Q8(d). SeO₂, NBS and Pb(OAc)₄ reactions
Identify the major product (X) formed in the following reactions:
(i) [structure] → (SeO₂) (X)
(ii) [structure] → (1) NBS, 2) hν/heat) (X)
(iii) [structure] → (Pb(OAc)₄) (X)
Answer in brief
(i) 2-naphthylglyoxal, ArCOCHO (SeO₂, α-oxidation). (ii) 3-bromocyclopentene (NBS, allylic radical bromination). (iii) cyclodecane-1,6-dione (Pb(OAc)₄ glycol cleavage, which works even on this trans-diol).
Key points
- SeO₂ oxidises a CH₂ or CH₃ group α to a carbonyl (Riley oxidation). The enol of the ketone attacks Se; the resulting α-seleninic ester rearranges and eliminates Se and water, leaving a C=O on the former α-carbon.
- X = 2-oxo-2-(naphthalen-2-yl)acetaldehyde (2-naphthylglyoxal), ArCOCHO, C₁₂H₈O₂: a 1,2-dicarbonyl compound. The aromatic ring is not attacked.
- Light or heat starts a radical chain. Br• abstracts an allylic H (the weakest C–H, because the allyl radical is resonance-stabilised).
- NBS reacts with the HBr formed to keep a very low concentration of Br₂; the allyl radical takes Br from Br₂ and regenerates Br•. At this low concentration addition of Br₂ to the C=C cannot compete (Wohl–Ziegler reaction).
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
(i) Riley oxidation
Scheme: The CH₃ next to C=O becomes CHO.
- SeO₂ oxidises a CH₂ or CH₃ group α to a carbonyl (Riley oxidation). The enol of the ketone attacks Se; the resulting α-seleninic ester rearranges and eliminates Se and water, leaving a C=O on the former α-carbon.
- X = 2-oxo-2-(naphthalen-2-yl)acetaldehyde (2-naphthylglyoxal), ArCOCHO, C₁₂H₈O₂: a 1,2-dicarbonyl compound. The aromatic ring is not attacked.
(ii) Allylic bromination with NBS
Scheme: Br replaces an allylic H; the C=C is kept.
- Light or heat starts a radical chain. Br• abstracts an allylic H (the weakest C–H, because the allyl radical is resonance-stabilised).
- NBS reacts with the HBr formed to keep a very low concentration of Br₂; the allyl radical takes Br from Br₂ and regenerates Br•. At this low concentration addition of Br₂ to the C=C cannot compete (Wohl–Ziegler reaction).
- In cyclopentene the two ends of the allyl radical are equivalent, so there is only one product: X = 3-bromocyclopentene (racemic).
(iii) Glycol cleavage of trans-decalin-4a,8a-diol
Scheme: The bond between the two carbinol carbons (the ring fusion) breaks; two C=O groups form in a ten-membered ring.
- Pb(OAc)₄ (like HIO₄) breaks the C–C bond of a 1,2-diol, giving two carbonyl groups. Both carbinol carbons are tertiary, so both become ketones.
- Cutting the shared C4a–C8a bond opens the bicyclic system into one large ring: X = cyclodecane-1,6-dione (C₁₀H₁₆O₂).
- The two OH groups of the trans-diol are anti-periplanar and cannot form the cyclic lead ester that cis-glycols form; it is still cleaved by Pb(OAc)₄ (more slowly, through an acyclic transition state), whereas HIO₄, which works through a cyclic periodate ester, does not cleave it readily. This is the classic case showing that the cyclic-intermediate mechanism does not explain all lead tetraacetate oxidations.
Conclusion. (i) 2-naphthylglyoxal, ArCOCHO (SeO₂, α-oxidation). (ii) 3-bromocyclopentene (NBS, allylic radical bromination). (iii) cyclodecane-1,6-dione (Pb(OAc)₄ glycol cleavage, which works even on this trans-diol).
Application. Glycol cleavage of bicyclic diols is used to make medium-sized rings (ten-membered rings for macrocyclic musks); NBS allylic bromination and SeO₂ oxidation are standard steps in terpene and steroid synthesis.
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All parts of Question 8
- 8(a) Woodward–Fieser rules: which structure absorbs at 229 nm15 marks
- 8(b) Mass spectrum: the M-1 peak of pentanal10 marks
- 8(c) Equivalent protons and number of 1H NMR signals10 marks
- 8(d) SeO₂, NBS and Pb(OAc)₄ reactions15 marks