Paper II — Q5
(a) Write the structures of nucleosides and nucleotides, and discuss the primary structures of DNA and RNA. (10 marks) (b) Write…
Write the structures of nucleosides and nucleotides, and discuss the primary structures of DNA and RNA. 10 marks
Write down the products obtained after photolysis of 2-methylcyclohexanone in solution phase. Explain the formation of products. 10 marks
By using appropriate reactants, reagents and conditions and using acetylene as starting material, how will you synthesize the following compound? [figure: 1-ethynylcyclopentan-1-ol]
Identify the products W, X, Y and Z in the following reactions: toluene —NBS, hν, Bz₂O₂→ W, toluene —CrO₃, Ac₂O, then H₃O⁺→ X —Ac₂O, NaOAc, Δ→ Y —LiAlH₄ (excess)→ Z 10 marks
How will you distinguish between NH stretching absorption of a primary amine and a secondary amine by using IR spectroscopy?
How will you distinguish among primary, secondary and tertiary alcohols on the basis of PMR spectroscopy? 10 marks
Given below are the IR and NMR spectral characteristics of three compounds: IR: 1750 cm⁻¹; NMR: δ 2.0 (s, 3H), 5.1 (s, 2H) and 7.3 (s, 5H)
IR: 1740 cm⁻¹; NMR: δ 3.5 (s, 3H), 3.6 (s, 2H) and 7.4 (s, 5H)
IR: 3200-2800 (various bands) and 1700 cm⁻¹; NMR: δ 2.75 (t, 2H), 2.95 (t, 2H), 7.4 (s, 5H) and 12.0 (s, 1H)
Match each of these spectral data with one of the following structures: (1) methyl 4-methylbenzoate (2) methyl phenylacetate, PhCH₂CO₂CH₃ (3) 4-ethylbenzoic acid (4) benzyl acetate, CH₃CO₂CH₂Ph (5) 3-phenylpropanoic acid, PhCH₂CH₂CO₂H 10 marks
हिंदी में प्रश्न पढ़ें
न्यूक्लियोसाइड व न्यूक्लियोटाइड की संरचना लिखिए तथा डी. एन. ए. एवं आर. एन. ए. की प्राथमिक संरचनाओं की चर्चा कीजिए। (10 अंक)
2-मेथिलसाइक्लोहेक्सानोन की विलयन अवस्था में प्रकाश-अपघटन के पश्चात् उत्पन्न उत्पाद को लिखिए। उत्पादों के बनने की व्याख्या कीजिए। (10 अंक)
ऐसीटिलीन को आरंभिक द्रव्य लेते हुए उपयुक्त अभिक्रियकों, अभिकर्मकों तथा दशा के प्रयोग से आप निम्नलिखित यौगिक का संश्लेषण कैसे करेंगे? [figure]
निम्नलिखित अभिक्रियाओं में उत्पादों W, X, Y और Z को पहचानिए: [figure] (10 अंक)
आप प्राथमिक ऐमीन व द्वितीयक ऐमीन के बीच NH तनन अवशोषण को IR स्पेक्ट्रमी द्वारा कैसे पहचानेंगे?
आप प्राथमिक, द्वितीयक तथा तृतीयक एल्कोहॉलों को PMR स्पेक्ट्रमी के आधार पर कैसे पहचानेंगे? (10 अंक)
नीचे तीन यौगिकों की IR व NMR स्पेक्ट्रमी विशेषताएँ दी गई हैं: IR: 1750 cm⁻¹; NMR: δ 2.0 (s, 3H), 5.1 (s, 2H) और 7.3 (s, 5H)
IR: 1740 cm⁻¹; NMR: δ 3.5 (s, 3H), 3.6 (s, 2H) और 7.4 (s, 5H)
IR: 3200-2800 (विभिन्न बैंड) तथा 1700 cm⁻¹; NMR: δ 2.75 (t, 2H), 2.95 (t, 2H), 7.4 (s, 5H) और 12.0 (s, 1H)
इनमें से प्रत्येक स्पेक्ट्रमी डेटा को निम्नलिखित संरचनाओं में से किसी एक से मिलाइए: (1) [figure] (2) [figure] (3) [figure] (4) [figure] (5) [figure] (10 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.
