Chemistry 2022 Paper II 50 marks Explain

Paper II — Q6

(a) Predict the products in the following reactions : (i) PhCO–CH₂CH₂CH₂CH₃ (valerophenone) + CH₃CO–CH₂CH₂–CD₂–CH₃…

(a)

Predict the products in the following reactions :

(i)

PhCO–CH₂CH₂CH₂CH₃ (valerophenone) + CH₃CO–CH₂CH₂–CD₂–CH₃ (hexan-2-one-5,5-d₂) --hv-->

Propose suitable mechanism to justify your answer.

(ii)

CH₃CO–CH₂–CH(CH₃)–CH(CH₃)–CH₂CH₃ (4,5-dimethylheptan-2-one) --hv-->

Giving justification, write the major and minor product(s). Comment upon the chirality of recovered reactant (if any). (5+10=15 marks)

(b)
(i)

How will you synthesize polypropylene (PP) by using Ziegler-Natta catalysis? Discuss the mechanism and its advantages over conventional polymerization. 10 marks

(ii)

How is Perlon synthesized from ε-Caprolactam? Give the mechanism of the reaction. 5 marks

(c)
(i)

Write the structures of the products X, Y and Z in the following reactions and indicate the mechanism for the formation of X :

tricyclic lactone (γ-lactone fused to a cyclopentane fused to a cyclopropane) whose cyclopropane carbon carries –CH=CH–CH₂CH₂CH₂Ph [figure] --(1) OsO4, Pyridine (2) H+, H2O--> X --HIO4--> Y + Z 10 marks

(ii)

Write the major and minor products in the following reaction. Discuss the stereochemistry along with reaction mechanism for the formation of the major product :

Ph–CH(CH₃)–CO–CH₃, single enantiomer (wedged CH₃) --m-CPBA--> Major product + Minor product 10 marks

हिंदी में प्रश्न पढ़ें
(a)

निम्नलिखित अभिक्रियाओं में उत्पादों का अनुमान लगाइए :

(i)

[figure] + [figure] --hv-->

अपने उत्तर का औचित्य सिद्ध करने के लिए उपयुक्त क्रियाविधि प्रस्तावित कीजिए।

(ii)

[figure] --hv-->

औचित्य प्रदान करते हुए मुख्य तथा अल्प उत्पाद/उत्पादों को लिखिए। पुनः-प्राप्त अभिकारक (अगर कोई है) की काइरलता पर टिप्पणी कीजिए। (5+10=15 अंक)

(b)
(i)

ज़िग्लर-नाट्टा उत्प्रेरण के प्रयोग से आप पॉलिप्रोपिलीन (PP) का संश्लेषण कैसे करेंगे? क्रियाविधि बताइए तथा पारंपरिक बहुलकरण (पॉलिमराइज़ेशन) की अपेक्षा फायदों की विवेचना कीजिए। (10 अंक)

(ii)

ε-कैप्रोलेक्टम से पर्लोन को कैसे संश्लेषित करते हैं? अभिक्रिया की क्रियाविधि दीजिए। (5 अंक)

(c)
(i)

निम्नलिखित अभिक्रियाओं में उत्पादों X, Y और Z की संरचना लिखिए तथा X के बनने की क्रियाविधि का उल्लेख कीजिए :

[figure] --(1) OsO4, पिरिडीन (Pyridine) (2) H+, H2O--> X --HIO4--> Y + Z (10 अंक)

(ii)

निम्नलिखित अभिक्रिया में मुख्य तथा अल्प उत्पादों को लिखिए। मुख्य उत्पाद के बनने की त्रिविम रसायन के साथ अभिक्रिया क्रियाविधि की विवेचना कीजिए :

[figure] --m-CPBA--> मुख्य उत्पाद + अल्प उत्पाद (10 अंक)

Q6 of the 2022 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2022 Chemistry paper

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) A chemical reaction scheme showing a starting material reacting with light (hv). The starting material is a ketone with the structure: a carbonyl group (C=O) attached to a methyl group on one side and a chiral carbon on the other. This chiral carbon is bonded to a hydrogen atom, a methyl group, and an isopropyl group. The reaction arrow points to the right, indicating the formation of products.

