Chemistry 2025 Paper II 50 marks Calculate

Paper II — Q4

(a) (i) After polymerization of p-hydroxybenzoic acid, IR analysis shows 0·2% unreacted —COOH. Calculate the molecular weight of…

(a)
(i)

After polymerization of p-hydroxybenzoic acid, IR analysis shows 0·2% unreacted —COOH. Calculate the molecular weight of the polymer and categorize it. 7 marks

(ii)

Write the structure of the polymer formed when p-hydroxybenzoic acid is polymerized. 3 marks

(iii)

What is the precursor of the main component of natural rubber? Draw the structures of both monomer and polymer. 5 marks

(b)
(i)

Compare the salient properties of Nylon 6 and Nylon 6,6. 5 marks

(ii)

How would you prepare syndiotactic polystyrene? 5 marks

(iii)

What are the salient features of alpha helix in regards to handedness, residues per turn and elongation per residue? What are the principal stabilizing factors for an alpha helix? 5 marks

(c)
(i)

Compounds 1 and 2 on reaction with NaN₃ yield same product but compound 1 reacts at room temperature while compound 2 reacts at 200 °C. Identify the product formed and explain the reasons for different reactivities of compounds 1 and 2: 10 marks

(ii)

Identify the major product in the following reactions and justify your answer: 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

p-हाइड्रॉक्सीबेंजोइक अम्ल के बहुलकन के बाद IR विश्लेषण 0·2% बचा हुआ —COOH दर्शाता है। बहुलक के आण्विक भार की गणना कीजिए और उसे वर्गीकृत कीजिए। 7 अंक

(ii)

p-हाइड्रॉक्सीबेंजोइक अम्ल के बहुलकन में बने बहुलक की संरचना लिखिए। 3 अंक

(iii)

प्राकृतिक रबर के मुख्य घटक का पूर्वगामी क्या है? एकलक तथा बहुलक, दोनों की संरचनाएं बनाइए। 5 अंक

(b)
(i)

नाइलॉन 6 तथा नाइलॉन 6,6 के मुख्य गुणधर्मों की तुलना कीजिए। 5 अंक

(ii)

सिंडियोटैक्टिक पॉलिस्टाइरीन का निर्माण कैसे करेंगे? 5 अंक

(iii)

ऐल्फा हेलिक्स के वर्तिता, प्रति मोड अवशिष्ट तथा प्रति अवशिष्ट दीर्घन के संबंध में मुख्य लक्षण क्या हैं? ऐल्फा हेलिक्स के प्रमुख स्थिरीकरण कारक क्या हैं? 5 अंक

(c)
(i)

यौगिक 1 तथा 2 की NaN₃ के साथ हुई अभिक्रिया में समान उत्पाद बनता है जबकि यौगिक 1 कमरे के तापमान पर अभिक्रिया करता है लेकिन यौगिक 2 की अभिक्रिया 200 °C पर होती है। बनने वाले उत्पाद की पहचान कीजिए तथा यौगिक 1 एवं 2 की विभिन्न अभिक्रियाशीलता के कारणों की व्याख्या कीजिए : 10 अंक

(ii)

निम्नलिखित अभिक्रिया में मुख्य उत्पाद की पहचान कीजिए तथा अपने उत्तर का औचित्य सिद्ध कीजिए : 10 अंक

Q4 of the 2025 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2025 Chemistry paper

The figure this question refers to, in words

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

(a) Two chemical structures labeled (1) and (2). Structure (1) is 2-chlorobutane. It shows a central carbon atom bonded to a methyl group (line to the left), a chlorine atom (Cl) with a solid wedge bond (pointing out of the page), a hydrogen atom (implied), and an ethyl group (line to the right). Structure (2) is 1,2-dimethylbenzene (o-xylene). It shows a benzene ring with two methyl groups (Me) attached to adjacent carbons. The bond to the top methyl group is a solid wedge, and the bond to the bottom methyl group is a solid wedge.

