Chemistry 2021 Paper II 50 marks Deduce

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

(b)

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.

(c)

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 ।

Q7 of the 2021 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2021 Chemistry paper
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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).

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.

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How this answer will be evaluated

Approach

(a) describe: define > structure or process in order > labelled diagram > significance | (b) discuss: intro > 3-4 dimensions > example > balanced close | (c) analyse: intro > causes > effects > stakeholders/linkages > way forward Full marks: All structures correct with mechanisms; reagents correctly matched with reasoning; structure deduced with full spectral assignment.

Key points expected

  • I: A is benzyllithium (C6H5-CHLi-CH3)
  • I: B is the Diels-Alder adduct (cyclohexene derivative)
  • II: A is the oxime of cyclohexanone
  • II: B is the nitrile (cyclohexanecarbonitrile)
  • LiAlH4 requires anhydrous ether solvents (e.g., Et2O)
  • NaBH4 is compatible with protic solvents (e.g., MeOH, H2O)
  • LiAlH4 is a stronger, less selective reducing agent
  • NaBH4 is a milder, more selective reducing agent

Evaluation rubric

Each sub-part is marked on its own, against the marks and word limit printed on the paper.

  1. (a) Draw structures of A, B, and C for both reaction sequences I and II. 20 marks

    describe— define → structure or process in order → labelled diagram → significance

    Must cover

    • I: A is benzyllithium (C6H5-CHLi-CH3)
    • I: B is the Diels-Alder adduct (cyclohexene derivative)
    • II: A is the oxime of cyclohexanone
    • II: B is the nitrile (cyclohexanecarbonitrile)

    Loses marks

    • Missing intermediate A or B in either sequence
    • Incorrect regiochemistry in Diels-Alder product

    Earns more

    • I: C is the alcohol product of hydration
    • II: C is the amine product of reduction
    • Correct stereochemistry shown for Diels-Alder product

    Extra mark

    • Mechanism of Beckmann rearrangement shown
  2. (b) Compare LiAlH4 and NaBH4 reactivity and select the correct reagent for the two transformations. 15 marks

    discuss— intro → 3-4 dimensions → example → balanced close

    Must cover

    • LiAlH4 requires anhydrous ether solvents (e.g., Et2O)
    • NaBH4 is compatible with protic solvents (e.g., MeOH, H2O)
    • LiAlH4 is a stronger, less selective reducing agent
    • NaBH4 is a milder, more selective reducing agent

    Loses marks

    • Suggesting LiAlH4 for the ketone reduction
    • Suggesting NaBH4 for the alkyne reduction

    Earns more

    • Identifies NaBH4 for the ketone to alcohol reduction
    • Identifies LiAlH4 for the alkyne to alkene reduction
    • Explains chemoselectivity regarding the amide group

    Extra mark

    • Mention of specific workup conditions for each reagent
  3. (c) Deduce the structure of the molecule C10H14O from the spectral data. 15 marks

    analyse— intro → causes → effects → stakeholders/linkages → way forward

    Must cover

    • Identifies the aromatic ring (1,3,5-trisubstituted pattern)
    • Identifies the isopropyl group (δ 1.3, d, 6H; δ 3.4, m, 1H)
    • Identifies the methyl group (δ 2.4, s, 3H)
    • Identifies the hydroxyl group (δ 4.6, s, D2O exchangeable)

    Loses marks

    • Incorrect substitution pattern on the benzene ring
    • Ignoring the D2O exchangeable signal

    Earns more

    • Correctly assigns the base peak at m/z 135 to loss of methyl
    • Draws the final structure: 2-isopropyl-4-methylphenol
    • Calculates degree of unsaturation correctly

    Extra mark

    • Explains the splitting pattern of the aromatic protons

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