Chemistry 2023 Paper II 50 marks Draw

Paper II — Q7

(a) (i) Draw the structure of 2'-deoxycytidine-3'-monophosphate. (ii) Why nucleotides and nucleic acids are acidic ? Mention the…

(a)
(i)

Draw the structure of 2'-deoxycytidine-3'-monophosphate.

(ii)

Why nucleotides and nucleic acids are acidic ? Mention the factor(s) that stabilize the DNA duplex.

(iii)

Show the hydrogen bonding in between cytosine and guanine base pair. 15 marks

(b)
(i)

Identify all the possible products in the above reaction showing the sequential pathways of their formation : [cyclopentanone, hν, vapour phase] 10

(ii)

1,3-Butadiene solution on irradiation with UV light follows path I. However, irradiating 1,3-butadiene in the presence of benzophenone at 366 nm follows path II. Give a suitable explanation. 5 marks

(c)
(i)

Write the structure of the compounds (A), (B), (C), (D) and (E). [(i) 2-methylpentan-3-ol, CS2/NaOH then MeI gives (A); SeO2 gives (B); (ii) a benzocycloheptene diene with two OMe groups and an isopropyl group, mCPBA, CH2Cl2, 0 °C gives (C); (iii) 2,2,3,3-tetramethylsuccinic acid, Pb(OAc)4, O2 gives (D); (iv) (R)-3-methylcyclohexene, CrO3-pyridine gives (E)] 10

(ii)

From the reaction sequence below, answer the following questions : [cyclohexene, OsO4/pyridine gives (A); (B) gives cis-cyclohexane-1,2-diol]

(I) Write structure of (A) and (B)

(II) Write the oxidation state of Osmium in the intermediate (A) and its geometry. 5 marks

(iii)

Write the structure of (A) and (B) in the above reactions. [D-mannitol (2R,3R,4R,5R), cyclohexanone, BF3·Et2O, (MeO)3CH gives (A); HIO4 gives (B)] 5

हिंदी में प्रश्न पढ़ें

(क) (i) 2'-डिऑक्सीसाइटिडिन-3'-मोनोफॉस्फेट की संरचना लिखिए।

(ii)

न्यूक्लिओटाइड तथा न्यूक्लिक अम्ल अम्लीय क्यों होते हैं ? DNA डबल हेलिक्स संरचना को स्थिर करने वाले कारकों का उल्लेख कीजिए।

(iii)

साइटोसीन व गुआनिन शार युग्म के मध्य हाइड्रोजन आबंधन को दिखाइए। 15

(ख)(i) निम्नलिखित अभिक्रिया में सभी संभावित उत्पादों की पहचान करें जो उनके गठन का अनुक्रमिक मार्ग दिखाते हैं। 10

(ख)(ii) 1,3-ब्यूटाडाइन विलयन पराबैंगनी किरण पर पथ I का अनुसरण करता है। हालांकि बेंजोफेनॉन की उपस्थिति में 366 nm पर 1,3-ब्यूटाडाइन का पराबैंगनी किरण पथ II का अनुसरण करता है। उपयुक्त स्पष्टीकरण दीजिए। 5

(ग)(i) यौगिकों (A), (B), (C), (D) तथा (E) की संरचना लिखिए। 10

(ग)(ii) नीचे दिए गए अभिक्रिया क्रम से, निम्नलिखित प्रश्नों का उत्तर दें :

(I) (A) तथा (B) की संरचना लिखिए ।

(II) मध्यवर्ती (A) में ऑस्मियम की ऑक्सीकरण अवस्था तथा ज्यामिति लिखिए । 5

(ग)(iii) निम्नलिखित अभिक्रियाओं में (A) तथा (B) की संरचना लिखिए । 5

Q7 of the 2023 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2023 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.

(b) A reaction scheme starting with 1,3-butadiene (a four-carbon chain with double bonds at positions 1 and 3). It branches into two pathways: Path I shows reaction with 'hv' to yield cyclobutene (a four-membered ring with one double bond) plus bicyclo[1.1.0]butane (a structure with two fused three-membered rings). Path II shows reaction with '(C6H5)2CO' (benzophenone) at '366 nm' to yield 3,4-dimethylenecyclobutane (a four-membered ring with two exocyclic double bonds) plus 1-ethylcyclohexene (a six-membered ring with one double bond and an ethyl group attached to a double-bonded carbon).

