[appamo]

anyone know how to explain the mechanism

[Mobius1988]

Im sorry I dont understand the question, what do you need explaining? The reasons for why it proceeds like this?

[appamo]

Im sorry I dont understand the question, what do you need explaining? The reasons for why it proceeds like this?

yes....can you help me in explaining the mechanism?
its a condensation between the o-phenylenediamine and formic acid.
but their electron transfer  i don't know

[Mobius1988]

Ok well firstly, in the last two molecules you seem to have lost one of the bonds in the benzene ring? Im going to assume this was a mistake and that the aromaticity is still there.

Do you know the conditions of the reaction at all? From a cursory glance I would immediately assume there is an acid catalyst in there and some 4A molecular sieves. This is why:

An acid catalyst would protonate the carbonyl on the carboxylic acid allowing the lone pair from the NH₂ to attack it. As they have drawn it they assume the NH₂ is already reactive enough to attack the carbonyl (as far as im aware this isnt true but I might be wrong). So you end up with the diol attached to the amine as shown in molecule 3.

Next I would have the acid catalyst protonating one of the OH groups, making it a good leaving group as H₂O. At this point either the lone pair from the other oxygen will form the carbonyl and kick out H₂O (which will be removed by the molecular sieves), or, the nitrogen lone pair will form an oxime and kick out H₂O but then tautomerism will occur. Either way you form the amide shown in molecule 4.

You have to remember that many of these steps will actually be in equilibrium so the presence of the acid catalyst and molecular sieves will push the equilibrium towards the products.

At this point the second NH₂ will get involved with the overall result being the formation of a second aromatic ring - this is highly favourable and is the driving force for the reaction!! Again I would assume the carbonyl is protonated to make it more reactive (since amides are normally relatively unreactive) and is attacked by the NH₂ followed by proton transfer to form molecule 6.

The final step is the actual formation of the aromatic ring and a lone pair from the nitrogen will push into the ring (forming the aromaticity) and H₂O will be lost as a byproduct. This step will be irreversible. Once the aromaticity is formed it will not be reversed just by water.

Thats quite a long and complicated explanation. If you want it made simpler let me know and Ill give it a go.

[appamo]

im not sure about the acid catalyst because while doing the experiment, only o-phenylenediamine and formic acid are mixed and heated on water bath for 2 hours. and then naoh added to make it alkaline(to remove the unreacted formic acid)...
what is 4A molecular sieves??
how come in your paragraph 3, i'll end up with diol attached to amine?

[appamo]

Mobius1988, i would like to have a simpler explanation.

[Mobius1988]

im not sure about the acid catalyst because while doing the experiment, only o-phenylenediamine and formic acid are mixed and heated on water bath for 2 hours. and then naoh added to make it alkaline(to remove the unreacted formic acid)...
what is 4A molecular sieves??
how come in your paragraph 3, i'll end up with diol attached to amine?

Sorry I explained that a bit poorly.

Ok, do you use excess formic acid? Since this is an acid, if it is used in excess it may act as the acid catalyst itself (one molecule protonates another to make it more reactive).

At what point do you add the NaOH? If it is at the end of the reaction as part of the work up you are probably just neutralising the acid and the NaOH doesnt take part in the reaction itself.

4A molecular sieves basically absorb water. So you add them to reactions where any water can cause side reactions or reverse your reaction. It absorbs the water and pushes the reaction towards the products. (If you have ever dried something with MgSO₄ it is the same principle.

The "diol attached to the amine" is the third molecule you have drawn, I wrote that poorly sorry. So you end up with the molecule with two hydroxyl groups on it that are bonded to the carbon, which is then bonded to the NH. That configuration of atoms is perfect for a protonated OH (ie water) to be lost as a leaving group and a double bond to be formed. You see that kind of mechanism very often in carbonyl chemistry.

Does that make a little more sense?

[appamo]

90% formic acid consider in excess?
i think naoh didnt involve in this reaction. just wana mention to you.
your explanation is very good. you help me a lot. actually im very weak in chemistry and always confuse with the mechanism. really thank you mobius:)

[Mobius1988]

Hmm, sounds like an excess of formic acid to me. Were you using an organic solvent for the reaction or was the formic acid the solvent?

Ill tell you the same thing all of my lecturers/supervisors have always told me. Once you learn the underyling principles of functional groups you can predict any reaction by drawing curly arrows.

Chemistry can be very difficult and when you see new reactions youve never seen before it can be intimidating. But in most cases you can always go back to your fundamental reactivity. Whats your electrophile, whats your nucleophile, whats the driving force for the reaction?? Focus on those things and itll help you improve very quickly.

[appamo]

agreed. i always get confuse with electrophile and nucleophile. haha
anyway...thank you very much.