Chemistry 2024 Paper II 50 marks Explain

Paper II — Q8

(a) Ethanolic solution of compound I on irradiation leads to the formation of compounds II, III and IV [structures in the paper…

(a)

Ethanolic solution of compound I on irradiation leads to the formation of compounds II, III and IV [structures in the paper: I = verbenone (4,6,6-trimethylbicyclo[3.1.1]hept-3-en-2-one); II = chrysanthenone (2,7,7-trimethylbicyclo[3.1.1]hept-2-en-6-one); III = ethyl 3,7-dimethylocta-2,6-dienoate; IV = ethyl 3,7-dimethylocta-3,6-dienoate]. The resulting reaction mixture exhibits bands at 1787, 1740, 1715 and 1685 cm⁻¹ for νC=O. Assign these C=O stretching frequencies to the corresponding compounds giving reasons : 15 marks

(b)
(i)

2,2-Dimethyl cyclopropanone undergoes ring opening when attacked by methoxide ion and the product obtained possesses the following spectral data : IR (ν, cm⁻¹) : 1740, 1160 ¹H NMR (δ) : 3·6 (3H, s), 1·2 (9H, s) Mass (m/z) : 116, 85, 59, 31 Deduce the structure of the product with reasons. Write down the structure of another possible product. 10 marks

(ii)

Arrange the following compounds in the order of increasing coupling constant values (J_Ha-Hb) : I: CH₃–C(Ha)=C(Hb)–COOCH₃, Ha and Hb trans [(E)-methyl crotonate] II: (Ha)(Hb)C=C(CH₃)–COOCH₃ [methyl methacrylate] III: CH₃–C(Hb)=C(Ha)–COOCH₃, Ha and Hb cis [(Z)-methyl crotonate] 5 marks

(c)
(i)

2-Chloro-2,3-dimethyl butane on dehydrohalogenation can lead to the formation of two products. Explain how the two can be distinguished using ¹H NMR and IR spectral data. 10 marks

(ii)

The mass spectral data of diethyl ether is as under : m/z 74, m/z 59, m/z 45, m/z 31, m/z 29 Explain the fragmentation pattern. 5 marks

(iii)

Using the following data, calculate the bond length of HCl : I = 2·70×10⁻⁴⁷ kg m² 1 a.m.u. = 1·661×10⁻²⁷ kg 5 marks

हिंदी में प्रश्न पढ़ें

(क) यौगिक I के एथेनॉलिक विलयन का किरणन करने पर यौगिक II, III तथा IV बनते हैं [प्रश्नपत्र में संरचनाएँ: I = वर्बीनोन (4,6,6-ट्राइमेथिलबाइसाइक्लो[3.1.1]हेप्ट-3-ईन-2-ओन); II = क्राइसैन्थिनोन (2,7,7-ट्राइमेथिलबाइसाइक्लो[3.1.1]हेप्ट-2-ईन-6-ओन); III = एथिल 3,7-डाइमेथिलऑक्टा-2,6-डाइईनोएट; IV = एथिल 3,7-डाइमेथिलऑक्टा-3,6-डाइईनोएट]। परिणामी अभिक्रिया मिश्रण 1787, 1740, 1715 तथा 1685 cm⁻¹ पर कार्बोनिल (νC=O) बैंड दर्शाता है। इन C=O तरंग आवृत्तियों को संबंधित यौगिकों में निर्दिष्ट कीजिए तथा उसका कारण दीजिए : (15 अंक)

(ख) (i) 2,2-डाइमेथिल साइक्लोप्रोपेनोन की मेथॉक्साइड आयन के साथ अभिक्रिया में वलय विवर्तन हो जाता है एवं बनने वाले उत्पाद का स्पेक्ट्रमी आँकड़ा निम्नलिखित है : IR (ν, cm⁻¹) : 1740, 1160 ¹H NMR (δ) : 3·6 (3H, s), 1·2 (9H, s) Mass (m/z) : 116, 85, 59, 31 कारणों के साथ उत्पाद की संरचना निकालिए। दूसरे संभावित उत्पाद की संरचना लिखिए। (10 अंक)