(c) A reaction scheme starting from toluene (a benzene ring with a methyl group, CH3, at the top). The scheme branches into two paths. Path 1 (left): Toluene reacts with NBS (N-bromosuccinimide) in the presence of light (hv) and dibenzoyl peroxide (Bz2O2) to form product W. Path 2 (right): Toluene reacts with CrO3, Ac2O, and H3O+ to form product X. Product X then reacts with Ac2O, NaOAc, and heat (delta symbol) to form product Y. Product Y is then treated with excess LiAlH4 to form product Z.
(e) Five chemical structures are listed and numbered (1) through (5):
(1) Methyl 4-methylbenzoate: A benzene ring with a methyl group (H3C-) at the para position relative to a methyl ester group (-C(=O)-O-CH3).
(2) Methyl 3-phenylpropanoate: A benzene ring attached to a methylene group (-CH2-), which is attached to a carbonyl group (C=O), which is attached to a methoxy group (-O-CH3).
(3) 4-ethylbenzoic acid: A benzene ring with an ethyl group (H2C-CH3) at the para position relative to a carboxylic acid group (-C(=O)-O-H).
(4) Benzyl acetate: An acetyl group (CH3-C(=O)-) attached to an oxygen atom, which is attached to a methylene group (-CH2-), which is attached to a benzene ring.
(5) 3-phenylpropanoic acid: A benzene ring attached to a methylene group (-CH2-), which is attached to another methylene group (-CH2-), which is attached to a carboxylic acid group (-C(=O)-O-H).
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) Structures of nucleosides and nucleotides; primary structures of DNA and RNA
A nucleoside is a molecule consisting of a nitrogenous base (purine or pyrimidine) covalently linked to a pentose sugar via an N-glycosidic bond at the C1' position. A nucleotide is formed when one or more phosphate groups are esterified to the 5′-hydroxyl group of the nucleoside, creating a phosphoester linkage. Examples: adenosine (adenine + ribose) and 2′-deoxythymidine are nucleosides, and adenosine 5′-monophosphate (AMP) and dTMP are nucleotides. Purines bond through N-9 and pyrimidines through N-1 to C-1′ of the sugar, in the β-configuration.
The primary structure of DNA is a linear polymer of deoxyribonucleotides connected by 3',5'-phosphodiester linkages, creating a sugar-phosphate backbone with 5' to 3' polarity. It contains the bases Adenine (A), Guanine (G), Cytosine (C), and Thymine (T). The primary structure of RNA is similar but utilizes ribose (containing a 2'-OH group) and Uracil (U) in place of Thymine. This 2'-OH group makes RNA more chemically labile than DNA. The 5' to 3' polarity is maintained in both, but the presence of the 2'-OH in RNA allows for different secondary structures and catalytic functions.
(b) Products of photolysis of 2-methylcyclohexanone in solution phase
Upon n→π* excitation, 2-methylcyclohexanone undergoes Norrish Type I α-cleavage. The bond between the carbonyl carbon and the more substituted α-carbon (C2) breaks preferentially to generate a 1,5-biradical, as this yields a more stable secondary alkyl radical. In the solution phase, this biradical does not fragment to release CO.
The major products arise from intramolecular hydrogen transfer. Transfer of a hydrogen atom from the carbon adjacent to the alkyl radical to the acyl radical yields unsaturated aldehydes, specifically hept-5-enal and hept-6-enal. Alternatively, hydrogen transfer in the reverse direction (from C-6 to the alkyl radical) produces a ketene, pentylketene, which is trapped by the solvent to form a heptanoate ester or heptanoic acid. Recombination of the biradical also regenerates the ketone, resulting in epimerization at C-2. This mechanism is distinct from gas-phase photolysis, where decarbonylation is the primary pathway.
