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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

Written 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.

Structure Elucidation of Acetone Derivative Treatment of acetone with dilute NaOH yields mesityl oxide (4-methylpent-3-en-2-one) via an aldol condensation. The IR spectrum confirms this structure: the peak at 1695 cm⁻¹ corresponds to the C=O stretch of a conjugated ketone (lowered from ~1715 cm⁻¹ due to resonance), while 1620 cm⁻¹ indicates the C=C stretch. The ¹H NMR data provides definitive evidence for the regiochemistry. The signal at δ 6.15 (s, 1H) is the vinylic proton. The methyl groups are non-equivalent: the acetyl methyl (CH₃CO-) appears as a singlet at δ 2.1 (3H), while the vinylic methyls show distinct chemical shifts due to their different positions relative to the carbonyl and the vinylic proton. The methyl group cis to the vinylic proton (C4-Me) is shielded, appearing at δ 1.9 (s, 3H), whereas the trans methyl (C5-Me) is slightly deshielded, appearing at δ 2.1 (s, 3H). The reported "2.1 (s, 6H)" in the prompt likely represents the overlap of the acetyl methyl and the C5-methyl, or a simplification, but the distinct δ 1.9 signal confirms the specific substitution pattern of mesityl oxide.

Comparative Carbonyl Frequencies In pair (a)(ii) I, the α,β-unsaturated lactone exhibits a higher carbonyl stretching frequency than the δ-valerolactone. Conjugation of the C=O with the C=C double bond reduces the bond order of the carbonyl, lowering its stretching frequency. The saturated lactone lacks this conjugation, retaining a higher C=O bond order and thus a higher ν_C=O. In pair II, ethyl acetate (CH₃COOC₂H₅) shows a higher frequency than ethyl benzoate (C₆H₅COOC₂H₅). The phenyl ring in ethyl benzoate conjugates with the carbonyl group, delocalizing the π-electrons and weakening the C=O bond, which lowers the stretching frequency compared to the aliphatic ester.

Protein Stabilization Interactions Protein tertiary structure is stabilized by specific inter-residue interactions. A salt bridge forms between the carboxylate group of Aspartate (Asp) or Glutamate (Glu) and the guanidinium group of Arginine (Arg) or ammonium group of Lysine (Lys), creating an electrostatic attraction. Hydrogen bonding occurs between the backbone amide (N-H) and carbonyl (C=O) groups, or between side chains like Serine (Ser) or Threonine (Thr) and backbone carbonyls, stabilizing secondary structures like α-helices. Van der Waals interactions arise from the close packing of non-polar side chains such as Alanine (Ala), Valine (Val), Leucine (Leu), and Isoleucine (Ile) in the protein core, driven by the hydrophobic effect. A disulfide bridge is a covalent bond formed between the thiol groups of two Cysteine (Cys) residues (Cys-S-S-Cys), providing strong structural rigidity, particularly in extracellular proteins.

Photochemical Mechanisms For reaction (c) I, assuming a standard alkene or aromatic substrate under hv, the mechanism involves excitation to the S₁ state followed by intersystem crossing to T₁. If the substrate is an alkene, [2+2] cycloaddition may occur to form a cyclobutane derivative. For reaction (c) II, the substrate is 1,3-diphenyl-2-butanone (PhCH₂COCH(Ph)₂). Under UV light (hv), the ketone undergoes Norrish Type I cleavage. The carbonyl group absorbs light to reach the excited singlet state (S₁), which undergoes intersystem crossing to the triplet state (T₁). In the T₁ state, the C-C bond adjacent to the carbonyl (α-cleavage) breaks homolytically. This generates an acyl radical and a carbon-centered radical. For this specific substrate, cleavage of the bond between the carbonyl and the CH(Ph)₂ group yields a benzoyl radical (PhCO•) and a 1,1-diphenylethyl radical (Ph₂C•CH₃). These radicals may then undergo decarboxylation (for the acyl radical) or hydrogen abstraction to form the final stable products, typically benzene (from decarboxylation of the acyl radical) and 1,1-diphenylethane (from hydrogen abstraction by the tertiary radical). The mechanism is driven by the stability of the resulting tertiary radical and the relief of steric strain.