(c) A reaction scheme: 3-methylcyclohexene (a six-membered ring with a double bond and a methyl group at the allylic position) reacts with 'CrO3-Pyridine' to form compound (E).

(c(ii)) A reaction scheme starting with a cyclohexene ring. An arrow points to the right, labeled above with 'OsO4' and below with 'Pyridine', leading to an intermediate labeled '(A)'. A second arrow points from '(A)' to the right, labeled above with '(B)', leading to a final product. The final product is a cyclohexane ring with two hydroxyl (OH) groups attached to adjacent carbons. Both OH groups are shown with solid wedges, indicating they are on the same side of the ring (cis-1,2-cyclohexanediol).

(c(iii)) A reaction scheme showing two reactants on the left. The first reactant is a complex sugar molecule (a polyhydroxylated chain with multiple chiral centers and OH groups). The second reactant is a cyclohexanone ring (a six-membered ring with a ketone group). An arrow points to the right, labeled above with 'BF3.Et2O' and below with '(MeO)3CH', leading to an intermediate labeled '(A)'. A vertical arrow points down from '(A)', labeled with 'HIO4', leading to a final product labeled '(B)'.

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) Nucleotides and nucleic acids

(i) 2′-Deoxycytidine-3′-monophosphate: cytosine is joined through N-1 to C-1′ of β-D-2-deoxyribofuranose by a β-N-glycosidic bond. C-2′ carries two hydrogens (no OH), C-5′ carries a free CH₂OH, and the phosphate is esterified to the 3′-OH.

Diagram: Structure of 2′-deoxycytidine-3′-monophosphate

(ii) Nucleotides and nucleic acids are acidic because of their phosphate groups. A phosphomonoester has pKₐ values of about 1 and 6, and the single remaining P–OH of each phosphodiester link in DNA or RNA has pKₐ near 1. At physiological pH the backbone is therefore fully ionised, a polyanion with one negative charge per nucleotide, and the conjugate base is resonance-stabilised over the P–O bonds.

The DNA duplex is stabilised by four factors. Watson–Crick hydrogen bonding (A=T two bonds, G≡C three) gives specificity. Base stacking, through van der Waals and π–π interactions between the planar bases, is the largest contribution. The hydrophobic effect buries the bases inside and exposes the sugar-phosphate backbone to water. Counter-ions (Na⁺, Mg²⁺, polyamines) screen the repulsion between the phosphates of the two strands.

(iii) Cytosine–guanine pairing has three hydrogen bonds: cytosine N4–H to guanine O6, guanine N1–H to cytosine N3, and guanine N2–H to cytosine O2.

Diagram: Watson–Crick C·G base pair (three hydrogen bonds)

(b)(i) Vapour-phase photolysis of cyclopentanone

Excitation (n→π*) is followed by Norrish Type I α-cleavage of a C(=O)–Cα bond, which gives the acyl–alkyl 1,5-biradical •C(=O)–CH₂CH₂CH₂CH₂•. This biradical can follow three paths. Recombination regenerates cyclopentanone. Intramolecular H-transfer, in which the acyl radical takes an H from the CH₂ next to the alkyl radical, gives pent-4-enal, CH₂=CH–CH₂CH₂–CHO. Loss of CO from the acyl radical gives the tetramethylene 1,4-biradical •CH₂CH₂CH₂CH₂•, which either closes to cyclobutane or fragments at its central C–C bond to two molecules of ethylene.

The products are therefore CO, ethylene, cyclobutane and pent-4-enal. In the vapour phase no solvent cage holds the radicals together, so decarbonylation (CO + ethylene + cyclobutane) is especially important.

Diagram: Norrish type I photolysis of cyclopentanone in the vapour phase and the products formed

(b)(ii) Direct versus sensitised photolysis of 1,3-butadiene

Path I, direct irradiation: butadiene absorbs only at short wavelength (π→π*, near 217 nm) and reacts from its singlet excited state. The s-cis conformer undergoes photochemically allowed disrotatory 4π electrocyclisation to cyclobutene, and the s-trans conformer gives bicyclo[1.1.0]butane. Both are unimolecular reactions of the singlet.

