UPSC Chemistry 2026 — Paper II

(a(i)) Two directives: (a) DRAW three possible geometrical (cis/trans) isomers of [10]annulene, and (b) EXPLAIN its aromaticity - i.e. show that although the 10 pi electrons satisfy Huckel's 4n+2 rule, none of the isomers is planar, so [10]annulene is non-aromatic. (a(ii)) Explain WHY each of two very different rings is aromatic: sydnone (a mesoionic 1,2,3-oxadiazol-5-one) and tropolone (2-hydroxycyclohepta-2,4,6-trien-1-one). Both compounds must be covered with structure, electron count and supporting evidence. (b) Explain, through mechanism, why the two printed tosylates solvolyse at very different rates in acetic acid: M is anti-7-norbornenyl tosylate (OTs on C7 pointing AWAY from the C2=C3 double bond) and N is the syn isomer (OTs on C7 over the double bond). The answer must give the mechanism of each, the rate comparison with its reason, and account for the retained product from M and the rearranged product from N. (c) Identify the five major products (2 marks each) of two linked sequences: benzene + benzoyl chloride/AlCl3 in excess benzene gives V; cyclopentyl chloride with PPh3 then BuLi gives W; W + V gives X; X with catalytic OsO4/NMO gives Y; Y with H2SO4 gives Z. Structures, not names alone, are expected. (d) Three linked demands on the HCl/Al2O3 addition to 1-phenylpropyne: (i) ASSIGN E/Z to the two printed geometrical isomers X (Cl and H on the same side; Ph and CH3 on the same side) and Y (Ph and H on the same side; Cl and CH3 on the same side); (ii) IDENTIFY which is the kinetic and which the thermodynamic product; (iii) PROPOSE a mechanism for the slow conversion of the kinetic into the thermodynamic isomer. (e(i)) Write the structures of the TWO cycloadducts formed when maleic anhydride is heated with the printed diene (o-quinodimethane, i.e. 5,6-bis(methylene)cyclohexa-1,3-diene) and state, with justification, which is major and which minor. (e(ii)) State HOW buta-1,3-diene can be converted into the two printed products - cyclobutene and bicyclo[1.1.0]butane - and DRAW the transition-state (closure) geometry leading to each.

• Three isomers drawn: all-Z, mono-E, di-E
• 10 π electrons satisfy 4n + 2 but the ring cannot be planar
• All-cis planar needs 144° angles (angle strain); trans forms have internal H···H clash
• Non-aromatic; contrast with 1,6-methano[10]annulene
• Sydnone: planar mesoionic ring with six π electrons, charge-separated
• Sydnone evidence: electrophilic substitution at C4, dipole moment
• Tropolone: seven-membered ring, tropylium-like 6 π electrons
• Tropolone evidence: acidity, substitution, chelation
• M (anti): backside π participation, non-classical bishomocyclopropenyl cation, rate about 10¹¹ faster
• M: attack anti to the bridging π bond gives retained product
• N (syn): no π assistance, slow; C1–C6 σ participation with rearrangement
• N: allyl cation in bicyclo[3.2.0] skeleton trapped to give bicyclo[3.2.0]hept-2-en-4-yl acetate
• V = benzophenone
• W = cyclopentylidene triphenylphosphorane and X = (diphenylmethylene)cyclopentane by Wittig reaction
• Y = cis diol, 1-(hydroxydiphenylmethyl)cyclopentanol
• Z = 2,2-diphenylcyclohexanone via pinacol ring expansion
• X = E, Y = Z with CIP reasoning
• X kinetic (syn addition), Y thermodynamic (less strain)
• Mechanism: reversible protonation to the α-chlorobenzyl cation
• C–C rotation then loss of H⁺ gives Y
• Major: adduct across exocyclic CH₂ termini, tetrahydronaphthalene anhydride
• Minor: adduct across the ring diene, bicyclo[2.2.2]octene with two exocyclic methylenes
• Justification: aromatisation gain versus loss of conjugation
• Endo or cis stereochemistry
• Photochemical (UV) conversion
• Cyclobutene: disrotatory closure of s-cis diene, C1–C4 bond
• Bicyclobutane: cross-bonding C1–C3 and C2–C4
• Transition-state structures drawn

