Sciencemadness Discussion Board - o-Phthalic Acid Ester Syntheses

o-Phthalic Acid Ester Syntheses

bfesser - 10-9-2008 at 19:54

I've been thinking of synthesizing the mono-methyl and mono-ethyl esters of o-phthalic acid. I have 99.5+% phthalic anhydride, reagent grade anhydrous methanol, and reagent grade 95% denatured ethanol. I plan to reflux a mixture of phthalic anhydride with an excess of the alcohol and then dilute the resultant solution with a large volume of ice cold distilled water to precipitate the products. The phthalic anhydride should be sufficiently reactive toward the alcohols to form the mono-esters without any (acid) catalyst, correct?

I've found a little bit of information regarding the solubilites of the products in water and ethanol, and the melting point of only the mono-ethyl ester (2°C; CRC Handbook of Chem. and Phys., 47th ed.).

Does anyone have any suggestions or warnings for conducting these syntheses--especially for the workup of the mono-ethyl ester? I've searched Org. Syn. with no results, and unfortunately just lost my access to my university's library temporarily (taking a semester off).

Nicodem - 10-9-2008 at 22:42

Phthalic anhydride might take a very long reflux in methanol or ethanol to form the corresponding monoesters as it its reactivity toward poor nucleophiles is understandingly poor so give it time. Examples from literature indicate about 8 hours or more reflux in methanol gives 80-90% isolated yield (an example using 36h reflux can be found on pages 3-4 of EP0043642; I guess you don't need other references since the procedure is simple enough).
I do know from experience that succinic and phthalic anhydrides react pretty fast with catalysis by trace amounts of acids even at room temperature (like 1 mol% of H2SO4 or even less), but acid catalysis will also promote the formation of more dialkyl ester so rather opt on long reflux.

panziandi - 11-9-2008 at 00:28

8hr reflux for a 80-90% yield would be quite acceptable. Phosphoric acid is used commonly for acetylation of phenols with acetic anhydride. Perhaps a small quantity of 85% phosphoric would be helpful? I expect the mono and dialky phthalates should be easily separated perhaps by shortpath distillation or perhaps extraction of the mono ester into bicarbonate, wash with AcOEt, then acidify with dilute acid and extract into more AcOEt then remove the solvent on a water bath etc. Just an idea but it shouldn't hydrolyse if the pH of the washes is not too high or low.

bfesser - 20-4-2009 at 06:16

I attempted to synthesize 2-(ethoxycarbonyl)benzoic acid by refluxing EtOH with phthalic anhydride, but was unable to isolate the product from the reaction mixture. Out of curiosity, I decided to do a computational analysis of the reaction to 'prove' that the product was indeed formed.

<img src='http://www.medievalsiege.net/bfesser/chem/mech.jpg' />

I've used MM2 and MOPAC (ChemBioDraw3D) to model the starting reactants, the tetrahedral bicyclic intermediate, the cleaved intermediate (before proton transfer), and the acid/ester product. I've recorded the MOPAC ΔHf for each, and am now at a bit of a loss as for the most appropriate or the correct way to draw the reaction energy diagram.

The Data:<ol type='I'>
<li>phthalic (cyclic) anhydride ΔHf = -68.99787 Kcal/mol</li>
<li>ethanol (1° alcohol) ΔHf = -62.70198 Kcal/mol</li>
<li>tetrahedral cyclic intermediate ΔHf = -135.74865 Kcal/mol</li>
<li>cleaved intermediateΔHf = -150.13136 Kcal/mol</li>
<li>2-(ethoxycarbonyl)benzoic acid (acid/ester product) ΔHf = -145.35263 Kcal/mol</li></ol>

My questions are these:<ol type='1'>
<li>Would it be correct to sum the ΔHf of the phthalic anhydride with the ethanol for the first energy level?</li>
<li>This reaction is exothermic overall, correct?</li>
<li>Am I supposed to just throw out the signs when drawing the reaction energy diagram?</li>
<li>Is the following reaction energy diagram correct?</li></ol>

<img src='http://www.medievalsiege.net/bfesser/chem/red.jpg' />

P.S. I know these are simple questions, but I've looked around in several of my books and online and have only become more confused.

[Edited on 4/20/09 by bfesser]

Nicodem - 21-4-2009 at 11:20

Quote: Originally posted by bfesser

My questions are these:<ol type='1'>
<li>Would it be correct to sum the ΔHf of the phthalic anhydride with the ethanol for the first energy level?</li>
<li>This reaction is exothermic overall, correct?</li>
<li>Am I supposed to just throw out the signs when drawing the reaction energy diagram?</li>
<li>Is the following reaction energy diagram correct?</li></ol>

I'm pretty clueless when it comes to physical chemistry so I'm not the most competent to answer your questions, but some things you obviously didn't do correctly.