Path II, benzophenone at 366 nm: butadiene does not absorb at 366 nm, but benzophenone does (n→π*). It crosses to its triplet by fast intersystem crossing (E_T about 69 kcal mol⁻¹) and transfers energy to butadiene (E_T about 60 kcal mol⁻¹), which is triplet sensitisation. Triplet butadiene lives long enough to meet a ground-state molecule and adds to it stepwise through a 1,4-biradical, giving 1,2-divinylcyclobutane and 4-vinylcyclohexene. The triplet state therefore gives bimolecular dimerisation instead of unimolecular electrocyclisation.

(c)(i) Structures of A–E

Compound A is the S-methyl xanthate of 2-methylpentan-3-ol: CS₂/NaOH forms the sodium xanthate and MeI S-methylates it, giving (CH₃)₂CH–CH(C₂H₅)–O–C(=S)–SCH₃.

Compound B: SeO₂ does not act on a xanthate, and the scheme shows no heating, so the Chugaev pyrolysis of A (a syn, six-membered cyclic elimination that releases COS and CH₃SH) is taken as implied before the SeO₂ step. Of the two β-carbons, the CH₂ with two hydrogens gives 4-methylpent-2-ene (E) as the main alkene, and the single tertiary H gives a little 2-methylpent-2-ene. SeO₂ then effects Riley allylic oxidation (ene reaction, [2,3]-shift, hydrolysis), attacking the allylic CH₃ in preference to the tertiary CH. B is 4-methylpent-2-en-1-ol (E); further oxidation to the enal is possible.

Compound C: m-CPBA epoxidises the more electron-rich, tetrasubstituted ring-fusion C=C selectively at 0 °C. The disubstituted C=C and the dimethoxy-arene survive. C is the racemic epoxide at that double bond.

Compound D: Pb(OAc)₄ causes oxidative bis-decarboxylation of the vicinal diacid 2,2,3,3-tetramethylsuccinic acid, losing two CO₂ and forming a C=C. D is 2,3-dimethylbut-2-ene, (CH₃)₂C=C(CH₃)₂.

Compound E: CrO₃·pyridine oxidises the allylic CH₂ (C-6) of (R)-3-methylcyclohexene to the enone. E is 4-methylcyclohex-2-en-1-one (R); the stereocentre is not touched and remains R.

Diagram: Structures of compounds A to E

(c)(ii) OsO₄ dihydroxylation of cyclohexene

(I) OsO₄ adds to the C=C in a concerted syn [3+2] cycloaddition. With pyridine, A is the cyclic osmate(VI) ester (the pyridine adduct of the five-membered Os–O–C–C–O ring). B is the reagent that cleaves the Os–O bonds, aqueous NaHSO₃ (or H₂S). It releases cis-cyclohexane-1,2-diol; both C–O bonds keep the syn stereochemistry of the addition.

(II) Osmium is in the +6 oxidation state in A (two oxo, −4, and the diolate, −2, balance Os⁺⁶). The osmate ester itself is four-coordinate, distorted tetrahedral; pyridine coordinates to give a five- or six-coordinate (pseudo-octahedral) adduct.

Diagram: Syn dihydroxylation of cyclohexene with OsO4 and pyridine: osmate(VI) ester and cis-diol

(c)(iii) Hexitol with cyclohexanone and then HIO₄

The hexitol drawn is D-mannitol, (2R,3R,4R,5R)-hexane-1,2,3,4,5,6-hexol. Compound A: cyclohexanone with BF₃·Et₂O, with HC(OMe)₃ removing the water, protects the two terminal 1,2-diols as five-membered cyclohexylidene ketals. A is 1,2:5,6-di-O-cyclohexylidene-D-mannitol, leaving the C-3/C-4 vicinal diol free.

