Paper II — Q8
8.(a) (i) Write the products X/Y in the above chemical transformations. I. MCPBA → X II. OsO₄ → X NaIO₄ → Y III. OH |…
8.(a) (i) Write the products X/Y in the above chemical transformations.
I. MCPBA → X
II. OsO₄ → X NaIO₄ → Y
III. OH | H₃C—CH—COOH BH₃/THF X
IV. CrO₃-Pyridine CH₂Cl₂ X
Predict the product in the reaction of 2-methyl-1-butene with diborane. Account for the regioselectivity observed in the reaction.
A molecule with molecular formula C₉H₁₈O exhibits only one signal at δ 1·2 ppm in ¹H NMR spectrum. This also exhibits a strong absorption at 1710 cm⁻¹ in IR spectrum. Propose structure for this molecule.
The UV spectrum of acetone exhibits two signals of different intensities, one at λₘₐₓ 280 nm and the other at λₘₐₓ 190 nm. Assign corresponding electronic transitions to the observed signals.
Predict the number of signals, nature of the signals (s/d/t/m) and approximate chemical shifts in ¹H NMR spectrum of methyl propenoate.
Acetone exhibits only one carbonyl stretching frequency in IR spectrum whereas chloroacetone exhibits two at 1725 and 1745 cm⁻¹. Explain why.
हिंदी में प्रश्न पढ़ें
८.(क) (i) निम्नलिखित रासायनिक रूपांतरणों में उत्पादित यौगिकों X/Y की संरचना लिखिए :
I. MCPBA → X
II. OsO₄ → X NaIO₄ → Y
III. OH | H₃C—CH—COOH BH₃/THF X
IV. CrO₃-पिरिडीन CH₂Cl₂ X
2-मेथिल-1-ब्यूटीन और डाइबोरेन की अभिक्रिया के उत्पाद का अनुमान लगाएं। इस अभिक्रिया में जो रीजियोसेलेक्टिविटी देखी गई उसका लेखा दें।
(ख) (i) एक अणु जिसका आणविक सूत्र C₉H₁₈O है वह ¹H NMR स्पेक्ट्रम में केवल एक सिग्नल δ 1·2 ppm पर देता है। यह IR स्पेक्ट्रम में भी एक प्रबल अवशोषण 1710 से.मी.⁻¹ पर देता है। इस अणु की संरचना अनुमानित करें।
ऐसीटोन का UV स्पेक्ट्रम विभिन्न तीव्रताओं के दो सिग्नल — एक λₘₐₓ 280 nm पर और दूसरा λₘₐₓ 190 nm पर दर्शाता है। इन सिग्नलों के संगत इलेक्ट्रॉनिक संक्रमण को अंकित करें।
(ग) (i) मेथिल प्रोपीनोएट के ¹H NMR स्पेक्ट्रम में सिग्नलों की संख्या, सिग्नलों का स्वरूप (s/d/t/m) और सन्निकट रासायनिक विस्थापन का अनुमान लगाएं।
IR स्पेक्ट्रम में ऐसीटोन केवल एक कार्बोनिल तनन आवृत्ति जबकि क्लोरोऐसीटोन दो 1725 से.मी.⁻¹ और 1745 से.मी.⁻¹ पर दर्शाती है। ऐसा क्यों, समझाइए।
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 II: 1,2-dihydronaphthalene (a benzene ring fused to a six-membered ring with one double bond). The first step shows reagent OsO4 above an arrow pointing to intermediate X. The second step shows reagent NaIO4 above an arrow pointing to product Y.
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) Products X and Y (15 marks)
I. The substrate is 1-phenyl-1,2-dihydronaphthalene. MCPBA epoxidises its C3=C4 double bond by concerted syn transfer of oxygen. X is the 3,4-epoxide, an oxirane fused to the dihydronaphthalene ring, formed on the face opposite the C1 phenyl group.
