Q5 50M Compulsory explain Organic spectroscopy and reaction mechanisms
(a) How many signals would you expect in the ¹H NMR spectrum of above compounds ? Mark these protons. 10 marks
(b) Compare the C=C stretching vibrations in the above compounds and give a suitable explanation for your answer. 10 marks
(c) (i) Identify A and B. The polymerisation of the rearranged product of B, unaided by any catalyst, gives rise to a 'synthetic rubber'. Name this rubber along with its structural formula.
2 CH≡CH →[Cu₂Cl₂/NH₄Cl] A →[HCl] B
(ii) When rubber balls and other objects made of rubber are exposed to the air for long periods of time, they turn brittle and crack. This does not happen to objects made of polyethylene. Explain. 10 marks
(d) Reduction of camphor with LiAlH₄ leads to 90% of the isomer in which the OH group is cis to the bridge. Give a suitable explanation of this observation. 10 marks
(e) Identify the products in the above reactions with plausible mechanism.
(i) CH₃—C(=O)—CH₃ →[hν/Vapour phase]
(ii) CH₃—C(=O)—CH₃ →[hν/Room temperature] 10 marks
हिंदी में पढ़ें
(a) आप निम्न यौगिकों के ¹H NMR स्पेक्ट्रम में कितने शीर्ष (सिग्नल) का अनुमान लगाते हैं ? इन प्रोटोनों को चिह्नित कीजिए । 10
(b) निम्न यौगिकों में C=C तनन आवृत्ति की तुलना कीजिए तथा अपने उत्तर का औचित्य सिद्ध करने के लिए उपयुक्त स्पष्टीकरण दीजिए । 10
(c) (i) निम्नलिखित अभिक्रिया अनुक्रम में A तथा B यौगिकों को पहचानिए । किसी भी उत्प्रेरक की सहायता के बिना B के पुनर्व्यवस्थित उत्पाद के बहुलकन से एक सिंथेटिक रबड़ बनता है । इस रबड़ का संरचनात्मक सूत्र सहित नाम लिखिए ।
2 CH≡CH →[Cu₂Cl₂/NH₄Cl] A →[HCl] B
(ii) जब रबड़ की गेंदों और रबड़ से बनी अन्य वस्तुओं को ज्यादा समय के लिए हवा के संपर्क में रखा जाता है, तो वे भंगुर हो जाती हैं और टूट जाती हैं । पॉलीथीन से बनी वस्तुओं के साथ ऐसा नहीं होता है । व्याख्या कीजिए । 10
(d) कपूर (camphor) को LiAlH₄ द्वारा अपचयन करने पर 90% मात्रा में एक ऐसा समावयव मिलता है जिसमें हाइड्रोक्सल (OH) सेतु के समीप है। इस अवलोकन के लिए एक उपयुक्त स्पष्टीकरण दें। 10
(e) निम्न अभिक्रियाओं में उत्पादों की पहचान कीजिए व क्रियाविधि लिखिए।
(i) CH₃—C(=O)—CH₃ →[hν/Vapour phase]
(ii) CH₃—C(=O)—CH₃ →[hν/Room temperature] 10
Answer approach & key points
(a) map: locate accurately > label > one line on why it matters | (b) compare: paired headings or table > key differences > significance > conclusion | (c) explain: definition/context > points in order > small example > short close | (d) explain: definition/context > points in order > small example > short close | (e) trace: start point > the stages in sequence > end point > what changed Full marks: All parts fully answered with correct structures, mechanisms, and reasoning; no errors in nomenclature or stereochemistry.
