Q5 50M Compulsory discuss Biochemistry, organic synthesis, and spectroscopy
(a) Write the structures of nucleosides and nucleotides, and discuss the primary structures of DNA and RNA. (10 marks)
(b) Write down the products obtained after photolysis of 2-methylcyclohexanone in solution phase. Explain the formation of products. (10 marks)
(c) (i) 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]
(ii) 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)
(d) (i) How will you distinguish between NH stretching absorption of a primary amine and a secondary amine by using IR spectroscopy?
(ii) How will you distinguish among primary, secondary and tertiary alcohols on the basis of PMR spectroscopy?
(10 marks)
(e) Given below are the IR and NMR spectral characteristics of three compounds:
(i) IR: 1750 cm⁻¹; NMR: δ 2.0 (s, 3H), 5.1 (s, 2H) and 7.3 (s, 5H)
(ii) IR: 1740 cm⁻¹; NMR: δ 3.5 (s, 3H), 3.6 (s, 2H) and 7.4 (s, 5H)
(iii) 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 पढ़ें
(a) न्यूक्लियोसाइड व न्यूक्लियोटाइड की संरचना लिखिए तथा डी. एन. ए. एवं आर. एन. ए. की प्राथमिक संरचनाओं की चर्चा कीजिए। (10 अंक)
(b) 2-मेथिलसाइक्लोहेक्सानोन की विलयन अवस्था में प्रकाश-अपघटन के पश्चात् उत्पन्न उत्पाद को लिखिए। उत्पादों के बनने की व्याख्या कीजिए। (10 अंक)
(c) (i) ऐसीटिलीन को आरंभिक द्रव्य लेते हुए उपयुक्त अभिक्रियकों, अभिकर्मकों तथा दशा के प्रयोग से आप निम्नलिखित यौगिक का संश्लेषण कैसे करेंगे?
[figure]
(ii) निम्नलिखित अभिक्रियाओं में उत्पादों W, X, Y और Z को पहचानिए:
[figure]
(10 अंक)
(d) (i) आप प्राथमिक ऐमीन व द्वितीयक ऐमीन के बीच NH तनन अवशोषण को IR स्पेक्ट्रमी द्वारा कैसे पहचानेंगे?
(ii) आप प्राथमिक, द्वितीयक तथा तृतीयक एल्कोहॉलों को PMR स्पेक्ट्रमी के आधार पर कैसे पहचानेंगे?
(10 अंक)
(e) नीचे तीन यौगिकों की IR व NMR स्पेक्ट्रमी विशेषताएँ दी गई हैं:
(i) IR: 1750 cm⁻¹; NMR: δ 2.0 (s, 3H), 5.1 (s, 2H) और 7.3 (s, 5H)
(ii) IR: 1740 cm⁻¹; NMR: δ 3.5 (s, 3H), 3.6 (s, 2H) और 7.4 (s, 5H)
(iii) 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)
Answer approach & key points
(a) discuss: intro > 3-4 dimensions > example > balanced close | (b) explain: definition/context > points in order > small example > short close | (c) trace: start point > the stages in sequence > end point > what changed | (d) compare: paired headings or table > key differences > significance > conclusion | (e) map: locate accurately > label > one line on why it matters Full marks: Precise structures, mechanisms, and spectral assignments with full reasoning.
- 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
- 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
- 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)
- 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
- 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
Q6 50M explain Organic photochemistry, polymerization, and stereochemistry
(a) Predict the products in the following reactions :
(i) PhCO–CH₂CH₂CH₂CH₃ (valerophenone) + CH₃CO–CH₂CH₂–CD₂–CH₃ (hexan-2-one-5,5-d₂) --hv-->
Propose suitable mechanism to justify your answer.