(a) Write the structures of five major products (3 marks each) from 4-bromonitrobenzene: M (Sn/HCl), N (M then NaNO2/HCl 0-5 C, then CuCN), O (cyanide ion), P (O then SOCl2, then NH3) and Q (P then Br2/NaOH with heat, then H2O). Structures, with the reaction type named, are expected. (b(i)) Benzaldehyde is treated with concentrated NaOD in D2O: write the products AND the mechanism. The point being tested is the Cannizzaro disproportionation and the fact that the hydride transferred comes from the substrate (intermolecular direct hydride transfer), not from the solvent, so no deuterium enters the benzylic carbon. (b(ii)) Decide which of the two printed diastereomers of 2-bromo-1,3-dimethylcyclohexane (Y: the two methyls on wedges, Br on a hash, i.e. Br trans to both methyls; Z: methyls and Br all on wedges, i.e. all-cis) can undergo E2 elimination, and explain using the anti-periplanar requirement and chair conformational analysis. (b(iii)) Write the major products of two conjugate-addition problems: M from but-3-en-2-one (methyl vinyl ketone) with NaCN/HCN at 5-10 C, and N from mesityl oxide (4-methylpent-3-en-2-one) with PhSH at 25 C. The contrast tested is 1,2- versus 1,4-addition (kinetic versus thermodynamic control and hard/soft nucleophiles). (c(i)) Identify the intermediate [A] generated when chloroform is treated with potassium tert-butoxide - one line, with the species named and its mode of formation. (c(ii)) Write, with correct stereochemistry, the products of dichlorocarbene addition to cis-2-butene (Reaction 1) and to trans-2-butene (Reaction 2), showing that the alkene geometry is preserved in each cyclopropane. (c(iii)) State whether the two dichlorocarbene additions of part (ii) are stereospecific or stereoselective, and define the term used. (c(iv)) Write the structure of the major product when phenol is treated with (i) NaOH, (ii) dichlorocarbene [A] at 70 C and (iii) H+ - that is, recognise and apply the Reimer-Tiemann reaction. (c(v)) Write the major product from one equivalent of dichlorocarbene with the printed hydrocarbon - a cyclopentane ring fused to a six-membered ring that contains two non-conjugated double bonds, one of them the tetrasubstituted ring-fusion bond. The test is chemoselectivity between the two alkenes.

• M = 4-bromoaniline (nitro reduction)
• N = 4-bromobenzonitrile (diazotisation, Sandmeyer)
• O = 4-nitrobenzonitrile (SNAr by cyanide)
• P = 4-nitrobenzamide (via acid chloride and NH₃)
• Q = 4-nitroaniline (Hofmann rearrangement)
• Cannizzaro: benzyl alcohol and benzoate
• OD⁻ adds to C=O; tetrahedral intermediate transfers hydride to a second molecule
• Hydride from substrate; no D at the benzylic carbon
• Z (all-cis) undergoes E2, Y cannot
• Anti-periplanar (trans-diaxial) H and Br needed
• Z: Br axial, methyls equatorial, H on C1 and C3 axial
• Y: Br axial only with axial methyls, so β-H equatorial
• M = 4-oxopentanenitrile by 1,4-addition
• N = 4-methyl-4-(phenylthio)pentan-2-one
• 1,2 (kinetic, reversible) versus 1,4 (thermodynamic); soft nucleophile
• Dichlorocarbene :CCl₂
• Formed by deprotonation to ⁻CCl₃ and α-elimination of Cl⁻
• cis-2-butene gives cis-1,1-dichloro-2,3-dimethylcyclopropane
• trans-2-butene gives trans-cyclopropane
• Concerted syn addition retains alkene geometry
• Stereospecific
• Definition: stereoisomeric reactants give stereoisomerically different products
• Stereoselective defined as preference for one stereoisomer
• Reimer-Tiemann reaction
• Phenoxide attacks :CCl₂ at ortho
• Hydrolysis of ArCHCl₂ to CHO
• Salicylaldehyde (major)
• Carbene adds to the more electron-rich tetrasubstituted ring-fusion C=C
• Dichlorocyclopropane at ring fusion, syn addition
• Disubstituted C=C untouched

