Chemistry 2025 Paper II 50 marks Compulsory Explain

Paper II — Q1

(a) (i) Tropolone is aromatic, but fulvene is non-aromatic. Why? (ii) Explain with example pseudo-aromaticity. (b) Identify the…

(a)
(i)

Tropolone is aromatic, but fulvene is non-aromatic. Why?

(ii)

Explain with example pseudo-aromaticity.

(b)

Identify the missing reagent and intermediates in the following chemical conversion :

(c)

Write the structure of the major product when neomenthyl chloride is reacted with sodium ethoxide in ethanol. Justify your answer :

(d)
(i)

Discuss in detail how the reaction of a carbene with cis-2-butene can be used to define the spin state (S/T) of carbene.

(ii)

para-Bromophenol on reaction with NaNH₂/NH₃ (l) followed by acidic workup yields one major product. Explain the reaction by writing the steps involved.

(e)

With example, elucidate the permanent and temporary denaturation of a protein.

हिंदी में प्रश्न पढ़ें
(a)
(i)

ट्रोपोलोन ऐरोमैटिक है, लेकिन फल्वीन नॉन-ऐरोमैटिक है। ऐसा क्यों है?

(ii)

छद-ऐरोमैटिकता को उदाहरण देकर समझाइए।

(b)

निम्नलिखित रासायनिक रूपांतरण में लुप्त अभिकर्मक तथा मध्यवर्तियों की पहचान कीजिए :

(c)

निओमेथिल क्लोराइड की सोडियम एथॉक्साइड के साथ एथेनॉल में अभिक्रिया करने पर बनने वाले मुख्य उत्पाद की संरचना लिखिए। अपने उत्तर का औचित्य सिद्ध कीजिए :

(d)
(i)

विस्तार से विवेचना कीजिए कि एक कार्बिन की सिस-2-ब्यूटीन के साथ हुई अभिक्रिया से कार्बिन की प्रचक्रण अवस्था (S/T) कैसे परिभाषित की जा सकती है।

(ii)

पैरा-ब्रोमोफिनोल की NaNH₂/NH₃ (l) के साथ हुई अभिक्रिया व तत्पश्चात् अम्लीय विवेचन से एक मुख्य उत्पाद प्राप्त होता है। इस अभिक्रिया की सम्मिलित चरणों का वर्णन करते हुए व्याख्या कीजिए।

(e)

एक प्रोटीन के स्थायी व अस्थायी विकृतीकरण को उदाहरण देते हुए स्पष्ट कीजिए।

Q1 of the 2025 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2025 Chemistry paper

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)

Tropolone is aromatic because its planar, cyclic, conjugated seven-membered ring behaves as a tropylium system with 6 π-electrons, satisfying Hückel’s 4n+2 rule (n = 1). Although it is a neutral molecule, its aromatic character is best understood through significant dipolar resonance contributors where the carbonyl oxygen bears a negative charge and the ring carbons bear a positive charge, creating a continuous cyclic delocalization of 6 π-electrons (a tropylium-type cation). In contrast, fulvene is non-aromatic. While it possesses 6 π-electrons in total (only 4 of them within the ring), the exocyclic double bond disrupts the cyclic conjugation required for aromaticity. A dipolar resonance structure places a positive charge on the exocyclic carbon and a negative charge on the ring, but it is only a minor contributor (fulvene's dipole moment is only about 0.4 D), and it does not result in a stable, closed-shell aromatic sextet within the ring itself. The ring in fulvene behaves more like a diene than an aromatic system, lacking the thermodynamic stability and bond-length equalization characteristic of aromatic compounds.

(a)(ii)

Pseudo-aromaticity is the property of certain cyclic, conjugated systems that show a part of the stabilisation characteristic of aromatics although they do not strictly satisfy Hückel's rule in a symmetric, fully delocalised ground state. Such molecules acquire aromatic character through charge-separated (dipolar) resonance forms, and they show pronounced bond-length alternation, low resonance energies and retention of some polyene (addition) chemistry. Classic example: cyclopropenone. One of its dipolar resonance forms contains the 2π-electron cyclopropenyl cation (Hückel-aromatic, n = 0) with the negative charge on oxygen; the carbonyl is strongly polarised, the C–C bonds have marked double-bond character, and the compound is far more stable than a strained cyclopropanone (it resists hydration in water) — which is why it is called pseudo-aromatic. Tropone and tropolone are likewise described as pseudo-aromatic: their 6π tropylium-cation resonance form (n = 1) gives partial aromatic stabilisation without complete bond equalisation. In some textbooks the term is also applied to aromatic transition states, e.g., the cyclic 6π-electron transition state of the Diels–Alder reaction.

