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Chemistry 2021 Paper II 50 marks Explain

Paper II — Q6

(a)(i) The molecule obtained on treatment of acetone with dilute sodium hydroxide exhibits the following spectral data. Propose…

(a)
(i)

The molecule obtained on treatment of acetone with dilute sodium hydroxide exhibits the following spectral data. Propose the structure of this molecule. IR : 1620 cm⁻¹ and 1695 cm⁻¹ ¹H NMR : δ 1·9(s, 3H), 2·1(s, 6H), 6·15(s, 1H) 15 marks

(ii)

Identify the compound in each of the following pairs, that can be expected to exhibit carbonyl stretching signal at higher frequency: I. [two structures] II. CH₃—C(=O)—O—C₂H₅ and C₆H₅—C(=O)—O—C₂H₅ 5 marks

(b)

Show salt bridge, hydrogen bond, van der Waals' interaction and disulfide bridge for stabilization of protein by choosing appropriate amino acid residues in the protein chain. 15 marks

(c)

Complete the following reactions by giving the suitable mechanisms: I. [diagram] hv ? (5 marks) II. [diagram] hv ? 10 marks

हिंदी में प्रश्न पढ़ें
(a)
(i)

ऐसीटोन और तनु सोडियम हाइड्रोक्साइड की अभिक्रिया में बनने वाला अणु निम्नलिखित स्पेक्ट्रमी आँकड़ा दर्शाता है। इस अणु की संरचना लिखें। IR : 1620 cm⁻¹ and 1695 cm⁻¹ ¹H NMR : δ 1·9(s, 3H), 2·1(s, 6H), 6·15(s, 1H) (15 अंक)

(ii)

निम्नलिखित प्रत्येक युगलों में उस यौगिक की पहचान करें जो उच्चतर आवृत्ति का कार्बोनिल तनन सिग्नल दिखाता है : I. [दो संरचनाएँ] II. CH₃—C—O—C₂H₅ और C₆H₅—C—O—C₂H₅ ‖ ‖ O O (5 अंक)

(b)

उपयुक्त एमीनो अम्ल का चयन करते हुए लवण सेतु, हाइड्रोजन आबंध, वान्डरवाल्स अन्योन्यक्रिया और डाइसल्फाइड सेतु द्वारा प्रोटीन के स्थायीकरण को दर्शाइए। (15 अंक)

(c)

निम्नलिखित अभिक्रियाओं को उनकी क्रियाविधि के साथ पूर्ण करें : I. [आरेख] hv ? (5 अंक) II. [आरेख] hv ? (10 अंक)

Q6 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) Pair I: Two six-membered cyclic ester (lactone) structures. The first structure is a six-membered ring containing one oxygen atom and one carbonyl group (C=O) adjacent to the ring oxygen, with a double bond between the two carbons opposite the oxygen (delta-valerolactone). The second structure is a six-membered ring containing one oxygen atom and one carbonyl group (C=O) adjacent to the ring oxygen, with a double bond between the carbon adjacent to the carbonyl and the next carbon (alpha,beta-unsaturated lactone). Pair II: Two ester structures. The first is CH3-C(=O)-O-C2H5 (ethyl acetate). The second is C6H5-C(=O)-O-C2H5 (ethyl benzoate).

(c) Reaction II: A ketone molecule with a central carbonyl group (C=O). On the left side of the carbonyl, there is a phenyl group (Ph) attached to a methylene group (CH2). On the right side of the carbonyl, there is a carbon atom bonded to two phenyl groups (Ph) and one hydrogen atom (implied). The reaction arrow points to the right with 'hv' (light) written above it, leading to a question mark.

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) Acetone + dilute NaOH: structure from spectra (15 marks)

Dilute NaOH converts acetone to diacetone alcohol (aldol addition), which dehydrates by E1cB to mesityl oxide, 4-methylpent-3-en-2-one, (CH₃)₂C=CH–CO–CH₃ (C₆H₁₀O).

Mechanism: the enolate of acetone adds to a second acetone C=O; the β-hydroxy ketone, (CH₃)₂C(OH)CH₂COCH₃, loses the acidic α-H to give an enolate, and OH⁻ is expelled to give the conjugated enone.

IR: 1695 cm⁻¹ is the conjugated ketone C=O (a saturated ketone absorbs near 1715 cm⁻¹; conjugation lowers it). 1620 cm⁻¹ is the conjugated C=C. The absence of a broad O–H band rules out diacetone alcohol.

¹H NMR (10 H):

  • δ 6.15 (s, 1H): the vinylic =CH–, a singlet because there is no vicinal H.
  • δ 1.9 (s, 3H): one of the two allylic (vinylic) CH₃ groups on the β-carbon.
  • δ 2.1 (s, 6H): the acetyl CH₃ (next to C=O) and the other vinylic CH₃, which fall at nearly the same shift and are integrated together.

(a)(ii) Higher C=O stretching frequency (5 marks)

Pair I: the enol lactone, in which the ring oxygen carries the C=C (O–CH=CH), absorbs higher (about 1760 cm⁻¹) than the lactone with a β,γ-double bond (about 1735 cm⁻¹). The ring-oxygen lone pair is delocalised into the C=C, so it donates less to the carbonyl and the C=O bond keeps more double-bond character.

Pair II: ethyl acetate (about 1740 cm⁻¹) absorbs higher than ethyl benzoate (about 1720 cm⁻¹), because conjugation of the C=O with the phenyl ring lowers its bond order.

