Chemistry 2022 Paper II 50 marks Explain

Paper II — Q3

(a) The compound A, acetal of acetaldehyde, on reaction with B, under given reaction conditions, yields C and the final product…

(a)

The compound A, acetal of acetaldehyde, on reaction with B, under given reaction conditions, yields C and the final product D:

A, CH₃CH(OMe)₂ + B, CH₂=CH–CH(OH)–R —propionic acid→ C —Δ→ D

(i)

Write down the structures of the products C and D. Propose a mechanism with suitable explanation.

(ii)

Write down the product(s) if ortho-ester E is used instead of the compound A under the similar reaction conditions:

E = CH₃C(OMe)₃

15 marks

(b)
(i)

Considering the structural features of ergosterol (A) and precalciferol (B), a precursor of vitamin D, giving mechanistic details, explain why vitamin D deficiency is endemic in those parts of the world where sunlight is scarce:

[figure]

10 marks

(ii)

How will you convert p-bromonitrobenzene to m-bromobenzoic acid? Give the mechanism.

5 marks

(c)
(i)

Assign the structure of the product with proper justification of kinetic aspects using energy profile diagram:

Cl2HC-CF3 --(EtO-)--> Product

10 marks

(ii)

Arrange the following compounds in order of their increasing reactivity towards nucleophilic substitution through SN2 mechanism. Assign the reason for the same also:

A = 1-bromobicyclo[2.2.1]heptane, B = 1-bromobicyclo[2.2.2]octane, C = tert-butyl bromide, D = 9-bromo-9-phenyl-9,10-dihydroanthracene [figure]

10 marks

हिंदी में प्रश्न पढ़ें
(a)

यौगिक A, ऐसीटाल्डिहाइड का ऐसिटैल, B के साथ दी हुई परिस्थितियों में अभिक्रिया कर C तथा अंतिम उत्पाद D बनाता है:

[figure]

(i)

उत्पादों C तथा D की संरचना लिखिए। उपयुक्त स्पष्टीकरण द्वारा क्रियाविधि प्रस्तावित कीजिए।

(ii)

यौगिक A के स्थान पर ऑर्थो-एस्टर E लेते हुए, उन्हीं अभिक्रिया परिस्थितियों में बने उत्पाद/उत्पादों को लिखिए:

[figure]

(15 अंक)

(b)
(i)

ऑर्गेस्टेरोल (A) व प्रीकैल्सिफेरोल (B), जो कि विटामिन D का अग्रसंरचक है, की संरचनात्मक विशेषताओं पर विचार कर, क्रियाविधि की विस्तृत जानकारी देते हुए बताइए कि क्यों विटामिन D की कमी संसार के उन हिस्सों में स्थानिक बीमारी है जहाँ सूर्य के प्रकाश की कमी है:

[figure]

(10 अंक)

(ii)

आप p-ब्रोमोनाइट्रोबेंजीन को m-ब्रोमोबेंजोइक अम्ल में कैसे रूपान्तरित करेंगे? क्रियाविधि दीजिए।

(5 अंक)

(c)
(i)

ऊर्जा प्रोफ़ाइल आरेख का उपयोग करते हुए गतिज पहलुओं के उचित औचित्य के साथ उत्पाद की संरचना निर्धारित कीजिए:

Cl2HC-CF3 --(EtO-)--> उत्पाद

(10 अंक)

(ii)

निम्नलिखित यौगिकों को SN2 क्रियाविधि द्वारा उनकी नाभिकरागी प्रतिस्थापन के प्रति बढ़ती अभिक्रियाशीलता के क्रम में व्यवस्थित कीजिए। उसके लिए कारण भी निर्धारित कीजिए:

[figure]

(10 अंक)

Q3 of the 2022 UPSC Mains Chemistry Paper II, as printed
The question as printed in the 2022 Chemistry paper

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) Claisen and Johnson–Claisen rearrangements

(i) Structures of C and D, with mechanism

A is acetaldehyde dimethyl acetal, CH₃CH(OMe)₂, and B is a secondary allylic alcohol, R–CH(OH)–CH=CH₂. Heating the two with a weak acid (propionic acid) is the standard way to make an allyl vinyl ether in situ, and the ether then undergoes a Claisen rearrangement.

  1. Trans-acetalisation. Propionic acid protonates one OMe of A, and MeOH leaves to give the oxocarbenium ion CH₃–CH=O⁺Me. The OH of B adds to this ion, and loss of H⁺ gives the mixed acetal CH₃CH(OMe)–O–CH(R)CH=CH₂.
  2. Elimination of MeOH. Protonation of the remaining OMe and loss of MeOH give a second oxocarbenium ion. Loss of an α-proton from its CH₃ then produces the enol ether. C = allyl vinyl ether, CH₂=CH–O–CH(R)–CH=CH₂.
  3. [3,3]-Sigmatropic (Claisen) rearrangement on heating. In a concerted, chair-like six-membered transition state, the allylic C–O σ-bond breaks while a new C–C bond forms between the terminal CH₂ of the vinyl group and the terminal CH₂ of the allyl group. The π bonds move to C=O and to a new internal C=C. Forming a strong C=O bond makes the step irreversible. D = γ,δ-unsaturated aldehyde, OHC–CH₂–CH₂–CH=CH–R, formed mainly as the (E)-alkene because R sits equatorial in the chair transition state.
C=COC(R)C=C  >>  O=CCC/C=C/R

(Example with R = CH₃: C = 3-(vinyloxy)but-1-ene and D = (E)-hex-4-enal, both C₆H₁₀O. The rearrangement is an isomerisation.)