(b)

The sequence converts acetylcyclohexane into (Z)-1-cyclohexylbut-1-ene by the standard ketone → alkyne → cis-alkene route:

M — 1-(1,1-dichloroethyl)cyclohexane (the gem-dichloride). PCl₅ converts the ketone into the gem-dichloride: chloride attacks the P(V) centre to give a chlorophosphorane-type intermediate, which loses POCl₃ to leave the CCl₂ group.

N — ethynylcyclohexane (1-cyclohexylethyne). Three equivalents of NaNH₂ are used: the first two effect the double dehydrohalogenation of the gem-dichloride (via 1-chloro-1-cyclohexylethylene) to give the alkyne as its sodium acetylide; the third equivalent removes the remaining acidic ≡C–H proton so that the mixture is completely deprotonated. The acid work-up (HA) then regenerates the terminal alkyne.

O — 1-cyclohexylbut-1-yne (the internal alkyne). NaNH₂ deprotonates the terminal alkyne to the acetylide, which is alkylated by C₂H₅Br in an SN2 reaction.

P — H₂ over Lindlar catalyst (Pd/CaCO₃ deactivated with quinoline, or poisoned with lead). Syn (cis) addition of hydrogen across the triple bond gives the (Z)-alkene, (Z)-1-cyclohexylbut-1-ene, as drawn.

(c)

The reaction of neomenthyl chloride with sodium ethoxide in ethanol proceeds by an E2 elimination. Neomenthyl chloride is the isomer of 3-chloro-1-methyl-4-isopropylcyclohexane in which the chlorine atom is axial in the most stable chair conformation. For E2 elimination the β-hydrogen must be anti-periplanar to the leaving group; in this chair the axial chlorine at C3 is anti-periplanar to the axial hydrogen at C2 and, most importantly, to the tertiary hydrogen at C4.

Elimination towards C2 gives 2-menthene (p-menth-2-ene, disubstituted), whereas elimination towards C4 gives 3-menthene (p-menth-3-ene, trisubstituted). The major product is 3-menthene (≈78%) together with about 22% of 2-menthene, in agreement with the Zaitsev rule: the more substituted — and therefore more stable — internal alkene is formed preferentially. The anti-periplanar geometry is readily satisfied for both the C2–H and the tertiary C4–H, but removal of the tertiary C4–H delivers the trisubstituted alkene, so 3-menthene predominates. Its epimer menthyl chloride must first ring-flip to place the chlorine axially and then eliminates much more slowly, giving mainly the non-Zaitsev alkene, 2-menthene; hence the striking difference between the two chlorides.

Diagram: E2 of neomenthyl chloride with NaOEt/EtOH: 3-menthene (78%) and 2-menthene (22%)

(d)(i)

The reaction of a carbene with cis-2-butene serves as a stereochemical probe to distinguish between singlet and triplet carbenes.

  • Singlet Carbene: A singlet carbene has a filled sp² hybrid orbital (lone pair) and an empty p-orbital. It reacts with alkenes in a concerted, stereospecific manner. The carbene adds to the double bond in a single step, preserving the stereochemistry of the alkene. When a singlet carbene (e.g., :CH₂) reacts with cis-2-butene, it yields exclusively cis-1,2-dimethylcyclopropane. The orbital interaction involves the overlap of the alkene π-orbital with the empty p-orbital of the carbene and the alkene π*-orbital with the filled sp² orbital of the carbene, leading to simultaneous bond formation on the same face.
  • Triplet Carbene: A triplet carbene has two unpaired electrons in orthogonal orbitals (one in an sp²-like orbital and one in a p-orbital). It reacts via a stepwise mechanism. First, one electron pairs with one electron from the alkene π-bond to form a single bond, generating a 1,3-diradical intermediate. This intermediate has a single bond between the former carbene carbon and the alkene carbon, allowing for free rotation around this C-C bond before the second bond forms. Consequently, the stereochemistry of the starting alkene is lost. The reaction of a triplet carbene with cis-2-butene yields a mixture of cis- and trans-1,2-dimethylcyclopropane. The lack of stereospecificity is the key diagnostic feature of triplet carbenes.

(d)(ii)

The reaction of para-bromophenol with NaNH₂ in liquid ammonia followed by acidic workup proceeds via a benzyne (aryne) mechanism.

