Is this reaction possible?

I've been trying to read up on ways to introduce a carboxylic functional group to a weakly deactivated aromatic ring, in pursuit of a method of synthesizing luminol from PET bottles that have been converted to benzene-1,4-dicarboxamide.

Using http://en.wikipedia.org/wiki/Nucleophilic_aromatic_substitut... and http://en.wikipedia.org/wiki/Deactivating_groups as references I've thought up the mechanism below.

Does it seem plausible? And if not, is there another method which I can use to introduce such functional groups and not risk reactions that would take place with the other functional groups on the aromatic ring, without the need for use of protecting groups?




Explanation of mechanism.......[From left to right]:-

1). Iodoform is deprotonated in an alcohol solution of equimolar sterically hindered, metal alkoxide which reacts to form the triiodomethyl anion and an alcohol molecule (meanwhile the metal cation acts as the counterion to the anion formed).....{If the molar ratio of iodoform to alkoxide is not equimolar then the iodoform should be present in the larger molar proportion}.

2). The triiodomethyl anion then attacks the carbon directly bonded to the halogen atom and forms a bond to the carbon. This causes electron density to be pushed from the carbon with the halogen substituent to the carbon in the ortho position with the electron withdrawing group. Due to the change in electron density in the electron withdrawing group it then rearranges to compensate for the change. In the process the aromatic ring loses its aromaticity and thus its stability.

3). To regain aromatic stability, the excess electron density in the EWG is withdrawn and is pushed into the better leaving group attached to the primary carbon atom, in this case the halogen atom iodine, forming the iodide ion. The iodide ion pairs with the lithium countercation and leaves the reaction mixture.

4). The resulting triiodomethyl-nitroarene molecule is then hydrolysed in base to the unstable orthoformic acid-nitorarene derivative and iodide byproduct. The orthoformic acid group then decomposes to produce the carboxylic acid-nitroarene and water. (The orthoformic acid-nitorarene derivative formation and decomposition was omitted).

Please reply telling me what you think of this reaction.

Thank You......

Revision of mechanism

Oh dang! I was hoping that it might have had at least a slight chance of working.

About the use of isopropoxides. I wasn't planning to use secondary alcohol alkoxides, just was doing something of a generalization. I was planning to synthesize a bulky alcohol alkoxide molecule like lithium dimethyl-phenyl-pentoxide which would hopefully, due to steric hindrance, be unable to form the orthoformate esters.

Please remember that I said I was dealing with deactivated aromatic rings, therefore the Reimer-Tiemann reaction most likely would not work in this case since a step in the mechanism involves pushing electron density from the activated aromatic ring to the dichlorocarbene. Since the aromatic rings will be deactivated then this important step would not be able to proceed.





Okay, so above is my revised reaction mechanism utilizing bromoform instead of iodoform. I was hoping that since bromine is more electronegative than iodine then that would mean that there would be a greater possibility for the formation of the tribromomethyl anion, due to the bromine's better electron withdrawing nature.

About the iodine. I really didn't think about the iodine's size and the size of the replacement substituent. I didn't realize that the sizes would play such important roles in the determination of the mechanism's feasibility.

Considering sizes, I think iodine would still be a good atom for replacement because it is a very good leaving group. However, since you highlighted that the S_N2 step of the mechanism in which both substituents are attached to the carbon could pose a problem, I'm wondering what would happen if I also utilized a stoichiometric amount of cuprous iodide in the reaction mixture.

I'm thinking that the CuI would form a complex or an adduct with the haloarene molecule which would make it easier for the iodine atom to be replaced be the tribromomethyl anion and in the process the CuI could react to form the complex anion [CuI₂]^- with the counterion being the alkali cation Li⁺. The conversion of the CuI to Li⁺[CuI₂]^- would be what helps in making the process go forward in favour of the formation of the trihalomethylarene derivative (maybe, just a thought).