Compound B: HIO₄ cleaves the free vicinal diol through a cyclic periodate ester, breaking the C3–C4 bond. B is two molecules of 2,3-O-cyclohexylidene-D-glyceraldehyde, (R)-1,4-dioxaspiro[4.5]decane-2-carbaldehyde.

Diagram: D-mannitol to 1,2:5,6-di-O-cyclohexylidene-D-mannitol and periodate cleavage to 2,3-O-cyclohexylidene-D-glyceraldehyde

How this answer will be evaluated

What earns full marks: All structures correct with mechanisms, stereochemistry, and named reactions identified.

Evaluation rubric

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

  1. (a) Answer (i)–(iii) on nucleotides: structure, acidity and duplex stability, C–G pairing. 15 marks

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

    Must cover

    • 2′-deoxycytidine-3′-monophosphate: 2′-deoxyribose, cytosine at C-1′, phosphate at 3′-O
    • Acidity from phosphate groups (polyanion at pH 7)
    • Duplex stability: H-bonds, base stacking, hydrophobic effect, counter-ions
    • C–G: three hydrogen bonds (N4–H···O6, N1–H···N3, N2–H···O2)

    Loses marks

    • Drawing a ribonucleotide (2′-OH) or a 5′-phosphate.

    Earns more

    • Partial credit for each sub-part correctly answered.

    Extra mark

    • Labels C-2′ as CH₂ and notes higher G-C raises Tm.
  2. (b)(i) Identify all products from vapour-phase photolysis of cyclopentanone and the pathway. 10 marks

    trace— start point → the stages in sequence → end point → what changed

    Must cover

    • Norrish Type I α-cleavage to acyl–alkyl biradical
    • Pent-4-enal by H-transfer
    • Decarbonylation to CO and tetramethylene biradical
    • Cyclobutane and two ethylene

    Loses marks

    • Naming Norrish Type II.

    Earns more

    • Partial credit for products without the biradical pathway.

    Extra mark

    • Notes cyclopentanone regenerated by recombination.
  3. (b)(ii) Explain the difference between direct and benzophenone-sensitised photolysis of butadiene. 5 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Direct: singlet, electrocyclisation
    • Products cyclobutene and bicyclobutane
    • Benzophenone absorbs at 366 nm; triplet sensitisation
    • Triplet adds stepwise: 1,2-divinylcyclobutane and 4-vinylcyclohexene

    Loses marks

    • Saying butadiene absorbs at 366 nm.

    Earns more

    • Partial credit for sensitisation without the products.

    Extra mark

    • Gives triplet energies.
  4. (c)(i) Write the structures of A–E. 10 marks

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

    Must cover

    • A S-methyl xanthate of 2-methylpentan-3-ol
    • B from Chugaev then SeO2: allylic alcohol/4-methylpent-2-ene route
    • C epoxide at the tetrasubstituted C=C; D 2,3-dimethylbut-2-ene
    • E (R)-4-methylcyclohex-2-en-1-one

    Loses marks

    • Changing the stereocentre of E.

    Earns more

    • Partial credit per compound correct.

    Extra mark

    • Notes Chugaev pyrolysis is implied between the steps.
  5. (c)(ii) Answer (I) and (II) on OsO4 dihydroxylation. 5 marks

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

    Must cover

    • A cyclic osmate(VI) ester
    • B aqueous NaHSO3 (or H2S) giving cis-diol
    • Os oxidation state +6
    • Geometry: tetrahedral ester; pyridine adduct pseudo-octahedral

    Loses marks

    • Giving +8 for Os in A.

    Earns more

    • Partial credit for A and B without oxidation state/geometry.

    Extra mark

    • Mentions syn [3+2] concerted addition.
  6. (c)(iii) Give the structures of A and B from the hexitol. 5 marks

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

    Must cover

    • Hexitol is D-mannitol
    • A 1,2:5,6-di-O-cyclohexylidene-D-mannitol
    • HIO4 cleaves the free C3–C4 diol
    • B two molecules of 2,3-O-cyclohexylidene-D-glyceraldehyde

    Loses marks

    • Protecting the internal diol.

    Earns more

    • Partial credit for A without B.

    Extra mark

    • Names (R)-1,4-dioxaspiro[4.5]decane-2-carbaldehyde.
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