UPSC Chemistry 2024 — Paper II

Framework: Concept > Structure or mechanism > Reasoning > Result. (a(i)) describe: define > structure or process in order > labelled diagram > significance | (a(ii)) explain: definition/context > points in order > small example > short close | (b) explain: definition/context > points in order > small example > short close | (c) explain: definition/context > points in order > small example > short close | (d) explain: definition/context > points in order > small example > short close | (e) derive: given > assumptions > stepwise derivation > result > check Full marks: Accurate mechanisms, correct stereochemical reasoning, and precise experimental evidence.

• Azulene and Pyridine identified as aromatic
• Cyclooctatetraene identified as nonaromatic
• Cyclopentadienyl cation identified as antiaromatic
• Sydnone classification with electronic justification
• Identify the compound as diphenylcyclopropenone
• Show that the polar C=O form gives an aromatic cyclopropenium oxide (2 pi electrons)
• Explain that nucleophilic addition would destroy the aromatic 2-pi system
• Support with the large dipole moment / O-protonation to the hydroxycyclopropenium ion
• Identification of benzyne (aryne) as the intermediate
• Mechanism involving elimination of HBr
• Description of an experimental proof (e.g., isotope labelling)
• Explanation of the evidence supporting the intermediate
• Identify SOCl2 without base and the S_Ni pathway
• Explain that (S)->(S) with unchanged CIP order means retention
• Show the alkyl chlorosulfite intermediate and intimate ion pair
• Contrast with S_N2 (pyridine added) which inverts to (R)
• Correct reagent for each of the five conversions
• Correct name for each reaction type
• Specific conditions where applicable (e.g., heat, light)
• Matching reagents to the specific structural change
• Prediction of conrotatory/disrotatory modes for each step
• Deduction of structure M based on reaction conditions
• Determination of cis/trans relationship of hydrogens
• Application of Woodward-Hoffmann rules

(a(i)) justify: claim > 3-4 reasons > evidence > conclusion | (a(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (b(i)) explain: definition/context > points in order > small example > short close | (b(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (c(i)) describe: define > structure or process in order > labelled diagram > significance | (c(ii)) explain: definition/context > points in order > small example > short close Full marks: Complete mechanisms with correct structures, proper justification, named reactions identified

• Identify E2 mechanism
• Cite primary kinetic isotope effect
• Explain C-H bond breaking in RDS
• Link isotope effect to transition state
• Identify kinetic vs thermodynamic control
• Explain lower activation energy for 1,2
• Explain greater stability of 1,4 product
• Link temperature to product distribution
• Identify X = diethyl benzylidenemalonate, PhCH=C(CO₂Et)₂
• Show enolate formation mechanism
• Show nucleophilic addition to carbonyl
• Name the reaction correctly
• Identify X = S-methyl xanthate, (CH₃)₂CH–C(CH₃)₂–O–C(=S)–SCH₃
• Name Chugaev elimination
• Explain syn-periplanar requirement
• Show cyclic transition state
• Product A: 2-substituted tetrahydrocarbazole, ethyl 2-cyano-2-(2,3,4,9-tetrahydro-1H-carbazol-2-yl)butanoate (4-isomer minor)
• Product B: new pyridine ring fused ortho to the former NH₂ (linear pyrido-fused azaxanthone)
• Product C: ring-opened CH₂=CH(CH₂)₄N(CH₃)₂ (N,N-dimethylhex-5-en-1-amine)
• Name each reaction: Fischer indole, Skraup, Hofmann elimination
• Disconnect to propanal + ethyl 2-bromopropanoate
• Show Zn insertion giving the BrZn ester enolate (Reformatsky reagent)
• Show addition to propanal giving ethyl 3-hydroxy-2-methylpentanoate
• Dehydrate and hydrolyse to CH₃CH₂CH=C(CH₃)COOH

(a(i)) explain: definition/context > points in order > small example > short close | (a(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (b(i)) explain: definition/context > points in order > small example > short close | (b(ii)) comment: context > arguments both sides > judgment > close | (b(iii)) describe: define > structure or process in order > labelled diagram > significance | (c(i)) explain: definition/context > points in order > small example > short close | (c(ii)) account for: state the phenomenon > the causes in order of weight > conclusion Full marks: Complete mechanisms with correct stereochemistry, orbital diagrams, and electronic reasoning throughout.

