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HOFMANN rearrangement

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Chembio    10

Can any one please how the HOFMANN rearrangement takes place in phthalimide ( isoindole-1,3-dione) in order to give 2-aminobenzoic acid? ( the reaction mecahanism)

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Mr Skeptic    1154

The Hofmann rearrangement requires a primary amide. So first you'd need to open up that N-containing ring.

http://en.wikipedia.org/wiki/Hofmann_rearrangement

http://en.wikipedia.org/wiki/Phthalimide

http://en.wikipedia.org/wiki/2-Aminobenzoic_acid

Edited by Mr Skeptic
can't tell the difference between a d and an n.

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Mr Skeptic    1154
Primary amiDe, quite a difference :D.

 

How much difference can one letter make? :D:doh:

 

It still needs to be primary though.

Edited by Mr Skeptic

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Chembio    10

phthalimide has two sites which are electrophilic, Carbonyl carbon and the nitrogen center with an acidic hydrogen. So when a nucleophile as OBr- or OH- is to attak where will it attack first??

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Mr Skeptic    1154

Well if you were to replace the NH with NBr, then hit one of the carbonyls with OH-, you'd end up at one of the steps shown in my link. I didn't study the Hofmann rearrangement, so don't take my word for it.

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Hello. I'm sorry I don't have a figure to link you to so I will describe it with words. :(

hopefully this will be of some use to you.

 

Step 1: hydroxide attacks one of the carbonyls - (it doesn't matter which because they are symmetrical) - and one of the bonds between the carbon of the carbonyl and the oxygen of the carbonyl moves to the oxygen to make a negatively charged oxygen.

 

Step 2: one of the lone pairs on the negative oxygen swings down to reform the carbonyl and the bond between that carbon and the nitrogen goes to the nitrogen (breaking open the five membered ring) - and the lone pair of the nitrogen removes the hydrogen from the "tertiary alcohol" formed in this intermediate.

*this isn't done by a water molecule because of pka(s)

 

Step 3: the lone pair of the primary amide now attacks a the Br2 and a Br- is kicked out.

 

Step 4: another hydroxide deprotonates the positively charged nitrogen.

 

Step 5: the Hofmann rearrangement occurs: another hydroxide attacks the reamaining hydrogen on the nitrogen, that bond goes to the carbon bearing the amide on the benzene ring, the bond between the benzene carbon and the amide goes to form a double between the nitrogen and the carbon of the amide carbonyl. Bromide is kicked out as well during this process.

 

Step 6: another hydroxide attacks the carbon of the isocyanate - (hopefully that is what it's called?) - and the bond between the carbon of the isocyanate and the nitrogen deprotonates a water molecule to give a hydrogen to the nitrogen.

 

Step 7: a decarboxylation occurs: the lone pair of the nitrogen attacks the hydrogen of the "tertiary alcohol" of the isocyanate, the bond between the hydrogen and the oxygen goes to the carbon of the isocyante creating an additional CO double bond, and the bond between the carbon of the isocynate goes to the nitrogen to produce a new lone pair.

 

step 8: you now have the primary amine and the carboxylate species in the 1 and 2 position of the benzene ring. H3O+ (will have to be added) - and is needed to protonate the carboxylate because water is not a strong enough acid, or you should draw your final product with just the carboxylate...

 

hopefully this was helpful.

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Chembio    10

thank you very much... it helped me a lot.. Got a one single question.. How about the acidic proton which is attached to the nitrogen atom. will it be stable in a basic medium...

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thank you very much... it helped me a lot.. Got a one single question.. How about the acidic proton which is attached to the nitrogen atom. will it be stable in a basic medium...

 

Hello, hopefully I'm understanding you correctly, if not let me know and I'll try to answer your question. :)

 

Are you talking about the amine protons in the final product? Those are stable to base they have a pka of approximately 38 or so in the presence of the carboxylate because making dianions becomes more difficult.

 

Here is a link to a good pka table. This is hands down the best pka table to have in my opinion. - http://docs.google.com/viewer?a=v&q=cache:UY6cBoMV640J:evans.harvard.edu/pdf/evans_pKa_table.pdf+evans+pka+table&hl=en&gl=us&pid=bl&srcid=ADGEESiNn7DVNlwVhRq6cGOxdfWQAHsjnCcQ6QZpJ4GvbQZB0Y7H_XIaxQXxllxBGCZ0d-Z3Dg0KCx7HlQq9zvrLmMkERPhGTn40Zp2ACSM2mXTTlFcjp2EA0jw6KUraN6GEmJ8pfcn5&sig=AHIEtbQ7OPFAjGYUC34QH5JZ5rH-m4vRlQ

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Chembio    10

no i'm not talking about the final product. My question is on the amide proton in phthalimide. Insted of attacking one of the carbonyls by the hydroxide ion won't it abstract the amide proton in phthalimide to initiate the reaction..

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Charlatan    2
Can any one please how the HOFMANN rearrangement takes place in phthalimide ( isoindole-1,3-dione) in order to give 2-aminobenzoic acid? ( the reaction mecahanism)

 

Well, if you were to take the phthalimide [organ or cell] filled with chemicals, and you know the answer is to produce acid for another organ, then it would serve you well to think that the other cell gets the mixture making the acid. If you were to take the mass of the acid you would see that they produce acid, or activate acid producing glands. Acids come from yuor bile thing and that means that they must be glands inside the bile stuff. Now, if you were to produce acid you would need to have some sort of hungry cells in it. These cells eat all the stuff that is dry - carbons without a liquid coating and composed of liquids, like foods and liquids.

 

But to make acid you need to take bacteria mixed with salt, because they will absorb the stuff as it gets eaten, making a 'dual destroyer'!

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no i'm not talking about the final product. My question is on the amide proton in phthalimide. Insted of attacking one of the carbonyls by the hydroxide ion won't it abstract the amide proton in phthalimide to initiate the reaction..

 

Yes you are right, the pka of the amide proton is 8.3 - so the first step is the deprotonation of the phthalimide hydrogen, and next the nitrogen anion attacks Bromine, and then attack of either of the carbonyls by hydroxide and then you will get to the same intermediate, you will still have to do the intramolecular proton transfer to avoid the carboxyllic acid formation... and then the rest of the mechanism should be the same...

 

hopefully this helps. :)

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