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All 7 posts Subject: MDP2Pol formate ester Please login to post Down

ylid
(Stranger)
04-16-03 12:31
No 427249
      MDP2Pol formate ester
(Rated as: excellent)

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.rhodium.ws/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).

Who do you believe, me or your own eyes?

Rhodium
(Chief Bee)
04-16-03 23:56
No 427422
User Picture       Addition of formic acid to allylbenzene

J. Org. Chem. 39, 2157-2166 (1974) (https://www.rhodium.ws/pdf/allylbenzene.hcooh.addition.pdf)

Lilienthal
(Moderator)
04-17-03 08:41
No 427528
      I would expect the rearrangement to be a ...

I would expect the rearrangement to be an unimolecular reaction, thereby independent of the concentration of the free alcohol. If so, the esterification doesn't really make sense... The real point is probably to find reaction conditions were the oxidation is way faster than the rearrangement.

ylid
(Stranger)
04-17-03 13:53
No 427575
      Correction to Rhodium's link

https://www.rhodium.ws/pdf/allylbenzene.hcooh.addition.pdf

Who do you believe, me or your own eyes?

Rhodium
(Chief Bee)
04-18-03 03:03
No 427718
User Picture       Gracias! I have now corrected the link in my...

Gracias! I have now corrected the link in my post above.

otto
(Hive Bee)
04-21-03 22:22
No 428486
      no full conversion

This method is intriguing, although otto is not optimistic about the NH4NO3 - oxydation.

otto put 1 mL 3,4-Dimethoxyallylbezene (1,036 g) and 30 mL HCOOH into a tube, sealed it and heated at 80°C for 20 hours. The liquid turned to a slightly pinkish color. The solution was then poured into 50 mL of destilled water and extracted twice using 15 mL DCM each time. The combined organic layers were washed with concentrated NaHCO3 - solution, dried over MgSO4 and then evaporated to give 1.07 g of a yellowish oil. GC analysis of this oil shows a 1:1 mixture of starting material and a heavier peak, supposingly 3,4-DMP2Pol.

Otto estimates the time needed for full conversion to be higher. Alternatively, one can raise the temperature.

otto
(Hive Bee)
04-27-03 00:29
No 429662
      3,4-DMP2Pol higher conversion
(Rated as: excellent)

Experimental

1 mL 3,4-Dimethoxyallylbenzene (1.036 g, 5.8 mmol) and 30 mL HCOOH were put into a pressure-tube and heated at 120°C over night. The liquid turned red [1]. The solution was then poured into 50 mL of destilled water and extracted twice using 15 mL DCM each time. The combined organic layers were washed with concentrated NaHCO3 - solution, dried over MgSO4 and then evaporated to give 1.11 g of a yellow viscous [2] oil. GC analysis of this oil shows a 1:10 mixture of starting material and a heavier peak, supposingly 3,4-DMP2Pol.

TLC (hexane/ethyl acetate ~3:1) shows the following spots

3,4-Dimethoxyallylbezene RF 0.75
3,4-DMP2Pol RF 0.5
Impurity RF 0.4

[1] supposingly oxidation of phenolic material
[2] this time the product is viscous, whereas in the first reaction was not

The weak Spot at RF 0.75 of this reaction in comparison to a TLC of the above and the higher intensity of the RF 0.5 spot clearly shows the higher conversion. This result is consistent with GC spec, a higher product mass and higher viscosity as one would expect for alcohols.

No NMR was available, to check the nature of the product there was undertaken a test for presence of a RCH(OH)CH3 group: the Iodoform reaction. As most of you propably know, 3,4-DMP2Pol will form CHI3 but 3,4-DMP1Pol would not.

RCH(OH)CH3 + I2 + 2OH- --> RCOCH3 + 2H2O + 2I-
R(C=O)CH3 + 3I2 + 3OH- --> R(C=O)CI3 + 3H2O + 3I-
R(C=O)CI3 + OH-    --> CHI3 + RCOO-

The test is easily carried out in a test tube by adding aqueous KI*I2 solution to a solution of one drop of substance in 10 mL water and 1 mL 2N NaOH. CHI3 will turn the solution cloudy and has a characteristic smell. The test was postive.

Conclusion

3,4-DMP2Pol can be prepared by the method stated above. The reported [3] moderate yield from a related substrate seems to arrise from insufficient conversion of the reactants. 120°C does not seem to be too high a temperature. However it remains still unclear wether rearrangement to the 3,4-DMP1Pol takes place or not. A product sample stored at room temperature not protected from light gives a positive Iodoform test still after one week.
The final yield can not exactly be told as the purity of the product is not known. The 1:10 peak ratio in GC and the absence of other peaks as well as an increase in weight near the calculated suggests a yield of 80 to 90 percent.

