Paper II — Q6
(a) (i) Elucidate the structure of the product and the intermediate (if any) in the following reactions: (A) [structure…
Elucidate the structure of the product and the intermediate (if any) in the following reactions: (A) [structure: H(Me)C(CO₂Et)(CH₂CO₂H)] 1) BH₃ 2) H⁺ → ? (B) [structure: cyclohexyl-CH=CH-CHO] NaBH₄ → ? (5+5=10 marks)
Describe the role of NMO during the dihydroxylation of an alkene using catalytic amount of OsO₄ in the presence of N-methylmorpholine N-oxide (NMO). 5 marks
Calculate the frequency of radiation required for a transition of J = 4 to J = 5 in the rotational spectrum of HCl. The rotation constant B = 10·6 cm⁻¹. 5 marks
The fundamental vibrational frequency of HCl is 2990 cm⁻¹. Calculate the fundamental vibrational frequency of DCl assuming the same bond strength. 5 marks
The molecular formula of a compound is C₃H₃N. The IR absorption frequencies are 1650 cm⁻¹, 2250 cm⁻¹ and 3100 cm⁻¹. Assign a structure for the compound. 5 marks
Write the mechanism for the following photochemical transformation: [structure not shown in transcript] 5 marks
Photobromination of cinnamic acid was carried out by using light of wavelength 480 nm with a light intensity of 1·5×10⁻³ J-s⁻¹. An exposure of 10 minutes showed a decrease of 0·05 millimole of Br₂. Calculate the quantum yield assuming that 80% of radiation is absorbed by cinnamic acid. (Planck's constant h = 6·627×10⁻³⁴ J-s and velocity of light c = 3×10⁸ m-s⁻¹) 5 marks
Predict the major and minor products for the following photoreaction. Give the logic: [structure not shown in transcript] 5 marks
Assign A, B and C in the following reaction: (Major) [reaction not fully shown in transcript] 5 marks
हिंदी में प्रश्न पढ़ें
निम्नलिखित अभिक्रियाओं में बनने वाले उत्पाद और मध्यवर्ती (यदि हो) की संरचना को स्पष्ट कीजिए: (A) [संरचना: H(Me)C(CO₂Et)(CH₂CO₂H)] 1) BH₃ 2) H⁺ → ? (B) [संरचना: साइक्लोहेक्सिल-CH=CH-CHO] NaBH₄ → ? (5+5=10 अंक)
N-मेथिलमॉर्फोलिन N-ऑक्साइड (NMO) की उपस्थिति में OsO₄ की उत्प्रेरक मात्रा का उपयोग करके एक ऐल्कीन के डाइहाइड्रॉक्सिलेशन अभिक्रिया में NMO की भूमिका का वर्णन कीजिए। (5 अंक)
HCl के घूर्णनात्मक स्पेक्ट्रम में J = 4 से J = 5 के संक्रमण के लिए आवश्यक विकिरण की आवृत्ति की गणना कीजिए। घूर्णक स्थिरांक B = 10·6 cm⁻¹। (5 अंक)
HCl की मूल कंपन आवृत्ति 2990 cm⁻¹ है। समान आबंध सामर्थ्य मानते हुए DCl की मूल कंपन आवृत्ति की गणना कीजिए। (5 अंक)
एक यौगिक का अणुसूत्र C₃H₃N है। इसकी IR अवशोषण आवृत्तियाँ 1650 cm⁻¹, 2250 cm⁻¹ और 3100 cm⁻¹ हैं। यौगिक के लिए एक संरचना निर्दिष्ट कीजिए। (5 अंक)
निम्नलिखित प्रकाश-रासायनिक रूपांतरण के लिए क्रियाविधि लिखिए: [संरचना प्रतिलिपि में नहीं दिखाई गई] (5 अंक)
सिनेमिक अम्ल का फोटोब्रोमिनेशन 480 nm तरंगदैर्घ्य के प्रकाश का उपयोग करके किया गया था, जिसकी प्रकाश तीव्रता 1·5×10⁻³ J-s⁻¹ थी। 10 मिनट के उद्भासन से Br₂ में 0·05 मिलीमोल की कमी देखी गई। यह मानते हुए क्वांटम लब्धि की गणना कीजिए कि सिनेमिक अम्ल द्वारा 80% विकिरण अवशोषित किया जाता है। (प्लांक स्थिरांक h = 6·627×10⁻³⁴ J-s और प्रकाश का वेग c = 3×10⁸ m-s⁻¹) (5 अंक)
निम्नलिखित प्रकाश-अभिक्रिया में बनने वाले प्रमुख तथा अल्प उत्पादों की प्रागुक्ति कीजिए। तर्क दीजिए: [संरचना प्रतिलिपि में नहीं दिखाई गई] (5 अंक)
निम्नलिखित अभिक्रिया में A, B और C निर्देश कीजिए: (Major) [अभिक्रिया प्रतिलिपि में पूर्ण रूप से नहीं दिखाई गई] (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) Reaction (B): A cyclohexane ring is bonded to a carbon-carbon double bond, which is bonded to an aldehyde group (CHO). The structure is 3-cyclohexylprop-2-enal. The reagent is NaBH4. The product is indicated by a question mark.
