Paper II — Q4
(a) (i) Heating of 3-deuterioindene causes scrambling of the deuterium. Explain with mechanism. (5 marks) (ii) Write the…
Heating of 3-deuterioindene causes scrambling of the deuterium. Explain with mechanism. 5 marks
Write the structure of product(s) formed during the thermal reaction of maleic anhydride with cyclopentadiene. Explain with mechanism. 20 marks
4.
Write the product(s) of the above reactions.
(CH3)3C-CH=CH2 --HCl--> ? 5 marks
CH2=CH-CH2-Br --HBr--> ? 10 marks
CH2=CH-CH2-Br --HBr, benzoyl peroxide--> ?
CH2=CH-CH2-Br --HBr, benzoyl peroxide--> ?
4.
Predict the product in the above chemical conversions and also identify the name reaction involved. 10 marks
I. PhCH2Cl --(1) (C6H5)3P, (2) RLi, (3) [cyclohexanone structure]--> ?
II. PhNHNH2 + [cyclohexanone structure] --Glacial acetic acid--> ?
Which of the following pair(s) gives α,β-unsaturated carbonyl compound in presence of base? Justify the answer. 5 marks
I. HCHO and PhCHO
II. PhCHO and Ph-CH-CHO (with CH3 below CH)
III. PhCHO and CH3CH2CHO
IV. HCHO and PhCOPh
SECTION 'B'
हिंदी में प्रश्न पढ़ें
3-ड्यूटेरोइंडीन को गर्म करने पर ड्यूटेरियम का व्यमिश्रण होता है। क्रियाविधि देते हुए समझाइए। (5 अंक)
मैलेइक एनहाइड्राइड की साइक्लोपेन्टाडाइन के साथ उष्मीय अभिक्रिया में बने उत्पादों की संरचना लिखें। क्रियाविधि के साथ समझाइए। (20 अंक)
4.
निम्नलिखित अभिक्रियाओं के उत्पाद/उत्पादों को लिखें:
(CH3)3C-CH=CH2 --HCl--> ? (5 अंक)
CH2=CH-CH2-Br --HBr--> ? (10 अंक)
CH2=CH-CH2-Br --HBr, बेंज़ॉयल परॉक्साइड--> ?
CH2=CH-CH2-Br --HBr, बेंज़ॉयल परॉक्साइड--> ?
4.
निम्नलिखित रासायनिक रूपान्तरणों का उत्पाद लिखें और संलिप्त अभिक्रिया को पहचानते हुए नाम लिखें: (10 अंक)
I. PhCH2Cl --(1) (C6H5)3P, (2) RLi, (3) [साइक्लोहेक्सेनोन संरचना]--> ?
II. PhNHNH2 + [साइक्लोहेक्सेनोन संरचना] --ग्लेशियल ऐसिटिक अम्ल--> ?
निम्नलिखित में से कौन सा युग्म धार की उपस्थिति में, α,β-असंतुप्त कार्बोनिल यौगिक देता है? उत्तर को सिद्ध करें: (5 अंक)
I. HCHO और PhCHO
II. PhCHO और Ph-CH-CHO (CH3 नीचे CH के नीचे)
III. PhCHO और CH3CH2CHO
IV. HCHO और PhCOPh
खण्ड 'B'
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 scheme for 4(b)(ii): The starting material is 3-bromoprop-1-ene (CH2=CH-CH2-Br). There are three reaction arrows originating from this structure. The first arrow points to the right with 'HBr' written above it, leading to a question mark. The second arrow points downwards with 'HBr' and 'benzoyl peroxide' written to the left of the arrow, leading to a question mark. The third arrow points diagonally down-right with 'HBr' and 'benzoyl peroxide' written to the right of the arrow, leading to a question mark.
(c) Chemical structure for pair II: A benzene ring attached to a CH group, which is attached to a CHO group. The CH group is also attached to a CH3 group below it.
