Q3(b). 2,6-Dimethylphenol: photo-Fries and para-Claisen rearrangement
Write the structures of the major products M, N, O and P in the following reaction sequence. Suggest a mechanism for the conversion of O to P.
Answer in brief
M = 2,6-dimethylphenyl acetate; N = 4-hydroxy-3,5-dimethylacetophenone (photo-Fries to para); O = the (E)-crotyl ether; P = (E)-4-(but-2-en-1-yl)-2,6-dimethylphenol, formed by a Claisen and then a Cope [3,3] shift.
Key points
- At 220°C the allyl aryl ether undergoes a Claisen [3,3]-sigmatropic rearrangement through a six-membered chair transition state: the O–C bond breaks while the terminal alkene carbon bonds to the ortho ring carbon. The allyl…
- The ortho carbon already carries a methyl group, so the 6,6-disubstituted dienone cannot tautomerise to a phenol.
- A second [3,3] shift (a Cope rearrangement) moves the allyl group from the ortho to the para carbon, inverting it again; the chain is now attached through its original CH₂.
- The para dienone has an H on the para carbon; it tautomerises to the aromatic phenol P.
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. 2,6-Dimethylphenol has both ortho positions blocked by methyl groups. Any group that migrates from oxygen to the ring must therefore end up at the para position. The left route is a photo-Fries rearrangement of the acetate; the right route is a para-Claisen rearrangement of an allylic ether.
Reading of the question: the alkenyl bromide is printed as CH₃CH=CHBr; a vinylic bromide does not alkylate a phenoxide, so it is read as (E)-crotyl bromide, CH₃CH=CHCH₂Br. (With allyl bromide the same steps give 4-allyl-2,6-dimethylphenol.)
M and N
Scheme: M = 2,6-dimethylphenyl acetate; N = 1-(4-hydroxy-3,5-dimethylphenyl)ethan-1-one
M. K₂CO₃ forms the phenoxide, which is acylated on oxygen: M = 2,6-dimethylphenyl acetate.
N. Light breaks the ester O–C(O) bond (Norrish type I), giving an acyl radical and a phenoxy radical held in a solvent cage. The acyl radical recombines with a ring carbon where the phenoxy radical has spin density, the ortho or para carbon. Both ortho carbons carry methyl groups, so it bonds at para; tautomerisation restores the phenol: N = 4-hydroxy-3,5-dimethylacetophenone. Some 2,6-dimethylphenol forms from radicals that escape the cage.
O and P
O. NaH forms the sodium phenoxide, which displaces bromide by SN2: O = (E)-2-(but-2-en-1-yloxy)-1,3-dimethylbenzene, the crotyl ether.
P (mechanism).
- At 220°C the allyl aryl ether undergoes a Claisen [3,3]-sigmatropic rearrangement through a six-membered chair transition state: the O–C bond breaks while the terminal alkene carbon bonds to the ortho ring carbon. The allyl group is inverted (it is now attached through the carbon that carried the CH₃).
- The ortho carbon already carries a methyl group, so the 6,6-disubstituted dienone cannot tautomerise to a phenol.
- A second [3,3] shift (a Cope rearrangement) moves the allyl group from the ortho to the para carbon, inverting it again; the chain is now attached through its original CH₂.
- The para dienone has an H on the para carbon; it tautomerises to the aromatic phenol P.
Scheme: Two [3,3] shifts invert the crotyl group twice: it arrives at para attached through CH₂, with the C=C still E.
Conclusion. M = 2,6-dimethylphenyl acetate; N = 4-hydroxy-3,5-dimethylacetophenone (photo-Fries to para); O = the (E)-crotyl ether; P = (E)-4-(but-2-en-1-yl)-2,6-dimethylphenol, formed by a Claisen and then a Cope [3,3] shift.
Application. The Claisen rearrangement is used to make allyl phenols such as eugenol analogues and the precursors of vitamin E (tocopherol); photo-Fries chemistry explains the yellowing of aromatic polyester films in sunlight.
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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