• Protonation of nitrile nitrogen
• Nucleophilic attack by water
• Tautomerization to amide intermediate
• Final product: acetic acid and ammonium chloride
• Anti-addition mechanism via bromonium ion
• Cis-2-butene yields racemic mixture
• Trans-2-butene yields meso compound
• Stereochemical reasoning for each
• Thermal 6π electrocyclic ring closure of the hexatriene
• Disrotatory motion (4n+2 electrons, thermal)
• Cyclohexadiene methyl enol ether intermediate
• Acid hydrolysis of the enol ether and C=C migration into conjugation give the enone
• Identify as a [3,3]-sigmatropic (Cope, divinylcyclopropane-type) rearrangement
• Identify the cyclopropane σ bond that breaks and the new C–C bond formed
• Suprafacial-suprafacial pathway
• Symmetry-allowed classification
• Butadiene is the diene and the quinone the dienophile
• Cyclohexene ring formation
• Addition at the methyl-bearing C=C (OMe-bearing C=C is deactivated)
• cis ring fusion: 2-methoxy-4a-methyl-4a,5,8,8a-tetrahydronaphthalene-1,4-dione
• Acyl azide formation from acid chloride
• Loss of N₂ gives the acyl nitrene (N with 6 valence electrons)
• Singlet nitrene orbital diagram
• Triplet nitrene orbital diagram
• Indole: 3-position carbocation stability
• Indole: aromaticity preservation
• Chlorobenzene: resonance gives partial C=Cl double bond (shorter, stronger bond)
• Chlorobenzene: SN2 backside attack blocked and phenyl cation too unstable for SN1

(a(i)) explain: definition/context > points in order > small example > short close | (a(ii)) explain: definition/context > points in order > small example > short close | (b(i)) justify: claim > 3-4 reasons > evidence > conclusion | (b(ii)) describe: define > structure or process in order > labelled diagram > significance | (c(i)) describe: define > structure or process in order > labelled diagram > significance | (c(ii)) explain: definition/context > points in order > small example > short close Full marks: Correct products, mechanisms, and diagrams with clear reasoning.

• Identify product X as 2-acetylfuran
• Show electrophilic attack of acylium ion on furan
• Indicate formation of sigma complex as RDS
• Show deprotonation to restore aromaticity
• (1) tert-butyl bromide in ethanol at 60 °C: unimolecular ionisation to the tert-butyl cation; on heating elimination dominates, so the major product is 2-methylpropene by E1, with tert-butyl ethyl ether (SN1) as the minor solvolysis product
• (2) 1-chloroprop-1-ene with NaNH₂: E2 dehydrohalogenation (H from C-2, Cl from C-1) gives propyne directly; excess NaNH₂ forms sodium propynide, and work-up returns propyne
• Mechanism labels stated for each: (1) E1 (with SN1 competing), (2) E2
• Reason: tertiary substrate + weak nucleophile + heat favour the cation and elimination; vinyl halide + very strong base favours E2
• Heat an equimolar mixture of two ethers labelled in both the ring and the allyl group
• Predict only the two non-crossover phenols if the reaction is intramolecular
• State absence of crossover products
• Conclude reaction is intramolecular
• Show two-step energy profile with intermediate
• Draw structure of sigma complex intermediate
• Draw structure of first transition state
• Draw structure of second transition state
• (A) 1,1′-bi(cyclopentyl)-1,1′-diol with conc. H2SO4: pinacol rearrangement with ring expansion; loss of water, migration of a ring C–C bond to the cation, giving the spiro ketone spiro[4.5]decan-6-one
• (B) 3-methylbutan-2-one: NH2OH gives the oxime; Al2O3 (Beckmann) migrates the group anti to the OH; the more substituted isopropyl group migrates from the major oxime, giving N-isopropylacetamide (N-methylisobutyramide from the minor oxime)
• (C) camphenilol (3,3-dimethylbicyclo[2.2.1]heptan-2-ol): ionisation to the 2-norbornyl cation, 1,2-methyl shift giving the tertiary cation, loss of H⁺ gives santene (2,3-dimethylbicyclo[2.2.1]hept-2-ene)
• Arrow-pushing shown for each rearrangement, with the reason the migrating bond migrates
• The ketone PhCO–CF2Cl has no α-hydrogen, so Favorskii and haloform paths are closed; Ac2O/NaOAc on heating supplies the acetate enolate (acetic anhydride enolised by acetate)
• Perkin-type condensation: the enolate of acetic anhydride adds to the highly electrophilic carbonyl (activated by CF2Cl), the mixed anhydride cyclises/eliminates on heating, and hydrolysis on work-up gives the α,β-unsaturated acid
• Product: (E)-4-chloro-4,4-difluoro-3-phenylbut-2-enoic acid, PhC(CF₂Cl)=CH–COOH (CF₂Cl trans to COOH, by analogy with CF₃ ketones)
• Steps: enolate formation, aldol-type addition, acetylation of the alkoxide, E1cB elimination, hydrolysis

(a) compare: paired headings or point-wise contrast > key differences > significance > conclusion | (b) map: locate accurately > label > one line on why it matters | (c) map: locate accurately > label > one line on why it matters | (d) compare: paired headings or point-wise contrast > key differences > significance > conclusion | (e) map: locate accurately > label > one line on why it matters Full marks: All structures correct, mechanisms shown, reasoning precise, no errors.