Possible experimental conditions and procedure:

An anhydrous polar aprotic solvent is selected which can dissolve all the chemicals. The compound to be carboxylated is dissolved in the solvent and the bromoform and cuprous iodide are added (with the bromoform in slight molar excess). The solution is thoroughly mixed and is gently warmed. A given amount of alkali alkoxide is dissolved in fresh solvent and is thoroughly mixed. With stirring, the alkoxide is added dropwise to the primary mixture. After complete addition of the alkoxide, the solution is refluxed for about an hour. An aqueous solution of base (in excess) is then added slowly to the boiled reaction mixture and after complete addition the mixture is then heated and stirred. The aqueous layer containing the carboxylate salt is separated and acidified. The resultant carboxylic acid precipitate is then filtered and purified.

To anyone who would like to help, your input would be greatly appreciated.

Thank You......

Exploratory experiment time

Well, I guess it is experiment time.

I've devised an experiment by which I can test the feasibility of this reaction, and I will test it on different deactivated halgenoarene substrates. If everything goes as planned the halogenoarene compound will be converted into a carboxylic acid.

Below is the proposed experiment, utilizing the substrates 1,4-diiodo-2-nitrobenzene and 2,4-diiodobenzoic acid, produced from poly(ethylene terephthalate) bottles.

If the reaction works then the first carboxylic acid to be synthesized will be 4-iodo-2-nitrobenzoic acid. The second carboxylic acid to form will be 4-iodobenzene-1,2-dicarboxylic acid. The 4-iodobenzene-1,2-dicarboxylic acid can then be dehydroxylated to phthalic acid, which can be used to produce luminol.

I will conduct these reactions and update this thread as such.

Thank You......

I can relate to where he's at. It's easy to get caught up in the end-goal from a certain way rather then the practicality and feasability. Sure PET bottles are cheap but do you really want to be doing all that? Especially considering a lot of those reactions even IF it worked wouldn't be very yield worthy.

There are a lot of flaws in this reaction mechanism. Hydroxylation of the iodobenzene functional group won't be going off without a hitch with NaOH. The second to last step, the removal of the phenol functionality with an acid will not work. Not sure what the Li-O- ligand is? Not sure if it really even matters.

I could try and sort through more of it but from a practical stand-point you might as well start from ortho-xylene and just oxidize to pthatlic acid. Then make the anhydride. Much cheaper and higher yielding with a whole lot less leg-work. Plus it's documented .

Just sayin'

Quote: Originally posted by AndersHoveland

I think your proposed reaction might work if you use chloroform instead. But then to transform the -CCl3 group into a carboxyl group, you would need to leave it for several months to react with NaOH solution, perhaps with a CuI catalyst. While refluxing it would speed the reaction up to only 4 hours, it would also lead to decarboxylation, and you would just get nitrobenzene. Another idea to speed up the reaction might be to use something like sodium methoxide, and then hydrolyse off the orthoesters. [Edited on 6-11-2012 by AndersHoveland]

Update #1

I have a bit of good news. The lab technician has agreed to give me a decent quantity of toluene, that I can use for research (and 30ml of benzene). So instead of synthesizing deactivated halo-arene compounds starting with poly(ethylene terephthalate), I'll be able to synthesize similar deactivated aromatic halides compounds using toluene.

@smaerd - I might not be able to access everything I want, but if I can't buy it or extract it, I will make it.

I can produce the nitrite from the carbon reduction of potassium nitrate (which I can simply buy from pharmacies), then further purification to remove oxide and hydroxide contamination. I can extract the lithium from energizer ultimate lithium batteries, with ease and then rinse the Li with some toluene to remove the electrolyte. I already have some nitric acid that I got a good while ago. Remember I did not specify what type of reduction I was going to use and I was not thinking of using palladium, carbon or platinum. But I was thinking of using nascent hydrogen generated in situ. But if on the topic of reduction aiding metals, I can get nickel from energizer nickel metal hydride batteries as well.

Below are three reactions dealing with the generation of deactivated halo-arene compounds and their subsequent dehalogenation\carboxylation to carboxylic acids.

Well, pertaining to the reaction scheme I think the first might be workable but the second and third might deviate due to steric effects however; as I said this thread is for testing the reaction's possibility.