1. Well yes, but you can not use the energy of formation in your diagram. Reaction coordinate diagrams represent the free energy levels (Gibbs energy). That is something quite different. To obtain the activation energy of transition states (the free energy difference between the starting material and transition state) you need to first determine what the transition states actually are by calculating the vibrational states during the transformation. Don't worry, I'm just as clueless as you on what this is and how to do this..., but I do know that to do this you will need one of the computational chemistry software packages. Unfortunately I can not help you here since I know just about nothing on this area. How about opening a thread in Computational chemistry section, with this specific reaction as a problem, and call for someone who knows how to use Gaussian to do a step by step description on how to solve the problem?

2. Yes, it is exothermic. The mechanism is different if non-catalysed or catalysed by acids or bases. Thus the activation energies, transition states and consequently also the reaction coordinate diagrams are also different (except for the energy levels of the starting and ending material).

3. These diagrams are about free energy differences with the position of the zero level unknown. Hence you can not have concrete/meaningful numerical values for each position unless you define the zero value first. But they do show the difference between energy levels and that is generally all what the chemists are interested in.

4. I'm afraid not. You can see that already from the first pair of states. Even though III is higher in energy than I+II, your diagram shows it as more stable than the starting material. If such a diagram would be real, then just dissolving phthalic anhydride in ethanol would immediately form III since there is no activation energy barrier and the I -> III is exothermic. Only later would the reaction from III to V proceed more slowly with IV as transition state and energy difference between III and IV as the activation energy. Of course, experience tells us this reaction does not happen this way since dissolving phthalic anhydride in ethanol initially only gives a solution in which phthalic anhydride can still be spectroscopically detected.

kmno4 - 22-4-2009 at 15:17

I think that monoesters can be prepared from diesters by partial saponification with stoichiometric amount of NaOH. I think I used to read papers about it, but I am not sure..... I will try to find them.

bfesser - 22-4-2009 at 16:56

Stoichiometric saponification would be an interesting route of synthesis. Unfortunately, it is not a viable route for me, as I have no strong acids or other catalysts to promote a Fisher Esterification required to produce the diesters from the anhydride to begin with.

I don't really have any planned use for the esters. I'm sort of just looking for something to do with my phthalic anhydride, and I thought esterification would be interesting.

Nicodem - 23-4-2009 at 00:15

Kmno4, the general procedure for selective alkaline hydrolysis of diesters to carboxy esters is described in the Organikum lab manual. As far as I remember it only works for diesters in which the carbonyls are mutually influenced by inductive effects, such as malonates. Since in phthalates the carbonyls are coupled by both, the inductive and mesomeric effects, it might also be applicable to them as well.

Bfesser, what is wrong with actually doing the monoalkyl ester of phthalic acid for the beginning? It is simple and you were already provided the reference needed. There are also a bunch of other stuff you can make starting from phthalic anhydride, various compound and polymeric resins...

grind - 23-4-2009 at 07:16

It is important to use absolute ethanol for the production of the monoethylester. Otherwise you get a very hard to separate mixture containing phthalic acid and the monoester. The diester can be separated as follows (no guarantee concerning solubilities):
Extract your crude product with cold (!) diluted NaOH-solution. All compounds except the diester and remaining anhydride are now in the aqueous solution (requirement: the sodium salt is soluble in water). Remove the insolubles with 2 extractions with ether or DCM or toluene or whatever you want. Apply vacuum to remove solvent residues. Now place the aqueous phase in an ice/salt-mixture and acidify slowly (good stirring) with HCl. Separate the oil and dry it. You obtain a monoester absolutely free of diester, with traces of phthalic acid.

bfesser - 23-4-2009 at 10:37

Quote: Originally posted by Nicodem

Bfesser, what is wrong with actually doing the monoalkyl ester of phthalic acid for the beginning?

I don't understand your question.

I don't have time right now to pursue this reaction further, at least at the bench. I do, however, have downtime between classes that I've been using to play around with computational models. I just thought it would be interesting to try to apply the models to analysis of an actual reaction I can carry out later on.

There are several reasons why I was unable to isolate the product. First and foremost, I didn't have enough time to let the reaction reflux for more than four hours. Second, I don't have any nonpolar or water immiscible solvents to perform extractions with. Finally, I don't have the acids needed to do acid-base extraction, anyways.

When I find the time and resources, I'll investigate this reaction further, and will be sure to post my results. Does anyone have any suggestions as far as computational chem is concerned?

[Edited on 4/23/09 by bfesser]

Polverone - 23-4-2009 at 11:14

I would suggest trying to get a feel for computational chemistry by reproducing older published computational work, preferably work where the authors provided experimental results for comparison with computation. There are enough details left out of papers that you should find it moderately (though not outrageously) challenging if you understand what the authors are doing. Since computers and software have improved you should be able to handle calculations on your PC that ran on supercomputers in the mid 1990s or earlier. Doing this will give you a much better sense of the applicability and problems of computational methods.