(ii)

निम्नलिखित यौगिकों को उनके युग्मन स्थिरांक मानों (J_Ha-Hb) के आरोही क्रमानुसार व्यवस्थित कीजिए : I: CH₃–C(Ha)=C(Hb)–COOCH₃, Ha तथा Hb ट्रांस [(E)-मेथिल क्रोटोनेट] II: (Ha)(Hb)C=C(CH₃)–COOCH₃ [मेथिल मेथैक्रिलेट] III: CH₃–C(Hb)=C(Ha)–COOCH₃, Ha तथा Hb सिस [(Z)-मेथिल क्रोटोनेट] (5 अंक)

(ग) (i) 2-क्लोरो-2,3-डाइमेथिल ब्यूटेन के विहाइड्रोहैलोजनन से दो उत्पाद बन सकते हैं। व्याख्या कीजिए कि ¹H NMR तथा IR स्पेक्ट्रमी आँकड़ों द्वारा दोनों उत्पादों का विभेदन कैसे कर सकते हैं। (10 अंक)

(ii)

डाइएथिल ईथर का द्रव्यमान स्पेक्ट्रमी आँकड़ा इस प्रकार दिया गया है : m/z 74, m/z 59, m/z 45, m/z 31, m/z 29 खंड प्रतिरूप की व्याख्या कीजिए। (5 अंक)

(iii)

निम्नलिखित आँकड़ों के द्वारा HCl की आबंध लंबाई की गणना कीजिए : I = 2·70×10⁻⁴⁷ kg m² 1 a.m.u. = 1·661×10⁻²⁷ kg (5 अंक)

Q8 of the 2024 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2024 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) A chemical reaction scheme showing the photochemical transformation of compound I into a mixture of compounds II, III, and IV. The reaction is indicated by an arrow labeled 'hv' (light) over 'EtOH' (ethanol). Compound I is a bicyclic ketone, specifically 2-methylbicyclo[2.2.1]hept-5-en-2-one (a norbornene derivative with a ketone group at C2 and a methyl group at C7). Compound II is a bicyclic ketone, specifically 7-methylbicyclo[2.2.1]hept-5-en-2-one (an isomer of I where the methyl group is at C7). Compound III is an acyclic ester, specifically ethyl (E)-2,6-dimethylhepta-2,5-dienoate, featuring a central single bond flanked by two double bonds, with a COOEt group at one end and a methyl group at the other. Compound IV is an acyclic ester, specifically ethyl (Z)-2,6-dimethylhepta-2,5-dienoate, which is the geometric isomer of III.

(b) Three chemical structures of methyl 2-methylbut-2-enoate isomers, labeled I, II, and III.

Structure I: An alkene with a double bond. On the left carbon of the double bond, a methyl group (CH3) is attached pointing down and a hydrogen atom labeled Ha is attached pointing up. On the right carbon of the double bond, a methoxycarbonyl group (COOCH3) is attached pointing up and a hydrogen atom labeled Hb is attached pointing down. This is the (E)-isomer.

Structure II: An alkene with a double bond. On the left carbon of the double bond, a hydrogen atom labeled Ha is attached pointing up and a hydrogen atom labeled Hb is attached pointing down. On the right carbon of the double bond, a methoxycarbonyl group (COOCH3) is attached pointing up and a methyl group (CH3) is attached pointing down. This is the (Z)-isomer.

Structure III: An alkene with a double bond. On the left carbon of the double bond, a methyl group (CH3) is attached pointing up and a hydrogen atom labeled Hb is attached pointing down. On the right carbon of the double bond, a methoxycarbonyl group (COOCH3) is attached pointing up and a hydrogen atom labeled Ha is attached pointing down. This is the (E)-isomer.