• Draw structures of A, T, G, C, U
• Identify 2'-OH in RNA vs 2'-H in DNA
• Link 2'-OH to RNA hydrolytic instability
• Mention DNA double helix stability
• P: (E)-2-methylhepta-2,5-diene (Na/NH₃ gives the trans alkene; the isolated C=C is untouched)
• Q: 2-methylheptane-2,3,5,6-tetrol by syn-dihydroxylation of both C=C
• R: acetone + malondialdehyde + acetaldehyde from HIO₄ cleavage of both vicinal diols
• Correct carbon skeleton in all
• (i) Photoreduction: the triplet keto ester abstracts the carbinol C–H of butan-2-ol
• (ii) Photoaddition of methanol via ketyl + •CH₂OH radicals
• Correct radical intermediates (ketyl and α-hydroxyalkyl radicals)
• Products: cross-coupled 1,2-diol esters (with pinacol/tartrate and butanone or HCHO by-products)
• Correct choice for all five pairs: A, B, B (marginal), A, B
• Woodward–Fieser values for (ii) 234 vs 244 nm and (v) 239 vs 254 nm
• Steric inhibition of resonance in (i) and charge-transfer band in (iv)
• Exocyclic C=C and α/β-substituent increments explained
• Correct number of signals: I 3, II 3, III 3
• Label distinct protons in each
• Order by increasing chemical shift
• Justify shift based on environment

(a(i)) explain: definition/context > points in order > small example > short close | (a(ii)) derive: given > assumptions > stepwise derivation > result > check | (b) explain: definition/context > points in order > small example > short close | (c) enumerate: list the items in order > one line each > no commentary Full marks: Complete mechanisms with correct intermediates and stereochemistry.

• Initiation: Benzoyl peroxide homolysis to benzoyloxy radicals
• Propagation: Radical addition to ethylene monomer
• Termination: Coupling or disproportionation of growing chains
• Arrow-pushing for each step
• Atactic: amorphous, transparent, glassy (Tg ≈ 100 °C), no sharp melting point, soluble
• Isotactic: Crystalline, high melting point, rigid
• Syndiotactic: Crystalline, high melting point, rigid
• Linkage of tacticity to crystallinity
• n→π* excitation of the cyclopentanone in A, then Norrish type I α-cleavage of the C(=O)–C bond at the quaternary ring-junction carbon, giving an acyl radical and a tertiary alkyl radical (biradical)
• Route to C: the alkyl radical passes a hydrogen from its adjacent methyl to the acyl radical (disproportionation), giving an aldehyde and the exocyclic methylene; the tertiary OH then adds intramolecularly to the aldehyde to give the cyclic hemiacetal (lactol) C
• Route to B: the acyl radical instead takes the hydrogen from the CH2 α to itself, giving a ketene; the tertiary OH adds intramolecularly to the ketene to give the spiro lactone B
• Draw the biradical and both H-transfer paths; note that the OH is what turns the open-chain products into the two rings
• A: cyclohexanone (NaBH4 in MeOH reduces it to cyclohexanol)
• B: cyclohexene, by acid-catalysed E1 dehydration of cyclohexanol
• C: the reagent NBS (with light or AIBN, CCl4): allylic bromination gives 3-bromocyclohexene
• D: mCPBA epoxidises the C=C: 3-bromocyclohexene oxide (2-bromo-7-oxabicyclo[4.1.0]heptane), mainly on the face anti to Br. Mechanism of the first step: hydride transfer from BH4⁻ to the carbonyl carbon through a four-centre transition state, giving the alkoxyborate, then protonation by methanol

Framework: Woodward-Fieser rules. (a) calculate: given > formula > substitution > result with units > interpretation | (b) trace: start point > the stages in sequence > end point > what changed | (c(i)) analyse: intro > causes > effects > stakeholders/linkages > way forward | (c(ii)) explain: definition/context > points in order > small example > short close Full marks: All parts answered with correct structures, mechanisms, and reasoning; no errors in calculations or assignments.