(c) A reaction scheme showing a chiral ketone reacting with m-CPBA. The starting material is 2-methyl-1-phenylpropan-1-one. The chiral center is at the carbon adjacent to the carbonyl, bonded to a phenyl group (Ph), a methyl group (CH3), and a hydrogen atom (shown with a dashed bond). The reaction arrow points to two products labeled '?' with 'Major product' and 'Minor product' underneath.

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) Norrish type II photolysis of the mixed ketones: a crossover test

Both ketones carry γ-hydrogens, so on n→π* excitation each undergoes a Norrish type II reaction. The excited C=O oxygen abstracts a γ-H through a six-membered (1,5-H shift) transition state, giving a 1,4-biradical (ketyl radical plus γ-alkyl radical). The biradical then either cleaves the Cα–Cβ bond (β-scission) to an enol and an alkene, or closes to a cyclobutanol (Yang cyclisation, minor).

  • Valerophenone, PhCO–CH₂CH₂CH₂CH₃: γ-H abstraction from C-4, then cleavage, gives the enol PhC(OH)=CH₂, which tautomerises to acetophenone, together with propene. A little 2-methyl-1-phenylcyclobutanol forms as the minor product.
  • Hexan-2-one-5,5-d₂, CH₃CO–CH₂CH₂–CD₂–CH₃: the γ-carbon carries only D, so the oxygen abstracts D. Cleavage gives the enol CH₃C(OD)=CH₂, which ketonises to acetone-d₁, CH₃COCH₂D, and the alkene CH₂=CD–CH₃ (propene-2-d).

The deuterium ends up only in the fragments of the labelled ketone. Acetophenone and propene from valerophenone contain no D, and the acetone carries exactly one D. There is no crossover, so the hydrogen transfer is intramolecular, through the six-membered cyclic transition state of each molecule, not by free radicals exchanging H between molecules. The C–D abstraction is also slower (primary isotope effect), so the labelled ketone reacts less efficiently.

(a)(ii) Photolysis of 4,5-dimethylheptan-2-one

The ketone is CH₃CO–CH₂(α)–CH(CH₃)(β, C-4)–CH(CH₃)(γ, C-5)–CH₂CH₃. It has two kinds of γ-H: one tertiary H at C-5 and three primary H on the C-4 methyl group. The tertiary C–H is weaker and gives the more stable radical, so abstraction occurs mainly there.

  • Major products (type II cleavage): the 1,4-biradical with the radical at C-5 cleaves C3–C4. This gives the enol of acetone, which tautomerises to acetone, and 3-methylpent-2-ene, CH₃CH=C(CH₃)CH₂CH₃ (E and Z).
  • Minor products: (1) Yang cyclisation of the same biradical gives 2-ethyl-1,2,3-trimethylcyclobutan-1-ol as a mixture of diastereomers. (2) Abstraction of a primary H from the C-4 methyl group gives acetone and 3-methylpent-1-ene.

Chirality of the recovered ketone. Some biradicals simply return the H from the ketyl O–H to C-5 and regenerate the ketone. In the biradical, C-5 is a planar radical, so the H can come back to either face. If a single stereoisomer is photolysed, the recovered ketone is therefore epimerised at C-5 (the γ-stereocentre). C-4 (β) never becomes sp², so its configuration is retained. The recovered material is a mixture of C-5 epimers that is still optically active because of C-4. This loss of configuration at the γ-carbon is evidence for the reversible 1,4-biradical.

(b)(i) Ziegler–Natta polypropylene

Catalyst: TiCl₄ supported on MgCl₂ (or TiCl₃) activated by AlEt₃, with an internal Lewis-base donor for stereocontrol. Polymerisation is run at about 60–80 °C and modest pressure.

Mechanism (Cossee–Arlman):

  1. AlEt₃ alkylates titanium, creating a Ti–C bond next to a vacant coordination site.
  2. Propylene coordinates to the vacant site through its π bond.
  3. Migratory insertion: the growing chain moves onto the coordinated alkene through a four-centre transition state, forming a new C–C bond and leaving a vacancy where the chain used to sit.
  4. Steps 2 and 3 repeat. Insertion is 1,2 (head-to-tail), and the chiral surface site makes each monomer present the same enantioface, so the product is isotactic PP.
  5. The chain ends by β-hydride elimination or by chain transfer to H₂, which is used to control molecular mass.