(c) A reaction scheme starting from a central molecule: 1-cyclohexylethanone (a cyclohexane ring attached to a carbonyl group, which is attached to a methyl group). The scheme branches into two paths. The left path shows reagents '1) LDA/THF' and '2) D2O' leading to product 'Q'. The right path shows reagents '1) Excess NaOH' and '2) D2O' leading to product 'P'.

Model answer

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

(a)(i)

Molecular weight from 0·2 % residual –COOH. *p*-Hydroxybenzoic acid (HBA) undergoes step‑growth condensation to a polyester, and every linear chain that is not cyclic retains one terminal –COOH group, which is exactly the end‑group the IR spectrum detects. If 0·2 % of the –COOH groups remain unreacted, the extent of reaction is p = 0·998, and the Carothers equation gives X̄ₙ = 1/(1 − p) = 1/0·002 = 500. The repeat unit –O–C₆H₄–CO– has M₀ = 120 g mol⁻¹, so M̄ₙ ≈ 500 × 120 = 6·0 × 10⁴ g mol⁻¹. This places the material firmly in the high‑molecular‑weight condensation‑polyester class, well above the fibre‑forming threshold.

(a)(ii)

Polymer structure. The product is the *para*‑linked aromatic polyester poly(4‑oxybenzoyl) (poly(p‑hydroxybenzoic acid), sold as Ekonol; HBA is also the main unit of liquid-crystalline Vectra copolyesters) with the repeating unit [–O–C₆H₄–CO–]ₙ, joined by ester (–CO–O–) linkages formed with loss of water as the condensation by‑product.

(a)(iii)

Precursor of natural rubber. The monomer (precursor) is isoprene, 2‑methylbuta‑1,3‑diene, CH₂=C(CH₃)–CH=CH₂. Natural rubber is cis‑1,4‑polyisoprene, in which the monomers add through the 1,4‑positions so that the residual C=C lies inside the chain: –[CH₂–C(CH₃)=CH–CH₂]–ₙ, with the two chain bonds on the same side of the double bond (cis configuration). The cis geometry is what gives the elastomer its coiled, elastic chain conformation.

(b)(i)

Nylon 6 vs Nylon 6,6.

  • Monomers and repeat unit: Nylon 6 is made from a single monomer, caprolactam, and has one C₆ unit, –[NH–(CH₂)₅–CO]–ₙ; Nylon 6,6 is made from hexamethylenediamine and adipic acid, –[NH–(CH₂)₆–NH–CO–(CH₂)₄–CO]–ₙ.
  • Route: Nylon 6 by ring-opening polymerisation of caprolactam (water-initiated or anionic); Nylon 6,6 by step-growth condensation of the nylon salt.
  • Melting point: Nylon 6 about 220 °C; Nylon 6,6 about 265 °C.
  • Mechanical and thermal behaviour: Nylon 6,6 has somewhat higher strength, stiffness and heat resistance; Nylon 6 has slightly lower crystallinity, better impact strength and surface finish.
  • Moisture and dye uptake: slightly higher for Nylon 6, which also dyes more easily; Nylon 6,6 absorbs slightly less water.
  • Uses: both are used for fibres (textiles, tyre cord, carpets) and engineering plastics; Nylon 6,6 is preferred where higher service temperature is needed.

Thus Nylon 6,6, with its more symmetric, better‑packing diad, is the stronger, higher‑melting fibre, while Nylon 6 is easier to process from one monomer.

(b)(ii)

Preparation of syndiotactic polystyrene. Syndiotactic PS (first made by Ishihara at Idemitsu in 1986) is made by coordinate (chain‑growth) polymerisation of styrene with a homogeneous single‑site catalyst — e.g. CpTiCl₃/MAO or Cp*Ti(OMe)₃/MAO (half-sandwich titanium catalysts) — under inert atmosphere at moderate temperature. Under chain-end control, each incoming styrene is inserted with the opposite face to the preceding unit, so the phenyl groups alternate sides of the backbone, giving the syndiotactic (r, racemo) placement. This contrasts with free‑radical polymerisation (atactic PS) and with TiCl₃/AlR₃ Ziegler catalysts (isotactic PS).