So since I got some toluene ( YAY!!! ) I'm going to first isolate some bromine for the bromination of the toluene to 2-bromotoluene and the reaction scheme should just fall into place.

Truly, what really interests me about this reaction is that if it works a trihalomethyl-arene will result initially. Using benzotrichloride as a suitable chemical analogue, this resulting trihalomethyl-arene will be a relatively reactive chemical intermediate, capable of hydrolysis to the carboxylic acid. However, if other nucleophiles are utilized a wide range of chemicals might result, depending on nucleophile, and reaction conditions etc...

Below is a proposed example of the compounds formed due to rxn of the trihalomethyl-arene with other nucleophiles.

@AndersHoveland - I think you might need to recheck whatever erroneous sources you got that reaction reference from. No offence but honestly I think that would be best. I don't think decarboxylation can result in aqueous solution. I think you have that confused with the dry distillation of carboxylates with calcium oxide or sodium hydroxide similar to the members publication http://www.sciencemadness.org/member_publications/benzene_pr... . The reaction described with the methoxide would most likely just result in the conversion of the haloform to methyl orthoesters or eventually to formic acid if water is present as well.

Thank You......

Update #2 ...... Understanding

@smaerd - I keep insisting especially to you that the thread is not primarily about making phthalic acid. The primary reason for the thread is to create a method by which deactivated halogenoarenes can be dehalogenated and carboxylated. Basically swapping a halogen atom for a carboxyl group on a deactivated aromatic molecule; and it wouldn't be bad to synthesize other compounds from the intermediates as well . I am asking persons opinions on the reactions I have worked out so I can adjust them so they have the best chance of working.

I pretty much think I had covered the reaction of the alkoxide and it's importance to the entire reaction, but I will elaborate.

According to my research, when chloroform is reacted with a strong base, a trichloromethyl anion results. However this molecule will easily undergo alpha elimination to produce a reactive chemical intermediate called dichlorocarbene. This intermediate is what is responsible for the formation of aldehydes from activated aromatic compounds (Reimer-Tiemann reaction), due to the ring's rich electron density. However, since I am working with deactivated aromatic compounds this would not work since the aromatic ring has less electron density. So even if dichlorocarbene was produced, a carbonyl group would not have been added, and I would not be able to oxidize it to a carboxylic acid.

So I had to devise my own method. It incorporates nucleophilic aromatic substitution, the Halex Reaction (Finkelstein Reaction), the Reimer-Tiemann reaction and effects such as the mesomeric effect and polar effects (electron withdrawing and donating effects).

Firstly, I researched chemical analogues to the trichloromethyl anion. Since treating chloroform with a strong base results in the formation of the trichloromethyl anion then bromoform reacted with strong base should result in the formation of the tribromomethyl anion. This is where the lithium alkoxide comes in, this will function as the base. A solution of the alkoxide (in an organic aprotic solvent) will be mixed with the bromofrom to create a solution of the alcohol (due to deprotonation of the bromoform) and an intermediate lithium salt.

Now comes the nucleophilic aromatic substitution:-

After forming the intermediate lithium salt the deactivated haloarene compound is then added. Hopefully, what would occur on a molecular level is that the nucleophilic tribromomethyl anion would attack the carbon attached to the halogen atom. Eventually what would occur is that the halogen being the better leaving group, (due to formation of an adduct with a compound added beforehand, which helps weaken the halogen-carbon bond), would be displaced from the molecule as halide and the halide salt would precipitate out of the organic mixture (halex reaction scheme). The precipitation would drive the reaction forward due to Le Chatlier's principle. The deactivated tribromomethylarene, which is similar to benzotrichloride in chemistry, would then show a great deal of increased reactivity; such in that it could be easily hydrolyzed to the carboxylic acid derivative or reacted with other nucleophiles to result in the formation of other compounds.

I hope this clears up any misconceptions (a bit of the details have been omitted).

Thank You......

Quote: Originally posted by Ionic Chemist

So I had to devise my own method.