II. The substrate is 1,4-dihydronaphthalene. OsO₄ adds syn to the isolated C2=C3 bond through a cyclic osmate ester, and hydrolysis gives X, the cis-1,2,3,4-tetrahydronaphthalene-2,3-diol. NaIO₄ cleaves the 1,2-diol through a cyclic periodate ester to give Y, o-phenylenediacetaldehyde, OHC–CH₂–C₆H₄–CH₂–CHO. Periodate stops at the dialdehyde.
III. The substrate is 3-hydroxybutanoic acid, CH₃CH(OH)CH₂COOH. BH₃/THF reduces a carboxylic acid faster than other carbonyl groups, through a triacyloxyborane. X is butane-1,3-diol, CH₃CH(OH)CH₂CH₂OH, and the secondary OH is untouched.
IV. CrO₃·pyridine (Collins reagent) in CH₂Cl₂ oxidises the allylic CH₂ of cyclohexene. X is cyclohex-2-en-1-one.
(a)(ii) 2-Methylbut-1-ene + diborane (5 marks)
Hydroboration followed by oxidation (H₂O₂/OH⁻) gives 2-methylbutan-1-ol, CH₃CH₂CH(CH₃)CH₂OH, the anti-Markovnikov alcohol.
Boron, the electrophilic end of B–H, bonds to the less hindered terminal CH₂ in a concerted four-centre transition state, while hydride goes to the more substituted carbon, which carries more of the developing positive charge. Both steric and electronic factors favour this orientation. Oxidation then replaces B by OH with retention.
(b)(i) C₉H₁₈O with one NMR signal and IR 1710 cm⁻¹ (10 marks)
The degree of unsaturation is 1, and the strong band at 1710 cm⁻¹ is a saturated ketone C=O, which accounts for it. One signal at δ 1.2 means all 18 H are equivalent, so there are two identical tert-butyl groups on the carbonyl.
The compound is 2,2,4,4-tetramethylpentan-3-one (di-tert-butyl ketone), (CH₃)₃C–CO–C(CH₃)₃. The singlet at δ 1.2 is the 18 equivalent methyl protons.
(b)(ii) UV spectrum of acetone (5 marks)
The weak band at 280 nm is the n→π* transition. It is symmetry-forbidden, so ε is small (about 15). The strong band at 190 nm is the π→π* transition, which is allowed and has a large ε. The non-bonding oxygen lone pair lies above the π level, so n→π* needs less energy and appears at the longer wavelength.
(c)(i) ¹H NMR of methyl propenoate, CH₂=CH–CO₂CH₃ (10 marks)
There are four signals, for the OCH₃ group and three non-equivalent vinylic protons, in the ratio 3:1:1:1.
- δ ≈ 3.7 (s, 3H): OCH₃.
- δ ≈ 6.4 (dd, 1H): the vinylic H trans to the other vinylic H and cis to the ester (J ≈ 17 and 1.5 Hz).
- δ ≈ 6.1 (dd, 1H): the α-H, =CH–CO (J ≈ 17 and 10 Hz).
- δ ≈ 5.8 (dd, 1H): the vinylic H cis to the α-H (J ≈ 10 and 1.5 Hz).
(c)(ii) Two C=O bands in chloroacetone (5 marks)
Chloroacetone exists as two rotational isomers (conformers) that are both present at room temperature and are seen separately on the IR time scale. In the cisoid form the C–Cl bond eclipses the C=O. The C–Cl and C=O dipoles repel through space (field effect), which stiffens the C=O and gives the higher band, about 1745 cm⁻¹. In the gauche form this repulsion is smaller, giving about 1725 cm⁻¹. Acetone has no such rotamers, so it shows one band.