Diagram: Claisen rearrangement of the allyl vinyl ether C through a chair transition state to the aldehyde D

(ii) With the ortho-ester E (trimethyl orthoacetate): the Johnson–Claisen rearrangement

E, CH₃C(OMe)₃, exchanges one OMe with the allylic alcohol under the same acid catalysis. It then loses MeOH to give a ketene acetal, CH₂=C(OMe)–O–CH(R)CH=CH₂, which carries one more OMe than C. The same [3,3] shift then produces an ester instead of an aldehyde.

  • Product: the γ,δ-unsaturated methyl ester MeO₂C–CH₂–CH₂–CH=CH–R (for R = CH₃, methyl (E)-hex-4-enoate), together with 2 MeOH.

(b)(i) Photochemical route to vitamin D and sunlight

Ergosterol (A) has a cyclohexa-1,3-diene unit (Δ⁵,⁷) in ring B, closed by the C9–C10 σ-bond. Precalciferol (B) has no ring B: the C9–C10 bond is broken, and the molecule is a (6Z)-hexatriene running C10=C5–C6=C7–C8=C9.

Mechanism

  1. The Δ⁵,⁷-diene absorbs UV-B light (about 290–315 nm). It is promoted to its excited state, where the HOMO is ψ₄ of the triene being formed.
  2. A 6π electrocyclic ring opening follows. For 4n + 2 electrons the photochemically allowed mode is conrotatory, and this mode opens ring B with correct stereochemistry to give precalciferol.
  3. Precalciferol then isomerises thermally at body temperature by a [1,7]-sigmatropic hydrogen shift, antarafacial and thermally allowed. An H moves from the C-19 methyl to C-9, which gives the 10(19)-exocyclic methylene triene of vitamin D₂ (ergocalciferol). In human skin the same sequence converts 7-dehydrocholesterol into vitamin D₃. Hydroxylation in the liver (C-25) and kidney (C-1) then gives the active hormone.

Why the deficiency is endemic where sunlight is scarce. The ring-opening step has no thermal alternative. The thermally allowed disrotatory opening of the fused, rigid ring is not accessible at body temperature, so without UV-B photons the provitamin is not converted. At high latitudes, in long winters, under persistent cloud, smog or full clothing, UV-B reaching the skin is too weak to drive the step. Few foods contain much vitamin D, so poor Ca²⁺ absorption leads to rickets in children and osteomalacia in adults. The deficiency is endemic where UV-B is scarce because the key step in making the vitamin is photochemical.

Diagram: Ergosterol to precalciferol (photochemical ring opening) to vitamin D2 ([1,7]-hydrogen shift)

(b)(ii) p-Bromonitrobenzene to m-bromobenzoic acid

The conversion takes one step, the von Richter reaction: p-bromonitrobenzene is heated with KCN in aqueous ethanol.

p-BrC₆H₄NO₂ + CN⁻ → m-BrC₆H₄COO⁻ + N₂, and acidification gives m-bromobenzoic acid.

Mechanism (cine substitution)

  1. CN⁻ adds to the ring carbon ortho to NO₂, which is meta to Br, to give a Meisenheimer-type σ-adduct.
  2. A nitro oxygen attacks the neighbouring nitrile carbon, and the cyclic adduct opens to an o-nitroso benzamide.
  3. The amide N and the nitroso N condense to form an N=N bond in a cyclic intermediate.
  4. Attack by hydroxide expels N₂, the carboxylate forms, and a proton ends up on the carbon that carried NO₂.

The COOH therefore ends up ortho to the old NO₂ position, which is meta to Br.

Diagram: von Richter reaction: p-bromonitrobenzene to m-bromobenzoic acid by cine substitution

(c)(i) Product of CHCl₂CF₃ with EtO⁻ and its energy profile

Product: CF₂=CCl₂ (1,1-dichloro-2,2-difluoroethene), formed by E1cB elimination of HF.

  • The C–H proton is unusually acidic. The carbanion ⁻CCl₂–CF₃ is stabilised by the strong –I effect of CF₃ and by the two polarisable chlorines on the carbanion carbon.
  • The only leaving groups β to that carbanion are fluorines, and C–F cleavage is difficult. Deprotonation is therefore fast and reversible, and loss of F⁻ is rate-limiting: an E1cB(reversible) mechanism.
  • Kinetic evidence. When the reaction is run with EtO⁻ in EtOD, the recovered substrate has exchanged its H for D faster than it forms alkene. The carbanion must therefore form and return to substrate many times before F⁻ is lost. For the same reason the elimination shows little primary H/D isotope effect.
Diagram: Energy profile of the E1cB elimination from CF₃CHCl₂

The energy profile shows a low TS1 for fast, reversible proton removal and a carbanion in a genuine potential well. TS2, where C–F breaks, is the highest point, so it controls the rate.

