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Chemistry 2021 Paper II 50 marks Explain

Paper II — Q3

(a) Complete the following reactions with mechanisms: (i) Conc. H₂SO₄ (10 marks) (ii) EtO⁻ (10 marks) (b) (i) Write the…

(a)
(i)

Complete the following reactions with mechanisms: Conc. H₂SO₄ 10 marks

(ii)

EtO⁻ 10 marks

(b)
(i)

Write the structure of the reaction product between 3-chlorocyclopropene and SbCl₅. What is the unique feature of this product in ¹H NMR spectrum?

(ii)

Which one of the above compounds is more acidic and why? (Structures A and B shown)

(iii)

Predict the aromaticity of tropolones and sydnones.

(c)
(i)

Complete the following reactions along with mechanisms: meso-2,3-dibromobutane →(I⁻) A →(OEt⁻) B

(ii)

CH₃—C—CHI—CH₃ →(AgNO₃, EtOH, Δ) ?

CH₃

हिंदी में प्रश्न पढ़ें
(a)
(i)

निम्नलिखित अभिक्रियाओं को उनकी क्रियाविधि दर्शाते हुए पूर्ण करें : Conc. H₂SO₄ 10 अंक

(ii)

EtO⁻ 10 अंक

(b)
(i)

3-क्लोरोसाइक्लोप्रोपीन और SbCl₅ की परस्पर अभिक्रिया के उत्पाद की संरचना लिखें। इस उत्पाद की ¹H NMR स्पेक्ट्रम में अनूठी विशेषता क्या है?

(ii)

निम्नलिखित यौगिकों में कौन सा ज्यादा अम्लीय है और क्यों? (A और B की संरचनाएं दिखाई गई हैं)

(iii)

ट्रोपोलोन्स और सिड्नोन्स की ऐरोमैटिकता का अनुमान लगाएं।

(c)
(i)

निम्नलिखित अभिक्रियाओं को उनकी क्रियाविधि के साथ पूर्ण करें : मेसो-2,3-डाइब्रोमोब्यूटेन →(I⁻) A →(OEt⁻) B

(ii)

CH₃—C—CHI—CH₃ →(AgNO₃, EtOH, Δ) ?

CH₃

Q3 of the 2021 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2021 Chemistry paper
Evaluate your answer to this question → See all 2021 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.

(a) (i) Reaction of a 1,2-diol: A cyclopentane ring where the top-right carbon is bonded to an -OH group and also bonded to an exocyclic carbon atom. The exocyclic carbon is bonded to an -OH group and two phenyl groups (-Ph). The reaction arrow is labeled with 'Conc. H2SO4' (सांद्र H2SO4) leading to a question mark (?).

(ii) Reaction of a substituted cyclohexane: A six-membered ring where the top carbon has a dashed wedge to a -CH3 group, and the bottom carbon has two dashed bonds: one going down and to the left to a -CH3OC group, and one going down and to the right to a -Cl atom. The reaction arrow is labeled with EtO^- leading to a question mark (?).

(a(i)) A chemical reaction scheme. The reactant is a cyclopentane ring. At the 1-position, there is a hydroxyl group (OH) and a carbon atom bonded to two phenyl groups (Ph) and a hydroxyl group (OH). The reagent is concentrated sulfuric acid (Conc. H2SO4). The product is indicated by a question mark.

(a(ii)) A chemical reaction scheme. The reactant is a cyclohexane ring. At the 1-position, there is a methyl group (CH3) attached with a dashed wedge bond (pointing away). At the 3-position, there is a methoxycarbonyl group (CH3OC=O) attached with a dashed wedge bond and a chlorine atom (Cl) attached with a solid wedge bond (pointing towards). The reagent is ethoxide ion (EtO-). The product is indicated by a question mark.

(a(i)) A reaction scheme showing a starting material reacting with 'Conc. H2SO4' to form a product indicated by a question mark. The starting material is a cyclopentane ring. At the 1-position of the ring, there is a quaternary carbon atom bonded to a hydroxyl group (-OH) and a carbon atom that is part of a gem-diol group (a carbon atom bonded to two hydroxyl groups and two phenyl groups, -C(OH)2(Ph)2).

(a(ii)) A reaction scheme showing a starting material reacting with 'EtO-' to form a product indicated by a question mark. The starting material is a cyclohexane ring. At the 1-position, there is a chlorine atom (Cl) on a wedge bond (pointing up). At the 4-position, there is a methoxymethyl group (-CH2OCH3) on a wedge bond (pointing up). At the 3-position, there is a methyl group (CH3) on a dashed bond (pointing down).