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) Deuterium scrambling in 3-deuterioindene Indene is a benzo-fused cyclopentadiene, so on heating it undergoes thermal [1,5]-sigmatropic hydrogen (deuterium) shifts around the five-membered ring, a suprafacial, six-electron, thermally allowed process under Woodward–Hoffmann rules. The sp³ carbon C1 or C3 carrying D migrates through a cyclic transition state to the adjacent carbon, giving 2-deuterio-2H-indene (isoindene, with a non-benzenoid o-quinodimethane structure), and a second shift moves D onwards. Repeated shifts scramble the label among C1, C2 and C3, so 3-D-indene is converted to a mixture of 1-D, 2-D and 3-D indenes. The isoindene is high in energy, so the process needs high temperature.
(a)(ii) Maleic anhydride with cyclopentadiene This is a thermal Diels–Alder [4+2] cycloaddition between the diene (cyclopentadiene, locked s-cis) and the electron-poor dienophile (maleic anhydride). It is a concerted, suprafacial, cycloaddition of 6 electrons, thermally allowed by orbital symmetry, so the cis geometry of the anhydride is retained: both carbonyl groups end up cis on the new ring. The product is endo-cis-bicyclo[2.2.1]hept-5-ene-2,3-dicarboxylic anhydride (cis-5-norbornene-endo-2,3-dicarboxylic anhydride), a bicyclic adduct with the anhydride bridge oriented towards the C5=C6 double bond. The endo rule applies: secondary orbital overlap between the anhydride carbonyl carbons and the diene C2/C3 stabilises the endo transition state, so the endo isomer is the kinetic product, formed faster. On prolonged heating, the reaction is reversible and the more stable exo isomer accumulates.
(b)(i) 3,3-Dimethyl-1-butene with HCl H⁺ adds to the terminal CH₂ (Markovnikov), giving the secondary cation (CH₃)₃C–CH⁺–CH₃. A 1,2-methyl shift transforms it into the more stable tertiary cation (CH₃)₂C⁺–CH(CH₃)₂. Chloride adds to it, giving 2-chloro-2,3-dimethylbutane as the major product, with a little unrearranged 3-chloro-2,2-dimethylbutane.
(b)(ii) Allyl bromide with HBr Ionic conditions (HBr alone): H⁺ adds to the terminal carbon to give the secondary cation (the CH₂Br group is destabilising but the secondary cation is still preferred), then Br⁻ gives 1,2-dibromopropane, CH₃CHBrCH₂Br (Markovnikov). Radical conditions (benzoyl peroxide, Kharasch peroxide effect): peroxide forms Br• from HBr; Br• adds to the terminal CH₂ to give the more stable secondary radical, which abstracts H from HBr (chain propagation), giving 1,3-dibromopropane, BrCH₂CH₂CH₂Br (anti-Markovnikov).
(c)(i) Name reactions I. Benzyl chloride with PPh₃ gives the phosphonium salt [PhCH₂PPh₃]⁺Cl⁻; RLi removes a benzylic proton to form the ylide PhCH=PPh₃, which adds to cyclohexanone through an oxaphosphetane, giving benzylidenecyclohexane (PhCH=C₆H₁₀) and Ph₃P=O: the Wittig reaction. II. Phenylhydrazine and cyclohexanone form the phenylhydrazone, which in glacial acetic acid undergoes [3,3]-sigmatropic rearrangement of its ene-hydrazine tautomer and loss of NH₃ to give 1,2,3,4-tetrahydrocarbazole: the Fischer indole synthesis.
(c)(ii) Pair giving an α,β-unsaturated carbonyl compound Only pair III: PhCHO and CH₃CH₂CHO, a crossed (Claisen–Schmidt) aldol condensation. PhCHO has no α-H and acts only as the electrophile; propanal forms the enolate, adds to PhCHO, and the β-hydroxy aldehyde dehydrates (E1cB) to the conjugated PhCH=C(CH₃)CHO. I (HCHO + PhCHO): neither has α-H, so no aldol (crossed Cannizzaro occurs). II: the only α-H of Ph–CH(CH₃)–CHO is on the tertiary carbon; after addition no α-H remains for dehydration. IV: neither has α-H.
