Q1(d). Baeyer–Villiger oxidation; Bischler–Napieralski synthesis
Identify the product (X) in the following reactions and write the name of the reaction involved:
(i) [structure] → (mCPBA) (X)(ii) [structure] → ((1) CH₃COCl(2) P₂O₅, (3) Pd/C100°C) (X)
Answer in brief
(i) Baeyer–Villiger oxidation: the ring-fusion carbon migrates to O, giving the seven-membered lactone X. (ii) Bischler–Napieralski reaction followed by Pd/C dehydrogenation: X = 1-methylisoquinoline.
Key points
- The peracid adds to the C=O to give the tetrahedral Criegee intermediate, C(OH)(O-O-COAr).
- In the rate-determining step one α-carbon migrates from C to O as ArCO₂⁻ leaves. Migratory aptitude: tertiary > secondary > primary. At C-1 the ring-fusion CH (C-8a, secondary) migrates rather than the CH₂ (C-2, primary), and…
- The C-6 ketone has two CH₂ neighbours, gives no regiochemical preference, and is the slower one.
- N-(2-phenylethyl)acetamide.
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) A peracid converts a ketone into an ester (a cyclic ketone into a lactone) by the Baeyer–Villiger oxidation; the more substituted carbon migrates. (ii) A β-arylethylamide closes to a dihydroisoquinoline by the Bischler–Napieralski reaction; Pd/C then dehydrogenates it to the aromatic isoquinoline.
(i) Baeyer–Villiger oxidation
Reading of the question: the 1,6-dione is read with one equivalent of mCPBA, which attacks the C-1 ketone; excess peracid would also convert the C-6 ketone into a lactone.
Scheme: O is inserted between C-1 and the ring-fusion carbon, which migrates with retention.
- The peracid adds to the C=O to give the tetrahedral Criegee intermediate, C(OH)(O-O-COAr).
- In the rate-determining step one α-carbon migrates from C to O as ArCO₂⁻ leaves. Migratory aptitude: tertiary > secondary > primary. At C-1 the ring-fusion CH (C-8a, secondary) migrates rather than the CH₂ (C-2, primary), and it moves with retention of configuration.
- The C-6 ketone has two CH₂ neighbours, gives no regiochemical preference, and is the slower one.
X is the ε-lactone: the original ring A becomes a seven-membered ring containing -O-CO⁻, fused to the cyclohexanone ring (an octahydro-1-benzoxepine-2,7-dione, the second C=O kept).
(ii) Bischler–Napieralski reaction, then dehydrogenation
Scheme: Acylation, cyclodehydration (SEAr on a nitrilium ion) and aromatisation.
- Acetyl chloride acylates the amine: N-(2-phenylethyl)acetamide.
- P₂O₅ activates the amide oxygen; loss of water gives an electrophilic nitrilium ion (or imidoyl phosphate), which attacks the ortho position of the ring (intramolecular electrophilic substitution, six-membered ring closure). Loss of H⁺ gives 1-methyl-3,4-dihydroisoquinoline.
- Pd/C at 100°C removes two H from C-3 and C-4: the ring becomes aromatic.
X = 1-methylisoquinoline.
Conclusion. (i) Baeyer–Villiger oxidation: the ring-fusion carbon migrates to O, giving the seven-membered lactone X. (ii) Bischler–Napieralski reaction followed by Pd/C dehydrogenation: X = 1-methylisoquinoline.
Application. Baeyer–Villiger oxidation makes caprolactone (for polycaprolactone) and steroid lactones; the Bischler–Napieralski route is the standard way into isoquinoline alkaloids such as papaverine.
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All parts of Question 1
- 1(a) Is [14]annulene aromatic?10 marks
- 1(b) One benzyne from two dichlorobenzenes10 marks
- 1(c) Neopentyl tosylate: 1,2-methyl shift10 marks
- 1(d) Baeyer–Villiger oxidation; Bischler–Napieralski synthesis10 marks
- 1(e) Thermal 6π electrocyclisation of a dienyl cyclohexene10 marks