Advantages over conventional free-radical polymerisation: propylene does not give useful high polymer by radical routes (allylic H abstraction stops the chain), whereas Ziegler–Natta gives high-molecular-mass, stereoregular, crystalline PP with Tₘ about 160–165 °C. It also works at low temperature and pressure, gives linear chains with little branching, and allows control of tacticity and molar mass.

(b)(ii) Perlon (nylon-6) from ε-caprolactam

Perlon is made by ring-opening polymerisation of ε-caprolactam, usually started with a little water at about 250 °C.

  1. Water hydrolyses some lactam to 6-aminohexanoic acid, H₂N(CH₂)₅COOH.
  2. The NH₂ group of this acid attacks the carbonyl carbon of another caprolactam. The tetrahedral intermediate collapses, the ring C–N bond breaks, and the chain grows by one unit while a new terminal NH₂ is formed.
  3. Repetition, together with condensation between chain ends, gives –[NH–(CH₂)₅–CO]ₙ–, nylon-6 (Perlon).

Anionic polymerisation with a strong base (NaH or Na) and an N-acyl-lactam activator is the faster alternative.

(c)(i) OsO₄ dihydroxylation, then periodate cleavage

The only C=C in the substrate is in the side chain, cyclopropyl–CH=CH–CH₂CH₂CH₂Ph. The lactone and the cyclopropane are untouched.

  • X: OsO₄ in pyridine adds to the alkene by a concerted [3+2] cycloaddition to give a cyclic osmate(VI) ester. Both C–O bonds form on the same face. Aqueous acid hydrolyses the osmate ester and releases the syn-1,2-diol, cyclopropyl–CH(OH)–CH(OH)–CH₂CH₂CH₂Ph. The two faces of the alkene give diastereomeric diols.
  • HIO₄ forms a cyclic periodate ester with the vicinal diol and cleaves the C–C bond between the two carbinol carbons.
  • Y: the tricyclic lactone carrying a cyclopropanecarbaldehyde (–CHO) at the former alkene carbon.
  • Z: 4-phenylbutanal, OHC–CH₂CH₂CH₂Ph.

(c)(ii) Baeyer–Villiger oxidation of 3-phenylbutan-2-one

The substrate is one enantiomer of Ph–CH(CH₃)–CO–CH₃.

Mechanism: m-CPBA adds to the protonated carbonyl to give the Criegee intermediate. The group anti-periplanar to the O–O bond migrates from carbon to oxygen in a concerted step while m-chlorobenzoate leaves. Migratory aptitude is tertiary > secondary (benzylic) > phenyl > primary > methyl.

  • Major product: the secondary benzylic group migrates, giving 1-phenylethyl acetate, CH₃CO–O–CH(CH₃)Ph. The migrating carbon keeps its bonds during the shift, so the reaction proceeds with retention of configuration. The (S)-ketone gives (S)-1-phenylethyl acetate, and the (R)-ketone gives the (R)-acetate.
  • Minor product: methyl migration gives methyl 2-phenylpropanoate, Ph–CH(CH₃)–CO–OCH₃, also with the stereocentre unchanged, since that carbon is not involved.
Diagram: Norrish II transition state, Cossee-Arlman insertion, osmate ester and Criegee intermediate

How this answer will be evaluated

Framework: UPSC Chemistry Paper 2

What earns full marks: Comprehensive mechanisms, correct stereochemistry, clear justification for all parts.