(b)(iii)

α‑Helix: features and stabilisation. The α‑helix is right‑handed, with 3·6 residues per turn and a rise (elongation) of 1·5 Å per residue (pitch ≈ 5·4 Å). It is stabilised principally by intra‑chain C=O(i)···H–N(i+4) hydrogen bonds running parallel to the helix axis, and by the outward orientation of the side chains, which packs favourably and shields the polar backbone. Proline residues and contiguous like‑charges disrupt the helix.

(c)(i)

Same azide from the two dichloro‑bicyclic sulfides. Compounds 1 and 2 are the two diastereomeric dichloro‑bicyclic thioethers (fused bicyclic sulfides) shown, differing in the relative orientation of the C–Cl bonds. The common product is the bicyclic sulfide azide in which N₃ has replaced the chlorine at the carbon β to sulfur (the assisted centre), the other C–Cl being retained. In 1 that C–Cl is antiperiplanar to the C–S bond, so the sulfur lone pair participates (neighbouring‑group / anchimeric assistance) to form a bridged three‑membered sulfonium (thiiranium) ion; this stabilised transition state lowers the activation energy, so N₃⁻ reacts at room temperature, then opens the sulfonium (second inversion → net retention). In 2 the same C–Cl is oriented syn/endo, not antiperiplanar to C–S, so sulfur cannot assist and displacement must proceed by a slow, high‑energy direct path, requiring 200 °C. Because the two isomers carry that Cl on opposite faces, the retention product from 1 and the inversion product from 2 place N₃ on the same face, hence the identical azide; steric/electronic effects and sulfonium stabilisation account for the rate difference.

(c)(ii)

Products P and Q from 1‑cyclohexylpropan‑1‑one. The substrate is the ethyl ketone C₆H₁₁–CO–CH₂–CH₃, with two distinct α‑sites: the less hindered ethyl α‑CH₂ and the more substituted ring α‑CH. Q (LDA/THF, 1 equiv, low T, then D₂O): LDA is strong, bulky and the enolate formation is irreversible, so the kinetic (less substituted) enolate forms at the ethyl CH₂; quenching with D₂O gives Q = C₆H₁₁–CO–CHD–CH₃ (D on the ethyl α‑carbon only). P (excess NaOH, then D₂O): hydroxide enolisation is reversible and thermodynamic, so every α‑hydrogen exchanges at both sites, giving P = (1‑D‑cyclohexyl)–CO–CD₂–CH₃ (D at the ring α‑CH and at both H of the ethyl α‑CH₂, i.e. three D). Thus Q reflects kinetic control at the less hindered site, while P reflects complete thermodynamic α‑deuteration.

Diagram: Chemistry 2025 Paper II Q4: poly(4-oxybenzoyl), isoprene and cis-1,4-polyisoprene, products P and Q

How this answer will be evaluated

What earns full marks: Complete mechanisms, correct structures, full working, precise terminology, all parts addressed thoroughly

Evaluation rubric

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

  1. (a)(i) Calculate molecular weight of polymer from 0.2% unreacted -COOH and categorize it. 7 marks

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

    Must cover

    • State formula relating % end groups to molecular weight
    • Substitute 0.2% and calculate MW
    • Categorize polymer (e.g., high molecular weight)
    • Show full working with units

    Loses marks

    • Missing formula or substitution
    • No categorization of polymer
    • Arithmetic errors in calculation

    Earns more

    • Mention degree of polymerization
    • Reference IR analysis for -COOH

    Extra mark

    • Mention specific polymer name (poly(p-hydroxybenzoic acid))
  2. (a)(ii) Write structure of polymer formed from p-hydroxybenzoic acid. 3 marks

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

    Must cover

    • Draw repeating unit structure
    • Show ester linkage formation
    • Indicate correct connectivity

    Loses marks

    • Wrong linkage type
    • Missing repeating unit notation
    • Incorrect connectivity

    Earns more

    • Label functional groups
    • Show condensation byproduct (water)

