Paper II — Q7
7.(a) Complete the above reaction sequence by writing the structures of A, B and C. I. C₆H₅—CH = CH₂ n-BuLi A 1,3-butadiene B…
7.(a) Complete the above reaction sequence by writing the structures of A, B and C.
I. C₆H₅—CH = CH₂ n-BuLi A 1,3-butadiene B H₂O C (excess)
II. [diagram] NH₂OH A 55% H₂SO₄ B Δ C
Discuss the solvent compatibility for LiAlH₄ and NaBH₄ reagents and the factors responsible for differential reactivity. Also suggest preferred reagent between the two for the above transformations.
A molecule with molecular formula C₁₀H₁₄O exhibits a broad band at 3464 cm⁻¹ in IR spectrum. Its mass spectrum exhibits base peak at m/z 135 and the ¹H NMR spectrum exhibits the following signals: δ 1·3 (d, 6H); 2·4 (s, 3H), 3·4 (m, 1H), 4·6 (s, D₂O exchangeable), 6·6 (s, 1H), 6·8 (d, 1H) and 7·1 (d, 1H). Deduce the structure.
हिंदी में प्रश्न पढ़ें
७.(क) निम्नलिखित अभिक्रियाओं के अनुक्रमों में A, B और C की संरचना लिखकर पूर्ण करें :
I. C₆H₅—CH = CH₂ n-BuLi A 1,3-ब्यूटाडाइीन B H₂O C (excess) 1,3-butadiene
II. [diagram] NH₂OH A 55% H₂SO₄ B Δ C
(ख) LiAlH₄ और NaBH₄ अभिकर्मकों की विलायक के प्रति अनुकूलता और उनकी अवकल अभिक्रियाशीलता के उत्तरदायी कारणों की विवेचना कीजिए । निम्नलिखित रूपांतरणों के लिए इनमें से कौन सा अभिकर्मक ज्यादा अच्छा है वह भी बताएं ।
(ग) एक अणु जिसका आण्विक सूत्र C₁₀H₁₄O है वह IR स्पेक्ट्रम में एक विस्तृत बैंड 3464 सेमी⁻¹ पर दर्शाता है । इसके मास स्पेक्ट्रम में m/z 135 पर आधार शिखर और ¹H NMR स्पेक्ट्रम में निम्नलिखित सिग्नल प्रदर्शित करता है : δ 1·3 (d, 6H); 2·4 (s, 3H), 3·4 (m, 1H), 4·6 (s, D₂O विनिमय (exchangeable)), 6·6 (s, 1H), 6·8 (d, 1H) और 7·1 (d, 1H). इस अणु की संरचना करें ।
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 Scheme II: A chemical structure of cyclohexanone (a six-membered ring with a double-bonded oxygen atom) is shown. An arrow points to the right with 'NH2OH' written above. The arrow points to a label 'A'. From 'A', an arrow points to the right with '55% H2SO4' written above. This arrow points to a label 'B'. From 'B', an arrow points to the right with a delta symbol (Δ) written above. This arrow points to a label 'C'.
(b) Reaction scheme 2: The reactant is 4-pentyn-2-ol, drawn as CH3-CH(OH)-C≡C-CH3. An arrow labeled with a question mark points to the product: 4-penten-2-ol, drawn as CH3-CH(OH)-CH=CH-CH3. The stereochemistry of the double bond is shown with the hydrogen on the first alkene carbon pointing down and the methyl group on the second alkene carbon pointing down (cis/Z configuration).
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 A, B and C (20 marks)
Sequence I: living anionic polymerisation. n-BuLi adds to the CH₂ end of styrene, putting the carbanion on the benzylic carbon. Because styrene is in excess and nothing terminates the chain, the benzylic anion keeps adding styrene. A is therefore the living polystyryl-lithium, Bu–(CH₂–CHPh)ₙ–CH₂–CH(Ph)⁻ Li⁺.