(a(i)) Write the structures of the major products of acetaldehyde with EXCESS formaldehyde in aqueous NaOH and give the mechanism - i.e. recognise the industrial pentaerythritol synthesis (three crossed aldol additions followed by a crossed Cannizzaro reaction). (a(ii)) From 5-oxohexanal (H-CO-CH2CH2CH2-CO-CH3): give X (aq. NaOH then CrO3/H2SO4, molecular formula C6H8O2) and Y (X with phenylhydrazine in acetic acid, heat), and write the mechanism of the X to Y conversion. (b) From 2,6-dimethylphenol give four products: M (K2CO3 then CH3COCl), N (M with hv), O (NaH then the printed 3-carbon halide) and P (O heated to 220 C), and supply a mechanism for the O to P conversion. The paper is testing the photo-Fries rearrangement on one branch and the para-Claisen rearrangement on the other. (c(i)) Write the products of (1) bromobenzene and (2) 2,6-dideuteriobromobenzene with NaNH2 in liquid NH3, decide which reaction is faster and justify the choice - i.e. the benzyne (elimination-addition) mechanism plus a primary kinetic isotope effect argument. (c(ii)) For the printed N-benzyl benzo-fused eight-membered cyclic amine (a 2-benzazocine, N flanked by a benzylic ring CH2 and a -(CH2)4- chain): give X (MeI) and Y (NaNH2 in liquid NH3) and write a suitable mechanism for the X to Y step - i.e. quaternisation followed by an ammonium-ylide rearrangement.

• Three crossed aldol additions at the three α-hydrogens
• Tris(hydroxymethyl)acetaldehyde intermediate (no α-H)
• Crossed Cannizzaro with HCHO as hydride donor
• Pentaerythritol and sodium formate
• Intramolecular aldol to 3-hydroxycyclohexanone; Jones oxidation to cyclohexane-1,3-dione (C₆H₈O₂)
• Y = 1,2,3,9-tetrahydro-4H-carbazol-4-one (Fischer indole)
• Mechanism: hydrazone or enehydrazine, [3,3] shift, rearomatisation, cyclisation, loss of NH₃
• Six-membered ring reasoning for the aldol
• M = 2,6-dimethylphenyl acetate
• N = 4-hydroxy-3,5-dimethylacetophenone (photo-Fries, para since ortho blocked)
• O = aryl allyl ether by Williamson synthesis
• P = 4-allyl-2,6-dimethylphenol
• Mechanism: Claisen [3,3] to ortho dienone then Cope to para, tautomerisation
• Benzyne mechanism (elimination-addition)
• Reaction 1 gives aniline
• Reaction 2 gives 2- and 3-deuterioaniline mixture
• Reaction 1 faster: primary kinetic isotope effect for C–H versus C–D in the rate-determining step
• X = N-methyl quaternary ammonium iodide
• NaNH₂ forms the ammonium ylide
• [2,3] Sommelet–Hauser rearrangement via a five-membered transition state
• Y = tertiary amine with a 2-methylbenzyl group on N; ring retained

(a(i)) Give the stereochemical product of the photochemical reaction of (2E,4E)-hexa-2,4-diene and EXPLAIN it by the frontier-molecular-orbital approach - i.e. the 4 pi electrocyclisation must be shown to be disrotatory in the excited state and the product identified as cis-3,4-dimethylcyclobutene. (a(ii)) Identify, with correct stereochemistry, the major product of heating the printed 3,4-dimethylhexa-1,5-diene (one methyl on a bold wedge, the other on a hash, i.e. the chiral R*,R* diastereomer) - a Cope [3,3]-sigmatropic rearrangement. (b(i)) Write the product(s) of the photochemical sigmatropic rearrangement of the printed alkene, 2-cyclopentylidenepropane (isopropylidenecyclopentane), and explain by the FMO approach - i.e. a suprafacial [1,3]-hydrogen shift, allowed photochemically and forbidden thermally. (b(ii)) Write the major product, with correct stereochemistry, of the thermal cycloaddition between 1-methoxycyclopenta-1,3-diene (OMe on a diene terminus) and methyl vinyl ketone - i.e. a Diels-Alder reaction whose regiochemistry ('ortho' rule) and endo selectivity must both be shown. (c(i)) From dicyclohexyl ketone: give E (Br2, 1 equivalent, in CH3COOH) and F (E with KOH) and write the mechanism of the E to F conversion - the Favorskii rearrangement via a cyclopropanone. (c(ii)) Arrange four alkenes - (1) Me2C=CH2, (2) trans-MeCH=CHMe, (3) Ph(Me)C=CH2 and (4) CH2=CH2 - in order of INCREASING reactivity towards Br2 in methanol (bromomethoxylation, in which MeO ends up on the more substituted carbon), and justify the order. (c(iii)) Write the major products of indole with (Reaction 1) n-BuLi then CO2 then H+, and (Reaction 2) n-BuMgBr then CO2 then H+ - the point being the N versus C-3 ambident reactivity of the indolyl anion as the counter-ion is changed.