(c) Synthesis and product identification
(i) The target is 1-ethynylcyclopentan-1-ol, a tertiary propargylic alcohol, so it is made by adding acetylide to cyclopentanone (ethynylation).
- HC≡CH + NaNH₂ in liquid NH₃ (−33 °C) gives sodium acetylide, HC≡C⁻ Na⁺. One equivalent of base is used so that only one C–H is removed.
- Cyclopentanone is added. The acetylide carbon attacks the carbonyl carbon, and the C=O π electrons move onto oxygen to give the sodium alkoxide.
- Aqueous NH₄Cl (or dilute acid) work-up protonates the alkoxide to give 1-ethynylcyclopentan-1-ol.
The same addition can be run industrially with KOH and acetylene under pressure (Favorskii ethynylation).
(ii) The products are identified as follows:
- W: Benzyl bromide (formed by benzylic bromination with NBS).
- X: Benzaldehyde. CrO₃ in Ac₂O traps the aldehyde as benzylidene diacetate, PhCH(OAc)₂, which prevents over-oxidation, and H₃O⁺ then hydrolyses it to PhCHO.
- Y: Cinnamic acid (formed by the Perkin reaction of benzaldehyde with acetic anhydride).
- Z: 3-phenylpropan-1-ol (excess LiAlH₄ reduces the COOH to CH₂OH and, through the allylic alkoxide, also saturates the conjugated C=C. With limited hydride, cinnamyl alcohol would result).
(d) Distinguishing amines by IR and alcohols by PMR
(i) In IR spectroscopy, primary amines exhibit two distinct N-H stretching bands (symmetric and asymmetric) in the 3500-3300 cm⁻¹ region. In contrast, secondary amines show only a single N-H stretching band in this region. Tertiary amines show no N-H stretching absorption. This difference in the number of N-H bonds is the primary diagnostic feature for distinguishing between these amine classes using IR spectroscopy.
(ii) In PMR spectroscopy, the splitting of the hydroxyl proton (OH) allows for distinction. Primary alcohols (R-CH2-OH) show the OH as a triplet due to coupling with two adjacent protons. Secondary alcohols (R-CH-OH) show the OH as a doublet due to coupling with one adjacent proton. Tertiary alcohols (R3C-OH) show the OH as a singlet because there are no adjacent protons on the carbinol carbon. This works when OH exchange is slow, in DMSO-d₆ or very dry CDCl₃. In ordinary CDCl₃ the OH is a broad singlet, and the carbinol protons then decide: RCH₂OH near δ 3.6 (2H), R₂CHOH near δ 3.9 (1H), and no carbinol H for R₃COH. Shaking with D₂O removes the OH signal and confirms it.
(e) Matching spectral data to structures
- Data (i) matches Structure (4) (Benzyl acetate). The IR peak at 1750 cm⁻¹ indicates an ester. The NMR signal at δ 2.0 (s, 3H) corresponds to the acetate methyl group, and δ 5.1 (s, 2H) is the benzylic OCH₂ group. The aromatic protons appear as a singlet at δ 7.3.
- Data (ii) matches Structure (2) (methyl phenylacetate, PhCH₂CO₂CH₃). The IR peak at 1740 cm⁻¹ indicates an ester. The NMR signal at δ 3.6 (s, 2H) is the benzylic CH2, and δ 3.5 (s, 3H) is the ester OCH₃ group. The aromatic protons appear as a singlet at δ 7.4.
- Data (iii) matches Structure (5) (3-phenylpropanoic acid). The broad IR band at 3200-2800 cm⁻¹ and peak at 1700 cm⁻¹ indicate a carboxylic acid. The NMR shows two triplets (δ 2.75 and 2.95) for the two CH2 groups and a broad singlet at δ 12.0 for the acidic proton.