(ii) CH₃CO–CH₂–CH(CH₃)–CH(CH₃)–CH₂CH₃ (4,5-dimethylheptan-2-one) --hv-->
Giving justification, write the major and minor product(s). Comment upon the chirality of recovered reactant (if any). (5+10=15 marks)
(b) (i) How will you synthesize polypropylene (PP) by using Ziegler-Natta catalysis? Discuss the mechanism and its advantages over conventional polymerization. (10 marks)
(ii) How is Perlon synthesized from ε-Caprolactam? Give the mechanism of the reaction. (5 marks)
(c) (i) Write the structures of the products X, Y and Z in the following reactions and indicate the mechanism for the formation of X :
tricyclic lactone (γ-lactone fused to a cyclopentane fused to a cyclopropane) whose cyclopropane carbon carries –CH=CH–CH₂CH₂CH₂Ph [figure] --(1) OsO4, Pyridine (2) H+, H2O--> X --HIO4--> Y + Z (10 marks)
(ii) Write the major and minor products in the following reaction. Discuss the stereochemistry along with reaction mechanism for the formation of the major product :
Ph–CH(CH₃)–CO–CH₃, single enantiomer (wedged CH₃) --m-CPBA--> Major product + Minor product (10 marks)
हिंदी में पढ़ें
(a) निम्नलिखित अभिक्रियाओं में उत्पादों का अनुमान लगाइए :
(i) [figure] + [figure] --hv-->
अपने उत्तर का औचित्य सिद्ध करने के लिए उपयुक्त क्रियाविधि प्रस्तावित कीजिए।
(ii) [figure] --hv-->
औचित्य प्रदान करते हुए मुख्य तथा अल्प उत्पाद/उत्पादों को लिखिए। पुनः-प्राप्त अभिकारक (अगर कोई है) की काइरलता पर टिप्पणी कीजिए। (5+10=15 अंक)
(b) (i) ज़िग्लर-नाट्टा उत्प्रेरण के प्रयोग से आप पॉलिप्रोपिलीन (PP) का संश्लेषण कैसे करेंगे? क्रियाविधि बताइए तथा पारंपरिक बहुलकरण (पॉलिमराइज़ेशन) की अपेक्षा फायदों की विवेचना कीजिए। (10 अंक)
(ii) ε-कैप्रोलेक्टम से पर्लोन को कैसे संश्लेषित करते हैं? अभिक्रिया की क्रियाविधि दीजिए। (5 अंक)
(c) (i) निम्नलिखित अभिक्रियाओं में उत्पादों X, Y और Z की संरचना लिखिए तथा X के बनने की क्रियाविधि का उल्लेख कीजिए :
[figure] --(1) OsO4, पिरिडीन (Pyridine) (2) H+, H2O--> X --HIO4--> Y + Z (10 अंक)
(ii) निम्नलिखित अभिक्रिया में मुख्य तथा अल्प उत्पादों को लिखिए। मुख्य उत्पाद के बनने की त्रिविम रसायन के साथ अभिक्रिया क्रियाविधि की विवेचना कीजिए :
[figure] --m-CPBA--> मुख्य उत्पाद + अल्प उत्पाद (10 अंक)
Answer approach & key points
Framework: UPSC Chemistry Paper 2. (a(i)) justify: claim > 3-4 reasons > evidence > conclusion | (a(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (b(i)) discuss: intro > 3-4 dimensions > example > balanced close | (b(ii)) explain: definition/context > points in order > small example > short close | (c(i)) explain: definition/context > points in order > small example > short close | (c(ii)) discuss: intro > 3-4 dimensions > example > balanced close Full marks: Comprehensive mechanisms, correct stereochemistry, clear justification for all parts.
- Identify the reaction as a Norrish Type II process
- Show the 1,5-hydrogen abstraction step
- Draw the resulting 1,4-biradical intermediate
- β-Scission products: acetophenone + propene from valerophenone, acetone-d₁ (CH₃COCH₂D) + propene-2-d from the labelled ketone
- Identifies the Norrish Type II reaction pathway
- Major: acetone + 3-methylpent-2-ene via the tertiary γ-H (C-5). Minor: cyclobutanol (Yang) and acetone + 3-methylpent-1-ene via the primary γ-H
- Explains the stereochemical outcome of the reaction
- Recovered ketone is epimerised at the γ-carbon C-5 by reverse H transfer, while C-4 keeps its configuration
- Identifies the Ziegler-Natta catalyst components (e.g., TiCl4, AlEt3)
- Describes the coordination-insertion mechanism of polymerization
- Explains the regio- and stereo-selectivity of the process
- Lists at least two advantages over conventional polymerization
- Identifies the reaction as ring-opening polymerization
- Shows the initiation step involving the catalyst
- Illustrates the propagation step of the polymer chain
- Names the final product as Perlon (nylon-6)
- X: syn-1,2-diol from the side-chain C=C via the cyclic osmate ester
- Y: the tricyclic lactone cyclopropanecarbaldehyde, Z: 4-phenylbutanal
- Shows the mechanism of OsO4-mediated dihydroxylation
- Explains the role of HIO4 in the cleavage reaction
- Identifies the reaction as a Baeyer-Villiger oxidation
- The secondary benzylic carbon migrates (not CH₃): major product 1-phenylethyl acetate
- Migration with retention of configuration at the migrating carbon
- Minor product: methyl 2-phenylpropanoate from methyl migration
Q7 50M discuss Protein structures and spectroscopic analysis
(a) Discuss the different types of secondary structures of proteins and compare these structures with tertiary structure of proteins. 15 marks
(b) Complete the following reactions and give suitable mechanisms for the formation of products:
(i) CH₃—CH₂—CH₂—C(=O)—Cl (1) 2NaBH₄ → ?