It is difficult to find a bridge between bench and computational chemistry that is accessible to the amateur. It is very difficult and computationally expensive to properly model solvent, but solvated chemistry is much different than gas-phase chemistry. It is comparatively easy to compute spectroscopic properties of gas phase molecules, but the amateur likely does not have the analytical instruments to compare theory and experiment (though I guess you can look at spectra that others have recorded).

Even though esterification of this sort is simple bench chemistry, producing a meaningful computational model is much more difficult than I think you realize. If you try such a hard problem right away I fear that you will not learn much before giving up in frustration.

kmno4 - 23-4-2009 at 15:20

From JACS article (ja01373a048):

.....Monomethyl Phthalate.—After a mixture of 25 g. of phthalic anhydride and 125 g. of absolute methyl alcohol (calcd. amt. X 23.1) had been refluxed on a water-bath for five hours, the excess methanol was distilled and the reaction product transferred to a beaker which was left in a desiccator over anhydrous calcium chloride for several hours.
The yield of monomethyl phthalate, m. p. 82-82.5°, was 24.4 g. (80.2%).....

It is one of the simplest procedures I have ever seen.

Another article is in German: Über Phtaläthylestersäure
[Monatshefte für Chemie (1915), 36 505-8]

Um Wasser mgglichst vollstiindig auszuschliet~en, trug der Kolben, der
das Phtalsiiureanhydrid enthielt, einen doppelt durchbohrten Stopfen. Durch
die eine Bohrung ging ein Kiihler, der als Riickfluflkiihler diente; an sein
oberes Ende war ein Kolben angesetzt, der zur Aufnahme des spiiter abzudestillierenden
Alkohols bestimmt war und dessen Inneres nur dutch ein
Chlorcalciumrohr mit der freien Luft in Verbindung stand. In die andere
Bohrung war ein Tropftrichter eingesetzt, in dessen obere Offnung ein VorstoI3
mit seitlichem, durch ein Chlorcalciumrohr verschlossenem Ansatzrohr
eingefiigt war. In den VorstoB war rnittels Kork ein Kiihler eingesetzt, durch den der vorher mit Kalk 6 Stunden gekochte und dann mit metallischem
Calcium versetzte Alkohol direkt in den Tropftrichter hineindestilliert wurde.
In der Hauptsache ist die Reaktion schon beendet, wenn nach LSsung
des Anhydrids noch 15 Minuten gekocht wird. Da abet bisweilen bei dieser
Kochdauer doch noch ein nicht unerheblicher Teil des Anhydrids unveriindert
blieb, wurde vorsichtshalber 2 Stunden gekocht. Noch l~ingeres
Kochen brachte keinen Vorteil.
(...) So wurde die Phtal/ithylesters~iure
in wohl ausgebildeten, zu grol3en Drusen vereinigten Krystallen
vom Schmelzpunkt 47 bis 48 ~ erhalten.
(sorry for copy/paste method)

I cannot read texts in German, but optically procedures look equally simple

grind - 23-4-2009 at 15:37

short translation of kmno4 procedure:

- working under anhydrous conditions, that means use of (really) absolute ethanol, use of a drying tube filled with CaCl2 connected with the condenser
- boil the phthalic anhydride with an excess of alcohol for 15 min after it went in solution (sometimes unreacted anhydride remains after this short reaction time) or boil for 2 hours (anhydride surely completely reacted)
- phthalic acid monoethyl ester obtained after distilling off unreacted alcohol in nice, big crystals with melting point 47-48 degree

[Edited on 23-4-2009 by grind]

[Edited on 24-4-2009 by grind]

Nicodem - 26-4-2009 at 12:04

Quote: Originally posted by bfesser

I don't have time right now to pursue this reaction further, at least at the bench. I do, however, have downtime between classes that I've been using to play around with computational models. I just thought it would be interesting to try to apply the models to analysis of an actual reaction I can carry out later on.

So... computing the Hartree-Fock energy levels of the starting and ending molecules as well as your intermediates using Gaussian is actually quite simple and in my opinion well within the reach of the average member here (assuming some familiarity with the annoying UNIX terminals and the ability to obtain Gaussian). I think even I could do that using this specific software package. But this is nearly not enough to actually build a computational model of this reaction. You also need the activation energies, the computation of which, so I was told, is not trivial at all. The other major problem is in that to my limited knowledge Gaussian only allow for dipolar solvent simulation models while it is unable to account for H-bonding. You might note that your anionic intermediates are all greatly stabilized by H-bond donating solvents (solvents such as excess ethanol) and that any gas phase or dipolar solvent model will give energies pretty much irrelevant to the real reaction. In this light, relying to the old fashioned kinetic measurements and mechanism investigating techniques might even sound a simpler solution (and more trustworthy).
Anyway, I would really like someone with the knowledge in the computational field to at least try to solve the problem. I'm interested in how this could be done.