- A has 7 signals (2 aromatic, COCH3, OCH2, =CH, two =CH2)
- Para ring symmetry makes ortho pairs equivalent
- B cis: 4 signals; trans: 3 signals
- Diastereotopic CH2 protons in cis
- Methylenecyclopropane higher (~1780 cm-1)
- Methylenecyclobutane lower (~1680 cm-1)
- Higher s-character of exocyclic bond from ring strain
- Frequency follows force constant, ν ∝ √(k/μ)
- A vinylacetylene
- B 4-chlorobuta-1,2-diene rearranging to chloroprene; neoprene
- Rubber has C=C and allylic C–H; autoxidation/ozone
- Polyethylene saturated, no allylic C–H
- Rigid bicyclic ketone
- Exo face blocked by syn C-7 methyl
- Hydride attacks endo face
- OH ends exo (cis to bridge): isoborneol major
- Norrish Type I α-cleavage to CH3• and CH3CO•
- Vapour phase: CO + ethane (minor methane)
- Room temperature: acetyl survives, biacetyl (+ ethane)
- Pinacol via H-abstraction/ketyl coupling in liquid
Q6 50M solve Vibrational energy, IR spectra, organic synthesis, NMR
(a) (i) The frequencies of vibration of the following molecules in their v = 0 states are HCl : 2885 cm⁻¹; D₂ : 2990 cm⁻¹; DCl : 1990 cm⁻¹ and HD : 3627 cm⁻¹. Calculate the energy change of the following reaction : HCl + D₂ → DCl + HD Determine whether energy is liberated or absorbed. [Given : h = 6.626 × 10⁻³⁴ Js Veloci c = 2.998 × 10⁸ ms⁻¹ N_A = 6.022 × 10²³ mol⁻¹] (10 marks)
(a) (ii) The IR spectra of butyric acid and ethyl butyrate show sharp strong singlet absorption at 1725 cm⁻¹ and 1740 cm⁻¹, respectively. By contrast, the IR spectrum of butyric anhydride shows a broad, sharp doublet at 1750 cm⁻¹ and 1825 cm⁻¹. Why are these so different? (5 marks)
(b) (i) Write the structure of product(s) in the above reactions : [(A) methyl 4-oxocyclohexane-1-carboxylate, (i) NaBH4, H2O, OH⁻ (ii) H2O, H⁺; (B) estrone, (i) HC≡CH, BuLi (ii) H⁺, H2O; (C) anisole, Na, NH3(l), Et2O, EtOH] (10 marks)
(b) (ii) What is meant by 'Tacticity' of a polymer ? Distinguish among isotactic, syndiotactic and atactic polymers. (5 marks)
(c) (i) An organic compound having molecular formula C₁₆H₂₅NO gave following IR and ¹H NMR data : IR(cm⁻¹) = 1690 ¹H NMR(CDCl₃, 400 MHz) : δ 1.11(t, J = 7Hz, 6H), 1.29(d, J = 7Hz 6H), 2.40 (g, J = 7Hz, 4H), 2.55(t, J = 7Hz, 2H), 2.65(t, J = 7 Hz, 2H), 3.12(septet, 1H), 7.21(d, J = 8Hz, 2H), 7.81(d, J = 8Hz, 2H) Determine the structure of the compound. (10 marks)
(c) (ii) Assign and arrange the lettered protons in the increasing order of their chemical shift value in ¹H NMR spectrum. [4H-oxepine ring: Ha and Hb on the O–CH=CH unit, He on the other O–CH=, Hc the CHO proton, Hd the COCH3 protons] (10 marks)
हिंदी में पढ़ें
(a) (i) निम्न यौगिकों की ऊर्जा v = 0 अवस्था में कंपनिक आवृत्ति इस प्रकार है : HCl : 2885 cm⁻¹; D₂ : 2990 cm⁻¹; DCl : 1990 cm⁻¹ व HD : 3627 cm⁻¹. निम्न अभिक्रिया के लिए ऊर्जा परिवर्तन की गणना करें । HCl + D₂ → DCl + HD निर्धारित करें कि इस अभिक्रिया में ऊर्जा मुक्त होगी या अवशोषित ? [Given : h = 6.626 × 10⁻³⁴ Js Veloci c = 2.998 × 10⁸ ms⁻¹ N_A = 6.022 × 10²³ mol⁻¹] (10 अंक)
