VKC ScienceChemistry for written exams Telegram

UPSC Civil Services (Mains)

Chemistry optional

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

Paper I covers physical and inorganic chemistry.

UPSC Civil Services (Mains), Chemistry optional · 2016 · Paper I

Q5(d). Oxidative addition against insertion

  • 10 marks
  • Answered
Question

What do you understand by an “oxidative addition” reaction? How does it differ from an “insertion-reaction”? Give one example in each case.

Answer in brief

In oxidative addition a molecule A–B is split and both A and B bond to the metal: oxidation state and coordination number each rise by 2. In (migratory) insertion an unsaturated ligand (CO, alkene) inserts into an existing M–X bond: no change in oxidation state, and a vacant site is created.

Key points

  • the A–B bond (H₂, HCl, CH₃I, Cl₂) is broken and two new M–A and M–B bonds form.
  • The metal's oxidation state rises by 2, its coordination number rises by 2, and the electron count by 2. It needs a coordinatively unsaturated, electron-rich metal, usually a 16-electron d⁸ (Ir(I), Rh(I), Pt(0)) or d¹⁰ complex…
  • trans-[IrCl(CO)(PPh₃)₂] + H₂ → [IrH₂Cl(CO)(PPh₃)₂] Ir(I), d⁸, CN 4, 16 e → Ir(III), d⁶, CN 6, 18 e. The two H atoms enter cis (concerted, three-centre addition).
  • The reverse, loss of A–B with a fall of 2 in oxidation state, is reductive elimination.

The full model answer

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

Oxidative addition

  1. LₙM + A-B → LₙM(A)(B): the A–B bond (H₂, HCl, CH₃I, Cl₂) is broken and two new M–A and M–B bonds form.
  2. The metal's oxidation state rises by 2, its coordination number rises by 2, and the electron count by 2. It needs a coordinatively unsaturated, electron-rich metal, usually a 16-electron d⁸ (Ir(I), Rh(I), Pt(0)) or d¹⁰ complex with a stable higher oxidation state.
  3. Example (Vaska's complex): trans-[IrCl(CO)(PPh₃)₂] + H₂ → [IrH₂Cl(CO)(PPh₃)₂] Ir(I), d⁸, CN 4, 16 e → Ir(III), d⁶, CN 6, 18 e. The two H atoms enter cis (concerted, three-centre addition).
  4. The reverse, loss of A–B with a fall of 2 in oxidation state, is reductive elimination.

Insertion (migratory insertion)

  1. An unsaturated ligand already on the metal (CO, alkene, SO₂) is inserted into an adjacent M–X bond (X = alkyl, H): LₙM(X)(CO) → LₙM(COX). In practice the X group migrates onto the cis CO.
  2. Example: [Mn(CH₃)(CO)₅] + CO → [Mn(COCH₃)(CO)₅] The methyl moves to a cis CO, giving an acyl and a vacant site, which the incoming CO fills. Labelling with ¹³CO shows the new CO is not the one in the acyl group.
  3. Alkene insertion into M–H gives an alkyl: M-H + CH₂=CH₂ → M-CH₂CH₃ (the key step of hydrogenation and of Ziegler–Natta polymerisation).
Oxidative additionInsertion
Oxidation stateincreases by 2unchanged
Coordination numberincreases by 2decreases by 1 (site opened)
Electron count+2 (16 → 18)−2 before the new ligand enters
BondsA–B broken; M–A, M–B formednew C–X bond formed within the ligand sphere
Reverse stepreductive eliminationβ-elimination / de-insertion

Application. both steps run in the Monsanto acetic acid process: oxidative addition of CH₃I to [Rh(CO)₂I₂]⁻, then CO insertion to the acetyl, then reductive elimination of CH₃COI.

Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

Model answer to Q5(d), 2016 Paper I: solution page 28
Solution page 28 of 46
Read the whole 2016 solution2016 question paper

All parts of Question 5