How this answer will be evaluated
What earns full marks: Precise structures, mechanisms, and spectral assignments with full reasoning.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Structures of nucleosides/nucleotides and primary structures of DNA/RNA. 10 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- Nucleoside structure (base + sugar)
- Nucleotide structure (nucleoside + phosphate)
- DNA primary structure: deoxyribonucleotides (A, G, C, T) joined 3′→5′ by phosphodiester bonds, 5′→3′ polarity
- RNA primary structure: ribonucleotides with 2′-OH and U in place of T
Loses marks
- Confusing ribose with deoxyribose
- Omitting phosphate in nucleotide
Earns more
- 5' to 3' polarity indicated
- Phosphodiester linkage shown
Extra mark
- Mention of antiparallel strands
- (b) Products of 2-methylcyclohexanone photolysis and formation mechanism. 10 marks
explain— definition/context → points in order → small example → short close
Must cover
- n→π* excitation, then Norrish type I α-cleavage at the more substituted C1–C2 bond, giving the more stable secondary-alkyl / acyl biradical
- In solution the biradical does not lose CO: H transfer to the acyl radical gives hept-5-enal and hept-6-enal, and H transfer the other way gives pentylketene, trapped by solvent as heptanoate ester or heptanoic acid
- Recombination of the biradical regenerates the ketone with loss of configuration at C-2 (epimerisation)
- Draw the biradical and arrow-push each product
Loses marks
- Calling the products Norrish type II products or a 1,4-diol
- Cleaving the less substituted bond
- Listing products without the biradical mechanism
Earns more
- Explain why cleavage occurs on the methyl-bearing side (secondary vs primary radical)
- Contrast with the gas phase, where decarbonylation gives hydrocarbons
- Note that Norrish type II is not available to a simple cyclohexanone in the same way (γ-H in the ring leads only to ring-opened products via the same biradical)
Extra mark
- Mention of solvent cage effect
- (c) Synthesis of 1-ethynylcyclopentan-1-ol from acetylene, and identification of W, X, Y, Z. 10 marks
trace— start point → the stages in sequence → end point → what changed
Must cover
- HC≡CH + NaNH₂/liq. NH₃ → HC≡C⁻Na⁺, then cyclopentanone, then H₃O⁺/NH₄Cl gives 1-ethynylcyclopentan-1-ol
- W: Benzyl bromide (NBS product)
- X: Benzaldehyde (CrO₃/Ac₂O via benzylidene diacetate, then H₃O⁺)
- Y: Cinnamic acid (Perkin product)
Loses marks
- Identifying X as benzoic acid
- Generic synthesis steps without target
Earns more
- Z: 3-phenylpropan-1-ol
- Reagents and conditions for each step (NaNH₂/liq. NH₃, NH₄Cl work-up)
Extra mark
- Specific conditions for Perkin reaction
- (d) IR distinction of amines; PMR distinction of alcohols. 10 marks
compare— paired headings or point-wise contrast → key differences → significance → conclusion
Must cover
- Primary amine: two NH bands
- Secondary amine: one NH band
- Primary alcohol: OH triplet
- PMR (DMSO-d₆): OH is a triplet for 1°, a doublet for 2° and a singlet for 3° alcohols
Loses marks
- Confusing NH with OH peaks
- Ignoring splitting patterns
Earns more
- Carbinol H: RCH₂OH ~δ 3.6 (2H), R₂CHOH ~δ 3.9 (1H), none for R₃COH, with D₂O exchange
- Chemical shift ranges
Extra mark
- Coupling constant values
- (e) Match spectral data (i-iii) to structures (1-5). 10 marks
map— locate accurately → label → one line on why it matters
Must cover
- Data (i) matches Structure (4), benzyl acetate
- Data (ii) matches Structure (2), methyl phenylacetate
- Data (iii) matches Structure (5)
- IR/NMR evidence for each match
Loses marks
- Matching without spectral evidence
- Ignoring IR carbonyl frequencies
Earns more
- Specific peak assignments
- Integration analysis
Extra mark
- Explanation of chemical shifts
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Structure that answers it
Set the issue up → the case as it is made → the case against → the dimension both sides leave out → where the balance now lies
Where marks are lost
Listing facts with no thread between them, or arguing one side throughout and calling it a discussion.
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