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Author: Subject: Sulfuric acid for Methyl Salicylate
Abromination
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[*] posted on 2-12-2018 at 20:36
Sulfuric acid for Methyl Salicylate


Very beginner question:
How much will the yeild of methyl salicylate be effected in the Fischer esterification of methanol and salicylic acid if we use sulfuric acid significantly less concentrated then 93%?




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[*] posted on 2-12-2018 at 21:27


How much less concentrated are we talking? The fischer esterification is an equilibrium reaction. Water will reduce the yield, but it won't quantitatively destroy product or anything.
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[*] posted on 2-12-2018 at 22:43


I havent been able to do a titration, but my sulfuric acid is between 30-60%



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[*] posted on 2-12-2018 at 23:51


Sulfuric acid is not too difficult to boil down. It does bump like crazy so use an oversize beaker. When it begins to fume profusely you have better than 70% which should be good for your purpose.
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[*] posted on 3-12-2018 at 03:03


Alternatively, if you can use a Dean-Stark trap (I DIY one from a handful of adapters) then it'll work just fine
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[*] posted on 3-12-2018 at 03:31


Quote: Originally posted by DavidJR  
Alternatively, if you can use a Dean-Stark trap (I DIY one from a handful of adapters) then it'll work just fine


mhhh how? the low boiling components are water and methanol, miscible in each other, methyl salicylate boils at 222°C, and is miscible with methanol, so something like a steam distillation (mostly methanol, some water and a bit of ester) wouldn't really work in my opinion, if it does could explain where i'm wrong?





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[*] posted on 3-12-2018 at 04:08


Quote: Originally posted by Ubya  
Quote: Originally posted by DavidJR  
Alternatively, if you can use a Dean-Stark trap (I DIY one from a handful of adapters) then it'll work just fine


mhhh how? the low boiling components are water and methanol, miscible in each other, methyl salicylate boils at 222°C, and is miscible with methanol, so something like a steam distillation (mostly methanol, some water and a bit of ester) wouldn't really work in my opinion, if it does could explain where i'm wrong?



Sorry, you're right, it won't work in this case. I was half asleep while writing that and was thinking about esterification generally...

In many cases though for esterification, it does work pretty well - as long as your ester forms a low-boiling azeotrope with water. I used this approach recently to prepare ethyl chloroacetate.
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[*] posted on 3-12-2018 at 10:56


Oh well...
I generally prefer not having to concentrate sulfuric acid but it doesn't lool like I'll have a choice if I want a decent yeild.




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[*] posted on 3-12-2018 at 22:21


I would look into using NaHSO4 as a catalyst. It is not terribly soluble, but it works. It might even give you better yields.

The reaction will proceed even without a catalyst, fyi. Whenever I have made methyl salicylate I recrystallize the SA from methanol, and it smells like wintergreen after it has dried.
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[*] posted on 4-12-2018 at 18:04


Update on this, my fumehood is not made for dense sulfuric acid fumes. Incredibly stupid of me to not just do it outside. I managed to put on a respirator and put it outside as it had just began to fume, but I could have put myself in an incredibly dangerous situation if I had not been present when it reached a high concentration.
Let this be a lesson to people like me.




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[*] posted on 5-12-2018 at 09:12


Quote: Originally posted by Abromination  
Update on this, my fumehood is not made for dense sulfuric acid fumes. Incredibly stupid of me to not just do it outside. I managed to put on a respirator and put it outside as it had just began to fume, but I could have put myself in an incredibly dangerous situation if I had not been present when it reached a high concentration.
Let this be a lesson to people like me.


Perhaps a suggestion to a safer path would be producing C6H4(OH)(CO2CH3) starting also with CH3OH and C6H4(OH)COOH, but just adding some light.

To be precise, I am proposing the simple UV photolysis of CH3OH in the presence of C6H4(OH)COOH. It may be a safer (than trying to conc H2SO4) and perhaps even better yield in what may be a new synthesis path!

Based on my prior research and very limited experience, a possible reaction scheme would be:

CH3OH + hv --> •CH3 + •OH (and other paths with associated products, see http://chemistry.emory.edu/faculty/widicusweaver/photolysis.... and do note the cool picture on the bottom of the web page!)

Next, an assumption (based on the side note below) as to which H atom would be most likely extracted by the hydroxyl radical:

•OH + C6H4(OH)COOH --> H2O + •C6H4(OH)COO

Followed by, to a limited degree based on random collisions:

•C6H4(OH)COO + •CH3 --> C6H4(OH)(CO2CH3)

However, as to when to terminate the UV treatment is a matter of experimenting. There are also other possible paths to products and even sensitivity of the target product itself to UV (which may reduce yield).