• Photochemical 4π electrocyclisation is disrotatory
• Excited-state HOMO is ψ3, with termini of the same phase
• Product is cis-3,4-dimethylcyclobutene (meso, 3R,4S)
• Diagram of ψ2 (thermal, conrotatory) versus ψ3 (photochemical, disrotatory)
• Cope rearrangement, a thermal [3,3]-sigmatropic shift via a chair transition state
• Both methyls equatorial in the chair for the chiral (R*,R*) isomer
• Product (2E,6E)-octa-2,6-diene
• Bond broken C3-C4, bond formed C1-C6
• Suprafacial [1,3]-hydrogen shift, photochemically allowed
• Products isopropenylcyclopentane and 1-isopropylcyclopentene
• Allyl fragment: thermal HOMO ψ2 has opposite-phase termini, so the suprafacial shift is forbidden
• Photoexcited HOMO ψ3 has same-phase termini, so the shift is allowed
• Thermal [4+2] cycloaddition giving a norbornene
• Ortho regiochemistry: acetyl on C2 adjacent to the C1-OMe bridgehead, by HOMO/LUMO coefficients
• Endo selectivity by secondary orbital overlap
• Product endo-2-acetyl-1-methoxybicyclo[2.2.1]hept-5-ene (racemic)
• E is (1-bromocyclohexyl)(cyclohexyl)ketone, by acid-catalysed α-bromination through the enol
• F is 1,1′-bicyclohexyl-1-carboxylic acid (Favorskii rearrangement)
• Enolate on the other α-carbon, intramolecular SN2 to a dispiro cyclopropanone
• OH⁻ addition, ring opening to the more stable carbanion, protonation
• Order: CH2=CH2 < trans-MeCH=CHMe < Me2C=CH2 < PhC(Me)=CH2
• Rate-determining bromonium ion with carbocation character
• Alkyl groups stabilise by hyperconjugation and induction
• Tertiary benzylic cation is stabilised by resonance, so it is fastest
• Both reagents deprotonate N-H to give an ambident indolyl anion
• Reaction 1: ionic N-Li, N-carboxylation gives indole-1-carboxylic acid (labile)
• Reaction 2: covalent N-Mg blocks N, C-3 attacks CO2 to give indole-3-carboxylic acid
• Site selectivity governed by the ionic versus covalent character of the metal-N bond

(a) Explain the directive 'Explain this phenomenon': show that a protein is not a rigid structure but samples conformations, and demonstrate it specifically through haemoglobin (T to R quaternary switch, cooperativity, allosteric effectors, local breathing motions). Both the general statement and the haemoglobin case study must be present. (b) For ONE substrate reacted in two different ways, draw the major product with correct stereochemistry and give the mechanism of each: (a) m-CPBA (arrow drawn to the left) and (b) (i) NBS then (ii) 30% NaOH. Both routes give epoxides, but of opposite (diastereomeric) configuration, and that contrast is the point of the question. (c) Two linked demands: (1) explain WHY 2-methylbenzophenone photo-enolises rapidly and reversibly (intramolecular gamma-H abstraction geometry, biradical, o-quinodimethane stabilisation, fast H-back-transfer); (2) discuss the stereoisomerism of the photo-enol - the two geometric isomers about the new enol C=C and their very different lifetimes. (d(i)) Distinguish TWO printed pairs using IR stretching frequencies: (a) ethyl benzoate C6H5CO-OCH2CH3 vs phenyl propanoate C6H5O-CO-CH2CH3 using nu(C=O); (b) 4-bromo-N-methylbenzamide 4-BrC6H4CO-NHCH3 vs 4'-bromoacetanilide 4-BrC6H4NH-COCH3 using nu(C=O) and nu(N-H). Numerical frequencies with reasons are required for all four compounds. (d(ii)) Apply the Woodward-Fieser rules numerically to the two printed tricyclic (phenalene-type) structures: (A) a methyl-substituted conjugated TRIENE containing a homoannular diene, and (B) a methyl-substituted cross-ring dienone. A stepwise additive calculation with a stated final lambda-max is required for each. (e(i)) Explain AND compare - give the actual 1H chemical shifts of ethene and ethyne, note that the order is the opposite of what hybridisation alone predicts, and explain the reversal by magnetic (diamagnetic) anisotropy of the pi systems. (e(ii)) Write the MECHANISM (stepwise, with structures) by which the benzyl cation (m/z 91) ring-expands to the tropylium cation in the mass spectrometer, and justify the driving force and the supporting evidence.