When 1,3-butadiene is added, the surviving chain end adds to the diene. B is the block copolymer polystyrene-*b*-polybutadienyl-lithium (mainly 1,4-units with some 1,2-units), still a living carbanion.
Water protonates the chain end and terminates the polymer. C is the polystyrene-*b*-polybutadiene (SB) diblock copolymer.
Sequence II: oxime, lactam, polyamide. Cyclohexanone condenses with NH₂OH to give A, cyclohexanone oxime. In 55% H₂SO₄ the oxime undergoes the Beckmann rearrangement: the ring C–C bond anti to the OH migrates to nitrogen as water leaves, expanding the ring to the seven-membered lactam. B is ε-caprolactam. On heating, caprolactam undergoes ring-opening polymerisation to C, nylon-6, [–NH–(CH₂)₅–CO–]ₙ.
(b) LiAlH₄ versus NaBH₄ (15 marks)
Solvents. LiAlH₄ reacts violently with water and alcohols, releasing H₂, so it needs dry ether or THF and is quenched only at the end. NaBH₄ is far milder and works in methanol, ethanol or water.
Reactivity. Aluminium is larger and less electronegative than boron, so the Al–H bond is weaker and more hydridic. LiAlH₄ is a stronger, less selective hydride donor, and Li⁺ also activates the carbonyl as a Lewis acid. It reduces aldehydes, ketones, esters, acids, amides and nitriles. NaBH₄ under normal conditions reduces only aldehydes, ketones and acyl halides.
First transformation (p-acetylbenzamide to the secondary alcohol, –CONH₂ untouched): use NaBH₄ in ethanol. LiAlH₄ would also reduce the amide to an amine.
Second transformation (CH₃CH(OH)C≡CCH₃ to the (E)-allylic alcohol): use LiAlH₄. The alkoxide binds aluminium, and intramolecular trans-hydroalumination of the alkyne gives the E-alkene after work-up. NaBH₄ does not reduce alkynes.
(c) Structure of C₁₀H₁₄O (15 marks)
The compound is thymol, 2-isopropyl-5-methylphenol.
- Degree of unsaturation = (2×10 + 2 − 14)/2 = 4, i.e. a benzene ring.
- IR 3464 cm⁻¹ (broad): hydrogen-bonded phenolic O–H.
- MS: M = 150, and the base peak m/z 135 is M − 15, loss of CH₃ from the isopropyl group to give a benzylic cation.
- δ 1.3 (d, 6H) with δ 3.4 (m, 1H): an isopropyl group on the ring. The methine at 3.4 is deshielded by the ortho OH.
- δ 2.4 (s, 3H): Ar–CH₃.
- δ 4.6 (s, exchanges with D₂O): phenolic OH.
- δ 6.6 (s, 1H, H-6), 6.8 (d, 1H, H-4) and 7.1 (d, 1H, H-3): three aromatic H with one isolated singlet, so the ring is 1,2,4-trisubstituted (OH at C-1, isopropyl at C-2, methyl at C-5).
How this answer will be evaluated
Framework: Concept > Structure or mechanism > Reasoning > Result
What earns full marks: Correct structures and reagents with clear reasoning; accurate spectral interpretation.
Evaluation rubric
Each sub-part is marked on its own, against the marks and word limit printed on the paper.