• Identify base value for each enone system
• Count and apply alkyl substituent increments
• Apply exocyclic double bond corrections
• Final values: A 357 nm, B 324 nm, C 317 nm
• X: CHO reduced to CH₂OH (Boc carbamate and C=C untouched at 0 °C)
• Y: anti-Markovnikov hydroboration–oxidation puts CH₂OH on C-3 (a 1,5-diol)
• Z: Jones oxidation of both primary OH to CO₂H (dicarboxylic acid; δ-lactone possible)
• Show correct regiochemistry for hydroboration
• Identify ester functional group from IR
• Assign NMR signals to specific protons
• Distinguish between the two isomers
• Justify assignments with coupling patterns
• m/z 58: [CH₂=C(OH)CH₃]⁺• from I by McLafferty rearrangement
• m/z 92: [C₇H₈]⁺• from II by γ-H transfer to the ring and loss of ethylene
• Show cleavage pathways for each ion
• Justify stability of fragment ions

(a) justify: claim > 3-4 reasons > evidence > conclusion | (b(i)) justify: claim > 3-4 reasons > evidence > conclusion | (b(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (c(i)) explain: definition/context > points in order > small example > short close | (c(ii)) explain: definition/context > points in order > small example > short close | (c(iii)) calculate: given > formula > substitution > result with units > interpretation Full marks: Accurate assignments, correct structures, clear reasoning, and precise calculations.

• 1685 cm⁻¹: compound I (verbenone), an α,β-unsaturated ketone in a six-membered ring; conjugation lowers the C=O bond order
• 1787 cm⁻¹: compound II (chrysanthenone), a cyclobutanone; ring strain raises the stretching frequency, and the C=C is β,γ, so no conjugation
• 1715 cm⁻¹: compound III, the α,β-unsaturated (conjugated) ethyl ester
• 1740 cm⁻¹: compound IV, the β,γ-unsaturated ester (C=C not conjugated with C=O), i.e. an ordinary ester carbonyl. Photochemistry: a [1,3]-shift of the allylic cyclobutane bond converts I into II; α-cleavage and β-scission of II give a dienyl ketene that ethanol traps as esters IV (β,γ) and III (conjugated)
• Identify product as methyl 2,2-dimethylpropanoate (pivalate)
• Link the 9H singlet (δ 1.2) to a tert-butyl group, C(CH₃)₃, with no neighbouring H
• Link the 3H singlet (δ 3.6) to the ester OCH₃
• Give the other possible product: methyl 3-methylbutanoate, (CH₃)₂CHCH₂COOCH₃ (cleavage toward the less stable tertiary carbanion)
• Read the geometry from the structures: in I, Ha and Hb are on different carbons and on opposite sides: trans (J ≈ 12–18 Hz)
• In II, Ha and Hb are on the same carbon: geminal (J ≈ 0–3 Hz)
• In III, Ha and Hb are on different carbons and on the same side: cis (J ≈ 6–12 Hz)
• Ascending order of J(Ha–Hb): II < III < I
• Identify the two alkenes: 2,3-dimethylbut-2-ene (Zaitsev, tetrasubstituted) and 2,3-dimethylbut-1-ene (Hofmann, terminal)
• ¹H NMR: the but-2-ene shows a single 12H singlet (δ ≈ 1.65) and no vinylic H; the but-1-ene shows =CH₂ signals at δ ≈ 4.7 (2H), a CH septet, an allylic CH₃ singlet and a 6H doublet
• IR: the but-1-ene shows =C–H stretch ≈ 3080, C=C ≈ 1645 and =CH₂ wag ≈ 890 cm⁻¹; the symmetrical but-2-ene has no =C–H band and a very weak or absent C=C stretch
• m/z 74: molecular ion M⁺• (loss of an electron from an O lone pair)
• m/z 59: α-cleavage with loss of CH₃•, giving CH₃CH₂O⁺=CH₂; m/z 45: loss of C₂H₅•, giving C₂H₅O⁺ (CH₃CH=OH⁺)
• m/z 31: CH₂=OH⁺ formed from m/z 59 by loss of ethene (H-transfer); it is the base peak
• m/z 29: ethyl cation C₂H₅⁺ from inductive C–O cleavage
• Reduced mass μ = m_H m_Cl/(m_H + m_Cl) = (1.008 × 35.45)/(36.46) ≈ 0.980 amu ≈ 1.63 × 10⁻²⁷ kg (0.973 amu with m_H = 1, m_Cl = 35.45)
• Moment of inertia I = μ r², so r = √(I/μ) = √(2.70 × 10⁻⁴⁷ / 1.63 × 10⁻²⁷) ≈ 1.29 × 10⁻¹⁰ m
• State the answer as ≈ 1.29 Å (129 pm); values 1.27–1.30 Å accepted depending on the masses used
• Units carried through each step