Q6(a). Chemoselective reductions
Write the structures of the products formed in the following reactions:
(i) [structure] → (NaBH₄/CeCl₃, MeOH) (ii) [structure] → (Na/NH₃ (l))
(iii) [structure] → ((i) B₂H₆, (ii) H⁺) (iv) [structure] → ((i) LiAlH₄, (ii) H⁺)
(v) [structure] → (B₂H₆, THF)
Answer in brief
(i) Cyclohex-2-en-1-ol; (ii) cyclohexanone; (iii) and (iv) 4-methyl-δ-valerolactone (borane reduces the acid, LiAlH₄ the ester); (v) methyl 1-benzylpyrrolidine-3-carboxylate.
Key points
- (iii) B₂H₆ is electrophilic: it reacts fastest with the electron-rich, acidic CO₂H (forming an acyloxyborane) and leaves the ester alone. The acid is reduced to CH₂OH: ethyl 5-hydroxy-3-methylpentanoate. On acidification the…
- (iv) LiAlH₄ first deprotonates the CO₂H; the carboxylate is negatively charged and is reduced only slowly. The nucleophilic hydride reduces the ester to CH₂OH: 5-hydroxy-3-methylpentanoic acid, which lactonises on acidification.
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. Each reagent reduces only one of the groups present. The choice depends on whether the reducing agent is a nucleophilic hydride (NaBH₄, LiAlH₄), an electrophilic, Lewis-acidic hydride (borane) or an electron donor (sodium in ammonia).
(i) Luche reduction
CeCl₃ is a hard Lewis acid; in methanol it converts NaBH₄ into NaBH(OMe)₃-type hard hydrides and activates the C=O. Hard hydride attacks the hard carbonyl carbon (1,2-addition) instead of the soft β-carbon. Product: cyclohex-2-en-1-ol (allylic alcohol), C=C kept.
(ii) Dissolving-metal reduction
Sodium gives an electron to the enone (radical anion); protonation at the β-carbon and a second electron give the enolate, which is protonated on work-up. The C=C is reduced (conjugate reduction) and the C=O survives as the enolate. Product: cyclohexanone.
Scheme: 1,2-reduction (Luche) and conjugate reduction (Na/NH₃) of the same enone.
(iii) and (iv) Borane and LiAlH₄ on the half-ester
- (iii) B₂H₆ is electrophilic: it reacts fastest with the electron-rich, acidic CO₂H (forming an acyloxyborane) and leaves the ester alone. The acid is reduced to CH₂OH: ethyl 5-hydroxy-3-methylpentanoate. On acidification the alcohol attacks the ester intramolecularly and closes the δ-lactone.
- (iv) LiAlH₄ first deprotonates the CO₂H; the carboxylate is negatively charged and is reduced only slowly. The nucleophilic hydride reduces the ester to CH₂OH: 5-hydroxy-3-methylpentanoic acid, which lactonises on acidification.
Scheme: (iii) via the hydroxy ester, (iv) via the hydroxy acid: the same δ-lactone.
Both give 4-methyltetrahydro-2H-pyran-2-one; the opposite ends of the chain are reduced. (If the half-ester were a single enantiomer, the two routes would give the two enantiomers of the lactone.)
(v) Borane with a lactam ester
Borane reduces amides (here the lactam C=O) quickly, because the nitrogen lone pair makes the oxygen basic enough to coordinate boron; esters react only slowly. The lactam C=O becomes CH₂, and the methyl ester is kept.
Scheme: Amide reduced to amine; ester untouched.
Conclusion. (i) Cyclohex-2-en-1-ol; (ii) cyclohexanone; (iii) and (iv) 4-methyl-δ-valerolactone (borane reduces the acid, LiAlH₄ the ester); (v) methyl 1-benzylpyrrolidine-3-carboxylate.
Application. Luche and borane chemoselectivity are routine in drug synthesis (prostaglandin allylic alcohols; amine drugs from amides without touching esters).
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All parts of Question 6
- 6(a) Chemoselective reductions10 marks
- 6(b) Nylons, combustion of polyethylene, polystyrene, Teflon and PVC20 marks
- 6(c) Natural rubber and gutta-percha; nucleic acids20 marks