Q3(c). Benzyne isotope effect; Stevens ring enlargement
(i) Write the structures of the products formed in the following reactions. Which one of these two reactions would proceed faster? Justify your answer.
Reaction 1 [structure] → (NaNH₂, NH₃ (l)) Reaction 2 [structure] → (NaNH₂, NH₃ (l))
(ii) Write the structures of the major products X and Y formed in the following reaction sequence. Write a suitable mechanism for the formation of Y from X.
[structure] → (MeI) X → (NaNH₂, NH₃ (l)) Y
Answer in brief
(i) Reaction 1 gives aniline; Reaction 2 gives aniline-2-d and aniline-3-d. Reaction 1 is faster (primary isotope effect, about 5), because breaking the ortho C–H bond to form benzyne is rate-determining. (ii) X = the N-benzyl-N-methyl ammonium salt; Y = the ring-enlarged 3-methyl-2-phenyl-3-benzazonine, by a Stevens [1,2] shift of the ammonium ylide.
Key points
- NH₂⁻ removes the most acidic proton, a benzylic H of the exocyclic N-CH₂Ph, giving the ammonium ylide N+-CH⁻Ph.
- The ring CH₂ that links N to the benzo ring (also benzylic, a good migrating group) moves from nitrogen to the carbanion carbon: a [1,2]-Stevens rearrangement. It goes by homolysis of this N–C bond to a radical pair inside the…
- The ylide carbon is inserted into the ring: the eight-membered ring grows to nine atoms, and nitrogen becomes a tertiary amine again.
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) is nucleophilic substitution of an unactivated aryl halide by NH₂⁻, which goes by elimination–addition through benzyne; deuterium at the ortho positions tests whether C–H breaking is rate-limiting. (ii) is a quaternary ammonium salt treated with a strong base: an ammonium ylide forms and rearranges.
(i) Products and relative rates
Reaction 1. NH₂⁻ removes the ortho H; bromide leaves to form benzyne; NH₂⁻ adds to either end and NH₃ protonates the aryl anion. Product: aniline.
Reaction 2. Here both ortho positions carry D, so NH₂⁻ must remove D⁺. The benzyne formed keeps one D, next to the triple bond. Addition at the former C–Br carbon gives aniline-2-d; addition at the other end gives aniline-3-d, in roughly equal amounts. (The proton that ends on the ring comes from NH₃.)
Scheme: Elimination–addition: the NH₂ enters at C-1 or at C-2 of the benzyne.
Which is faster? Reaction 1. The slow step is removal of the ortho hydrogen (together with, or just before, loss of Br⁻). A C–D bond has a lower zero-point energy than C–H, so it needs more energy to break: a primary kinetic isotope effect, k_H/k_D≈ 5 (Roberts' value for bromobenzene-2-d is 5.5). A large k_H/k_D is itself evidence that benzyne formation, not addition of NH₂⁻, is rate-determining.
(ii) X and Y: Stevens ring enlargement
X. Methyl iodide quaternises the tertiary amine: X = 2-benzyl-2-methyl-1,2,3,4,5,6-hexahydro-2-benzazocinium iodide.
Y (mechanism).
- NH₂⁻ removes the most acidic proton, a benzylic H of the exocyclic N-CH₂Ph, giving the ammonium ylide N+-CH⁻Ph.
- The ring CH₂ that links N to the benzo ring (also benzylic, a good migrating group) moves from nitrogen to the carbanion carbon: a [1,2]-Stevens rearrangement. It goes by homolysis of this N–C bond to a radical pair inside the solvent cage and fast recombination, so the migrating carbon keeps its configuration.
- The ylide carbon is inserted into the ring: the eight-membered ring grows to nine atoms, and nitrogen becomes a tertiary amine again.
Scheme: Y = 3-methyl-2-phenyl-2,3,4,5,6,7-hexahydro-1H-3-benzazonine (one-carbon ring enlargement)
A Sommelet–Hauser [2,3] shift would need an ylide carbon that can reach the ortho carbon of a benzyl group; with this ring the Stevens [1,2] product is formed as the sole product.
Conclusion. (i) Reaction 1 gives aniline; Reaction 2 gives aniline-2-d and aniline-3-d. Reaction 1 is faster (primary isotope effect, about 5), because breaking the ortho C–H bond to form benzyne is rate-determining. (ii) X = the N-benzyl-N-methyl ammonium salt; Y = the ring-enlarged 3-methyl-2-phenyl-3-benzazonine, by a Stevens [1,2] shift of the ammonium ylide.
Application. Benzyne routes are used to make substituted anilines and natural-product skeletons; ammonium-ylide ring enlargements give medium-ring amines found in alkaloids and in some CNS drugs.
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All parts of Question 3
- 3(a) Tollens condensation; aldol, Jones oxidation and Fischer indole15 marks
- 3(b) 2,6-Dimethylphenol: photo-Fries and para-Claisen rearrangement15 marks
- 3(c) Benzyne isotope effect; Stevens ring enlargement20 marks