• General statement: proteins sample interconnected conformations over fs to ms, with small energy gaps shifted by ligands, pH and temperature
• T to R quaternary switch: salt bridges broken, Fe moves into the porphyrin plane (about 0.6 Å), about 15° rotation of one αβ dimer
• Cooperativity: sigmoidal curve (Hill n about 2.8) against hyperbolic myoglobin; MWC and KNF models
• Allosteric effectors (H⁺/CO₂ Bohr effect, Cl⁻, 2,3-BPG in the central cavity) stabilise T; local breathing motions let O₂ enter the haem pocket
• m-CPBA: concerted syn delivery of O to the less hindered face, opposite to the allylic methyl, giving the epoxide trans to the substituents
• NBS: bromonium ion on the less hindered face, anti attack of water from the substituent face, giving the trans-bromohydrin
• NaOH: alkoxide intramolecular SN2 (Williamson) on the C-Br carbon, anti-periplanar, giving the epoxide cis to the substituents
• Clear statement that the two epoxides are diastereomers, with a labelled drawing of each
• n,π* triplet by fast intersystem crossing; syn o-methyl gives a 1,5 (six-membered) intramolecular γ-H abstraction
• 1,4-Biradical stabilised as an o-quinodimethane, closing to the photoenol
• Reversibility: [1,5]-H shift or proton transfer regenerates the ketone, so the compound is photostable
• Z-enol short-lived (ns to μs) by direct [1,5]-H return; E-enol long-lived (ms to s), trapped by dienophiles
• Ethyl benzoate about 1720 cm⁻¹, lowered by conjugation with the ring
• Phenyl propanoate about 1760 cm⁻¹, higher because the O lone pair is drawn into the phenyl ring and there is no conjugation of C=O
• 4-Bromo-N-methylbenzamide C=O about 1640 cm⁻¹ (conjugated, full N donation); 4′-bromoacetanilide about 1665 cm⁻¹ (N lone pair shared with ring)
• N-H stretch about 3300 cm⁻¹ for both (about 3450 in dilute solution), so the pair is told apart mainly by amide I
• A: base 253 (homoannular) + 30 (one extra conjugated C=C) + 10 (two exocyclic) + 30 (six alkyl/ring residues) = 323 nm
• B: base 215 + 30 (one extra conjugated C=C) + 10 (two exocyclic) = 255 before substituents
• B substituents: α ring residue +10, β +12, γ +18, δ two groups +36, total 331 nm
• Correct identification of the chromophore: homoannular diene plus one extra double bond in A; linear O=C-C=C-C=C in B
• Ethene δ about 5.3 and ethyne δ about 1.8-1.9
• Hybridisation alone (sp C-H more s-character) would predict ethyne downfield
• Ethene: in-plane H in the deshielding region of the π circulation
• Ethyne: aligned along B₀; π circulation about the axis gives a shielding cone at the H
• Benzyl cation formed by benzylic cleavage of an alkylbenzene molecular ion
• Ring C-C migration to CH₂⁺, giving a bicyclic norcaradienyl-type cation, then electrocyclic ring opening to the seven-membered ion
• Driving force: aromatic tropylium, 6 π electrons in seven p orbitals, Hückel 4n + 2
• Evidence: ¹³C/D labelling scrambling, loss of C₂H₂ to m/z 65, and the same ion from cycloheptatriene

