Another go at this old canard...
It seems likely that the oxidation of MDP2Pol to MDP2P suffers from the rearrangement of the alcohol to MDP1Pol at room temperature[
https://www.thevespiary.org/rhodium/Rhodium/chemistry/mdp2pol.html
]. If this is true, it might be possible to perform the oxidation at low temps without having the alcohol hang around in the reaction mixture, by using the formate ester instead.
It looks like formate esters of 1-phenyl-propan-2-ols can be prepared in fair yield by addition of 97% formic acid at 77.5°C
1. With a bit of luck, formic acid can be added to safrole in a similar way.
What I am wondering is whether it is possible to prepare MDP2P directly from this ester, without first hydrolysing it to isolate MDP2Pol. Perhaps the necessary hydrolysis/oxidation might work using Cu(NO3)2 and NH4NO3 in dilute HNO3. Acid conditions would suit the hydrolysis of the ester, and with a suitable excess of Cu++ and cool enough temperatures it might just be possible to oxidise the alcohol that results before it rearranges to MDP1Pol.
I guess the hydrolysis would proceed rather slowly at low temps. But maybe this way of doing the oxidation would be an improvement because the alcohol only comes into solution as quickly as the ester is hydrolysed, so that at any one time only small concentrations of alcohol are present in solution. Am I right that this would improve the reaction dynamics by favouring the oxidation over the rearrangement?
On the other hand, Terbium's posts on this topic suggest that MDP1Pol is formed by rearrangement of the intermediate carbocation when the C=C double bond is protonated. In which case, it doesn't matter whether H2SO4 or HCOOH is used as the acid, the method is doomed.
The oxidation (properly dehydrogenation) reactions would presumably look like this:
Ar-CH2CH(OOCH)CH3 + H2O
__> Ar-CH2CH(OH)CH3 + HCOOH
Ar-CH2CH(OH)CH3 + Cu++
__> Ar-CH2CH(O)CH3 + Cu+ + H+
2 Cu+ + NH4NO3 + 2 H+
__> 2 Cu++ + N2 + 3 H2O
1 Kwart H, Drayer D (1974)
J.Org.Chem. 39:2157-2166.
Addition of 97% HCOOH to allylbenzene. A solution of 5.00g (42.3mmol) of allylbenzene in 120ml of 97% HCOOH when heated for 40h at 77.5°C yielded 4.00g (58%) of alpha-methylphenyl alcohol formate, a colorless liquid, bp 58-59°C (0.50 mm).
Addition of 97% HCOOH to o-allylanisole. After a solution of 3.30g (22.3mmol) of o-allylanisole in 120ml of 97% HCOOH when heated for 9.5h at 77.5°C yielded 2.1g (48%) of o-methoxy-alpha-methylphenyl alcohol formate, a colorless liquid, bp 76-78°C (0.40 mm).