Chemistry 2023 Paper II 50 marks Solve

Paper II — Q6

(a)(i) The frequencies of vibration of the following molecules in their v = 0 states are HCl : 2885 cm⁻¹; D₂ : 2990 cm⁻¹; DCl …

(a)
(i)

The frequencies of vibration of the following molecules in their v = 0 states are HCl : 2885 cm⁻¹; D₂ : 2990 cm⁻¹; DCl : 1990 cm⁻¹ and HD : 3627 cm⁻¹. Calculate the energy change of the following reaction : HCl + D₂ → DCl + HD. Determine whether energy is liberated or absorbed. [Given : h = 6·626 × 10⁻³⁴ Js, c = 2·998 × 10⁸ ms⁻¹, Nₐ = 6·022 × 10²³ mol⁻¹] 10 marks

(ii)

The IR spectra of butyric acid and ethyl butyrate show sharp strong singlet absorption at 1725 cm⁻¹ and 1740 cm⁻¹, respectively. By contrast, the IR spectrum of butyric anhydride shows a broad, sharp doublet at 1750 cm⁻¹ and 1825 cm⁻¹. Why are these so different ? 5 marks

(b)
(i)

Write the structure of product(s) in the above reactions : 10 marks

(ii)

What is meant by 'Tacticity' of a polymer ? Distinguish among isotactic, syndiotactic and atactic polymers. 5 marks

(c)
(i)

An organic compound having molecular formula C₁₆H₂₅NO gave following IR and ¹H NMR data : IR(cm⁻¹) = 1690; ¹H NMR(CDCl₃, 400 MHz) : δ 1·11(t, J = 7Hz, 6H), 1·29(d, J = 7Hz 6H), 2·40 (q, J = 7Hz, 4H), 2·55(t, J = 7Hz, 2H), 2·65(t, J = 7 Hz, 2H), 3·12(septet, 1H), 7·21(d, J = 8Hz, 2H), 7·81(d, J = 8Hz, 2H). Determine the structure of the compound. 10 marks

(ii)

Assign and arrange the lettered protons in the increasing order of their chemical shift value in ¹H NMR spectrum. 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

ब्यूटिरिक एसिड व एथिल ब्यूटिरेट अवरक्त (IR) स्पेक्ट्रम में क्रमशः: 1725 cm⁻¹ व 1740 cm⁻¹ पर एक प्रबल अवशोषण दिखाते हैं। इसके विपरीत, ब्यूटिरिक एनहाइड्राइड अवरक्त (IR) स्पेक्ट्रम में 1750 cm⁻¹ व 1825 cm⁻¹ पर द्विक अवशोषण दिखाता है। ये इतने अलग क्यों हैं ? 5 marks

(b)
(i)

निम्नलिखित अभिक्रियाओं में उत्पाद(ओं) की संरचना लिखिए : 10 marks

(ii)

बहुलक की टैक्टिसिटी (व्यवस्था) से क्या अभिप्राय है ? समव्यवस्थ (आइसोटैक्टिक), एकान्तर व्यवस्थ (सिन्डियोटैक्टिक) तथा अव्यवस्थ (एटैक्टिक) बहुलकों के बीच अंतर बताएं। 5

(c)
(i)

एक कार्बनिक यौगिक जिसका आण्विक सूत्र C₁₆H₂₅NO है वह निम्नलिखित IR व ¹H NMR आँकड़ा देता है : IR(cm⁻¹) = 1690; ¹H NMR(CDCl₃, 400 MHz) : δ 1·11(t, J = 7Hz, 6H), 1·29(d, J = 7Hz 6H), 2·40 (q, J = 7Hz, 4H), 2·55(t, J = 7Hz, 2H), 2·65(t, J = 7 Hz, 2H), 3·12(septet, 1H), 7·21(d, J = 8Hz, 2H), 7·81(d, J = 8Hz, 2H). यौगिक की संरचना निर्धारित कीजिए। 10

(ii)

अक्षरों से चिह्नित प्रोटोनों को उनके ¹H NMR स्पेक्ट्रम में रासायनिक सृति मानों के अनुसार आरोही क्रम में लिखें। 10

Q6 of the 2023 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2023 Chemistry paper
Model answer coming soon See all 2023 Chemistry questions

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) Reaction 3: Anisole (a benzene ring with a methoxy group, -OMe). The reaction conditions are Na, NH3(l) in Et2O, EtOH. The product is labeled C.

(c) A chemical structure of a bicyclic molecule consisting of a six-membered ring fused to a five-membered ring. The six-membered ring contains an oxygen atom and two double bonds. The five-membered ring contains a double bond. Specific protons are labeled with subscripts: Ha is attached to a carbon in the six-membered ring; Hb is attached to a carbon in the six-membered ring; Hc appears twice, once on a carbon in the five-membered ring and once on a methyl group attached to a carbonyl group; Hd is attached to a methyl group (CH3d) which is part of an acetyl group (C=O) attached to the five-membered ring. Another acetyl group (C=O) is attached to the six-membered ring.