(a) Draw only the STRUCTURES of the products of five printed reactions (2 marks each); no mechanism is asked. The examiner is testing chemoselectivity of hydride/borane reductions: (i) Luche reduction of an enone, (ii) dissolving-metal reduction of the same enone, (iii) B2H6 on an acid-ester, (iv) LiAlH4 on the same acid-ester, (v) BH3-THF on a lactam bearing an ester. (b(i)) State the preparation of nylon 6 and of nylon 6,6 - monomers, reaction type and conditions - and write the repeat unit of each. Both polymers must be covered; a note on only one caps the answer at half marks. (b(ii)) Calculate the number of gram-molecules (moles) of CO2 formed on complete combustion of 20 g of polyethylene in excess air. Requires recognising the repeat unit -CH2- (14 g/mol) or -CH2CH2- (28 g/mol) and writing a balanced combustion equation. (b(iii)) Explain the contrast in mechanical behaviour: why polystyrene is brittle whereas polyethylene is flexible. The explanation must rest on chain structure (pendant phenyl vs bare -CH2-), glass transition temperature relative to room temperature, and crystallinity. (b(iv)) Explain why Teflon (PTFE) is chemically inert at high temperature while PVC is reactive. The answer must compare C-F with C-Cl bond strength, the fluorine shielding/steric sheath in PTFE, and the availability of beta-hydrogen in PVC that permits dehydrochlorination. (c(i)) Compare the PHYSICAL properties of natural rubber and gutta-percha, explain the difference in terms of cis versus trans 1,4-polyisoprene geometry and the resulting chain packing/crystallinity, AND draw the structures (repeat units) of both polymers - the structures are explicitly asked for and carry marks. (c(ii)) Two demands: define/describe nucleic acids (composition, the nucleoside and nucleotide units, the two classes DNA and RNA), and DRAW a schematic diagram of the primary structure of a nucleic acid showing the 3'-5' phosphodiester backbone. The diagram is explicitly asked and carries marks.

• (i) cyclohex-2-en-1-ol (1,2-reduction); (ii) cyclohexanone (C=C reduced)
• (iii) ethyl 5-hydroxy-3-methylpentanoate (acid reduced, ester intact)
• (iv) 3-methylpentane-1,5-diol (both reduced)
• (v) methyl 1-benzylpyrrolidine-3-carboxylate (lactam C=O reduced to CH2, ester intact)
• Nylon 6: caprolactam, ring-opening polymerisation, about 250-270 °C with a little water
• Repeat unit -[NH-(CH₂)₅-CO]-
• Nylon 6,6: adipic acid and hexamethylenediamine, condensation with loss of water, via the nylon salt, about 250-280 °C under pressure
• Repeat unit -[NH-(CH₂)₆-NH-CO-(CH₂)₄-CO]-
• Repeat unit -CH₂CH₂- = 28 g/mol (or CH₂ = 14 g/mol)
• Balanced equation C₂H₄ + 3O₂ → 2CO₂ + 2H₂O
• 20/28 = 0.714 mol of repeat units
• CO₂ = 2 × 0.714 = 1.43 mol
• Polyethylene: bare -CH₂- chain, flexible, packs closely, partly crystalline
• Polyethylene Tg far below room temperature, so amorphous regions are rubbery
• Polystyrene: pendant phenyl groups restrict rotation and stiffen chains; atactic, so amorphous
• Polystyrene Tg about 100 °C, above room temperature, so glassy and brittle
• C-F about 485 kJ/mol against C-Cl about 330 kJ/mol
• Fluorine sheath shields the carbon backbone and is non-polarisable
• PTFE has no hydrogen, so there is no elimination route
• PVC has β-hydrogens; dehydrochlorination (HCl loss, zip reaction) gives polyenes above about 100-150 °C
• Rubber: soft, elastic, amorphous, low Tg (about -70 °C); gutta-percha: hard, tough, inelastic, crystalline (about 60 °C melting point)
• Both are 1,4-polyisoprene; rubber is cis (Z), gutta-percha is trans (E)
• Cis gives kinked, poorly packed chains, so elasticity; trans gives straight, closely packed, crystalline chains
• Repeat units -[CH₂-C(CH₃)=CH-CH₂]- drawn correctly with the cis and trans geometries
• Nucleic acids are polymers of nucleotides; nucleotide = sugar + base + phosphate; nucleoside = sugar + base
• DNA: deoxyribose, A G C T; RNA: ribose, A G C U; purines and pyrimidines named
• Diagram: alternating sugar-phosphate backbone with bases at C1′
• Phosphodiester bond links 3′-OH of one sugar to 5′-OH of the next; 5′ and 3′ ends labelled