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 is verbenone (4,6,6-trimethylbicyclo[3.1.1]hept-3-en-2-one). Irradiation in ethanol first causes a photochemical [1,3]-shift: the excited enone breaks the allylic cyclobutane C–C bond to the CMe₂ bridge, giving a tertiary radical and an allyl radical delocalised onto the carbon α to C=O; recombination at that terminus moves the CMe₂ bridge and gives chrysanthenone II, in which the carbonyl sits in a four-membered ring and the C=C is no longer conjugated with it. II absorbs further: α-cleavage of the cyclobutanone followed by β-scission opens the skeleton to a dienyl ketene, which ethanol traps (EtOH adds across C=C=O) to give ethyl 3,7-dimethylocta-3,6-dienoate IV (β,γ-unsaturated) and its conjugated isomer, ethyl 3,7-dimethylocta-2,6-dienoate III. The irradiated mixture therefore contains residual I plus II, III and IV, each giving one C=O band.

Reference values: saturated open-chain or six-ring ketone ≈ 1715 cm⁻¹; saturated ester ≈ 1735–1750 cm⁻¹; conjugation lowers ν by ≈ 20–35 cm⁻¹; a four-membered ring raises a ketone by ≈ 65–70 cm⁻¹.

  • 1787 cm⁻¹ → II (chrysanthenone). A cyclobutanone C=O (cyclobutanone itself ≈ 1780–1785). The small internal C–C(O)–C angle puts more p-character into the ring bonds and more s-character into the C=O σ bond, stiffening it; the C=C is β,γ, so there is no conjugative lowering.
  • 1740 cm⁻¹ → IV. An ester whose C=C is β,γ (a CH₂ separates it from C=O), so it behaves as an ordinary saturated ester: the −I effect of the alkoxy oxygen outweighs its +M donation and raises ν above that of a ketone.
  • 1715 cm⁻¹ → III. The same ester group, now conjugated with the α,β-C=C; delocalisation (C=C–C=O ↔ ⁺C–C=C–O⁻) gives the C=O more single-bond character and lowers ν by ≈ 25 cm⁻¹ (α,β-unsaturated esters absorb at 1715–1730 cm⁻¹).
  • 1685 cm⁻¹ → I (unreacted verbenone). A conjugated cyclohexenone (≈ 1675–1690 cm⁻¹); conjugation lowers ν from the 1715 cm⁻¹ of a saturated six-ring ketone.

Order of frequency: II (strained, unconjugated ketone) > IV (unconjugated ester) > III (conjugated ester) > I (conjugated ketone).

Diagram: Photochemistry of verbenone and C=O stretching frequencies

(b)(i) Methoxide attacks the very electrophilic carbonyl of 2,2-dimethylcyclopropanone (an sp² carbon forced into a 60° ring) to give a tetrahedral hemiketal alkoxide. The alkoxide re-forms C=O and expels one ring carbon as a carbanion, relieving the ring strain (Favorskii-type ring opening). Two bonds can break:

  • Path a (C1–C3): the leaving carbon is the unsubstituted CH₂, a primary (more stable) carbanion; protonation by methanol gives methyl 2,2-dimethylpropanoate (methyl pivalate), (CH₃)₃C–COOCH₃.
  • Path b (C1–C2): the leaving carbon is the CMe₂, a less stable tertiary carbanion; this would give methyl 3-methylbutanoate, (CH₃)₂CH–CH₂–COOCH₃, the other possible product.

The data fit methyl pivalate (C₆H₁₂O₂, M = 116):

  • IR: 1740 cm⁻¹ is a saturated ester C=O; 1160 cm⁻¹ is the C–O–C stretch.
  • ¹H NMR: δ 3.6 (3H, s) is the ester OCH₃; δ 1.2 (9H, s) is a tert-butyl group, nine equivalent protons with no neighbouring H. There is no CH₂ or CH signal, which rules out methyl 3-methylbutanoate (that would show δ ≈ 2.2, d, 2H; ≈ 2.1, m, 1H; ≈ 0.95, d, 6H).
  • MS: m/z 116 is M⁺•; 85 = M − •OCH₃ (31), the acylium ion (CH₃)₃C–C≡O⁺; 59 = M − •C(CH₃)₃ (57), ⁺CO₂CH₃; 31 = CH₃O⁺. (m/z 57, (CH₃)₃C⁺, is also strong.)