(a) (ii) ब्यूटिरिक एसिड व एथिल ब्यूटिरेट अवरक्त (IR) स्पेक्ट्रम में क्रमश: 1725 cm⁻¹ व 1740 cm⁻¹ पर एक प्रबल अवशोषण दिखाते हैं । इसके विपरीत, ब्यूटिरिक एनहाइड्राइड अवरक्त (IR) स्पेक्ट्रम में 1750 cm⁻¹ व 1825 cm⁻¹ पर द्विक अवशोषण दिखाता है । ये इतने अलग क्यों हैं ? (5 अंक)
(b) (i) निम्नलिखित अभिक्रियाओं में उत्पाद(ओं) की संरचना लिखिए : [(A) मेथिल 4-ऑक्सोसाइक्लोहेक्सेन-1-कार्बोक्सिलेट, (i) NaBH4, H2O, OH⁻ (ii) H2O, H⁺; (B) एस्ट्रोन, (i) HC≡CH, BuLi (ii) H⁺, H2O; (C) ऐनिसोल, Na, NH3(द्रव), Et2O, EtOH] (10 अंक)
(b) (ii) बहुलक की टैक्टिसिटी (व्यवस्था) से क्या अभिप्राय है ? समव्यवस्थ (आइसोटैक्टिक), एकान्त व्यवस्थ (सिंडियोटैक्टिक) तथा अव्यवस्थ (एटैक्टिक) बहुलकों के बीच अंतर बताएं । (5 अंक)
(c) (i) एक कार्बनिक यौगिक जिसका आणविक सूत्र C₁₆H₂₅NO है वह निम्नलिखित IR व ¹H NMR आंकड़ा देता है : IR(cm⁻¹) = 1690 ¹H NMR(CDCl₃, 400 MHz) : δ 1.11(t, J = 7Hz, 6H), 1.29(d, J = 7Hz 6H), 2.40 (g, J = 7Hz, 4H), 2.55(t, J = 7Hz, 2H), 2.65(t, J = 7 Hz, 2H), 3.12(septet, 1H), 7.21(d, J = 8Hz, 2H), 7.81(d, J = 8Hz, 2H) यौगिक की संरचना निर्धारित कीजिए । (10 अंक)
(c) (ii) अक्षरों से चिह्नित प्रोटोनों को उनके ¹H NMR स्पेक्ट्रम में रासायनिक सुति मानों के अनुसार आरोही क्रम में लिखें । [4H-ऑक्सेपीन वलय: Ha और Hb O–CH=CH इकाई पर, He दूसरे O–CH= पर, Hc CHO प्रोटॉन, Hd COCH3 प्रोटॉन] (10 अंक)
Answer approach & key points
(a(i)) calculate: given > formula > substitution > result with units > interpretation | (a(ii)) explain: definition/context > points in order > small example > short close | (b(i)) map: locate accurately > label > one line on why it matters | (b(ii)) define: precise definition > the distinguishing feature > one example | (c(i)) map: locate accurately > label > one line on why it matters | (c(ii)) map: locate accurately > label > one line on why it matters Full marks: All parts fully answered with correct reasoning, mechanisms, and structures. Numerical work shown completely. NMR assignments justified with electronic effects.
- E = ½hcν̃ (zero-point energy)
- Δν̃ = (1990+3627) − (2885+2990) = −258 cm-1
- N_A h c = 11.96 J mol-1 cm; ΔE = −1.54 kJ mol-1
- Negative, so energy liberated
- Acid and ester: one C=O, single band
- Acid lower due to hydrogen-bonded dimer
- Anhydride: two coupled C=O, symmetric and asymmetric stretches
- Shared O lone pair gives higher frequency than ester
- A methyl 4-hydroxycyclohexanecarboxylate (ester intact)
- B 17α-ethynylestradiol (acetylide addition at C-17)
- C 1-methoxycyclohexa-1,4-diene (Birch)
- Excess BuLi for the phenol / OMe stays on sp2 carbon
- Tacticity: regular stereochemical arrangement of substituents
- Isotactic: same side
- Syndiotactic: alternating sides
- Atactic: random; crystalline vs amorphous
- 5 degrees of unsaturation; IR 1690 aryl ketone
- Para-disubstituted ring (7.21, 7.81 d)
- Isopropyl (1.29 d, 3.12 septet) and NEt2 (1.11 t, 2.40 q)
- CH2CH2 linker: 3-(diethylamino)-1-(4-isopropylphenyl)propan-1-one
- Hd acetyl CH3 lowest (~2.2)
- Hb enol-ether β-H (~5)
- Ha (~6.3) then He (~7.5)
- Hc aldehyde highest (~9.7); order Hd<Hb<Ha<He<Hc
Q7 50M draw Nucleic acids, photochemistry and organic spectroscopy
(a) (i) Draw the structure of 2'-deoxycytidine-3'-monophosphate.