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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How this answer will be evaluated

Approach

Framework: Concept > Structure or mechanism > Reasoning > Result. (a(i)) justify: claim > 3-4 reasons > evidence > conclusion | (a(ii)) justify: claim > 3-4 reasons > evidence > conclusion | (b) describe: define > structure or process in order > labelled diagram > significance | (c(i)) justify: claim > 3-4 reasons > evidence > conclusion | (c(ii)) justify: claim > 3-4 reasons > evidence > conclusion Full marks: Correct structures, mechanisms, and reasoning for all parts with clear diagrams.

Key points expected

  • Identify product as mesityl oxide (4-methylpent-3-en-2-one)
  • Correlate IR 1695 cm⁻¹ with conjugated C=O stretch
  • Correlate IR 1620 cm⁻¹ with C=C stretch
  • Assign NMR 1.9(s, 3H) to methyl ketone group
  • Select the non-conjugated lactone in pair I
  • Select ethyl benzoate in pair II
  • Explain that conjugation lowers C=O stretching frequency
  • Explain that electron-withdrawing groups increase C=O frequency

Evaluation rubric

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

  1. (a(i)) Propose the structure of the molecule formed from acetone and dilute NaOH based on spectral data. 15 marks

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

    Must cover

    • Identify product as mesityl oxide (4-methylpent-3-en-2-one)
    • Correlate IR 1695 cm⁻¹ with conjugated C=O stretch
    • Correlate IR 1620 cm⁻¹ with C=C stretch
    • Assign NMR 1.9(s, 3H) to methyl ketone group

    Loses marks

    • Proposing diacetone alcohol (aldol) instead of mesityl oxide
    • Ignoring the C=C stretch in IR data

    Earns more

    • Assign NMR 2.1(s, 6H) to two equivalent methyls
    • Assign NMR 6.15(s, 1H) to vinylic proton
    • Mention aldol condensation mechanism briefly

    Extra mark

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

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

    Must cover

    • Select the non-conjugated lactone in pair I
    • Select ethyl benzoate in pair II
    • Explain that conjugation lowers C=O stretching frequency
    • Explain that electron-withdrawing groups increase C=O frequency

    Loses marks

    • Selecting the conjugated compound as higher frequency
    • Failing to distinguish between inductive and resonance effects

    Earns more

    • Mention specific inductive effects of phenyl group

    Extra mark

    • Cite approximate wavenumber values for comparison
  3. (b) Show salt bridge, hydrogen bond, van der Waals' interaction, and disulfide bridge for protein stabilization. 15 marks

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

    Must cover

    • Draw salt bridge between charged residues (e.g., Lys and Asp)
    • Draw hydrogen bond between polar side chains or backbone
    • Draw disulfide bridge between two Cysteine residues
    • Illustrate van der Waals interaction between non-polar side chains

    Loses marks

    • Using incorrect amino acids for specific interactions (e.g., Cys for salt bridge)
    • Drawing 2D structures without showing spatial proximity

    Earns more

    • Labeling specific amino acid residues used
    • Showing the 3D spatial arrangement of the interactions

    Extra mark

    • Mentioning the role of these interactions in tertiary structure
  4. (c(i)) Complete the photochemical reaction of the cyclohexanone derivative and give the mechanism. 5 marks

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

    Must cover

    • Identify the reaction as Norrish Type II cleavage
    • Show the 1,5-hydrogen abstraction step
    • Show the $eta$-cleavage of the 1,4-diradical
    • Identify the products as an aldehyde and an alkene

    Loses marks

    • Proposing Norrish Type I cleavage instead
    • Failing to show the diradical intermediate

    Earns more

    • Drawing the intermediate 1,4-diradical structure

    Extra mark

    • Mentioning the specific wavelength of light required
  5. (c(ii)) Complete the photochemical reaction of the benzoyl derivative and give the mechanism. 10 marks

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

    Must cover

    • Identify the reaction as Norrish Type I cleavage
    • Show the $eta$-cleavage of the excited carbonyl
    • Show the formation of an acyl radical and a benzyl radical
    • Show the decarboxylation of the acyl radical

    Loses marks

    • Proposing Norrish Type II cleavage instead
    • Failing to show the decarboxylation step

    Earns more

    • Showing the recombination of radicals to form the final product
    • Identifying the specific radical intermediates

    Extra mark

    • Comparing the stability of the radicals formed

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