A side note, the hydrolysis of Methyl salicylate in the present of a strong acid or base returns Salicylic acid:

C6H4(OH)(CO2CH3) + H2O ---Strong Base or Acid---> C6H4(OH)COOH + Product (see https://en.wikipedia.org/wiki/Methyl_salicylate )

[Edited on 5-12-2018 by AJKOER]
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[*] posted on 5-12-2018 at 09:21


Quote: Originally posted by AJKOER  
Quote: Originally posted by Abromination  
Update on this, my fumehood is not made for dense sulfuric acid fumes. Incredibly stupid of me to not just do it outside. I managed to put on a respirator and put it outside as it had just began to fume, but I could have put myself in an incredibly dangerous situation if I had not been present when it reached a high concentration.
Let this be a lesson to people like me.


Perhaps a suggestion to a safer path would be producing C6H4(OH)(CO2CH3) starting also with CH3OH and C6H4(OH)COOH, but just adding some UV light.

I am proposing the simple UV photolysis of CH3OH in the presence of C6H4(OH)COOH. It may be a safer (than trying to conc H2SO4) and perhaps even better yield in what may be a new synthesis path!

Based on my prior research and very limited experience, a possible reaction scheme would be:

CH3OH + hv --> •CH3 + •OH (and other paths with associated products, see http://chemistry.emory.edu/faculty/widicusweaver/photolysis.... and do note the cool picture on the bottom of the web page!)

Next, an assumption (based on the side note below) as to which H atom would be most likely extracted by the hydroxyl radical:

•OH + C6H4(OH)COOH --> H2O + •C6H4(OH)COO

Followed by, to a limited degree based on random collisions:

•C6H4(OH)COO + •CH3 --> C6H4(OH)(CO2CH3)

However, as to when to terminate the UV treatment is a matter of experimenting. There are also other possible paths to products and even sensitivity of the target product itself to UV (which may reduce yield).

A side note, the hydrolysis of Methyl salicylate in the present of a strong acid or base returns Salicylic acid:

C6H4(OH)(CO2CH3) + H2O ---Strong Base or Acid---> C6H4(OH)COOH + Product (see https://en.wikipedia.org/wiki/Methyl_salicylate )

[Edited on 5-12-2018 by AJKOER]


Do you ever stop spouting such unsubstantiated rubbish?


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[*] posted on 5-12-2018 at 10:30


Quote: Originally posted by DavidJR  
Quote: Originally posted by AJKOER  
Quote: Originally posted by Abromination  
Update on this, my fumehood is not made for dense sulfuric acid fumes. Incredibly stupid of me to not just do it outside. I managed to put on a respirator and put it outside as it had just began to fume, but I could have put myself in an incredibly dangerous situation if I had not been present when it reached a high concentration.
Let this be a lesson to people like me.


Perhaps a suggestion to a safer path would be producing C6H4(OH)(CO2CH3) starting also with CH3OH and C6H4(OH)COOH, but just adding some UV light.

I am proposing the simple UV photolysis of CH3OH in the presence of C6H4(OH)COOH. It may be a safer (than trying to conc H2SO4) and perhaps even better yield in what may be a new synthesis path!

Based on my prior research and very limited experience, a possible reaction scheme would be:

CH3OH + hv --> •CH3 + •OH (and other paths with associated products, see http://chemistry.emory.edu/faculty/widicusweaver/photolysis.... and do note the cool picture on the bottom of the web page!)
.......
[Edited on 5-12-2018 by AJKOER]


Do you ever stop spouting such unsubstantiated rubbish?



By the way, without any drama, I used photolysis on several occasions on this forum (with success and safely) to produce chlorates and other products. My source cited above is hardly 'unsubstantiated' (but you can share your opinion directly with Emory University Chemistry Dept) and, I would add, a very honest picture of a significant issue associated with photolysis relating to the array of possible products, with some paths and products, however, being selectively preferred.

I suspect you have not been exposed to any of the recent literature or have significant knowledge of the underlying radical chemistry, so please comment again when you have more knowledge/experience to share.

Also, I would ask you to read, yet again, the comment below so that you understand the context under which my suggestion was made.

Quote: Originally posted by Abromination  
Update on this, my fumehood is not made for dense sulfuric acid fumes. Incredibly stupid of me to not just do it outside. I managed to put on a respirator and put it outside as it had just began to fume, but I could have put myself in an incredibly dangerous situation if I had not been present when it reached a high concentration.
Let this be a lesson to people like me.