(a(i)) Name and briefly explain the photophysical phenomena responsible for the emitted light having a LONGER wavelength (lower energy) than the exciting light - i.e. explain the Stokes shift and the processes that create it. (a(ii)) Explain why phosphorescence is a slower (longer-lived) emission than fluorescence. The explanation must rest on the spin-selection rule: T1 -> S0 is spin-forbidden and occurs only through weak spin-orbit coupling, so its radiative rate constant is orders of magnitude smaller. (b(i)) Complete the printed Norrish Type I photoreaction of the unsymmetrical dibenzyl ketone (C6H5)2CH-CO-CH(4-CH3C6H4)2, i.e. give the products and the radical pathway: alpha-cleavage, decarbonylation and radical recombination. (b(ii)) Answer whether Norrish Type II reactions proceed from the singlet state, the triplet state, or both, and JUSTIFY with two different examples. The examiner expects the answer 'both, depending on the ketone' supported by diene-quenching (Stern-Volmer) evidence: aliphatic ketones react from S1 and T1, aryl alkyl ketones only from T1. (c) For each of five printed reactions, give the structure of the MAJOR product AND the mechanism (4 marks each): (i) NaBH4/EtOH on methyl 3-formylbenzoate, (ii) LiAlH4 then H2O on 4-tert-butylcyclohexanone, (iii) Na/NH3(l)-ether-ethanol on anisole, (iv) SeO2 on the trisubstituted alkene RCH=C(CH3)2, (v) Pb(OAc)4 on 2-cyanohydroquinone. Product without mechanism, or mechanism without product, earns only part marks.

• Stokes shift: emission at lower energy than absorption
• Vibrational relaxation and internal conversion within S₁ before emission (Kasha's rule)
• Solvent reorganisation and excited-state geometry change
• Intersystem crossing to T₁ giving phosphorescence at longer wavelength; excimer/exciplex emission
• Fluorescence S₁ → S₀ is spin-allowed, k about 10⁶-10⁹ s⁻¹, ns lifetime
• Phosphorescence T₁ → S₀ is spin-forbidden (ΔS ≠ 0)
• It occurs only by weak spin-orbit coupling, enhanced by heavy atoms
• Radiative rate constant is small, lifetime ms to s; observed mainly at 77 K in rigid media
• α-Cleavage of the C-CO bond gives an acyl radical and a diarylmethyl radical
• Decarbonylation of the acyl radical to a diarylmethyl radical and CO
• Radical pair Ph₂CH• and Tol₂CH• recombines
• Products: Ph₂CH-CH(Tol)₂ (cross), Ph₂CH-CHPh₂ and Tol₂CH-CHTol₂, about 1:2:1 statistically; CO released
• Both, depending on the ketone
• Mechanism: six-membered γ-H abstraction, 1,4-biradical, cleavage (enol + alkene) or Yang cyclisation
• Aliphatic ketone (2-pentanone/2-hexanone): S₁ and T₁ both react; diene quenching is incomplete
• Aryl alkyl ketone (valerophenone): only T₁ because of very fast ISC; diene quenches completely (Stern-Volmer)
• (i) methyl 3-(hydroxymethyl)benzoate, hydride on CHO, ester untouched; (ii) trans-4-tert-butylcyclohexanol by axial hydride attack, OH equatorial (about 90%)
• (iii) 1-methoxycyclohexa-1,4-diene by the Birch mechanism: radical anion, protonation ortho, second electron, protonation para
• (iv) (E)-RCH=C(CH₃)CH₂OH by ene reaction, [2,3]-sigmatropic shift, hydrolysis; attack at the E-methyl
• (v) 2-cyano-1,4-benzoquinone; ArO-Pb(OAc)₃, then electron flow with Pb(IV) → Pb(II) and 2 AcOH

(a) Three independent 5-mark parts: (i) CALCULATE the rotational constant of a diatomic with I = 13.97 x 10^-47 kg m^2 using the given h and c; (ii) ARRANGE HF, HI, NO and CO in increasing order of vibrational frequency in cm^-1 with justification; (iii) EXPLAIN the specific C=O and O-H vibrational frequencies of pentane-2,4-dione and of 2-hydroxypropiophenone. (b(i)) Discuss the spin-spin interactions and coupling constants in the 1H NMR spectrum of furfuraldehyde (furan-2-carbaldehyde): assign the four protons, give their multiplicities and the three ring coupling constants with values. (b(ii)) Two separate 1H NMR distinctions must be made: (A) anthracene versus phenanthrene, and (B) pure ethanol versus ethanol containing a trace of acidic impurity. Both halves must be answered; the argument must be based on the number/intensity/position of signals and, for (B), on exchange averaging of the OH coupling. (b(iii)) Differentiate 2-nitroacetophenone (ortho) from 3-nitroacetophenone (meta) using approximate 1H chemical shifts and the aromatic splitting pattern, considering ONLY ortho-coupling (J approximately 8 Hz). The splitting patterns of the four aromatic protons in each isomer must be spelled out. (c(i)) Explain how conjugation in 3-buten-2-one (methyl vinyl ketone) changes BOTH the lambda(max) and the INTENSITY of the n-pi* and pi-pi* bands relative to acetone. Values for both compounds and both transitions are expected. (c(ii)) Explain, using electronic (UV-visible) spectroscopy, the formation of charge-transfer complexes in two specific pairs: (A) picric acid with anthracene and (B) tetracyanoethylene (TCNE) with aniline. The distinguishing spectroscopic signature of a CT complex and the donor/acceptor roles must be identified in each case. (c(iii)) For ethyl 4-chlorobenzoate, identify (a) the molecular ion PEAKS - note the plural, because chlorine gives an M/M+2 isotope pair - and (b) the peak(s) arising from McLafferty rearrangement, with the structure of the rearranged ion. (c(iv)) Write the STRUCTURES (not merely the formulae) of the five ions at m/z 108, 93, 78, 77 and 65 in the mass spectrum of anisole, i.e. give the fragmentation pathway of methoxybenzene under electron impact.