(c) A chemical reaction scheme showing the photolysis of two different ketones. The reactants are benzyl 2-benzyl-2-oxoacetate (Ph2CH-C(=O)-CHPh2) and benzyl 2-benzyl-2-oxoacetate (PhCH2-C(=O)-CH2Ph). They react under light (hv) to form products A, B, C, and CO. Product C is labeled as the major product.
(c(i)) A chemical reaction scheme showing a benzene ring (six-membered aromatic ring with alternating double bonds) on the left, an arrow pointing to the right labeled with 'hv' (indicating photochemical conditions), and a cyclopentadiene ring (five-membered ring with two double bonds) on the right. The cyclopentadiene ring has a double bond extending from one of its carbons to a terminal carbon (an exocyclic double bond), forming a methylene group (=CH2).
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)
In (A), BH₃ is selective for the free carboxylic acid. The acid first forms a monoacyloxyborane, RCOO–BH₂, with evolution of H₂; hydride transfer to the acyl carbon and aqueous work-up give the primary alcohol, while the ester is unchanged. This γ-hydroxy ester, CH₃CH(CO₂Et)CH₂CH₂OH, is the intermediate; with H⁺ it cyclises with loss of EtOH to the product, α-methyl-γ-butyrolactone (3-methyloxolan-2-one). In (B), NaBH₄ reduces the aldehyde faster than the C=C bond. The immediate intermediate is an alkoxyborohydride adduct, RCH₂O–BH₃⁻, which hydrolyses to the allylic alcohol cyclohexylCH=CHCH₂OH.
(a)(ii)
OsO₄ adds to the alkene to form a cyclic osmate(VI) ester. NMO is the co-oxidant: it oxidises Os(VI) back to Os(VIII), releasing the cis-diol and regenerating OsO₄. NMO is reduced to N-methylmorpholine, so only catalytic OsO₄ is needed and stoichiometric osmium is avoided.
(b)(i)
For a rigid rotor, Δν~ = 2B(J+1). For J=4 to 5, Δν~ = 2×10.6×5 = 106 cm⁻¹. Frequency = cν~ = 3×10¹⁰ cm s⁻¹ × 106 = 3.18×10¹² s⁻¹.
(b)(ii)
ν is proportional to the square root of k/μ. With the same force constant, ν(DCl)/ν(HCl) = square root of μ(HCl)/μ(DCl). μ(HCl) = 35.5×1/36.5 and μ(DCl) = 35.5×2/37.5, so the ratio is 0.717. Hence ν(DCl) = 2990×0.717 ≈ 2144 cm⁻¹.
(b)(iii)
C₃H₃N has three degrees of unsaturation. 3100 cm⁻¹ is sp² =C–H stretch, 2250 cm⁻¹ is C≡N stretch, and 1650 cm⁻¹ is C=C stretch. The compound is CH₂=CH–CN, acrylonitrile.
(c)(i)
Benzene absorbs hν to the S₁ singlet state. Bonding between C-1 and C-3 (meta bridging) gives the bicyclo[3.1.0]hexenyl 1,3-diradical ('prefulvene'). A 1,2-H shift with cleavage of the three-membered-ring bond converts this diradical to methylenecyclopentadiene (fulvene); closure of the same diradical gives benzvalene.