Evaluation rubric

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

  1. (a)(i) Products of photolysing valerophenone with hexan-2-one-5,5-d₂, and the mechanism. 5 marks

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

    Must cover

    • Identify the reaction as a Norrish Type II process
    • Show the 1,5-hydrogen abstraction step
    • Draw the resulting 1,4-biradical intermediate
    • β-Scission products: acetophenone + propene from valerophenone, acetone-d₁ (CH₃COCH₂D) + propene-2-d from the labelled ketone

    Loses marks

    • Fails to show the biradical intermediate
    • Incorrect hydrogen abstraction site

    Earns more

    • No D in the valerophenone fragments: no crossover, so the H transfer is intramolecular
    • Mentions the role of the excited carbonyl state

    Extra mark

    • Mentions the primary isotope effect for C–D abstraction
  2. (a)(ii) Major and minor products of photolysing 4,5-dimethylheptan-2-one, and the chirality of recovered ketone. 10 marks

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

    Must cover

    • Identifies the Norrish Type II reaction pathway
    • Major: acetone + 3-methylpent-2-ene via the tertiary γ-H (C-5). Minor: cyclobutanol (Yang) and acetone + 3-methylpent-1-ene via the primary γ-H
    • Explains the stereochemical outcome of the reaction
    • Recovered ketone is epimerised at the γ-carbon C-5 by reverse H transfer, while C-4 keeps its configuration

    Loses marks

    • Fails to distinguish between major and minor products
    • Ignores the chirality of the recovered reactant

    Earns more

    • Provides a clear mechanism for the formation of each product
    • Justifies product distribution using electronic/steric factors

    Extra mark

    • Uses the epimerisation as evidence for a reversible 1,4-biradical
  3. (b)(i) Synthesize polypropylene using Ziegler-Natta catalysis, discussing mechanism and advantages. 10 marks

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

    Must cover

    • Identifies the Ziegler-Natta catalyst components (e.g., TiCl4, AlEt3)
    • Describes the coordination-insertion mechanism of polymerization
    • Explains the regio- and stereo-selectivity of the process
    • Lists at least two advantages over conventional polymerization

    Loses marks

    • Fails to describe the mechanism of polymerization
    • Does not mention the advantages over conventional methods

    Earns more

    • Draws the active site of the catalyst
    • Mentions the specific type of polypropylene produced (e.g., isotactic)

    Extra mark

    • Mentions the specific industrial conditions (temperature, pressure)
  4. (b)(ii) Explain the synthesis of Perlon from ε-Caprolactam and give the mechanism. 5 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • Identifies the reaction as ring-opening polymerization
    • Shows the initiation step involving the catalyst
    • Illustrates the propagation step of the polymer chain
    • Names the final product as Perlon (nylon-6)

    Loses marks

    • Fails to show the mechanism of ring-opening
    • Does not name the final product

    Earns more

    • Mentions the specific catalyst used (e.g., water, acid, base)
    • Shows the structure of the repeating unit of the polymer

    Extra mark

    • Mentions the specific conditions for the reaction
  5. (c)(i) Write structures of X, Y, Z and indicate the mechanism for the formation of X. 10 marks

    explain— definition/context → points in order → small example → short close

    Must cover

    • X: syn-1,2-diol from the side-chain C=C via the cyclic osmate ester
    • Y: the tricyclic lactone cyclopropanecarbaldehyde, Z: 4-phenylbutanal
    • Shows the mechanism of OsO4-mediated dihydroxylation
    • Explains the role of HIO4 in the cleavage reaction

    Loses marks

    • Fails to show the mechanism for the formation of X
    • Incorrect structures for Y or Z

    Earns more

    • Correctly assigns the stereochemistry of the diol (X)
    • Shows the mechanism of the oxidative cleavage step

    Extra mark

    • Mentions the specific stereochemical outcome of the dihydroxylation
  6. (c)(ii) Write major/minor products and discuss stereochemistry and mechanism for the major product. 10 marks

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

    Must cover

    • Identifies the reaction as a Baeyer-Villiger oxidation
    • The secondary benzylic carbon migrates (not CH₃): major product 1-phenylethyl acetate
    • Migration with retention of configuration at the migrating carbon
    • Minor product: methyl 2-phenylpropanoate from methyl migration

    Loses marks

    • Fails to show the mechanism for the major product
    • Incorrect assignment of major/minor products

    Earns more

    • Provides a clear mechanism for the Baeyer-Villiger oxidation
    • Justifies the regioselectivity based on migratory aptitude

    Extra mark

    • Shows the anti-periplanar Criegee intermediate
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

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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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