(ii) Why nucleotides and nucleic acids are acidic ? Mention the factor(s) that stabilize the DNA duplex.
(iii) Show the hydrogen bonding in between cytosine and guanine base pair. 15
(b)(i) Identify all the possible products in the above reaction showing the sequential pathways of their formation : [cyclopentanone, hν, vapour phase] 10
(b)(ii) 1,3-Butadiene solution on irradiation with UV light follows path I. However, irradiating 1,3-butadiene in the presence of benzophenone at 366 nm follows path II. Give a suitable explanation. 5
(c)(i) Write the structure of the compounds (A), (B), (C), (D) and (E). [(i) 2-methylpentan-3-ol, CS2/NaOH then MeI gives (A); SeO2 gives (B); (ii) a benzocycloheptene diene with two OMe groups and an isopropyl group, mCPBA, CH2Cl2, 0 °C gives (C); (iii) 2,2,3,3-tetramethylsuccinic acid, Pb(OAc)4, O2 gives (D); (iv) (R)-3-methylcyclohexene, CrO3-pyridine gives (E)] 10
(c)(ii) From the reaction sequence below, answer the following questions : [cyclohexene, OsO4/pyridine gives (A); (B) gives cis-cyclohexane-1,2-diol]
(I) Write structure of (A) and (B)
(II) Write the oxidation state of Osmium in the intermediate (A) and its geometry. 5
(c)(iii) Write the structure of (A) and (B) in the above reactions. [D-mannitol (2R,3R,4R,5R), cyclohexanone, BF3·Et2O, (MeO)3CH gives (A); HIO4 gives (B)] 5
हिंदी में पढ़ें
(क) (i) 2'-डिऑक्सीसाइटिडिन-3'-मोनोफॉस्फेट की संरचना लिखिए।
(ii) न्यूक्लिओटाइड तथा न्यूक्लिक अम्ल अम्लीय क्यों होते हैं ? DNA डबल हेलिक्स संरचना को स्थिर करने वाले कारकों का उल्लेख कीजिए।
(iii) साइटोसीन व गुआनिन शार युग्म के मध्य हाइड्रोजन आबंधन को दिखाइए। 15
(ख)(i) निम्नलिखित अभिक्रिया में सभी संभावित उत्पादों की पहचान करें जो उनके गठन का अनुक्रमिक मार्ग दिखाते हैं। 10
(ख)(ii) 1,3-ब्यूटाडाइन विलयन पराबैंगनी किरण पर पथ I का अनुसरण करता है। हालांकि बेंजोफेनॉन की उपस्थिति में 366 nm पर 1,3-ब्यूटाडाइन का पराबैंगनी किरण पथ II का अनुसरण करता है। उपयुक्त स्पष्टीकरण दीजिए। 5
(ग)(i) यौगिकों (A), (B), (C), (D) तथा (E) की संरचना लिखिए। 10
(ग)(ii) नीचे दिए गए अभिक्रिया क्रम से, निम्नलिखित प्रश्नों का उत्तर दें :
(I) (A) तथा (B) की संरचना लिखिए ।
(II) मध्यवर्ती (A) में ऑस्मियम की ऑक्सीकरण अवस्था तथा ज्यामिति लिखिए । 5
(ग)(iii) निम्नलिखित अभिक्रियाओं में (A) तथा (B) की संरचना लिखिए । 5
Answer approach & key points
(a) describe: define > structure or process in order > labelled diagram > significance | (b(i)) trace: start point > the stages in sequence > end point > what changed | (b(ii)) explain: definition/context > points in order > small example > short close | (c(i)) describe: define > structure or process in order > labelled diagram > significance | (c(ii)) describe: define > structure or process in order > labelled diagram > significance | (c(iii)) describe: define > structure or process in order > labelled diagram > significance Full marks: All structures correct with mechanisms, stereochemistry, and named reactions identified.