(c)(ii)
E per photon = hc/λ = 6.627×10⁻³⁴×3×10⁸/(480×10⁻⁹) = 4.14×10⁻¹⁹ J. The 600 s exposure at 1.5×10⁻³ J s⁻¹ with 80% absorption corresponds to 0.72 J absorbed, i.e. 0.72/4.14×10⁻¹⁹ = 1.74×10¹⁸ photons ≈ 2.89×10⁻⁶ mol photons. Br₂ consumed = 0.05 mmol = 5×10⁻⁵ mol. Quantum yield φ = moles Br₂ reacted/moles photons absorbed = 5×10⁻⁵/2.89×10⁻⁶ ≈ 17.3.
(c)(iii)
The ketone is 4,5-dimethylhexan-2-one, CH₃–CO–CH₂–CH(CH₃)–CH(CH₃)₂, and this is a classic Norrish type II reaction. The excited C=O abstracts a γ-hydrogen through a six-membered cyclic transition state. Two γ-C–H sites are available: the tertiary C5–H and the primary C–H of the methyl on C4; the weaker tertiary γ-C–H is abstracted preferentially, giving the 1,4-biradical. β-Scission of the Cα–Cβ bond then gives the enol of acetone together with the alkene. Thus the major products are acetone and 2-methylbut-2-ene ((CH₃)₂C=CHCH₃, from the tertiary γ-H), while abstraction of a primary γ-methyl hydrogen gives acetone and 3-methylbut-1-ene as the minor pair; closure of the biradical to a cyclobutanol is a further minor pathway. Logic: Norrish II requires a γ-H, the weakest (tertiary) γ-C–H is removed most readily, and the resulting radical prefers fragmentation to cyclisation.
(c)(iv)
Both ketones undergo Norrish type I α-cleavage under hν. Ph₂CH–CO–CHPh₂ gives Ph₂CH• and Ph₂CHCO•; PhCH₂–CO–CH₂Ph gives PhCH₂• and PhCH₂CO•. The acyl radicals rapidly lose CO, leaving a radical pool of Ph₂CH• and PhCH₂•. Recombination then gives A = 1,1,2,2-tetraphenylethane, Ph₂CH–CHPh₂; B = bibenzyl, PhCH₂–CH₂Ph; and C (major) = 1,1,2-triphenylethane, Ph₂CH–CH₂Ph, because cross-combination of the two different (both resonance-stabilised) radicals is statistically and kinetically favoured over homodimerisation. Carbon monoxide is released in the acyl-radical step.
How this answer will be evaluated
Framework: Concept > Structure or mechanism > Reasoning > Result
What earns full marks: Correct structures, mechanisms, and calculations with full reasoning.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i) Deduce product and intermediate structures for reactions (A) and (B). 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- BH₃ selectively reduces the free –COOH (via the acyloxyborane, H₂ evolution, then hydride transfer); the ester is untouched
- Intermediate: the γ-hydroxy ester CH₃CH(CO₂Et)CH₂CH₂OH; H⁺ cyclises it with loss of EtOH to the 5-membered lactone
- Product (A): α-methyl-γ-butyrolactone (3-methyloxolan-2-one), stereochemistry retained
- In (B), NaBH₄ gives selective 1,2-reduction of the aldehyde (no C=C reduction); product: the allylic alcohol cyclohexylCH=CHCH₂OH
Loses marks
- Missing intermediate structure
- Wrong product connectivity
Earns more
- Correct stereochemistry in (A)
- Mechanism arrows for hydroboration
Extra mark
- Mention of regioselectivity rules
- (a)(ii) Explain the role of NMO in catalytic dihydroxylation. 5 marks
describe— define → structure or process in order → labelled diagram → significance
Must cover
- NMO is the co-oxidant; it re-oxidises Os(VI) to Os(VIII)
- Cyclic osmate(VI) ester intermediate; cis-diol released and OsO₄ regenerated
- Catalytic OsO₄ is sufficient; NMO is reduced to N-methylmorpholine
Loses marks
- Calling NMO the primary oxidant of the alkene (it only re-oxidises osmium)
- Missing catalytic aspect
Earns more
- Structure of NMO
- Byproduct N-methylmorpholine
Extra mark
- Upjohn dihydroxylation name
- (b)(i) Calculate frequency for J=4 to J=5 transition in HCl. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- Δν̃ = 2B(J+1); J = 4→5 with B = 10.6 cm⁻¹ gives 106 cm⁻¹
- Convert to frequency: ν = cν̃ = 3.18×10¹² s⁻¹ (≈3.18×10¹² Hz)
Loses marks
- Wrong formula for transition
- Missing unit conversion
Earns more
- Correct unit conversion
- Step-by-step calculation
Extra mark
- Mention of rigid rotor model