How this answer will be evaluated
Framework: Concept > Structure or mechanism > Reasoning > Result
What earns full marks: All products/structures correct with mechanisms and full reasoning
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a)(i) Draw structures of products X/Y for the four specified transformations. 15 marks
explain— definition/context → points in order → small example → short close
Must cover
- I: MCPBA epoxidises the ring C3=C4 double bond of 1-phenyl-1,2-dihydronaphthalene: X is the 3,4-epoxide (oxirane fused to the dihydronaphthalene), formed on the face anti to the phenyl group
- II: OsO4 gives syn dihydroxylation of 1,4-dihydronaphthalene: X is the cis-2,3-diol of tetralin; NaIO4 cleaves the diol: Y is o-phenylenediacetaldehyde (2,2′-(1,2-phenylene)diacetaldehyde)
- III: BH3/THF reduces the carboxylic acid selectively: 3-hydroxybutanoic acid gives butane-1,3-diol; the secondary OH survives
- IV: CrO3·pyridine (Collins reagent) in CH2Cl2 performs allylic oxidation of cyclohexene: X is cyclohex-2-en-1-one
Loses marks
- Writing an epoxide for IV or cyclohexanone
- Writing a hydroboration product for III (no C=C present)
- Leaving Y as a dicarboxylic acid: NaIO4 stops at the dialdehyde
Earns more
- Explain the stereochemical outcome of I (anti to Ph) and II (syn diol)
- Mechanism of periodate cleavage via the cyclic iodate ester
- Why borane, unlike NaBH4, reduces acids: formation of the triacyloxyborane
Extra mark
- Mechanism arrows for one transformation
- (a)(ii) Predict product of 2-methyl-1-butene with diborane and explain regioselectivity. 5 marks
account for— state the phenomenon → the causes in order of weight → conclusion
Must cover
- Product is 2-methyl-1-butanol
- Anti-Markovnikov addition explained
- Boron attaches to less substituted carbon
- Steric and electronic factors discussed
Loses marks
- Product without mechanism
- Wrong regioselectivity explanation
Earns more
- Mechanism of hydroboration shown
Extra mark
- Comparison with Markovnikov addition
- (b)(i) Propose structure for C9H18O with one NMR signal and 1710 cm-1 IR. 10 marks
suggest— the problem in one line → implementable measures → who acts → conclusion
Must cover
- Structure is 2,2,4,4-tetramethyl-3-pentanone
- Formula C9H18O verified
- Single NMR signal explained by symmetry
- 1710 cm-1 assigned to ketone C=O
Loses marks
- Proposing C9H16O structure
- Ignoring symmetry requirement
Earns more
- Degree of unsaturation calculation
Extra mark
- Alternative isomers ruled out
- (b)(ii) Assign electronic transitions for acetone UV signals at 280 and 190 nm. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- 280 nm assigned to n→π* transition
- 190 nm assigned to π→π* transition
- Relative intensities explained
- Orbital energy levels referenced
Loses marks
- Swapping transition assignments
- No explanation of intensity difference
Earns more
- Molecular orbital diagram
Extra mark
- Comparison with other carbonyls
- (c)(i) Predict NMR signals, nature, and shifts for methyl propenoate. 10 marks
derive— given → assumptions → stepwise derivation → result → check
Must cover
- Four signals: OCH3 plus three non-equivalent vinylic H
- Splitting correct: OCH3 singlet; each vinylic H a doublet of doublets
- Chemical shifts approximate and reasonable
- Integration ratio 3:1:1:1
Loses marks
- Giving three signals (treating the vinylic H as fewer than three types)
- Incorrect splitting patterns
Earns more
- Coupling constants estimated
Extra mark
- Assignment to specific protons
- (c)(ii) Explain why chloroacetone shows two IR bands while acetone shows one. 5 marks
explain— definition/context → points in order → small example → short close
Must cover
- Chloroacetone exists as two rotational isomers (conformers) seen separately in IR
- C-Cl eclipsing C=O (cisoid) gives the higher band (about 1745) by dipolar/field effect; the gauche form gives the lower (about 1725)
- Both conformers coexist, so two bands appear
- Acetone has no such rotamers, so it shows one band
Loses marks
- Attributing the two bands to Fermi resonance or to two different carbonyl groups
- Not linking the band positions to the C-Cl/C=O dipole orientation
Earns more
- Newman or rotamer sketch of the two conformers
Extra mark
- Comparison with other halogenated ketones
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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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