By the way, what is your opinion on the photo of the UV synthesis I alluded too on the right at the bottom of the web page?

Cool I think, not cool is a short clip of me trying to escape a hot sulfuric acid shower!
-------------------------------------

Here is a better source perhaps (a bit dated, from 2010, from some school called Harvard, see http://adsbit.harvard.edu/cgi-bin/nph-iarticle_query?bibcode... ) showing only 3 primary paths for the UV photolysis of methanol:

CH3OH + hv --> •CH3 + •OH (1a)

CH3OH + hv --> •CH2OH + •H (1b)

CH3OH + hv --> •CH3O + •H (1c)

But, no cool picture.:(

Note, radical reactions involving •H are usually much slower than •OH. Also, to quote a source on the stability of Methyl salicylate under UV exposure (see https://toxnet.nlm.nih.gov/cgi-bin/sis/search/a?dbs+hsdb:@te... ):

“Methyl salicylate will react with photochemically-produced hydroxyl radicals in the atmosphere resulting in as estimated half-life of 1.4 days.”

Also:
"The UV absorption maximum of a methanol solution of methyl salicylate is 305 nm(1) which indicates that methyl salicylate can undergo direct photolysis. One photolysis study was performed which yielded a half-life of methyl salicylate in solution of about 48 min(2). The exact medium was not identified, but the authors stated that compounds with low water solubilities were dissolved in a 10% ethanol-water mixture. Based on the concentration of methyl salicylate which was used (3.5 g/L), and the water solubility of the ester of 7.4 g/L(3), it is likely that the result above was obtained using an ethanol-water mixture or absolute ethanol. The UV dose used in the study was 0.0077 erg/sq cm/min, and the UV wavelengths ranged from 300 nm to 400 nm with a maximum at 350 nm. Direct photolysis may, therefore, be an important degradative process in the environment; no available data, however, were collected under environmental conditions."

[Edited on 5-12-2018 by AJKOER]
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[*] posted on 5-12-2018 at 12:48


I am not disputing that UV photolysis of methanol can occur. What I am disputing is the idea that it's in any way useful for making esters. If you can provide experimental evidence to the contrary then I apologise. Until then, I disagree.


[Edited on 5-12-2018 by DavidJR]
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[*] posted on 5-12-2018 at 14:11


Quote: Originally posted by DavidJR  
I am not disputing that UV photolysis of methanol can occur. What I am disputing is the idea that it's in any way useful for making esters. If you can provide experimental evidence to the contrary then I apologise. Until then, I disagree.

[Edited on 5-12-2018 by DavidJR]


On p. 642 of Photochemistry, Volume 6, edited by D Bryce-Smith, I did find some reagent plus methanol and warming under photo irradiation forming a methyl salicylate product. See https://books.google.com/books?id=s3UoDwAAQBAJ&pg=PA642&... .

As the prior page is not available, unclear on the reagent used in the photolysis with methanol and warming.

So, potentially, I am but half mad and you are just half right.;)

[Edited on 5-12-2018 by AJKOER]
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[*] posted on 5-12-2018 at 15:04


This got heated quickly...
I might try the photolysis method for fun, but I have a few kilos of sodium metabisulfite coming in the mail for an SO2 generator coming in the mail, Ill just make my acid that way. Way cheaper than what I was doing before for my acid, and hydrogen peroxide is cheap as dirt. I think Ill just stick to the acid.:P




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[*] posted on 6-12-2018 at 04:14


Quote: Originally posted by AJKOER  

On p. 642 of Photochemistry, Volume 6, edited by D Bryce-Smith, I did find some reagent plus methanol and warming under photo irradiation forming a methyl salicylate product. See https://books.google.com/books?id=s3UoDwAAQBAJ&pg=PA642&... .

As the prior page is not available, unclear on the reagent used in the photolysis with methanol and warming.

So, potentially, I am but half mad and you are just half right.;)

[Edited on 5-12-2018 by AJKOER]


That talks about a peroxide which salicylic acid is not. So I don't really think that supports your statements.

And frankly, even if it is possible (which I don't believe), then that still doesn't mean it would be in any way preferable to Fischer-Speier esterification. And I also debate that it would be safer than using sulphuric acid...
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[*] posted on 6-12-2018 at 05:39


Quote: Originally posted by DavidJR  

......................
That talks about a peroxide which salicylic acid is not. So I don't really think that supports your statements.

And frankly, even if it is possible (which I don't believe), then that still doesn't mean it would be in any way preferable to Fischer-Speier esterification. And I also debate that it would be safer than using sulphuric acid...