(b)(ii) I is (E)-methyl crotonate (Ha and Hb on adjacent alkene carbons, trans); II is methyl methacrylate (Ha and Hb on the same =CH₂ carbon, geminal); III is (Z)-methyl crotonate (Ha and Hb on adjacent carbons, cis).

  • Geminal =CH₂ coupling (²J) is small, ≈ 0–3 Hz (≈ 1–2 Hz in methyl methacrylate).
  • Vicinal ³J depends on the H–C=C–H dihedral angle (Karplus relation): 0° (cis) gives ≈ 6–12 Hz (≈ 11–12 Hz in the Z-ester); 180° (trans) gives the best orbital overlap and the largest value, ≈ 12–18 Hz (≈ 15–16 Hz in the E-ester).

Increasing J(Ha–Hb): II < III < I.

(c)(i) Loss of HCl from (CH₃)₂CCl–CH(CH₃)₂ can remove a β-H from the CH (Zaitsev) or from a CH₃ (Hofmann):

  • 2,3-Dimethylbut-2-ene, (CH₃)₂C=C(CH₃)₂: tetrasubstituted, the major product with a small base such as EtO⁻.
  • 2,3-Dimethylbut-1-ene, CH₂=C(CH₃)–CH(CH₃)₂: terminal, favoured with a bulky base such as t-BuO⁻.

¹H NMR: the but-2-ene has four equivalent methyls and no vinylic H, so it shows one singlet at δ ≈ 1.65 (12H). The but-1-ene shows two vinylic =CH₂ signals at δ ≈ 4.65 and 4.70 (1H each), an allylic CH septet at δ ≈ 2.25 (1H), a vinylic CH₃ singlet at δ ≈ 1.72 (3H) and an isopropyl doublet at δ ≈ 1.03 (6H).

IR: the but-1-ene shows =C–H stretch at ≈ 3080 cm⁻¹, a medium C=C stretch at ≈ 1645 cm⁻¹ and the strong =CH₂ out-of-plane wag at ≈ 890 cm⁻¹. The but-2-ene has no =C–H band above 3000 cm⁻¹; because it is symmetrical, its C=C stretch (≈ 1670 cm⁻¹) causes almost no dipole change and is very weak or absent.

(c)(ii) Diethyl ether, CH₃CH₂–O–CH₂CH₃, M = 74:

  • m/z 74: molecular ion M⁺•, formed by removing a non-bonding electron from oxygen.
  • m/z 59: α-cleavage (of the C–C bond next to O⁺•) expels CH₃•, giving the resonance-stabilised oxonium ion CH₃CH₂–O⁺=CH₂.
  • m/z 45: cleavage of a C–O bond with loss of C₂H₅• gives C₂H₅O⁺, which rearranges to the stable CH₃CH=OH⁺.
  • m/z 31: the m/z 59 ion transfers an H from its ethyl group to O and loses ethene (C₂H₄, 28) to give CH₂=OH⁺, the base peak.
  • m/z 29: inductive (heterolytic) C–O cleavage with the charge on carbon gives the ethyl cation C₂H₅⁺.

(c)(iii) For H³⁵Cl, the reduced mass μ = m_H·m_Cl/(m_H + m_Cl) = (1.008 × 34.97)/(35.98) amu = 0.980 amu = 0.980 × 1.661 × 10⁻²⁷ kg = 1.627 × 10⁻²⁷ kg.

Since I = μr², r = √(I/μ) = √(2.70 × 10⁻⁴⁷ / 1.627 × 10⁻²⁷) m = √(1.660 × 10⁻²⁰) m = 1.29 × 10⁻¹⁰ m.