- 2′-deoxycytidine-3′-monophosphate: 2′-deoxyribose, cytosine at C-1′, phosphate at 3′-O
- Acidity from phosphate groups (polyanion at pH 7)
- Duplex stability: H-bonds, base stacking, hydrophobic effect, counter-ions
- C–G: three hydrogen bonds (N4–H···O6, N1–H···N3, N2–H···O2)
- Norrish Type I α-cleavage to acyl–alkyl biradical
- Pent-4-enal by H-transfer
- Decarbonylation to CO and tetramethylene biradical
- Cyclobutane and two ethylene
- Direct: singlet, electrocyclisation
- Products cyclobutene and bicyclobutane
- Benzophenone absorbs at 366 nm; triplet sensitisation
- Triplet adds stepwise: 1,2-divinylcyclobutane and 4-vinylcyclohexene
- A S-methyl xanthate of 2-methylpentan-3-ol
- B from Chugaev then SeO2: allylic alcohol/4-methylpent-2-ene route
- C epoxide at the tetrasubstituted C=C; D 2,3-dimethylbut-2-ene
- E (R)-4-methylcyclohex-2-en-1-one
- A cyclic osmate(VI) ester
- B aqueous NaHSO3 (or H2S) giving cis-diol
- Os oxidation state +6
- Geometry: tetrahedral ester; pyridine adduct pseudo-octahedral
- Hexitol is D-mannitol
- A 1,2:5,6-di-O-cyclohexylidene-D-mannitol
- HIO4 cleaves the free C3–C4 diol
- B two molecules of 2,3-O-cyclohexylidene-D-glyceraldehyde
Q8 50M solve Spectroscopy: mass, UV, rotational, NMR and IR
(a)(i) For the following compound :
(i) Identify the site of initial ionization under EI conditions.
(ii) Draw the structure of ion having m/z = 58 value.
(iii) Calculate the m/z value of metastable ion formed during the formation of above ion. 10
(a)(ii) Draw the structural formula of a ketone with MF C₇H₁₂O that absorbs in the UV with λₘₐₓ = 249 nm. 5
(b)(i) (A) Which of the above molecule(s) will show a microwave rotational spectrum ? Justify your answer.
H₂, HCl, BrF, CH₄ and H₂O
(B) The first rotational absorption of ¹²C¹⁶O was observed at 3·84235 cm⁻¹ while that of ¹³C¹⁶O was observed at 3·67337 cm⁻¹. Find their rotational constants. 10
(b)(ii) Estimate the expected splitting (J in Hz) for the lettered protons in the above compounds : 5
(c)(i) In the ¹H NMR spectrum of acetyl acetone the height of the integration curve at methylene (–CH₂) signal was found to be 10 mm and that of methine (=CH) signal was 22 mm. Calculate the percentage of keto and enol form in the sample. 10
(c)(ii) How could the above pair of isomeric compounds be differentiated by mass spectrometry ? 5
(c)(iii) How many fundamental vibrational frequencies would be expected to be observed in the IR spectrum of CO₂ ? Sketch them. Identify the IR-active vibrations with justification. 5
हिंदी में पढ़ें
(क)(i) निम्नलिखित यौगिक में :
(i) ई आई (EI) अवस्था के तहत प्रारंभिक आयनन स्थल की पहचान करें ।
(ii) m/z = 58 मान वाले आयन की संरचना बनाइए ।
(iii) उपरोक्त आयन के निर्माण के दौरान बनने वाले (metastable) मितस्थायी आयन के m/z मान की गणना करें । 10
