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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2026 · Paper II

Q2(c). Dichlorocarbene and its reactions

  • 20 marks
  • Answered
Question

Intermediate [A] is formed in the following reaction: CHCl₃ + ^tBuOK → [A]
(i) Identify the intermediate [A].
(ii) Write the products of the following reactions with the correct stereochemistry.
Reaction 1 [structure] → ([A]) Reaction 2 [structure] → ([A])
(iii) Are reactions 1 and 2 stereospecific or stereoselective?
(iv) Write the structure of the major product formed in the following reaction: [structure] → ((i) NaOH, (ii) [A], 70°C(iii) H⁺)
(v) Write the major product formed in the following reaction: [structure] → ([A] (one equiv.))

Answer in brief

[A] = dichlorocarbene. It adds to cis- and trans-but-2-ene stereospecifically (cis and trans dichlorodimethylcyclopropanes), converts phenoxide into salicylaldehyde (Reimer–Tiemann), and with one equivalent attacks only the tetrasubstituted fusion C=C of 4,7-dihydroindane.

Key points

  • Introduction. A strong, hindered base removes the acidic proton of chloroform (the three Cl atoms stabilise the carbanion); CCl₃⁻ then loses Cl⁻ by α-elimination. The neutral, six-electron carbon that results is an…
  • CHCl₃ + ^tBuO⁻ ⇌ ⁻CCl₃ + ^tBuOH ⁻CCl₃ →:CCl₂ + Cl⁻ [A] = dichlorocarbene,:CCl₂, a singlet carbene: the two non-bonding electrons are paired in an sp² orbital and the p orbital is empty. Back-donation from the chlorine lone…

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. A strong, hindered base removes the acidic proton of chloroform (the three Cl atoms stabilise the carbanion); CCl₃⁻ then loses Cl⁻ by α-elimination. The neutral, six-electron carbon that results is an electrophile that adds to C=C bonds and to electron-rich arenes.

(i) Identity of [A]

CHCl₃ + ^tBuO⁻ ⇌ ⁻CCl₃ + ^tBuOH ⁻CCl₃ →:CCl₂ + Cl⁻ [A] = dichlorocarbene,:CCl₂, a singlet carbene: the two non-bonding electrons are paired in an sp² orbital and the p orbital is empty. Back-donation from the chlorine lone pairs into the empty p orbital makes it more stable and more selective than:CH₂.

(ii) Products with stereochemistry

Scheme: 1,1-Dichloro-2,3-dimethylcyclopropanes: the alkene geometry is kept.

Singlet:CCl₂ forms both new C–C bonds in one concerted step (a cheletropic [2+1] addition, its empty p orbital taking the π electrons while its filled sp² orbital interacts with π^*). There is no intermediate in which the C–C bond could rotate. Reaction 1 gives cis-1,1-dichloro-2,3-dimethylcyclopropane (meso, both CH₃ on one face); Reaction 2 gives trans-1,1-dichloro-2,3-dimethylcyclopropane as a racemate.

(iii) Stereospecific or stereoselective?

Both reactions are stereospecific: stereoisomeric starting materials (cis and trans alkene) give stereoisomeric products, each with complete retention of the alkene geometry. A stereoselective reaction is one that merely prefers one stereoisomer when several could form from the same starting material.

(iv) Phenol: Reimer–Tiemann reaction

Scheme: The phenoxide attacks:CCl₂ through its ortho carbon; the CHCl₂ group is hydrolysed to CHO.

The phenoxide ion is the nucleophile; it attacks the electrophilic carbene with its ortho carbon, helped by the close approach of:CCl₂ to the O⁻/Na⁺ ion pair. Re-aromatisation gives an ortho CHCl₂ group, which base hydrolyses through the gem-diol to the aldehyde. Major product: 2-hydroxybenzaldehyde (salicylaldehyde), with a little of the para isomer.

(v) One equivalent of [A] with 4,7-dihydroindane

The diene has a tetrasubstituted C=C (the ring-fusion bond) and a disubstituted C=C. Electrophilic:CCl₂ adds to the more electron-rich, more highly alkylated double bond. With one equivalent only that bond reacts; the cyclopropane spans the two fusion carbons and gives a [4.3.1]propellane.

Figure: 10,10-Dichlorotricyclo[4.3.1.01,6]dec-3-ene (a propellane).

Conclusion. [A] = dichlorocarbene. It adds to cis- and trans-but-2-ene stereospecifically (cis and trans dichlorodimethylcyclopropanes), converts phenoxide into salicylaldehyde (Reimer–Tiemann), and with one equivalent attacks only the tetrasubstituted fusion C=C of 4,7-dihydroindane.

Application. Dichlorocarbene addition followed by ring opening is used to make allenes and ring-expanded tropones; the Reimer–Tiemann reaction gives salicylaldehyde for coumarin (perfumery) and aspirin-related chemistry.

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Model answer to Q2(c), 2026 Paper II: solution page 10
Solution page 10 of 26

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