(2) H₃O⁺
(ii) CH₃—C≡C—CH₃ Na in liq. NH₃ → ?
10+5=15 marks
(c) (i) A compound (A) containing C, H and O has molecular weight 102 and displays two signals in the ¹H NMR spectrum at δ 1·1 (d) and 3·55 (septet) in the integral ratio of 6:1. Treatment of A with 1 mole of HI gives rise to B and C. In the IR spectrum, B gives a strong absorption band at 3330 cm⁻¹, whereas its ¹H NMR spectrum shows signals at δ 1·05(d, 6H), 3·6 (septet, 1H) and 4·4 (s, 1H, disappeared with D₂O). The ¹H NMR spectrum of C gives signals at δ 1·9(d, 6H) and 4·25 (septet, 1H). Reaction of A with excess of HI gives only C. Identify A, B and C. Write all the reactions involved. 10 marks
(ii) Three isomeric compounds having MF C₅H₁₀O give positive 2,4-DNP test and display the following NMR spectral characteristics. Identify the compounds. Among them, one isomeric compound on treatment with KOH (concentrated) gives two products. Write their structures also:
(1) A triplet at δ1.05 and a quartet at δ2.47
(2) Two singlets
(3) A doublet at δ1.0, a singlet at δ2.1 and a septet at δ2.2
10 marks
हिंदी में पढ़ें
(a) प्रोटीन के भिन्न प्रकारों की द्वितीयक संरचनाओं की विवेचना कीजिए तथा इन संरचनाओं की तुलना प्रोटीन के तृतीयक संरचना से कीजिए। 15 अंक
(b) निम्नलिखित अभिक्रियाओं को पूर्ण कीजिए तथा उत्पादों के बनने की उपयुक्त क्रियाविधि दीजिए :
(i) CH₃—CH₂—CH₂—C(=O)—Cl (1) 2NaBH₄ → ?
(2) H₃O⁺
(ii) CH₃—C≡C—CH₃ तरल NH₃ में Na → ?
10+5=15 अंक
(c) (i) C, H तथा O के एक यौगिक (A) का आणविक भार 102 है और ¹H NMR स्पेक्ट्रम में दो सिग्नल δ 1·1 (d) तथा 3·55 (सेप्टेट) पर अभिन्न अनुपात 6:1 में देता है। A का विवेचन जब 1 मोल HI के साथ किया जाता है, तो B तथा C उत्पाद बनते हैं। B के IR स्पेक्ट्रम में 3330 cm⁻¹ पर एक प्रबल अवशोषण बैंड है, जबकि ¹H NMR स्पेक्ट्रम में δ 1·05(d, 6H), 3·6 (सेप्टेट, 1H) और 4·4 (s, 1H, D₂O में लुप्त) पर सिग्नल दिखाता है। C के ¹H NMR स्पेक्ट्रम में δ 1·9(d, 6H) तथा 4·25 (सेप्टेट, 1H) पर सिग्नल देता है। यदि A की अभिक्रिया अतिरिक्त HI से की जाती है, तो केवल C बनता है। A, B तथा C को पहचानिए। सभी संबंधित अभिक्रियाओं को लिखिए। 10 अंक
(ii) तीन समावयवी यौगिकों, जिनका आणविक सूत्र (MF) C₅H₁₀O है, 2,4-DNP परीक्षण सकारात्मक देते हैं तथा निम्नलिखित NMR स्पेक्ट्रमी अभिलक्षण देते हैं। यौगिकों को पहचानिए। इनमें एक समावयवी यौगिक KOH (संद्रित) के साथ विवेचन करने पर दो उत्पाद देता है। उत्पादों की संरचना भी बताइए :
(1) δ1.05 पर एक ट्रिप्लेट और δ2.47 पर एक क्वार्टेट
(2) दो सिंगलेट
(3) δ1.0 पर एक डबलेट, δ2.1 पर एक सिंगलेट और δ2.2 पर एक सेप्टेट
10 अंक
Answer approach & key points
Framework: Concept > Structure or mechanism > Reasoning > Result. (a) discuss: intro > 3-4 dimensions > example > balanced close | (b(i)) explain: definition/context > points in order > small example > short close | (b(ii)) explain: definition/context > points in order > small example > short close | (c(i)) analyse: intro > causes > effects > stakeholders/linkages > way forward | (c(ii)) analyse: intro > causes > effects > stakeholders/linkages > way forward Full marks: Accurate structures, mechanisms, and spectral analysis with clear reasoning.
