Paper II — Q2
(a) Involving the stereochemical concept, compare the elimination behaviour of compounds A and B in the presence of base. Also…
Involving the stereochemical concept, compare the elimination behaviour of compounds A and B in the presence of base. Also give the product(s) of the reactions. 20 marks
Comment upon the structure and stability of cyclopropylmethyl carbocation. (ii) Which one of the above compound is more acidic and why ? (iii) Write the product(s) in the above reaction. 15 marks
Indicating the stereochemistry of the product, complete the above reaction with mechanism. 10 marks (ii) Write down the products in the above reactions. 5 marks
हिंदी में प्रश्न पढ़ें
त्रिविम रासायनिक संकल्पना को सम्मिलित करते हुए क्षार की उपस्थिति में यौगिक A और B के उन्मूलन व्यवहार की तुलना करें । अभिक्रिया के उत्पादों को भी लिखें । 20 अंक
साइक्लोप्रोपाइलमेथिल कार्बोकैटायन की संरचना और स्थायित्व पर टिप्पणी करें । (ii) निम्नलिखित यौगिकों में कौन सा ज़्यादा अम्लीय है और क्यों ? (iii) निम्नलिखित अभिक्रिया में उत्पाद/उत्पादों को लिखें : 15 अंक
उत्पाद की त्रिविम रसायन को दर्शाते हुए निम्नलिखित अभिक्रिया को क्रियाविधि देते हुए पूर्ण करें : 10 अंक (ii) निम्नलिखित अभिक्रियाओं के उत्पादों को लिखें : 5 अंक
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 A and B. Both are substituted cyclohexane rings. Structure A shows a cyclohexane ring with a chlorine atom (Cl) on a dashed bond (pointing away) and a methyl group (CH3) on a solid wedge bond (pointing towards) on adjacent carbons. Structure B shows a cyclohexane ring with a chlorine atom (Cl) on a solid wedge bond and a methyl group (CH3) on a solid wedge bond on adjacent carbons.
(b) Two chemical structures labeled A and B. Structure A is 1H-indene, a bicyclic aromatic compound 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.
(c) Reaction scheme: Cyclohexanone (a six-membered ring with a ketone group) reacts with NaH in THF to form an intermediate labeled [P]. From [P], two reaction paths diverge: one path reacts with CH3I to form product Q; the other path reacts with (CH3)3SiCl to form product R.
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) E2 elimination of A and B A is trans-1-chloro-2-methylcyclohexane and B is cis-1-chloro-2-methylcyclohexane. E2 with a base (e.g., NaOEt) requires the leaving group and the β-hydrogen to be anti-periplanar, i.e., both axial (trans-diaxial) on the chair. B (cis): the stable chair has Cl axial and CH₃ equatorial. The C2–H is then axial and anti to Cl, so elimination proceeds easily from the preferred conformer, giving mainly the more substituted alkene, 1-methylcyclohexene (Zaitsev, major); the axial H on C6 also gives a little 3-methylcyclohexene. B reacts fast. A (trans): in the stable conformer both Cl and CH₃ are equatorial, so no axial Cl and no E2 is possible. Ring-flip puts Cl and CH₃ both axial (higher energy, with 1,3-diaxial strain). Now the C2–H is equatorial (gauche, not anti) because CH₃ is axial, so the only anti hydrogen is the axial H on C6. Elimination gives only 3-methylcyclohexene (the less substituted, non-Zaitsev product) and is slower, since it has to pass through the strained diaxial conformer. Thus stereochemistry, not alkene stability, controls the product: B gives 1-methylcyclohexene (major), A gives 3-methylcyclohexene exclusively, and B eliminates faster than A.
(b)(i) Cyclopropylmethyl carbocation It is a primary cation, c-C₃H₅–CH₂⁺, yet is unusually stable, more than isopropyl and comparable to benzyl. The strained cyclopropane C–C bonds are bent (“banana”) bonds with high p-character (Walsh orbitals), and overlap sideways with the vacant p orbital of the CH₂⁺ in the bisected conformation (σ–π conjugation, sometimes called dancing resonance), delocalising the positive charge into the ring. It can equilibrate with cyclobutyl and homoallyl cations.
(b)(ii) Acidity Cyclohexane-1,3-dione is far more acidic (pKa ≈ 5). Its C2–H₂ lies between two carbonyls, and the planar enolate is stabilised by delocalisation over both oxygens. In the bridged bicyclic dione, the α-hydrogen at the bridgehead cannot enolise, since the enolate double bond would be at a bridgehead (Bredt’s rule), and the other α-CH₂ is activated by only one carbonyl. Its acidity is that of a simple ketone (pKa ≈ 19–20).
(b)(iii) Büchner reaction Heating tert-butyl diazoacetate loses N₂ to form the carbene :CH–CO₂C(CH₃)₃, which adds to a C=C of benzene (the hexagon read as benzene) to give the norcaradiene. This bicyclic cyclopropane undergoes electrocyclic ring-opening to tert-butyl cyclohepta-2,4,6-triene-1-carboxylate (ring expansion). If the ring were saturated cyclohexane, only unselective C–H insertion could occur.