(b) Stabilising interactions in a protein (15 marks)

  • Salt bridge: the Lys side-chain –NH₃⁺ and the Asp (or Glu) –COO⁻ form an ion pair.
  • Hydrogen bond: the Ser or Thr side-chain –OH donates to a backbone C=O (as does the backbone N–H···O=C pairing of an α-helix or β-sheet).
  • van der Waals (hydrophobic) packing: Leu, Ile, Val or Phe side chains pack together in the core.
  • Disulfide bridge: two Cys residues are oxidised to Cys–S–S–Cys, a covalent cross-link.
Diagram: Interactions that stabilise a folded protein

(c)(i) 2-Cyclopropylcyclohexanone + hv (5 marks)

This is a Norrish Type I α-cleavage. The n→π* excited ketone breaks the C1–C2 bond to give an acyl radical and a cyclopropylcarbinyl radical. The cyclopropylcarbinyl radical opens rapidly (a radical-clock ring opening) to a homoallylic radical, CH₂=CH–CH₂–CH₂•. The biradical then ends by H-transfer, giving an unsaturated aldehyde (a nona-dienal), or by recombination, giving a ring-expanded unsaturated ketone. The aldehyde is the expected product.

(c)(ii) PhCH₂–CO–CHPh₂ + hv (10 marks)

This is also Norrish Type I (there is no γ-H for Type II). The α-cleavage occurs on the side that gives the more stable radical, Ph₂CH• (diphenylmethyl) plus PhCH₂–C•=O (phenylacetyl). The phenylacetyl radical loses CO to give PhCH₂• (benzyl).

Recombination of the cage radicals gives 1,1,2-triphenylethane, Ph₂CH–CH₂Ph, as the cross-product. Radical pairs that escape the cage also give bibenzyl and 1,1,2,2-tetraphenylethane.

How this answer will be evaluated

What earns full marks: Correct structures, complete mechanisms, precise spectral assignments, clear diagrams

Evaluation rubric

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

  1. (a)(i) Propose structure of acetone dimer and justify with spectral data. 15 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Identify product as mesityl oxide (4-methylpent-3-en-2-one)
    • Assign IR 1695 cm⁻¹ to C=O and 1620 cm⁻¹ to C=C
    • Assign NMR 1.9 (s, 3H) to one vinylic (allylic) CH3
    • Assign NMR 2.1 (s, 6H) to the acetyl CH3 plus the other vinylic CH3

    Loses marks

    • Proposing mesityl oxide without spectral justification
    • Incorrect assignment of NMR chemical shifts

    Earns more

    • Assign NMR 6.15(s, 1H) to vinylic proton
    • Mention aldol condensation mechanism

    Extra mark

    • Draw the aldol condensation mechanism
  2. (a)(ii) Identify higher frequency carbonyl in each pair with reasoning. 5 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Pair I: the enol lactone (O-CH=CH in the ring) has the higher C=O frequency than the beta,gamma-unsaturated lactone
    • Pair II: ethyl acetate has the higher C=O frequency than ethyl benzoate
    • Explain conjugation lowers C=O stretching frequency
    • Explain that phenyl conjugation lowers the ester C=O bond order

    Loses marks

    • Selecting the beta,gamma-unsaturated lactone as higher in pair I
    • Selecting ethyl benzoate in pair II

    Earns more

    • Mention inductive effect of alkyl groups

    Extra mark

    • Cite specific frequency ranges for each
  3. (b) Show four types of protein stabilization interactions with residues. 15 marks

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

    Must cover

    • Show salt bridge (e.g., Lys-Asp)
    • Show hydrogen bond (e.g., Ser-Gly)
    • Show van der Waals interaction (e.g., Leu-Ile)
    • Show disulfide bridge (Cys-Cys)

    Loses marks

    • Missing any of the four interaction types
    • Using incorrect amino acid pairs

    Earns more

    • Label specific amino acid residues
    • Draw clear structural diagrams

    Extra mark

    • Mention role in tertiary structure
  4. (c)(i) Complete photochemical reaction of 2-cyclopropylcyclohexanone with mechanism. 5 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Identify the reaction as Norrish Type I alpha-cleavage of the C1-C2 bond
    • Show opening of the cyclopropylcarbinyl radical to a homoallylic radical
    • Give an unsaturated aldehyde or ring-expanded ketone as the end product
    • Show the radical steps

    Loses marks

    • Giving bicyclo[4.3.0]nonan-2-one by a non-Norrish path
    • Missing the ring expansion product

    Earns more

    • Mention the cyclopropylcarbinyl radical-clock ring opening

    Extra mark

    • Draw intermediate biradical
  5. (c)(ii) Complete the photochemical reaction of PhCH2COCHPh2 with mechanism. 10 marks

    justify— claim → 3-4 reasons → evidence → conclusion

    Must cover

    • Identify Norrish Type I alpha-cleavage
    • Show acyl-alkyl bond cleavage
    • Draw the Ph2CH and PhCH2CO (phenylacetyl) radicals, with loss of CO to give benzyl
    • Show radical recombination to 1,1,2-triphenylethane

    Loses marks

    • Identifying as Norrish Type II
    • Missing radical intermediates

    Earns more

    • Mention Norrish Type I mechanism
    • Show radical intermediates clearly

    Extra mark

    • Mention possible side products
What "Explain" is asking you to do

Make the working of something clear — what sets it off, what follows from what, and what it produces. Explain is the Commission's mechanism word: it dominates the technical papers and the “explain why” stems, where the marks sit in the causal chain and not in the label.

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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