[3] Kwart H, Drayer D (1974) J.Org.Chem. 39:2157-2166. See first post in this threat

otto

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	All 7 posts		Subject: MDP2Pol formate ester		Please login to post		Down


		ylid (Stranger) 04-16-03 12:31 No 427249				MDP2Pol formate ester (Rated as: excellent)
						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.rhodium.ws/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). Who do you believe, me or your own eyes?



		Rhodium (Chief Bee) 04-16-03 23:56 No 427422				Addition of formic acid to allylbenzene
						J. Org. Chem. 39, 2157-2166 (1974) (https://www.rhodium.ws/pdf/allylbenzene.hcooh.addition.pdf)



		Lilienthal (Moderator) 04-17-03 08:41 No 427528				I would expect the rearrangement to be a ...
						I would expect the rearrangement to be an unimolecular reaction, thereby independent of the concentration of the free alcohol. If so, the esterification doesn't really make sense... The real point is probably to find reaction conditions were the oxidation is way faster than the rearrangement.



		ylid (Stranger) 04-17-03 13:53 No 427575				Correction to Rhodium's link
						https://www.rhodium.ws/pdf/allylbenzene.hcooh.addition.pdf Who do you believe, me or your own eyes?



		Rhodium (Chief Bee) 04-18-03 03:03 No 427718				Gracias! I have now corrected the link in my...
						Gracias! I have now corrected the link in my post above.



		otto (Hive Bee) 04-21-03 22:22 No 428486				no full conversion
						This method is intriguing, although otto is not optimistic about the NH4NO3 - oxydation. otto put 1 mL 3,4-Dimethoxyallylbezene (1,036 g) and 30 mL HCOOH into a tube, sealed it and heated at 80°C for 20 hours. The liquid turned to a slightly pinkish color. The solution was then poured into 50 mL of destilled water and extracted twice using 15 mL DCM each time. The combined organic layers were washed with concentrated NaHCO3 - solution, dried over MgSO4 and then evaporated to give 1.07 g of a yellowish oil. GC analysis of this oil shows a 1:1 mixture of starting material and a heavier peak, supposingly 3,4-DMP2Pol. Otto estimates the time needed for full conversion to be higher. Alternatively, one can raise the temperature.



		otto (Hive Bee) 04-27-03 00:29 No 429662				3,4-DMP2Pol higher conversion (Rated as: excellent)
						Experimental 1 mL 3,4-Dimethoxyallylbenzene (1.036 g, 5.8 mmol) and 30 mL HCOOH were put into a pressure-tube and heated at 120°C over night. The liquid turned red [1]. The solution was then poured into 50 mL of destilled water and extracted twice using 15 mL DCM each time. The combined organic layers were washed with concentrated NaHCO3 - solution, dried over MgSO4 and then evaporated to give 1.11 g of a yellow viscous [2] oil. GC analysis of this oil shows a 1:10 mixture of starting material and a heavier peak, supposingly 3,4-DMP2Pol. TLC (hexane/ethyl acetate ~3:1) shows the following spots 3,4-Dimethoxyallylbezene RF 0.75 3,4-DMP2Pol RF 0.5 Impurity RF 0.4 [1] supposingly oxidation of phenolic material [2] this time the product is viscous, whereas in the first reaction was not The weak Spot at RF 0.75 of this reaction in comparison to a TLC of the above and the higher intensity of the RF 0.5 spot clearly shows the higher conversion. This result is consistent with GC spec, a higher product mass and higher viscosity as one would expect for alcohols. No NMR was available, to check the nature of the product there was undertaken a test for presence of a RCH(OH)CH3 group: the Iodoform reaction. As most of you propably know, 3,4-DMP2Pol will form CHI3 but 3,4-DMP1Pol would not. RCH(OH)CH3 + I2 + 2OH- --> RCOCH3 + 2H2O + 2I- R(C=O)CH3 + 3I2 + 3OH- --> R(C=O)CI3 + 3H2O + 3I- R(C=O)CI3 + OH- --> CHI3 + RCOO- The test is easily carried out in a test tube by adding aqueous KII2 solution to a solution of one drop of substance in 10 mL water and 1 mL 2N NaOH. CHI3 will turn the solution cloudy and has a characteristic smell. The test was postive. Conclusion* 3,4-DMP2Pol can be prepared by the method stated above. The reported [3] moderate yield from a related substrate seems to arrise from insufficient conversion of the reactants. 120°C does not seem to be too high a temperature. However it remains still unclear wether rearrangement to the 3,4-DMP1Pol takes place or not. A product sample stored at room temperature not protected from light gives a positive Iodoform test still after one week. The final yield can not exactly be told as the purity of the product is not known. The 1:10 peak ratio in GC and the absence of other peaks as well as an increase in weight near the calculated suggests a yield of 80 to 90 percent. [3] Kwart H, Drayer D (1974) J.Org.Chem. 39:2157-2166. See first post in this threat otto

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