(c)(i) SOCl₂/pyridine on (S)-1-phenylethanol The alcohol attacks SOCl₂ to give the chlorosulfite ROS(O)Cl (HCl lost to pyridine). Pyridine takes up HCl, and free Cl⁻ then attacks the benzylic carbon from the back (SN2), expelling SO₂ and Cl⁻, with inversion of configuration. The product is (R)-(1-chloroethyl)benzene (the alcohol, read as S from the drawing). Without pyridine, the chlorosulfite collapses in an SNi step with retention.
(c)(ii) Cyclohexanone enolate NaH in THF deprotonates the α-carbon to the sodium enolate P. With CH₃I the enolate carbon attacks by SN2 to give Q = 2-methylcyclohexanone (C-alkylation, soft electrophile). With (CH₃)₃SiCl, the hard oxygen attacks the strongly oxophilic silicon, giving R = 1-(trimethylsilyloxy)cyclohexene (silyl enol ether).
How this answer will be evaluated
What earns full marks: Complete mechanisms with correct stereochemistry, all products drawn accurately, clear reasoning throughout.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Compare elimination behaviour of A and B with stereochemical reasoning and products. 20 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- Identify A as trans and B as cis isomer
- Apply anti-periplanar requirement for E2 elimination
- Show B (cis) eliminates easily from the stable chair (axial Cl, axial C2-H) to give mainly 1-methylcyclohexene (Zaitsev), minor 3-methylcyclohexene
- Show A (trans) needs diaxial ring flip and, with C2-H equatorial, gives only 3-methylcyclohexene (slower)
Loses marks
- Ignores stereochemical requirement for E2
- Confuses cis and trans configurations
- Fails to show anti-periplanar geometry
Earns more
- Draws Newman projections for both isomers
- Mentions Zaitsev product as major
- Notes steric hindrance in A's transition state
- Uses wedge-dash notation correctly
Extra mark
- Mentions E1 pathway possibility
- Draws chair conformations explicitly
- (b) Comment on cyclopropylmethyl carbocation stability, compare acidity of two compounds, and write reaction products. 15 marks
comment— context → arguments both sides → judgment → close
Must cover
- Explain cyclopropylmethyl cation stability by sigma-pi (Walsh, bent-bond) conjugation in the bisected conformation
- Identify cyclohexane-1,3-dione as more acidic: delocalised planar enolate; Bredt's rule prevents enolisation of the bridged dione
- (iii) Buchner ring expansion: carbene from the diazoester adds to benzene to give norcaradiene in equilibrium with cycloheptatriene
- Draw correct product: tert-butyl cyclohepta-2,4,6-triene-1-carboxylate
Loses marks
- Fails to explain carbocation stabilization
- Wrong acidity assignment without reasoning
- Incorrect product structures
Earns more
- Draws resonance structures for carbocation
- Mentions pKa values for acidity comparison
- Shows carbene formation by loss of N2, cyclopropanation, then electrocyclic ring opening
- Names the reaction as Buchner ring expansion
Extra mark
- Mentions Walsh orbitals for cyclopropane
- Mention of cyclopropylmethyl-cyclobutyl-homoallyl cation equilibration
- (c)(i) Complete reaction with SOCl2/pyridine showing stereochemistry and mechanism. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Shows SN2 mechanism with inversion
- Draws product with inverted stereochemistry
- Shows pyridine as base removing HCl
- Indicates backside attack on carbon
Loses marks
- Shows retention instead of inversion
- Fails to show mechanism
- Wrong stereochemistry in product
Earns more
- Draws intermediate chlorosulfite
- Shows leaving group departure
- Uses correct arrow pushing
- Labels stereochemistry clearly
Extra mark
- Mentions retention with SOCl2 alone
- Draws transition state
- (c)(ii) Write products P, Q, and R from cyclohexanone reactions. 5 marks
enumerate— list the items in order → one line each → no commentary
Must cover
- Identifies P as enolate ion
- Shows Q as alpha-methylcyclohexanone
- Shows R as silyl enol ether
- Draws correct structures for all three
Loses marks
- Wrong product structures
- Fails to show enolate formation
- Confuses alkylation and silylation products
Earns more
- Shows NaH deprotonation step
- Indicates SN2 alkylation for Q
- Shows silylation mechanism for R
- Uses correct reagent conditions
Extra mark
- Mentions kinetic vs thermodynamic enolate
- Draws resonance of enolate
What "Compare" is asking you to do
Set the items against each other on named dimensions. In UPSC practice compare already carries both halves — likeness and difference — and where the stem names the dimensions, as in region, nature and climatic impact, those are the headings the examiner expects to see.
Structure that answers it
Dimensions named → both items on dimension 1 → dimension 2 → dimension 3 → where they converge and where they part
Where marks are lost
Two self-contained descriptive blocks with the comparison left for the reader to make. Marks here sit on the dimensions, so an answer that names none of them gives the examiner nothing to award.
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.
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