Per Wikipedia on Organic peroxide (https://en.wikipedia.org/wiki/Organic_peroxide ):

“Organic peroxides are organic compounds containing the peroxide functional group (ROOR′). If the R′ is hydrogen, the compounds are called organic hydroperoxides. Peresters have general structure RC(O)OOR. The O−O bond easily breaks, producing free radicals of the form RO•.

So, the final synthesis equation is, starting with the irradiation of the peroxide compound (177):

.RO + .CH3 = C6H4(OH)(COOCH3)

So, .RO is a radical of the form .C6H4(OH)COO, which is precisely the radical created by the action of a hydroxyl radical (or the hydrogen atom radical as well yielding H2 and not H2O) on C6H4(OH)COOH as I detailed originally:

.OH + C6H4(OH)COOH --> H2O + .C6H4(OH)COO

.H + C6H4(OH)COOH --> H2 + .C6H4(OH)COO

Now, there could be cost or yield advantages in employing the peroxide (177) for the photosynthesis (to be determined) as, per my reaction equations above, some water is introduced.

[Edited on 6-12-2018 by AJKOER]
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[*] posted on 6-12-2018 at 06:25


Quote: Originally posted by AJKOER  
Quote: Originally posted by DavidJR  

......................
That talks about a peroxide which salicylic acid is not. So I don't really think that supports your statements.

And frankly, even if it is possible (which I don't believe), then that still doesn't mean it would be in any way preferable to Fischer-Speier esterification. And I also debate that it would be safer than using sulphuric acid...


Per Wikipedia on Organic peroxide (https://en.wikipedia.org/wiki/Organic_peroxide ):

“Organic peroxides are organic compounds containing the peroxide functional group (ROOR′). If the R′ is hydrogen, the compounds are called organic hydroperoxides. Peresters have general structure RC(O)OOR. The O−O bond easily breaks, producing free radicals of the form RO•.

So, the final synthesis equation is, starting with the irradiation of the peroxide compound (177):

.RO + .CH3 = C6H4(OH)(COOCH3)

So, .RO is a radical of the form .C6H4(OH)COO, which is precisely the radical created by the action of a hydroxyl radical (or the hydrogen atom radical as well yielding H2 and not H2O) on C6H4(OH)COOH as I detailed originally:

.OH + C6H4(OH)COOH --> H2O + .C6H4(OH)COO

.H + C6H4(OH)COOH --> H2 + .C6H4(OH)COO

Now, there could be cost or yield advantages in employing the peroxide (177) for the photosynthesis (to be determined) as, per my reaction equations above, some water is introduced.

[Edited on 6-12-2018 by AJKOER]


Ok, I'm willing to accept that it's one possible product. But that doesn't mean it's a practical synthetic route. I believe there will also be many other products too.
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[*] posted on 6-12-2018 at 07:51


Quote: Originally posted by DavidJR  
.....Ok, I'm willing to accept that it's one possible product. But that doesn't mean it's a practical synthetic route. I believe there will also be many other products too.


True, but sometimes the attraction/stability of two radicals (referred to as selectivity) is so high that the yield is quite surprising, see K. Ogura, "Photolysis of CH4, NH3, H2O mixture: formation of methylamine and ethylenediamine", in Journal of Photochemistry and Photobiology A: Chemistry, Volume 49, Issues 1–2, September 1989, Pages 53-61, https://doi.org/10.1016/1010-6030(89)87105-9, link: http://www.sciencedirect.com/science/article/pii/10106030898... . Here is the abstract:

"The photolysis of the mixture CH4, NH3, H2O was performed with a low pressure mercury lamp at 100 °C and atmospheric pressure. Of the products, the major nitrogen-containing compounds were methylamine and ethylenediamine; the maximum selectivity of the two amines exceeded 99% in the mol per cent of N-containing products. The other products were methanol, ethane and hydrogen, but the formation of methanol and ethane decreased rapidly with an increase in the amount of added ammonia. This was attributed to the preferential reactivity of methyl radicals with NH2 rather than with OH and/or CH3. The formation rates of NH2, CH3NH2 and NH2C2H4NH2 are discussed."
-------------------------------------------

The proposed photolysis path to methyl salicylate may, as of yet, have an unmentioned favorable feature as the starting reagents are both colorless, but methyl salicylate can be yellow to red in appearance and has a distinctive odor. This could assist in determining the stopping point of the UV/warming treatment together with knowledge of its half-life under irradiation. An example, see my picture following a photolysis treatment at http://www.sciencemadness.org/talk/viewthread.php?tid=77186&... .

[Edited on 7-12-2018 by AJKOER]
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