(b) Two chemical structures labeled A and B. Structure A is 1H-indene, consisting of a benzene ring fused to a five-membered ring containing one double bond. Structure B is cyclopentadiene, a five-membered ring containing two double bonds.

(b(ii)) Two chemical structures labeled A and B. Structure A is 2,4-dihydroxybenzaldehyde (a benzene ring with an aldehyde group at position 1, and hydroxyl groups at positions 2 and 4). Structure B is 3,5-dihydroxybenzaldehyde (a benzene ring with an aldehyde group at position 1, and hydroxyl groups at positions 3 and 5).

(b(iii)) Two chemical structures labeled A and B. Structure A is tropolone (a seven-membered ring with three double bonds, one ketone group, and one hydroxyl group). Structure B is sydnone (a five-membered ring containing a nitrogen atom, a carbonyl group, and an N-oxide group, fused to a benzene ring).

(c) A reaction scheme showing a reactant structure: a central carbon atom bonded to a methyl group (CH3) on the left, a methyl group (CH3) on top, a methyl group (CH3) on the bottom, and a CHI group on the right, which is further bonded to a CH3 group. An arrow points to the right labeled with AgNO3 over EtOH, delta (heat symbol), leading to a question mark.

(c(i)) A reaction scheme. The starting material is meso-2,3-dibromobutane (CH3-CHBr-CHBr-CH3 with specific stereochemistry). It reacts with I- to form product A. Product A then reacts with OEt- to form product B.

(c(ii)) A reaction scheme. The starting material is 2-iodo-2-methylbutane (CH3-C(CH3)(I)-CH2-CH3). It reacts with AgNO3 in EtOH with heat (delta symbol) to form a product indicated by a question mark.

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) 1-(Hydroxydiphenylmethyl)cyclopentanol with conc. H₂SO₄ This is a pinacol rearrangement. Protonation of the two hydroxyls gives two possible cations, and the one on the diphenylcarbinol carbon is far more stable (diphenylmethyl cation, stabilised by resonance into both rings), so water leaves from that carbon: Ph₂C⁺–C(OH)(cyclopentane). A C–C bond of the neighbouring ring carbon now migrates to the cation (1,2-alkyl shift), and the ring expands from five to six, which also relieves strain. The positive charge moves to the oxygen-bearing carbon, giving the protonated ketone (oxocarbenium), and loss of H⁺ gives the product, 2,2-diphenylcyclohexanone.

(a)(ii) α-Chloro ketone with ethoxide: Favorskii rearrangement The substrate is 1-acetyl-1-chloro-4-methylcyclohexane, a tertiary α-halo ketone with enolisable α′-hydrogens (the acetyl CH₃). Ethoxide removes an α′-proton to give the enolate, which displaces chloride from the ring carbon by intramolecular SN2 (the carbon is tertiary, but the three-ring closes), forming a spiro-cyclopropanone intermediate. Ethoxide attacks the strained carbonyl to give a tetrahedral alkoxide, which opens the three-membered ring by cleaving the bond that gives the more stable carbanion, the primary CH₂⁻ rather than the tertiary ring carbon. Protonation by ethanol converts CH₂⁻ to CH₃. The product is ethyl 1,4-dimethylcyclohexane-1-carboxylate. Because the symmetric cyclopropanone has the two carbons on the same footing, the original stereochemistry at the quaternary carbon is not retained, and both cis and trans diastereomers can form.

(b)(i) 3-Chlorocyclopropene with SbCl₅ SbCl₅ abstracts chloride (a Lewis acid), giving the cyclopropenyl cation, C₃H₃⁺ SbCl₆⁻. It is a planar cyclic conjugated cation with 2 π electrons (Hückel 4n + 2, n = 0), so it is aromatic and stable. All three CH groups are equivalent, so its ¹H NMR shows a single sharp singlet (about δ 11).

(b)(ii) Indene (A) and cyclopentadiene (B) Cyclopentadiene (pKa ≈ 16) is more acidic than indene (pKa ≈ 20). Loss of H⁺ gives the cyclopentadienyl anion, which has 6 π electrons delocalised over five carbons (aromatic). In the indenyl anion, the negative charge must be shared with the fused benzene ring, whose aromaticity is partly sacrificed, so the stabilisation is smaller.

(b)(iii) Tropolone and sydnone Tropolone is aromatic: the seven-membered ring with C=O polarised as C⁺–O⁻ gives a tropylium-like 6π-electron delocalised system, with an intramolecular hydrogen bond with the adjacent OH. Sydnones are mesoionic and aromatic: the five-membered ring (O1, N2, N3, C4, C5) holds six delocalised π electrons, with the exocyclic oxygen carrying negative charge.