Bond length r(H–Cl) ≈ 1.29 Å (129 pm). Using m_H = 1 and m_Cl = 35.5 amu gives μ = 1.616 × 10⁻²⁷ kg and the same 1.29 Å; the accepted value is 1.275 Å.

Diagram: Cyclopropanone ring opening, alkene coupling constants, alkene isomers, ether fragmentation and HCl bond length

How this answer will be evaluated

What earns full marks: Accurate assignments, correct structures, clear reasoning, and precise calculations.

Evaluation rubric

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

  1. (a) Assign the four C=O stretching frequencies to the specific compounds I, II, III, and IV. 15 marks

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

    Must cover

    • 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)

    Loses marks

    • Assigning 1715 to IV or 1740 to III (the conjugated ester is the lower one)
    • Attributing 1787 to conjugation
    • Assigning without reasons

    Earns more

    • Quote the base values (ketone 1715, ester 1735-1750, conjugation −25 to −30, four-membered ring +70)
    • Mechanism of the 1,3-acyl shift and of ketene trapping

    Extra mark

    • References specific bond order changes
  2. (b)(i) Deduce the structure of the product from spectral data and propose an alternative. 10 marks

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

    Must cover

    • 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)

    Loses marks

    • Identifying product as a ketone (e.g., 4,4-dimethyl-3-oxobutyl methyl ether)
    • Ignoring the 9H singlet in structure deduction

    Earns more

    • Explains the ring-opening mechanism
    • Interprets the mass-spectral fragments (m/z 85 acylium, 59 ⁺CO₂CH₃, 31 CH₃O⁺)

    Extra mark

    • Draws the mechanism with curved arrows
  3. (b)(ii) Arrange the compounds in order of increasing coupling constant values. 5 marks

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

    Must cover

    • 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

    Loses marks

    • Swapping cis and geminal
    • Ordering by substituent electronegativity instead of geometry
    • Placing I (trans) below III (cis)

    Earns more

    • Karplus/dihedral reasoning for cis vs trans (180° gives the largest J)
    • Note that the ester substituent does not change the order

    Extra mark

    • Mention of Karplus relationship
  4. (c)(i) Explain how to distinguish the two dehydrohalogenation products using ¹H NMR and IR. 10 marks

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

    Must cover

    • 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

    Loses marks

    • Failing to identify the specific products
    • Vague description of NMR/IR differences

    Earns more

    • Mentions specific chemical shifts
    • Mentions specific wavenumbers for C=C or C-H

    Extra mark

    • Draws the structures of both products
  5. (c)(ii) Explain the fragmentation pattern of diethyl ether. 5 marks

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

    Must cover

    • 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

    Loses marks

    • Incorrect assignment of major fragments
    • Failing to explain the mechanism

    Earns more

    • Draws the fragmentation mechanism
    • Explains the loss of ethyl radical

    Extra mark

    • Mentions rearrangement if applicable
  6. (c)(iii) Calculate the bond length of HCl using the given data. 5 marks

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

    Must cover

    • 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

    Loses marks

    • Using the mass of H or of Cl instead of the reduced mass
    • Forgetting to convert amu to kg
    • Answer without units

    Earns more

    • Compare with the literature value 1.275 Å
    • Note that I comes from the rotational constant B = h/8π²cI if asked

    Extra mark

    • Converts the final answer to picometers
What "Explain" is asking you to do

Make the working of something clear — what sets it off, what follows from what, and what it produces. Explain is the Commission's mechanism word: it dominates the technical papers and the “explain why” stems, where the marks sit in the causal chain and not in the label.

Structure that answers it

State what it is → the initiating condition → the chain of cause, step by step → an instance where it plays out → what the chain produces

Where marks are lost

Describing what something looks like instead of why it works that way. Naming the stages without linking them reads as description too.

All UPSC directive words, compared →

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Chemistry 2024 Paper II

Write this answer and get it marked in about 20 seconds

Free account: 5 AI evaluations a month, marked against the rubric above.

Start free

Already have an account? Log in