Q6(c). NBS, KMnO₄, HIO₄; hydroboration, dehydration, OsO₄ and Pb(OAc)₄
Identify the compounds A, B and C in the following conversions:
(i) [structure] + NBS → A → (1) OH⁻, 2) KMnO₄, 0°C) B → (HIO₄) C
(ii) CH₃CH₂CH=CH₂ → (A) CH₃CH₂CH₂CH₂OH → (B) CH₃-CH=CH-CH₃ → (1) OsO₄, 2) Pb(OAc)₄) C
Answer in brief
(i) A = 3-bromocyclohexene, B = cyclohexane-1,2,3-triol, C = pentanedial (OHC(CH₂)₃CHO) with HCOOH. (ii) A = BH₃/THF then H₂O₂/OH⁻; B = conc. H₂SO₄/heat; C = CH₃CHO (two molecules).
Key points
- A = 3-bromocyclohexene. NBS keeps a very low Br₂ concentration, so instead of addition a radical chain allylic substitution occurs: Br^ takes an allylic H (weak C–H, the radical is allylic and resonance-stabilised), and the…
- Cyclohex-2-en-1-ol. OH⁻ replaces Br (SN2, or SN1 through the allylic cation).
- B = cyclohexane-1,2,3-triol. Cold, dilute (alkaline) KMnO₄ adds two OH groups syn through a cyclic manganate ester to the C=C.
- C = pentanedial (glutaraldehyde) + HCOOH. HIO₄ cleaves each C–C bond between two OH-bearing carbons through a cyclic periodate ester. A 1,2,3-triol is cut twice (2 HIO₄): the two end carbons become CHO groups, still linked by…
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. (i) uses allylic bromination, substitution, syn-dihydroxylation and periodate cleavage of a 1,2,3-triol. (ii) uses anti-Markovnikov hydration, acid-catalysed dehydration with rearrangement, and glycol cleavage.
(i) From cyclohexene
Scheme: Allylic bromide → allylic alcohol → 1,2,3-triol → pentanedial and formic acid.
- A = 3-bromocyclohexene. NBS keeps a very low Br₂ concentration, so instead of addition a radical chain allylic substitution occurs: Br^ takes an allylic H (weak C–H, the radical is allylic and resonance-stabilised), and the radical takes Br from Br₂.
- Cyclohex-2-en-1-ol. OH⁻ replaces Br (SN2, or SN1 through the allylic cation).
- B = cyclohexane-1,2,3-triol. Cold, dilute (alkaline) KMnO₄ adds two OH groups syn through a cyclic manganate ester to the C=C.
- C = pentanedial (glutaraldehyde) + HCOOH. HIO₄ cleaves each C–C bond between two OH-bearing carbons through a cyclic periodate ester. A 1,2,3-triol is cut twice (2 HIO₄): the two end carbons become CHO groups, still linked by the three CH₂ groups, and the middle carbon leaves as formic acid.
(ii) From but-1-ene
- A = 1) BH₃·THF (or B₂H₆); 2) H₂O₂, NaOH. Hydroboration puts boron on the terminal carbon (steric and electronic control, syn addition), and oxidation replaces B by OH with retention: anti-Markovnikov hydration to butan-1-ol.
- B = conc. H₂SO₄, heat (or H₃PO₄, or Al₂O₃ at high temperature). The protonated alcohol loses water with a 1,2-hydride shift to the secondary carbon (or by E2-like loss); the proton is removed to give the more substituted Zaitsev alkene, but-2-ene (mainly trans).
- C = ethanal (acetaldehyde), 2 CH₃CHO. OsO₄ gives butane-2,3-diol (syn); Pb(OAc)₄ cleaves the 1,2-diol through a cyclic lead(IV) ester into two molecules of CH₃CHO.
Scheme: The symmetrical diol is split into two identical aldehydes.
Conclusion. (i) A = 3-bromocyclohexene, B = cyclohexane-1,2,3-triol, C = pentanedial (OHC(CH₂)₃CHO) with HCOOH. (ii) A = BH₃/THF then H₂O₂/OH⁻; B = conc. H₂SO₄/heat; C = CH₃CHO (two molecules).
Application. Glutaraldehyde is the cross-linking agent used to fix tissues and to sterilise medical equipment; periodate and lead tetra-acetate cleavage are standard tests for 1,2-diols in carbohydrate structure work.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram

All parts of Question 6
- 6(a) Photochemical ring expansion of cyclobutanone (Norrish type I)15 marks
- 6(b) o-Xylylene from a tetrabromide; photochemical ring closure15 marks
- 6(c) NBS, KMnO₄, HIO₄; hydroboration, dehydration, OsO₄ and Pb(OAc)₄20 marks