(c)(i) meso-2,3-Dibromobutane With I⁻, the reaction is a debromination (anti E2: I⁻ attacks one Br, the other leaves), through the anti-periplanar conformer in which the methyls are anti; the product is A = (E)-but-2-ene (trans). With ethoxide, E2 elimination of HBr needs H and Br anti-periplanar; in the meso isomer, this makes the methyls end up on the same side, so B = (E)-2-bromobut-2-ene (the two CH₃ groups cis; Br and the other CH₃ trans).

(c)(ii) 2-Iodo-3,3-dimethylbutane with AgNO₃/EtOH, Δ Ag⁺ pulls out iodide (AgI precipitates), giving a secondary cation, (CH₃)₃C–CH⁺–CH₃. A 1,2-methyl shift converts it to the more stable tertiary cation, (CH₃)₂C⁺–CH(CH₃)₂. Ethanol (the solvent) attacks it and loses H⁺ (SN1) to give the ether 2-ethoxy-2,3-dimethylbutane as the major product; E1 loss of H⁺ gives some 2,3-dimethylbut-2-ene and 2,3-dimethylbut-1-ene.

How this answer will be evaluated

What earns full marks: Complete mechanisms with correct stereochemistry and reasoning

Evaluation rubric

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

  1. (a)(i) Product and mechanism for acid-catalyzed dehydration of the diol. 10 marks

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

    Must cover

    • Protonation of hydroxyl group
    • Loss of water from the diphenylcarbinol carbon to give the more stable diphenylmethyl cation
    • Stabilization by phenyl groups
    • 1,2-alkyl (ring) migration to the cation with ring expansion, then loss of H+ to give 2,2-diphenylcyclohexanone

    Loses marks

    • Product without mechanism
    • Incorrect carbocation intermediate

    Earns more

    • Mention of resonance stabilization
    • Correct stereochemistry of product

    Extra mark

    • Named reaction (e.g., Pinacol rearrangement if applicable)
  2. (a)(ii) Product and mechanism for reaction with ethoxide. 10 marks

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

    Must cover

    • Identify tertiary alpha-chloroketone with alpha-prime hydrogens (CH3CO)
    • Enolate formation and intramolecular displacement of Cl to the spiro-cyclopropanone
    • Ethoxide attack on C=O and ring opening to the more stable (primary) carbanion
    • Final product ethyl 1,4-dimethylcyclohexane-1-carboxylate; cis/trans mixture

    Loses marks

    • Ignoring stereochemistry
    • Wrong product structure

    Earns more

    • Explanation of steric hindrance
    • Correct arrow pushing

    Extra mark

    • Name the reaction: Favorskii rearrangement
  3. (b) Product of 3-chlorocyclopropene + SbCl5, NMR feature, acidity comparison, and aromaticity of tropolones/sydnones. 15 marks

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

    Must cover

    • Structure of cyclopropenyl cation product
    • Single peak in 1H NMR due to symmetry
    • Comparison of acidity based on conjugate base stability
    • Aromaticity prediction for tropolones and sydnones

    Loses marks

    • Incorrect cation structure
    • Confusing aromaticity with anti-aromaticity

    Earns more

    • Explanation of aromaticity (4n+2 pi electrons)
    • Resonance structures for acidity

    Extra mark

    • Specific chemical shift values for NMR
  4. (c)(i) Products A and B from meso-2,3-dibromobutane reactions. 5 marks

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

    Must cover

    • A: (E)-but-2-ene by anti-elimination (debromination) with iodide
    • B: (E)-2-bromobut-2-ene by anti E2 with ethoxide
    • Anti-periplanar geometry shown for the meso isomer (Newman projection)
    • Stereochemical outcome: meso gives E alkene (A) and E bromoalkene (B)

    Loses marks

    • Wrong intermediate A
    • Ignoring stereochemistry

    Earns more

    • Explanation of E2 (not SN1/SN2) mechanism
    • Correct stereochemistry of products

    Extra mark

    • Contrast with the dl-isomer giving the opposite geometry
  5. (c)(ii) Product and mechanism for reaction with AgNO3 in ethanol. 10 marks

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

    Must cover

    • Ag+ abstracts iodide giving secondary carbocation
    • 1,2-methyl shift gives tertiary carbocation, which ethanol attacks
    • Final ether 2-ethoxy-2,3-dimethylbutane (E1 alkenes minor)
    • Mechanism steps

    Loses marks

    • Product without mechanism
    • Wrong carbocation intermediate

    Earns more

    • Explanation of silver ion role
    • Correct product structure

    Extra mark

    • Mention of SN1 mechanism
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

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

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