- Define alpha-helix and beta-pleated sheet structures
- Identify hydrogen bonding as the stabilizing force
- Distinguish secondary from tertiary structure (R-group interactions)
- Mention specific interactions in tertiary (disulfide, ionic, hydrophobic)
- Identify product as butan-1-ol
- Show nucleophilic attack of hydride on carbonyl carbon
- Show elimination of chloride ion
- Show protonation of the alkoxide intermediate
- Identify product as trans-2-butene
- Show radical anion formation
- Explain anti-addition of protons
- Justify trans stereochemistry via steric repulsion
- Identify A as diisopropyl ether
- Identify B as isopropanol (2-propanol)
- Identify C as isopropyl iodide (2-iodopropane)
- Write the reaction of A with HI to form B and C
- Identify (1) as pentan-3-one (triplet + quartet only)
- Identify (2) as 2,2-dimethylpropanal (two singlets: 9H t-Bu, 1H CHO)
- Identify (3) as 3-methylbutan-2-one
- 2,2-Dimethylpropanal (no α-H) undergoes the Cannizzaro reaction with conc. KOH: 2,2-dimethylpropan-1-ol + potassium 2,2-dimethylpropanoate
Q8 50M explain Spectroscopic techniques in organic chemistry
(a) (i) Among the following compounds I and II, which has more carbonyl stretching frequency in IR spectra? Explain:
I = 1-acetylcyclohexene
II = 1-acetyl-2-methylcyclohexene
5 marks
(ii) Complete the following reaction. Write the structures of A, B and C, and explain how IR spectroscopy is helpful to distinguish among them:
1,2-dimethylcyclohexan-1-ol —H⁺→ A + B + C
10 marks
(b) (i) Calculate the λ_max values of the following compounds:
(1) steroid tetraene with C3=C4, C5=C6, C7=C8 and C9=C11 double bonds (C₉H₁₉ side chain at C-17)
(2) 1-vinyl-octahydronaphthalene with C1=C8a and C4a=C5 double bonds
10 marks
(ii) Explain why 1,3-butadiene exhibits a lower λ_max for π → π* transitions compared to that of 1,3,5-hexatriene.
5 marks
(c) (i) What is McLafferty rearrangement? Discuss the mass spectral fragmentation of butyl butyrate with the following given data of ions:
m/z 101, m/z 73, m/z 71 and m/z 56
Write the structures of fragment ions.
10 marks
(ii) (1) A compound with MF C₂H₂BrCl exhibits two doublets (J = 16 Hz) in its PMR spectrum. Suggest a suitable structure along with other possible structures.
(2) How can the structures of A (β-ionone, 4-(2,6,6-trimethylcyclohex-1-enyl)but-3-en-2-one) and B (4-(cyclohex-1-enyl)but-3-en-2-one) be decided based on their UV spectral data?