How this answer will be evaluated
What earns full marks: Accurate mechanisms with correct stereochemistry and clear justification for all parts.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Mechanism for deuterium scrambling in 3-deuterioindene and product of maleic anhydride/cyclopentadiene reaction. 20 marks
explain— definition/context → points in order → small example → short close
Must cover
- Concerted [1,5]-sigmatropic shift mechanism for deuterium scrambling
- Diels-Alder reaction mechanism with correct arrow pushing
- Endo product structure for the Diels-Alder adduct
- Explanation of thermal conditions favoring the reaction
Loses marks
- Proposing a stepwise cationic mechanism for sigmatropic shift
- Ignoring stereochemistry in the Diels-Alder product
Earns more
- Mention of orbital symmetry conservation
- Discussion of secondary orbital interactions for endo preference
Extra mark
- Drawing of transition state geometry
- (b)(i) Product of the reaction between 3,3-dimethyl-1-butene and HCl. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Markovnikov protonation gives the secondary carbocation (CH3)3C-CH(+)-CH3
- 1,2-methyl shift gives the more stable tertiary carbocation (CH3)2C(+)-CH(CH3)2
- Final product: 2-chloro-2,3-dimethylbutane
Loses marks
- Direct addition without carbocation rearrangement
- Incorrect placement of the chlorine atom
Earns more
- Explicit drawing of the carbocation rearrangement step
Extra mark
- Mention of Markovnikov's rule as the initial driving force
- (b)(ii) Products of allyl bromide reacting with HBr under ionic and radical conditions. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- Ionic product: 1,2-dibromopropane (Markovnikov addition)
- Radical product: 1,3-dibromopropane (Anti-Markovnikov addition)
- Mechanism for ionic addition via secondary carbocation
- Mechanism for radical addition via bromine radical attack
Loses marks
- Confusing the ionic and radical products
- Omitting the role of benzoyl peroxide in the radical mechanism
Earns more
- Comparison of the two pathways in a single scheme
Extra mark
- Mention of the Kharasch effect for the radical pathway
- (c)(i) Products and names of reactions for PhCH2Cl and cyclohexanone conversions. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Reaction I: Wittig reaction yielding an alkene
- Reaction II: Fischer indole synthesis (phenylhydrazone, acid, [3,3] shift) giving 1,2,3,4-tetrahydrocarbazole
- Correct structure of the ylide in the Wittig reaction
- Identification of the Fischer indole synthesis (ene-hydrazine, loss of NH3)
Loses marks
- Identifying the second reaction as a simple hydrazone formation or Wolff-Kishner reduction
- Incorrect structure of the indole product
Earns more
- Step-by-step mechanism for the Wittig reaction
- Mention of the role of RLi in ylide formation
Extra mark
- Discussion of E/Z isomerism in the Wittig product
- (c)(ii) Identification of pairs yielding α,β-unsaturated carbonyl compounds via base-catalyzed reaction. 5 marks
justify— claim → 3-4 reasons → evidence → conclusion
Must cover
- Identification of pairs with at least one α-hydrogen donor
- Justification based on the requirement for enolate formation
- Correct selection of only pair III (PhCHO + CH3CH2CHO)
- Explanation of why HCHO + PhCHO and HCHO + PhCOPh have no alpha-H, and why pair II cannot dehydrate
Loses marks
- Selecting pairs where neither component has α-hydrogens
- Failing to explain the necessity of the α-hydrogen
Earns more
- Drawing of the enolate intermediate for the selected pairs
Extra mark
- Mention of the crossed aldol condensation specifically
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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.
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