(c)(ii) SN2 reactivity order of A–D

Order of increasing S_N2 reactivity: A ≈ B < D < C, where A = 1-bromobicyclo[2.2.1]heptane, B = 1-bromobicyclo[2.2.2]octane, C = tert-butyl bromide and D = 9-bromo-9-phenyl-9,10-dihydroanthracene.

  • A and B (bridgehead bromides): no S_N2 at all. Backside attack would have to come from inside the cage, where the C–Br σ* lobe points. The rigid framework also cannot flatten into a trigonal-bipyramidal transition state or allow Walden inversion. The more compact norbornyl cage (A) is the more rigid of the two.
  • D: extremely slow. The reacting carbon is tertiary and carries three aryl groups: the phenyl and the two fused rings. They screen the back face more heavily than the three methyls of C.
  • C: the "fastest" of the four. It is also tertiary, but its small methyl groups still allow some backside approach, so a very slow S_N2 is possible.

All four are tertiary halides, and in practice none reacts usefully by S_N2. In polar solvents C and D react by S_N1 instead, D very rapidly because its cation is stabilised by resonance over three rings. A and B resist even S_N1, because a bridgehead cation cannot become planar (Bredt-type strain). If the examiner intends ionisation (S_N1) reactivity, the order is A < B < C < D: the norbornyl cation is the most strained, the bicyclo[2.2.2]octyl cation somewhat less so, the tert-butyl cation is freely planar, and the cation from D is both tertiary and benzylic.

How this answer will be evaluated

What earns full marks: Complete mechanisms, correct structures, clear reasoning, energy diagrams where required

Evaluation rubric

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

  1. (a) Structures of C and D, mechanism, and product from ortho-ester E 15 marks

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

    Must cover

    • C = allyl vinyl ether CH₂=CH–O–CH(R)–CH=CH₂ (via trans-acetalisation and loss of MeOH)
    • D = γ,δ-unsaturated aldehyde OHC–CH₂–CH₂–CH=CH–R by [3,3] Claisen rearrangement
    • Mechanism with arrow-pushing
    • With E: ketene acetal then Johnson–Claisen gives the γ,δ-unsaturated methyl ester MeO₂C–CH₂–CH₂–CH=CH–R

    Loses marks

    • Product without mechanism
    • Calling the Claisen step ionic or giving the wrong (Z) geometry without a chair-TS argument

    Earns more

    • Acetal hydrolysis step shown
    • Acid-catalyzed condensation logic
    • Ketene acetal formation from the ortho ester

    Extra mark

    • Named reaction identification
  2. (b)(i) Mechanistic explanation of Vitamin D synthesis from ergosterol 10 marks

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

    Must cover

    • UV light role in bond cleavage
    • Ergosterol to precalciferol conversion
    • Structural features of A and B
    • Link to sunlight scarcity

    Loses marks

    • No mechanism provided
    • Confusing A and B structures

    Earns more

    • Electron pushing for ring opening
    • Thermal isomerization to calciferol

    Extra mark

    • Specific wavelength of UV mentioned
  3. (b)(ii) Conversion of p-bromonitrobenzene to m-bromobenzoic acid 5 marks

    trace— start point → the stages in sequence → end point → what changed

    Must cover

    • Stepwise reaction sequence
    • Mechanism of key steps
    • Correct regioselectivity

    Loses marks

    • Missing mechanism
    • Wrong product structure

    Earns more

    • Intermediate structures drawn
    • Reagents and conditions specified

    Extra mark

    • Alternative pathway mentioned
  4. (c)(i) Product structure with kinetic justification and energy profile 10 marks

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

    Must cover

    • Product CF₂=CCl₂ by elimination of HF
    • Energy profile diagram
    • E1cB(reversible): fast reversible deprotonation, rate-limiting C–F cleavage (H/D exchange faster than elimination)
    • Role of EtO- base

    Loses marks

    • No energy diagram
    • Wrong product isomer

    Earns more

    • Transition state analysis
    • Steric vs electronic factors

    Extra mark

    • Specific energy values if known
  5. (c)(ii) Order of SN2 reactivity with reasons 10 marks

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

    Must cover

    • Correct increasing order
    • Steric hindrance analysis
    • Electronic effects
    • Reasoning for each compound

    Loses marks

    • Wrong order
    • No reasoning provided

    Earns more

    • Transition state comparison
    • Specific steric interactions

    Extra mark

    • Solvent effects mentioned
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.

All UPSC directive words, compared →

Practice this exact question

Write your answer and it is marked point by point against the model answer above — what you covered, what you missed, what you got wrong.

Evaluate my answer →

More from Chemistry 2022 Paper II

Write this answer and get it marked in about 20 seconds

Free account: 5 AI evaluations a month, marked against the rubric above.

Start free

Already have an account? Log in