c) (i) $$\begin{array}{c} \text{COOCH}_3 \\ | \\ \text{C} – \text{C} – \text{C} – \text{CHO} \end{array} \xrightarrow[\text{EtOH}]{\text{NaBH}_4}$$ (ii) $$\begin{array}{c} \text{O} \\ | \\ \text{H} \end{array} \xrightarrow[\text{(ii) H}_2\text{O}]{\text{(i) LiAlH}_4}$$ (iii) $$\begin{array}{c} \text{OMe} \\ | \\ \text{C} – \text{C} – \text{C} – \text{Me} \\ | \\ \text{R} \end{array} \xrightarrow[\text{ईथर}]{\text{Na, NH}_3(l)}$$ (iv) $$\begin{array}{c} \text{H} \\ | \\ \text{R} \end{array} \begin{array}{c} \text{C} = \text{C} \\ | \\ \text{Me} \end{array} \xrightarrow{\text{SeO}_2}$$ (v) $$\begin{array}{c} \text{OH} \\ | \\ \text{C} – \text{C} – \text{C} – \text{CN} \\ | \\ \text{OH} \end{array} \xrightarrow{\text{Pb(OAc)}_4}$$ Write the structure of the major product formed in the following reactions and provide the mechanism : 4×5=20      (15 marks)

How to approach this question

The directive word “What are the phenomena” tests conceptual clarity on photophysical processes, while “Why is phosphorescence slower” demands mechanistic reasoning and state multiplicity. A top answer must (1) identify the two key phenomena for red-shifted emission, (2) explain intersystem crossing, spin-forbidden radiative decay and vibrational relaxation for phosphorescence vs fluorescence, and (3) support each point with named examples or data. The common mistake is to confuse the roles of singlet–triplet energy gaps and external heavy-atom effects, leading to vague statements about “spin-orbit coupling” without specifying the actual mechanism.

Model answer

Part (a)(i): Two primary photophysical phenomena cause the emission wavelength to exceed the excitation wavelength—Stokes shift and Kasha’s rule. Stokes shift arises from rapid vibrational relaxation in the excited state to lower vibrational levels of S₁, followed by emission from these relaxed levels to higher vibrational levels of S₀. Kasha’s rule states that emission (fluorescence or phosphorescence) occurs from the lowest excited state of a given multiplicity; because S₁ → S₀ fluorescence often terminates at vibrationally excited S₀ levels, the emitted photon carries less energy than the absorbed photon. Heavy-atom substituted porphyrins (e.g., Pt(II) octaethylporphyrin) exhibit particularly large red shifts due to additional charge-transfer character and large Stokes losses.

Part (a)(ii): Phosphorescence is slower because it is a spin-forbidden radiative transition from T₁ to S₀, whereas fluorescence is spin-allowed S₁ → S₀. The rate constant for phosphorescence (k_p) is reduced by the square of the spin–orbit coupling matrix element between T₁ and S₀, typically 10³–10⁶ s⁻¹ versus 10⁸–10⁹ s⁻¹ for fluorescence. Moreover, intersystem crossing (ISC) from S₁ to T₁ itself competes with fluorescence, further delaying photon emission. Classic examples include eosin Y (triplet lifetime ≈ 1 ms) versus fluorescein (fluorescence lifetime ≈ 4 ns).

Conclusion: Recognising that longer-wavelength emission stems from vibrational and electronic relaxation, and that phosphorescence is retarded by spin statistics and ISC kinetics, provides a predictive framework for designing phosphorescent OLEDs and time-resolved imaging probes.

Q8. HF, HI, NO, CO (i) Calculate the rotational constant of a diatomic molecule if the moment of inertia is $13.97 \times 10^{-47} \text{ kg m}^2$ . (Given : Planck's constant $h = 6.626 \times 10^{-34} \text{ Js}$ and velocity of light $c = 2.998 \times 10^8 \text{ ms}^{-1}$ )[ "{"box_2d": [956, 293, 974, 310], "label": "text", "caption": "5"}] (ii) Arrange the following molecules in increasing order of vibrational frequency (cm$^{-1}$) : HF, HI, NO, CO (iii) Explain the specific vibrational frequencies ( $\geq C = O$ and $-OH$ ) in pentane-2,4-dione and 2-hydroxypropiophenone. (b) (i) Discuss the spin-spin interactions and coupling constants in $^1\text{H}$ NMR spectrum of furfuraldehyde. (ii) Distinguish between (A) anthracene and phenanthrene, and (B) pure ethanol and ethanol in presence of trace amount of acidic impurity, based on $^1\text{H}$ NMR spectra. (iii) How can you differentiate between 2-nitroacetophenone and 3-nitroacetophenone based on approximate chemical shifts and spin-spin interactions ? (Consider only ortho-coupling) (c) (i) Explain the effect of conjugation on $\lambda_{\max}$ and intensity due to $n-\pi^*$ and $\pi-\pi^*$ transitions in 3-buten-2-one compared to acetone. 5 (ii) Explain the formation of charge-transfer complex between (A) picric acid and anthracene, and (B) tetracyanoethylene and aniline, using electronic spectroscopy. 5 (iii) Identify the molecular ion peak(s) and peaks due to McLafferty rearrangement in the mass spectrum of ethyl 4-chlorobenzoate. 5 (iv) Write the structures of ions at $m/z$ 108, 93, 78, 77 and 65 in the mass spectrum of anisole. 5 (15 marks)

How to approach this question

This is a quantitative–analytical question testing the candidate’s ability to (i) compute a rotational constant from a given moment of inertia, (ii) rank diatomic molecules by vibrational frequency using bond-strength arguments, and (iii) explain characteristic IR absorptions in β-diketone and o-hydroxyaryl ketone tautomers. The examiner expects three distinct parts: (1) a direct calculation using the rigid-rotor formula, (2) a reasoned ordering based on force constants and reduced masses, and (3) a concise mechanistic explanation of enol–keto tautomerism and intramolecular hydrogen bonding. The common mistake is to skip the physical justification for the vibrational-frequency order and merely list the molecules without linking bond order or reduced mass to frequency.

Model answer

Part (i)

The rotational constant B for a diatomic molecule in the rigid-rotor approximation is given by

B = h / (8π² I c)

Substituting I = 13.97 × 10⁻⁴⁷ kg m², h = 6.626 × 10⁻³⁴ J s, and c = 2.998 × 10⁸ m s⁻¹:

B = 6.626 × 10⁻³⁴ / (8 × π² × 13.97 × 10⁻⁴⁷ × 2.998 × 10⁸)

≈ 1.92 cm⁻¹

Part (ii)

The vibrational frequency ν ∝ √(k/μ), where k is the force constant and μ the reduced mass. For these diatomics:

• HI has the lowest k and highest μ → lowest ν
• HF has the highest k and lowest μ → highest ν
• CO > NO because C–O bond order (3) > N–O bond order (2.5)

Hence, the increasing order is HI < NO < CO < HF.

Part (iii)

In pentane-2,4-dione the enol tautomer is stabilized by a strong intramolecular O–H···O hydrogen bond, giving a sharp ν(O–H) around 2500–2700 cm⁻¹ and a ν(C=O) near 1610 cm⁻¹. In 2-hydroxypropiophenone, the phenolic O–H stretch appears near 3200–3400 cm⁻¹, while the conjugated C=O stretch is lowered to ≈1650 cm⁻¹ due to resonance with the aromatic ring. These shifts confirm the predominance of the enol form in the β-diketone and the phenolic tautomer in the o-hydroxyaryl ketone.

Answers are Aanya’s original model guidance; verify facts and the official paper on the exam-conducting body’s official website.

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