Q6(b). NBS (radical and ionic), KMnO₄/NaIO₄, OsO₄/H₂O₂
Identify (X) and (Y) in the following:
(i)(X) ← (NBS, φ. OOH) [structure] → (NBS, (CH₃)₄N+Br⁻) (Y)
(ii) [structure] → (Dil. KMnO₄) (X) → (NaIO₄) (Y) + HCHO
(iii)₆H₆ → (H₂O₂, OsO₄ in t-C₄H₇OH) (X)
Answer in brief
(i) X = 3-bromocyclohexene (radical allylic bromination), Y = trans-1,2-dibromocyclohexane (ionic anti addition). (ii) X = 1-(hydroxymethyl)cyclobutan-1-ol, Y = cyclobutanone. (iii) X = cyclohexanehexol (inositol) by repeated syn hydroxylation.
Key points
- X (Wohl–Ziegler bromination). The peroxide starts a chain: Br. abstracts an allylic H (weakest C–H, resonance-stabilised allyl radical); the radical takes Br from the low, steady concentration of Br₂ made from NBS and HBr. Low…
- Y. Bromide ion reacts with NBS to release Br₂ (or Br₃⁻) in high local concentration. Electrophilic addition goes through a bromonium ion, opened by Br⁻ from the back face: anti addition. Y = trans-1,2-dibromocyclohexane (racemic).
- H₂O₂ with catalytic OsO₄ in tert-butanol (Milas hydroxylation) adds two OH syn to a C=C through an osmate ester; H₂O₂ reoxidises osmium, so only a trace of OsO₄ is needed.
- Benzene resists addition because of its aromatic stabilisation, so it reacts only very slowly; this is not a preparative reaction, and the expected answer is the product of complete hydroxylation. Once the first syn…
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. The same reagent can act by different mechanisms: NBS gives a bromine radical with a peroxide initiator but Br₂/Br⁺ in the presence of bromide ion; KMnO₄ and OsO₄ add two OH syn; NaIO₄ cleaves 1,2-diols.
Reading of the question: “φ.OOH” is taken as a peroxide radical initiator (benzoyl peroxide or perbenzoic acid) and “t-C₄H₇OH” as tert-butanol (C₄H₉OH).
(i) NBS on cyclohexene
Scheme: One reagent, two mechanisms.
- X (Wohl–Ziegler bromination). The peroxide starts a chain: Br. abstracts an allylic H (weakest C–H, resonance-stabilised allyl radical); the radical takes Br from the low, steady concentration of Br₂ made from NBS and HBr. Low Br₂ concentration prevents addition to C=C. X = 3-bromocyclohexene.
- Y. Bromide ion reacts with NBS to release Br₂ (or Br₃⁻) in high local concentration. Electrophilic addition goes through a bromonium ion, opened by Br⁻ from the back face: anti addition. Y = trans-1,2-dibromocyclohexane (racemic).
(ii) Methylenecyclobutane
Scheme: Dihydroxylation then glycol cleavage.
Cold, dilute (alkaline) KMnO₄ adds two OH groups through a cyclic manganate ester: X = 1-(hydroxymethyl)cyclobutan-1-ol. Periodate forms a cyclic periodate ester with the 1,2-diol and cleaves the C–C bond between the two carbinol carbons: the ring carbon becomes C=O and the CH₂OH becomes HCHO. Y = cyclobutanone.
(iii) Benzene with H₂O₂/OsO₄ (Milas reagent)
- H₂O₂ with catalytic OsO₄ in tert-butanol (Milas hydroxylation) adds two OH syn to a C=C through an osmate ester; H₂O₂ reoxidises osmium, so only a trace of OsO₄ is needed.
- Benzene resists addition because of its aromatic stabilisation, so it reacts only very slowly; this is not a preparative reaction, and the expected answer is the product of complete hydroxylation. Once the first syn dihydroxylation has occurred, the product, cis-cyclohexa-3,5-diene-1,2-diol, is no longer aromatic, and its two remaining C=C bonds are hydroxylated in the same way.
- X = cyclohexane-1,2,3,4,5,6-hexol (an inositol), C₆H₆(OH)₆, with each pair of OH added together on one face (syn).
Scheme: Three syn dihydroxylations.
Conclusion. (i) X = 3-bromocyclohexene (radical allylic bromination), Y = trans-1,2-dibromocyclohexane (ionic anti addition). (ii) X = 1-(hydroxymethyl)cyclobutan-1-ol, Y = cyclobutanone. (iii) X = cyclohexanehexol (inositol) by repeated syn hydroxylation.
Application. Allylic bromination with NBS is the standard route to allylic functionalisation in steroid and terpene chemistry; dihydroxylation followed by periodate cleavage (Lemieux–von Rudloff with KMnO₄, Lemieux–Johnson with OsO₄) converts an exocyclic methylene group into a ring ketone.
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All parts of Question 6
- 6(a) Structure of C₇H₉N from UV, IR and NMR: p-toluidine15 marks
- 6(b) NBS (radical and ionic), KMnO₄/NaIO₄, OsO₄/H₂O₂15 marks
- 6(c) Nylon-6 from cyclohexanone: Beckmann rearrangement20 marks