    Extra mark

    • Mention polymer name (poly(p-hydroxybenzoic acid))
  3. (a)(iii) Identify precursor of natural rubber and draw monomer and polymer structures. 5 marks

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

    Must cover

    • Name precursor (isoprene/2-methyl-1,3-butadiene)
    • Draw monomer structure
    • Draw polymer structure (polyisoprene)
    • Show correct stereochemistry if applicable

    Loses marks

    • Wrong precursor identification
    • Missing monomer or polymer structure
    • Incorrect stereochemistry

    Earns more

    • Mention cis/trans configuration
    • Label double bonds

    Extra mark

    • Mention natural rubber source (Hevea brasiliensis)
  4. (b)(i) Compare salient properties of Nylon 6 and Nylon 6,6. 5 marks

    compare— paired headings or point-wise contrast → key differences → significance → conclusion

    Must cover

    • Compare monomer composition
    • Compare repeat-unit structure (one C₆ unit vs C₆ diamine + C₆ diacid)
    • Compare thermal properties
    • Compare mechanical properties

    Loses marks

    • No structured comparison
    • Missing key property categories
    • Inaccurate property values

    Earns more

    • Mention synthesis method differences
    • Reference specific property values

    Extra mark

    • Mention industrial applications
  5. (b)(ii) Explain preparation of syndiotactic polystyrene. 5 marks

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

    Must cover

    • Name catalyst system (half-sandwich titanium single-site catalyst with MAO)
    • Describe polymerization mechanism
    • Explain tacticity control
    • Mention reaction conditions

    Loses marks

    • No catalyst identification
    • Missing mechanism explanation
    • Incorrect tacticity description

    Earns more

    • Draw catalyst structure
    • Show monomer insertion mechanism

    Extra mark

    • Mention specific catalyst (e.g., CpTiCl₃/MAO, Ishihara 1986)
  6. (b)(iii) Highlight alpha helix features and stabilizing factors. 5 marks

    highlight— name the salient points → one line of substance each → close

    Must cover

    • State handedness (right-handed)
    • Give residues per turn (3.6)
    • State elongation per residue (1.5 Å)
    • List stabilizing factors (H-bonds, side chains)

    Loses marks

    • Wrong handedness
    • Incorrect residues per turn
    • Missing stabilizing factors

    Earns more

    • Mention H-bond pattern (i to i+4)
    • Reference side chain orientation

    Extra mark

    • Mention specific amino acid preferences
  7. (c)(i) Identify product and explain different reactivities of compounds 1 and 2 with NaN₃. 10 marks

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

    Must cover

    • Identify product structure
    • Explain compound 1 reactivity (room temp)
    • Explain compound 2 reactivity (200°C)
    • Provide mechanistic reasoning

    Loses marks

    • Wrong product identification
    • No mechanistic explanation
    • Missing reactivity comparison

    Earns more

    • Draw reaction mechanism
    • Mention steric/electronic factors
    • Reference transition state stability

    Extra mark

    • Mention specific orbital interactions
    • Reference computational data
  8. (c)(ii) Identify major products P and Q and justify the answer. 10 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Identify product P structure
    • Identify product Q structure
    • Justify P formation (NaOH conditions)
    • Justify Q formation (LDA/THF conditions)

    Loses marks

    • Wrong product identification
    • No mechanistic justification
    • Missing condition-specific reasoning

    Earns more

    • Draw reaction mechanisms
    • Mention kinetic vs thermodynamic control
    • Reference base strength differences

    Extra mark

    • Mention specific enolate formation
    • Reference solvent effects
What "Calculate" is asking you to do

Apply the standard formula or schedule to data the question has already supplied — a table of readings, cost records, a balance sheet — and produce the number. The method is rarely in doubt; the marks sit in the named intermediate quantities, each of which has to appear as a labelled line.

Structure that answers it

Data as given → formula or standard treatment, named → substitution → each intermediate, labelled → result with units

Where marks are lost

Omitting an intermediate the marking scheme pays for separately, or rounding at an intermediate line so the final figure drifts. In commerce and accountancy, any figure in a statement that no numbered working note supports is treated as unearned.

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