- (a) Complete reaction sequences I and II by writing structures of A, B, and C. 20 marks
trace— start point → the stages in sequence → end point → what changed
Must cover
- Sequence I is living anionic polymerisation: n-BuLi adds to styrene and, with excess styrene, A is polystyryl-lithium, Bu–(CH2–CHPh)n–CH2–CH(Ph)⁻ Li⁺ (a living carbanion chain end)
- B: the living chain adds 1,3-butadiene to give the block copolymer polystyrene-b-polybutadienyl-lithium (1,4- and 1,2-units)
- C: water protonates the chain end, terminating the living polymer: the styrene–butadiene diblock copolymer
- Sequence II: A cyclohexanone oxime; B ε-caprolactam by Beckmann rearrangement in 55% H2SO4; C nylon-6 by ring-opening polymerisation of caprolactam on heating
Loses marks
- Treating A as a small-molecule product (an alkene or a single carbanion) and ignoring the excess styrene
- Stopping II at caprolactam: C is the polymer
- Writing a Diels–Alder adduct for B
Earns more
- Explain living character: no termination or chain transfer, block copolymer possible only because the anion survives
- Mechanism arrows for initiation and propagation; Beckmann migration of the anti group
- Note industrial relevance: SB block copolymers, nylon-6
Extra mark
- I: Mechanism of n-BuLi deprotonation
- II: Beckmann rearrangement mechanism
- (b) Discuss solvent compatibility and reactivity of LiAlH4 vs NaBH4; suggest reagents. 15 marks
discuss— intro → 3-4 dimensions → example → balanced close
Must cover
- LiAlH4: reacts violently with protic solvents, needs dry ether or THF, reduces esters, acids, amides, nitriles and, via the aluminate, propargylic alcohols; NaBH4: usable in water or alcohols, reduces only aldehydes and ketones (and acyl halides) under normal conditions
- Reason for the difference: Al–H is more hydridic (Al less electronegative, larger, weaker Al–H bond) than B–H, so LiAlH4 is the stronger and less selective donor
- First transformation (aryl ketone reduced with the primary amide untouched): NaBH4 in methanol or ethanol, because LiAlH4 would also reduce the amide to the amine
- Second transformation (propargylic alcohol to the (E)-allylic alcohol): LiAlH4 (or Red-Al) — the alkoxide directs intramolecular trans hydroalumination, giving the trans alkene after work-up; NaBH4 does not reduce alkynes
Loses marks
- Choosing LiAlH4 for the first transformation (amide would be reduced)
- Choosing NaBH4 for the alkyne (no reaction)
- Discussing solvent compatibility without giving the reason (hydride donor strength)
Earns more
- Mention Lewis-acid assistance from Li⁺ in carbonyl activation
- Contrast with Lindlar or Na/NH3 for simple alkynes
- Note chemoselectivity examples for each reagent
Extra mark
- Mechanism of hydride transfer
- Steric hindrance effects on reactivity
- (c) Deduce the structure of C10H14O from IR, MS, and 1H NMR data. 15 marks
analyse— intro → causes → effects → stakeholders/linkages → way forward
Must cover
- Identify the structure as thymol (2-isopropyl-5-methylphenol), a 1,2,4-trisubstituted phenol
- IR 3464 cm-1 (broad) is a hydrogen-bonded phenolic O-H
- MS base peak m/z 135 is M-15 (150-15), loss of CH3 from the isopropyl group
- NMR 1.3 (d, 6H) with 3.4 (m, 1H) is an isopropyl group on the ring
Loses marks
- Proposing a para-disubstituted ring, which cannot show a singlet plus two doublets for three aromatic H
- Proposing an alcohol, ketone or ether that does not fit C10H14O and the D2O-exchangeable OH
- Assigning 3.4 (m, 1H) and 1.3 (d, 6H) to anything other than an isopropyl group
Earns more
- NMR 2.4 (s, 3H) is an aryl CH3
- NMR 4.6 (s, D2O exchangeable) is the phenolic OH
- NMR 6.6 (s), 6.8 (d) and 7.1 (d) show three aromatic H, i.e. a 1,2,4-trisubstituted ring
Extra mark
- Degree of unsaturation = 4 for C10H14O
- Methine at 3.4 (ortho to OH) favours thymol over its isomer carvacrol
What "Deduce" is asking you to do
Work backwards from the evidence given — a spectrum, a reaction sequence, a set of premises — to the structure or conclusion it forces, showing which datum rules out each alternative.
Structure that answers it
Evidence listed → what each datum indicates → alternatives eliminated → the conclusion → consistency check against every datum
Where marks are lost
Naming the right answer without showing what compels it. A conclusion that fits only part of the evidence, with one band or one premise left unaccounted for, is marked as a guess.
Practice this exact question
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