- (b)(ii) Calculate fundamental vibrational frequency of DCl. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- ν ∝ √(k/μ) at constant force constant
- Reduced masses of HCl and DCl; ratio 0.717
- ν(DCl) ≈ 2990 × 0.717 ≈ 2144 cm⁻¹
Loses marks
- Using atomic mass instead of reduced mass
- Wrong proportionality
Earns more
- Correct reduced mass calculation
- Final value in cm⁻¹
Extra mark
- Mention of harmonic oscillator
- (b)(iii) Assign structure for C3H3N based on IR data. 5 marks
map— locate accurately → label → one line on why it matters
Must cover
- 1650 cm⁻¹ = C=C stretch; 2250 cm⁻¹ = C≡N stretch; 3100 cm⁻¹ = sp² =C–H stretch
- Structure: acrylonitrile CH₂=CH–CN (prop-2-enenitrile), consistent with C₃H₃N
Loses marks
- Ignoring C≡N peak
- Structure not matching formula
Earns more
- Correct IUPAC name
- Consistency with molecular formula
Extra mark
- Alternative isomer discussion
- (c)(i) Write mechanism for photochemical transformation of benzene. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Excitation of benzene to S₁; meta-bridging gives the bicyclo[3.1.0]hexenyl 1,3-diradical (prefulvene)
- 1,2-H shift with cleavage of the three-membered ring gives fulvene (methylenecyclopentadiene), the product drawn
- The same diradical can also close to benzvalene
Loses marks
- Missing intermediate
- Wrong product structure
Earns more
- Arrow pushing for electron movement
- Mention of valence isomerization
Extra mark
- Energy diagram
- (c)(ii) Calculate quantum yield for photobromination of cinnamic acid. 5 marks
calculate— given → formula → substitution → result with units → interpretation
Must cover
- E = hc/λ = 4.14×10⁻¹⁹ J per photon (480 nm)
- Absorbed energy = 1.5×10⁻³ × 600 × 0.8 = 0.72 J → 1.74×10¹⁸ photons = 2.89×10⁻⁶ mol
- φ = 5×10⁻⁵/2.89×10⁻⁶ ≈ 17.3
Loses marks
- Ignoring 80% absorption factor
- Wrong time conversion
Earns more
- Correct unit handling
- Final numerical value
Extra mark
- Definition of quantum yield
- (c)(iii) Predict major and minor products for the photoreaction. 5 marks
map— locate accurately → label → one line on why it matters
Must cover
- Norrish type II: γ-H abstraction through a six-membered cyclic TS → 1,4-biradical
- β-Scission gives acetone (as the enol) + 2-methylbut-2-ene — major, via the tertiary γ-C5–H
- Minor: acetone + 3-methylbut-1-ene via a primary γ-methyl H; cyclobutanol closure is a further minor path
- Logic: weakest tertiary γ-C–H abstracted; fragmentation preferred
Loses marks
- Wrong cleavage site
- Missing logic for major/minor
Earns more
- Logic based on radical stability
- Correct product structures
Extra mark
- Mention of decarbonylation
- (c)(iv) Assign structures A, B, and C in the reaction. 5 marks
map— locate accurately → label → one line on why it matters
Must cover
- Norrish type I α-cleavage of both ketones; acyl radicals lose CO
- Radical pool Ph₂CH• + PhCH₂•
- A = (Ph₂CH)₂ (1,1,2,2-tetraphenylethane); B = bibenzyl (PhCH₂CH₂Ph)
- C (major) = Ph₂CH–CH₂Ph (1,1,2-triphenylethane) — cross-combination favoured; CO released
Loses marks
- Wrong product assignment
- Missing CO byproduct
Earns more
- Note that no stereocentres are formed
- Logic for major product
Extra mark
- Statistical 1:2:1 ratio (A:C:B) for equal radical populations
What "Elucidate" is asking you to do
Make a stated proposition plain and then prove it with instances. Elucidate stems almost always carry a claim or a named concept, and very often the words “with examples” or “with suitable diagrams” — the illustration is part of the directive, not decoration.
Structure that answers it
Plain-language statement of what the proposition means → the part that is obscure, resolved → first illustration → second illustration → why the proposition holds
Where marks are lost
Adding terminology; elucidate rewards removing it. The commoner loss is a clean explanation with no example, when the stem asked for examples.
Practice this exact question
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