  1. Deprotonation: The strong base NaNH₂ deprotonates the phenolic -OH group to form a phenoxide ion.
  2. Elimination: A second NH₂⁻ removes the ring proton ortho to bromine, and the resulting aryl anion expels the bromide ion from the para-position. This elimination is driven by the formation of a highly strained triple bond in the ring, known as 4-oxidobenzyne (or 3,4-dehydrophenoxide). The intermediate is a benzyne with a triple bond between C3 and C4 (numbering the oxygen as position 1, or C4 and C5 if numbering the ring carbons 1-6 with OH at 1).
  3. Nucleophilic Attack: The amide ion (NH₂⁻) attacks the benzyne intermediate. The attack is regioselective. The negatively charged phenoxide oxygen is electron-releasing and destabilises a carbanion close to it. Attack at C3 (meta to O) places the developing negative charge at C4, para to and farther from O⁻, whereas attack at C4 (para to O) would place it at C3, nearer to O⁻. The aryl carbanion lies in an sp² orbital in the ring plane, so it cannot be delocalised onto the oxygen; this field effect decides the regiochemistry. Thus, NH₂⁻ attacks at C3.
  4. Protonation: The resulting anion is protonated by ammonia to form the amine. Acidic workup protonates the phenoxide back to the phenol. The major product is m-aminophenol (3-aminophenol). The reaction demonstrates the elimination-addition mechanism characteristic of aryl halides with strong bases.

(e)

Protein denaturation is the loss of the native secondary, tertiary and quaternary structure of a protein, and hence of its biological activity, while the primary structure (the peptide bonds) remains intact. It is not hydrolysis.

  • Temporary (reversible) denaturation: the unfolding agent disrupts hydrogen bonds, hydrophobic and ionic interactions and, sometimes, disulfide bonds, but the chain does not aggregate or become chemically modified, so the protein refolds when the agent is removed. Example: Anfinsen's classic experiment on ribonuclease A, which was unfolded by 8 M urea and β-mercaptoethanol (which reduces the four disulfide bonds). When urea and the thiol were removed slowly by dialysis and the cysteines re-oxidised in air, the enzyme regained nearly full activity with the correct disulfides, showing that the amino acid sequence contains all the information for folding. Mild changes of pH or temperature that unfold small proteins such as RNase or lysozyme are also reversible.
  • Permanent (irreversible) denaturation: the unfolded chains aggregate or are chemically modified, so they cannot refold. Example: heating (boiling or frying) egg white: ovalbumin unfolds, exposes hydrophobic groups and forms new intermolecular disulfide bonds, giving an opaque, insoluble gel that does not revert on cooling. Other examples are the coagulation of milk proteins by heat and acid, precipitation by heavy-metal ions (Hg²⁺, Pb²⁺, Ag⁺) that bind thiol groups, and strong acids or alcohol used as disinfectants. In the RNase experiment, removing urea while the thiols were already oxidised produced "scrambled" disulfides and inactive enzyme, which shows how incorrect covalent cross-links can make denaturation effectively irreversible.

How this answer will be evaluated

What earns full marks: Accurate mechanisms, correct structures, and clear reasoning for all parts.

Evaluation rubric

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

  1. (a) Contrast the aromaticity of tropolone and fulvene using Hückel's rule. | Define pseudo-aromaticity and provide a valid example. 10 marks

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

    Must cover

    • Tropolone is aromatic through its 6π tropylium-type (n = 1) dipolar resonance form; the six π-electrons satisfy Hückel's rule
    • Zwitterionic resonance of tropolone (O⁻ and ring cation) giving continuous cyclic delocalisation
    • Fulvene is non-aromatic: only 4 π-electrons in the ring; the exocyclic C=C interrupts cyclic conjugation (minor dipolar contributor, dipole moment ≈ 0.4 D)
    • Pseudo-aromaticity defined: partial aromatic character of systems not strictly obeying Hückel's rule (charge-separated resonance forms; bond-length alternation; low resonance energy)
    • Examples: cyclopropenone (2π cyclopropenyl-cation character) and tropone/tropolone (6π tropylium character)
    • Why 'pseudo': incomplete delocalisation; aromatic transition states (e.g., Diels–Alder 6π TS) are also called pseudo-aromatic

    Loses marks

    • Miscounting tropolone's ring π-electrons (e.g., counting the carbonyl oxygen lone pairs to get 8π)
    • Defining pseudo-aromaticity as a non-cyclic system
    • Ignoring the role of the oxygen atom in tropolone
    • Using 1,3-butadiene as an example

    Earns more

    • Comparison with true aromaticity
    • Drawing of tropolone resonance structures
    • Explicit statement of Hückel's rule (4n+2)
    • Mention of dipole moment in cyclopropenone

    Extra mark

    • Mention of bond length equalization in tropolone
    • Mention of specific bond lengths in the example
  2. (b) Identify intermediates M, N, O, and reagent P in the reaction sequence. 10 marks