(क)(ii) आणविक सूत्र C₇H₁₂O वाले एक कीटोन का संरचनात्मक सूत्र बनाइए जो पराबैंगनी प्रकाश में λₘₐₓ = 249 nm पर अवशोषित करता है । 5
(ख)(i) (A) निम्नलिखित में से कौन सा अणु सूक्ष्मतरंग घूर्णनात्मक स्पेक्ट्रम दिखाएगा ? अपने उत्तर का औचित्य दीजिए ।
H₂, HCl, BrF, CH₄ and H₂O
(B) ¹²C¹⁶O का प्रथम घूर्णनात्मक अवशोषण 3·84235 cm⁻¹ पर देखा गया जबकि ¹³C¹⁶O का 3·67337 cm⁻¹ पर दिखा । उनका घूर्णनात्मक स्थिरांक ज्ञात कीजिए । 10
(ख)(ii) निम्नलिखित यौगिकों में अक्षरित प्रोटोनों के लिए अपेक्षित विपाटन (J in Hz) का अनुमान लगाइए । 5
(ग)(i) एसिटाइल एसीटोन के ¹H NMR स्पेक्ट्रम की समाकलन वक्र रेखा में मेथिलीन (–CH₂) सिग्नल की ऊँचाई 10 mm और मेथाइन (=CH) सिग्नल की ऊँचाई 22 mm पाई गई है । प्रतिदर्श में कीटो और एनोल रूप के प्रतिशत की गणना करें । 10
(ग)(ii) मास स्पेक्ट्रोमेट्री द्वारा समावयवी यौगिकों की निम्नलिखित जोड़ी को कैसे विभेदित किया जा सकता है ? 5
(ग)(iii) CO₂ के अवरक्त स्पेक्ट्रम में कितनी मौलिक कंपन आवृत्तियों की अपेक्षा की जा सकती है ? उनका आरेख (स्केच) बनाइए । ओचित्य के साथ अवरक्त सक्रिय कंपनों की पहचान करें । 5
Answer approach & key points
(a(i)) map: locate accurately > label > one line on why it matters | (a(ii)) describe: define > structure or process in order > labelled diagram > significance | (b(i)) justify: claim > 3-4 reasons > evidence > conclusion | (b(ii)) calculate: given > formula > substitution > result with units > interpretation | (c(i)) calculate: given > formula > substitution > result with units > interpretation | (c(ii)) compare: paired headings or table > key differences > significance > conclusion | (c(iii)) describe: define > structure or process in order > labelled diagram > significance Full marks: All parts answered with correct structures, mechanisms, and calculations; full working shown
- Initial ionisation at the nitrogen lone pair
- m/z 58 is CH2=N+(CH3)2 by α-cleavage
- Metastable ion m* = 58²/279
- m* = 12.06
- Degrees of unsaturation 2: C=O + C=C, acyclic
- Woodward–Fieser: 215 + 10 (α) + 24 (2β)
- 3,4-dimethylpent-3-en-2-one
- Permanent dipole needed
- HCl, BrF, H2O active; H2, CH4 inactive
- B = ν̃/2 for the J=0→1 line
- B(12CO) = 1.92118 and B(13CO) = 1.83669 cm-1
- A: equivalent geminal protons, no splitting observed
- B: geminal =CH2, J about 0–3 Hz
- C: vicinal cis ring alkene, J about 10 Hz
- Reason from H–C–H angle and dihedral
- Integration proportional to number of protons
- Keto 10/2 = 5; enol 22/1 = 22
- Total 27
- Enol 81.5%; keto 18.5%
- M+• = 101 for both
- McLafferty ion m/z 73 only in N-methylbutanamide
- α-cleavage ions: I 72 and 57; II 58 and 71
- Conclusion distinguishing I and II
- Linear: 3N−5 = 4 modes, 3 distinct frequencies
- Symmetric stretch (~1340) IR inactive
- Antisymmetric stretch (~2349) IR active
- Bend (~667, degenerate) IR active