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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 · 2020 · Paper II

Q4(b). Rates of HCl addition: an enol ether reacts fastest

  • 15 marks
  • Answered
Question

Consider the following reactions X, Y and Z, and answer the given questions:
1.3tabular@l@
X: propene + HCl → 2-chloropropane + 1-chloropropane
Y: 2-methoxypropene + HCl → 2-chloro-2-methoxypropane + 1-chloro-2-methoxypropane
Z: 3,3-dimethylbut-1-ene + HCl → 2-chloro-3,3-dimethylbutane + 1-chloro-3,3-dimethylbutane
tabular
(i) Which reaction is the fastest? Briefly explain.
(ii) What is the major product of the fastest reaction? Explain briefly.
(iii) Write a detailed curved arrow mechanism of the fastest identified reaction and identify the rate-determining step of your mechanism.
(iv) Offer an explanation for your choice of rate-determining step.

Answer in brief

(i) Y is fastest: its cation is an oxygen-stabilised oxocarbenium ion. (ii) 2-chloro-2-methoxypropane (Markovnikov). (iii)–(iv) Protonation of the C=C to give the oxocarbenium ion is the slow, rate-determining step; capture by Cl⁻ is fast.

Key points

  • (i) Fastest: Y. Protonation of 2-methoxypropene gives a tertiary cation that is also stabilised by the lone pair of the OMe oxygen (+M): the C=O⁺ resonance form has an octet on every atom. X (propene) and Z…
  • (ii) Major product: 2-chloro-2-methoxypropane. H⁺ goes to the terminal CH₂ (Markovnikov), because that puts the positive charge on C-2, next to oxygen. Protonation at C-2 would give a primary cation with no oxygen help, so…
  • (iii) Mechanism. Step 1: the C=C π electrons attack the H of H–Cl; the H–Cl bond breaks, giving the oxocarbenium ion and Cl⁻ (rate-determining). Step 2: Cl⁻ attacks the electrophilic C-2 (fast). The product is an α-chloro ether.
  • (iv) Why step 1 is the RDS. Step 1 breaks a π bond and the strong H–Cl bond and creates two charged species from neutral molecules: it is strongly endergonic, and by the Hammond postulate its transition state resembles the…

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. All three are electrophilic additions of HCl that pass through a carbocation. The rate is set by the stability of the cation formed in the first step; a cation with an adjacent oxygen lone pair (an oxocarbenium ion) is far more stable than any alkyl cation.

Scheme: Step 1: H⁺ adds to the CH₂ end (slow). Step 2: Cl⁻ adds to the oxocarbenium carbon (fast).

  1. (i) Fastest: Y. Protonation of 2-methoxypropene gives a tertiary cation that is also stabilised by the lone pair of the OMe oxygen (+M): the C=O⁺ resonance form has an octet on every atom. X (propene) and Z (3,3-dimethylbut-1-ene) give only secondary alkyl cations, stabilised by hyperconjugation alone. Enol ethers are electron-rich alkenes and add acids many orders of magnitude faster.
  2. (ii) Major product: 2-chloro-2-methoxypropane. H⁺ goes to the terminal CH₂ (Markovnikov), because that puts the positive charge on C-2, next to oxygen. Protonation at C-2 would give a primary cation with no oxygen help, so 1-chloro-2-methoxypropane is negligible.
  3. (iii) Mechanism. Step 1: the C=C π electrons attack the H of H–Cl; the H–Cl bond breaks, giving the oxocarbenium ion and Cl⁻ (rate-determining). Step 2: Cl⁻ attacks the electrophilic C-2 (fast). The product is an α-chloro ether.
  4. (iv) Why step 1 is the RDS. Step 1 breaks a π bond and the strong H–Cl bond and creates two charged species from neutral molecules: it is strongly endergonic, and by the Hammond postulate its transition state resembles the cation, high in energy. Step 2 combines a cation with an anion, is strongly exergonic and has almost no barrier. Rate =k[alkene][HCl], and the rate correlates with cation stability, as found (Y≫ X, Z).
  5. Note on Z. The secondary cation from 3,3-dimethylbut-1-ene undergoes a 1,2-methyl shift to a tertiary cation, so 2-chloro-2,3-dimethylbutane is also formed; Z is still much slower than Y.

Conclusion. (i) Y is fastest: its cation is an oxygen-stabilised oxocarbenium ion. (ii) 2-chloro-2-methoxypropane (Markovnikov). (iii)–(iv) Protonation of the C=C to give the oxocarbenium ion is the slow, rate-determining step; capture by Cl⁻ is fast.

Application. The same fast protonation of enol ethers is used to protect alcohols as THP and MOM-type acetals (dihydropyran or 2-methoxypropene with an acid catalyst) and explains why enol ethers hydrolyse rapidly in dilute acid.

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Model answer to Q4(b), 2020 Paper II: book edition page 15
Book edition, page 15
Model answer to Q4(b), 2020 Paper II: book edition page 16
Book edition, page 16

All parts of Question 4