• (i) B = h/8π²cI = 200.4 m⁻¹ = 2.00 cm⁻¹, with the substitution shown
• (ii) NO < CO < HI < HF, from ν̃ = (1/2πc)√(k/μ), with k and μ compared
• (iii) pentane-2,4-dione enol: intramolecular chelated H-bond, C=O about 1600-1640, very broad O-H about 2500-3200; keto form 1725/1707
• (iii) 2-hydroxypropiophenone: chelated, C=O lowered to about 1630-1640, O-H broad about 3000-3200 and concentration-independent
• Four protons: CHO δ about 9.65, H5 about 7.70, H3 about 7.26, H4 about 6.62
• Ring couplings J₃,₄ about 3.6, J₄,₅ about 1.7 and J₃,₅ about 0.8 Hz
• Multiplicities: H5 dd, H3 d (fine splitting), H4 dd, CHO small doublet or singlet
• Assignment logic: α-H to oxygen most downfield; H4 most upfield
• (A) Anthracene: three signals (H9/10 singlet about 8.4, 4H about 8.0, 4H about 7.45), ratio 2:4:4
• (A) Phenanthrene: five signals of 2H each, H4/H5 about 8.7 (bay region)
• (B) Pure ethanol: OH triplet, CH₂ doublet of quartets, CH₃ triplet
• (B) Trace acid: fast proton exchange removes the OH coupling; OH singlet, CH₂ quartet, CH₃ triplet
• 2-Nitro: d, t, t, d at about 8.1, 7.7, 7.6 and 7.4
• 3-Nitro: s (H2 about 8.75), d (about 8.4), d (about 8.25), t (about 7.7)
• Acetyl CH₃ about 2.55 (ortho, twisted) against about 2.68 (meta)
• Decisive test: singlet near 8.75 and a single triplet in the meta isomer; two triplets and no singlet in the ortho
• Acetone: n-π* about 279 nm (ε about 15), π-π* about 187-190 nm (ε about 900)
• 3-Buten-2-one: n-π* about 324 nm (ε about 25), π-π* about 219 nm (ε about 3,600)
• Conjugation narrows the π-π* gap: raised HOMO, lowered LUMO, shifts to longer wavelength and hyperchromic
• n level nearly unchanged, π* lowered, so n-π* also red-shifts, with only slight intensity gain (still forbidden)
• CT band: new, broad, intense visible absorption, belonging to neither component
• (A) Picric acid is the π-acceptor, anthracene the π-donor; coloured (red-orange) complex
• (B) TCNE is the acceptor, aniline the donor (n/π); coloured complex, CT band in the visible
• Mulliken relation hν_CT ≈ I_D - E_A - C: lower donor ionisation energy gives longer wavelength
• M⁺• at m/z 184 (³⁵Cl) and M+2 at 186 (³⁷Cl), intensity about 3:1
• McLafferty: γ-H transfer through a six-membered state with loss of ethene (28)
• Rearranged ion 4-chlorobenzoic acid radical cation at m/z 156/158
• Other ions: 139/141 (ArCO⁺) and 111/113 (ClC₆H₄⁺)
• 108: C₆H₅-O⁺•-CH₃ (molecular ion)
• 93: C₆H₅O⁺ (phenoxy cation) by loss of •CH₃; 65: C₅H₅⁺ by loss of CO from 93
• 78: C₆H₆⁺• by loss of CH₂O with H transfer
• 77: C₆H₅⁺ (phenyl cation) by loss of •OCH₃