What "Solve" is asking you to do

Choose the method, then carry it through to a final answer. Identifying what kind of problem this is and why that method applies is the first thing marked; a correct figure arrived at invisibly earns almost nothing.

Structure that answers it

Given data and what is required → method chosen, with the reason it applies → set-up (equation, circuit, free body, trial balance) → working, step by step → answer with units and any condition of validity

Where marks are lost

Doing the middle steps mentally and writing only the result. In mathematics papers, a further loss comes from giving a decimal where the exact value in surds or fractions was wanted, or from skipping the justification a part explicitly asks for.

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How this answer will be evaluated

Approach

(a(i)) calculate: given > formula > substitution > result with units > interpretation | (a(ii)) explain: definition/context > points in order > small example > short close | (b(i)) map: locate accurately > label > one line on why it matters | (b(ii)) define: precise definition > the distinguishing feature > one example | (c(i)) map: locate accurately > label > one line on why it matters | (c(ii)) map: locate accurately > label > one line on why it matters Full marks: All parts answered with correct mechanisms, structures, and full working.

Key points expected

  • Energy formula E = hν or E = hc/λ
  • Calculation of ΔE = ΣE(products) - ΣE(reactants)
  • Correct sign indicating energy liberated or absorbed
  • Final value in J/mol with correct units
  • Identification of symmetric and asymmetric C=O stretching modes
  • Explanation of coupling between two carbonyl groups
  • Contrast with single carbonyl in acid/ester
  • Product A: Reduction of ketone to alcohol, ester intact

Evaluation rubric

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

  1. (a(i)) Energy change of the reaction and its sign (liberated/absorbed). 10 marks

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

    Must cover

    • Energy formula E = hν or E = hc/λ
    • Calculation of ΔE = ΣE(products) - ΣE(reactants)
    • Correct sign indicating energy liberated or absorbed
    • Final value in J/mol with correct units

    Loses marks

    • Omitting Avogadro's number for molar energy
    • Incorrect sign convention for ΔE

    Earns more

    • Explicit conversion of wavenumbers to frequency
    • Step-by-step substitution of constants

    Extra mark

    • Comparison of bond strengths qualitatively
  2. (a(ii)) Reason for the doublet in anhydride vs singlet in acid/ester. 5 marks

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

    Must cover

    • Identification of symmetric and asymmetric C=O stretching modes
    • Explanation of coupling between two carbonyl groups
    • Contrast with single carbonyl in acid/ester

    Loses marks

    • Attributing doublet to impurities
    • Ignoring the coupling mechanism

    Earns more

    • Mention of Fermi resonance if applicable
    • Specific wavenumber assignment to modes

    Extra mark

    • Diagram of normal modes
  3. (b(i)) Structures of products A, B, and C. 10 marks

    map— locate accurately → label → one line on why it matters

    Must cover

    • Product A: Reduction of ketone to alcohol, ester intact
    • Product B: Alkyne addition to ketone forming propargylic alcohol
    • Product C: Birch reduction of anisole to 1-methoxy-1,4-cyclohexadiene
    • Correct regiochemistry and stereochemistry where applicable

    Loses marks

    • Reducing the ester in product A
    • Wrong regioisomer for Birch reduction

    Earns more

    • Mechanism for Birch reduction electron transfer
    • Explanation of chemoselectivity in NaBH4 reaction

    Extra mark

    • IUPAC names of products
  4. (b(ii)) Definition of tacticity and distinction between isotactic, syndiotactic, atactic. 5 marks

    define— precise definition → the distinguishing feature → one example

    Must cover

    • Definition of tacticity as spatial arrangement of substituents
    • Isotactic: all substituents on same side
    • Syndiotactic: alternating sides
    • Atactic: random arrangement

    Loses marks

    • Confusing tacticity with regiochemistry
    • Vague descriptions without spatial reference

    Earns more

    • Schematic representation of the three types
    • Mention of crystallinity differences

    Extra mark

    • Example of a specific polymer for each type
  5. (c(i)) Structure of the organic compound C16H25NO. 10 marks

    map— locate accurately → label → one line on why it matters

    Must cover

    • Identification of para-substituted aromatic ring from NMR
    • Assignment of isopropyl group from septet and doublet
    • Assignment of ethyl group from triplet and quartet
    • Identification of amide linkage from IR and NMR

    Loses marks

    • Ignoring the integration values
    • Incorrect placement of substituents on ring

    Earns more

    • Calculation of degree of unsaturation
    • Logical assembly of fragments

    Extra mark

    • IUPAC name of the compound
  6. (c(ii)) Order of protons Ha, Hb, Hc, Hd by increasing chemical shift. 10 marks

    map— locate accurately → label → one line on why it matters

    Must cover

    • Identification of deshielding effects of carbonyl groups
    • Assignment of Hc as most downfield (alpha to C=O)
    • Assignment of Ha/Hb based on alkene position
    • Correct ordering of all four protons

    Loses marks

    • Ignoring the effect of the ring oxygen
    • Incorrect relative order of Ha and Hb

    Earns more

    • Explanation of anisotropic effects
    • Comparison of alpha vs beta protons

    Extra mark

    • Approximate chemical shift values

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