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

    Must cover

    • M: 1-(1,1-dichloroethyl)cyclohexane (gem-dichloride); PCl₅ converts C=O to CCl₂ with loss of POCl₃
    • N: ethynylcyclohexane (1-cyclohexylethyne); two dehydrohalogenations by NaNH₂, the third equivalent forms the acetylide, HA work-up liberates the terminal alkyne
    • O: 1-cyclohexylbut-1-yne; the acetylide is alkylated by C₂H₅Br (SN2)
    • P: H₂/Lindlar catalyst (Pd/CaCO₃ with quinoline or Pb poison) — syn addition giving (Z)-1-cyclohexylbut-1-ene as drawn

    Loses marks

    • Calling M an α-chloro ketone or N an enone
    • Omitting the acid work-up after the third NaNH2
    • Choosing a reagent for P that gives the (E)-alkene or the alkane

    Earns more

    • Explain why three equivalents of NaNH2 are needed
    • Note that Na/NH3(l) in place of Lindlar would give the (E)-alkene, so the drawn geometry fixes P
    • Mechanism arrows for the E2 steps

    Extra mark

    • Detailed arrow-pushing for the elimination steps
  3. (c) Determine the major product of neomenthyl chloride with sodium ethoxide and justify. 10 marks

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

    Must cover

    • E2 requires anti-periplanar (trans-diaxial) H–Cl; the axial Cl at C3 is anti-periplanar to the axial H at C2 and the tertiary H at C4
    • Major product 3-menthene (p-menth-3-ene), the trisubstituted (Zaitsev) alkene ≈78%; minor 2-menthene (disubstituted) ≈22%
    • Structure of 3-menthene drawn (labelled structure), double bond C3=C4 with the isopropyl group on the alkene
    • Contrast: menthyl chloride must ring-flip, eliminates much more slowly and gives mainly the non-Zaitsev 2-menthene

    Loses marks

    • Claiming the major product is the Hofmann product
    • Ignoring the stereochemical requirements of E2 elimination

    Earns more

    • Drawing of the chair conformation of neomenthyl chloride
    • Explicit statement of the E2 mechanism

    Extra mark

    • Comparison with the menthyl chloride reaction
  4. (d) Explain how carbene reaction with cis-2-butene defines spin state. | Explain the reaction of para-bromophenol with NaNH2/NH3 followed by acid workup. 10 marks

    discuss— intro → 3-4 dimensions → example → balanced close

    Must cover

    • Benzyne mechanism: deprotonation of the phenol, then elimination of HBr (ortho-H with Br) to give 4-oxidobenzyne (3,4-didehydrophenoxide)
    • NH₂⁻ attacks the benzyne; the carbanion lies in an in-plane sp² orbital and cannot be resonance-stabilised by O⁻ — the inductive field effect directs attack to C3
    • Major product m-aminophenol (3-aminophenol); acidic work-up restores the phenol
    • Singlet carbene (filled sp² + empty p): concerted syn addition — stereospecific; cis-2-butene gives only cis-1,2-dimethylcyclopropane
    • Triplet carbene (two unpaired electrons): stepwise diradical; rotation about the new C–C bond gives a cis + trans mixture (stereochemistry lost)
    • Spin state/multiplicity thus diagnosed from the stereochemical outcome

    Loses marks

    • Confusing the stereochemistry of the products
    • Failing to link the mechanism to the spin state
    • Failing to show the benzyne intermediate
    • Ignoring the role of the phenol group in the reaction

    Earns more

    • Drawing of the benzyne intermediate
    • Drawing of the transition states for singlet and triplet
    • Explanation of the orbital interactions
    • Explanation of the regioselectivity of the amide attack

    Extra mark

    • Mention of specific examples of singlet and triplet carbenes
    • Mention of the final product structure (m-aminophenol)
  5. (e) Explain permanent and temporary denaturation of proteins with examples. 10 marks

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

    Must cover

    • Definition: loss of the native secondary/tertiary structure and of biological activity with the primary structure intact
    • Temporary (reversible): e.g., Anfinsen's ribonuclease A unfolded by 8 M urea + β-mercaptoethanol refolds with full activity on slow removal; mild pH/heat changes
    • Permanent (irreversible): boiling egg albumin, heat/acid coagulation of milk, heavy-metal ions, strong acid — aggregation or wrong covalent links prevent refolding

    Loses marks

    • Confusing denaturation with hydrolysis
    • Failing to provide examples for both types

    Earns more

    • Explanation of the structural changes in each case
    • Mention of the reversibility of temporary denaturation

    Extra mark

    • Mention of specific proteins used in the examples
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