[ λ_max = 296 nm (ε_max = 10700) and
λ_max = 281 nm (ε_max = 20800) ]
10 marks
हिंदी में पढ़ें
(a) (i) निम्नलिखित यौगिकों I तथा II में किसकी कार्बनिल तनन आवृत्ति IR स्पेक्ट्रा में अधिक है? व्याख्या कीजिए :
I
II
5 अंक
(ii) निम्नलिखित अभिक्रिया को पूर्ण कीजिए। A, B और C की संरचना लिखिए तथा व्याख्या कीजिए किस प्रकार से IR स्पेक्ट्रमिकी इनमें अंतर बताने में सहायक है :
H₃C OH CH₃
H⁺
A + B + C
10 अंक
(b) (i) निम्नलिखित यौगिकों की λ_max मानों का परिकलन कीजिए :
(1)
(2)
10 अंक
(ii) 1,3-ब्यूटाडाइन के π → π* संक्रमण का λ_max मान 1,3,5-हेक्साट्राइन की तुलना में कम क्यों है? व्याख्या कीजिए।
5 अंक
(c) (i) मैकलफर्टी पुनर्विन्यास क्या है? ब्यूटिल ब्यूटिरेट के द्रव्यमान स्पेक्ट्रमी खंडन में निम्नलिखित आयन प्राप्त हुए हैं :
m/z 101, m/z 73, m/z 71 और m/z 56
इनकी विवेचना कीजिए। खंड आयनों की संरचना लिखिए।
10 अंक
(ii) (1) एक यौगिक, जिसका आण्विक सूत्र (MF) C₂H₂BrCl है, PMR स्पेक्ट्रम में दो डबलेट (J = 16 Hz) देता है। एक उपयुक्त संरचना का सुझाव दीजिए तथा संभावित दूसरी संरचनाएं भी दीजिए।
(2) A तथा B की संरचना को UV स्पेक्ट्रमी आँकड़ों के आधार पर कैसे तय कर सकते हैं?
[ λ_max = 296 nm (ε_max = 10700) और
λ_max = 281 nm (ε_max = 20800) ]
10 अंक
Answer approach & key points
(a(i)) explain: definition/context > points in order > small example > short close | (a(ii)) explain: definition/context > points in order > small example > short close | (b(i)) calculate: given > formula > substitution > result with units > interpretation | (b(ii)) explain: definition/context > points in order > small example > short close | (c(i)) discuss: intro > 3-4 dimensions > example > balanced close | (c(ii)) suggest: the problem in one line > implementable measures > who acts > conclusion Full marks: Precise mechanisms, correct Woodward-Fieser application, and accurate spectral assignments.
- Identify Compound II (1-acetyl-2-methylcyclohexene) as having the higher frequency
- I is fully conjugated (planar enone, ~1685 cm⁻¹)
- In II the 2-methyl group twists the acetyl out of plane (allylic 1,3-strain)
- Steric inhibition of resonance restores C=O double-bond character in II, raising ν(C=O)
- A: 1,2-dimethylcyclohexene (tetrasubstituted, Zaitsev major)
- C: 2-methyl-1-methylidenecyclohexane (exocyclic =CH₂)
- B: 1,6-dimethylcyclohexene (trisubstituted ring alkene)
- IR: no =C–H and weak C=C for A, =C–H ~3020 and oop 800–840 for B, =C–H ~3080 and strong 890 cm⁻¹ wag for C
- Compound (1): homoannular diene in ring B, base 253 nm
- (1): + 2 × 30 extended conjugation + 5 ring residues (25) + 3 exocyclic C=C (15) = 353 nm
- Compound (2): heteroannular, base 214 nm
- (2): + 30 extended conjugation + 4 ring residues (20) + 2 exocyclic C=C (10) = 274 nm
- State that 1,3,5-hexatriene has a longer conjugated system
- Explain that increased conjugation decreases the HOMO-LUMO gap
- Link the smaller energy gap to a longer wavelength (λmax)
- McLafferty rearrangement: in a carbonyl compound with a γ-hydrogen, the γ-H transfers to the carbonyl oxygen through a six-membered transition state and the β,γ bond breaks, expelling a neutral alkene and leaving an enol radical cation (or, when the charge stays with the alkene, an alkene radical cation)
- Butyl butyrate, C8H16O2, M = 144. m/z 56: butene radical cation C4H8⁺• from McLafferty transfer of a γ-H of the butyl chain (the complementary neutral is butanoic acid, 88)
- m/z 71: butanoyl acylium CH3CH2CH2C≡O⁺ from α-cleavage of the C–O bond (loss of •OC4H9, 73); m/z 73: the complementary C4H9O⁺ oxonium ion (CH2=O⁺–C3H7)
- m/z 101: [M − C₃H₇]⁺, loss of a propyl radical (⁺O≡C–OC₄H₉, or C₃H₇COOCH₂⁺ from the butyl chain)
- trans-1-Bromo-2-chloroethene, with cis isomer and 1-bromo-1-chloroethene as the other structures
- J = 16 Hz is trans vicinal coupling (cis smaller, geminal 0–3 Hz)
- A (β-ionone): 215 + 30 + 18 + 2 × 18 = 299 nm calc, assigned 296 nm
- B: 215 + 30 + 18 + 18 = 281 nm, assigned 281 nm, with lower ε of A explained by steric twisting