Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: kanonsviel on September 08, 2010, 02:08:54 AM
-
Is it possible to undergo a nucleophilic aromatic substitution reaction?
But, without a EDG attaching to the para or ortho position, it's unlikely to be able to form a Meisenheimer complex...
So, what kind of reactions could occur in this reaction other than a nucleophilic aromatic substitution?
-
It could undergo the benzyne mechanism, I forget the name, but without some intense heat I don't know if it occurs appreciably.
-
Is it possible to undergo a nucleophilic aromatic substitution reaction?
But, without a EDG attaching to the para or ortho position, it's unlikely to be able to form a Meisenheimer complex...
So, what kind of reactions could occur in this reaction other than a nucleophilic aromatic substitution?
You will form a deuterated benzyne which will polymerise under the reaction conditions.
-
Humm...seems like benzyne is the final answer.
but, what's the fate of these two deuteriums? does one of them remain intact if it was to form a polymer?
and the mechanism leading to a benzyne...is this an one step reaction(-OH's attack towards the deuterium, whereupon the Iodo group gets expelled??) or there is an intermediate? ???
-
Humm...seems like benzyne is the final answer.
but, what's the fate of these two deuteriums? does one of them remain intact if it was to form a polymer?
and the mechanism leading to a benzyne...is this an one step reaction(-OH's attack towards the deuterium, whereupon the Iodo group gets expelled??) or there is an intermediate? ???
No it seems to be concerted, de-protonation and immediate elimination of DI to give the benzyne. Benzynes can be trapped by dienophiles at very low temperature (-78°C)
-
I'm sorry to ask this persistently...but, please forgive my crappy English, what did you mean by "concerted", I have referred to a dictionary, yet still can't understand what it supposed to mean... :'(
may I take it for "Meanwhile"??
you were suggesting that this is an one-step reaction, do I get it precisely?
and about the deuterium, was my conjecture (one would remain intact) right or wrong? ???
-
concerted means the entire reaction occurs at once. All bonds are formed and broken simultaneously.
-
I'm sorry to ask this persistently...but, please forgive my crappy English, what did you mean by "concerted", I have referred to a dictionary, yet still can't understand what it supposed to mean... :'(
may I take it for "Meanwhile"??
you were suggesting that this is an one-step reaction, do I get it precisely?
and about the deuterium, was my conjecture (one would remain intact) right or wrong? ???
One deuterium should remain
-
I see... thanks for your scrupulous explanation :)
-
I see... thanks for your scrupulous explanation :)
This is why I am not a chemistry teacher, what explanation do you require exactly?
-
As written, I too expect a benzyne intermediate. It should then react with -OH, to give after protonation, a mixture of ortho and meta-deuterophenol.
Concerted means the electron movements occur all at the same time so no additional intermediate is thought to be present.
-
This is why I am not a chemistry teacher, what explanation do you require exactly?
I hope the sentence I have written is not misleading...I just translated it directly from Japanese, in order to express my gratitude. did it seem that I was requiring a further explanation? ???
As written, I too expect a benzyne intermediate. It should then react with -OH, to give after protonation, a mixture of ortho and meta-deuterophenol.
Concerted means the electron movements occur all at the same time so no additional intermediate is thought to be present.
According to my text book, it seems that this is the industrial route to produce phenol from halogenobenzene, but as discodermolide said, this reaction only occurs under a high temperature and pressure condition :)
-
Without further discussion, I suggest it would give the mixture of phenols or no reaction. If as suggested these conditions are not harsh enough (though I have no indication of temperature or pressure), then it should be no reaction.
-
Without further discussion, I suggest it would give the mixture of phenols or no reaction. If as suggested these conditions are not harsh enough (though I have no indication of temperature or pressure), then it should be no reaction.
Without further discussion!!!
Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.
-
Without further discussion!!!
Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.
Certainly, you could write that as your solution. If you could verify this as well, it would be even more convincing. I have neither. Being unaware of the benzyne polymerization rate and if I were attempting to perform this reaction and I was getting polymerization, I would doubtlessly slowly add the iodobenzene to the KOH to reduce the benzyne concentrations and increase the concentration of hydroxide. That should increase the bimolecular reaction rate.
This looks like a reaction scheme taken from an example of a benzyne reaction in which the objective is to predict the benzyne products. I presumed that was the point of using a dideuterated starting material.
-
Without further discussion!!!
Depending on the KOH concentration and temperature these are conditions that will produce benzyne, water may re-add but benzyne is so reactive that it will poylmerise rapidly. So at best you will get a mess.
Certainly, you could write that as your solution. If you could verify this as well, it would be even more convincing. I have neither. Being unaware of the benzyne polymerization rate and if I were attempting to perform this reaction and I was getting polymerization, I would doubtlessly slowly add the iodobenzene to the KOH to reduce the benzyne concentrations and increase the concentration of hydroxide. That should increase the bimolecular reaction rate.
This looks like a reaction scheme taken from an example of a benzyne reaction in which the objective is to predict the benzyne products. I presumed that was the point of using a dideuterated starting material.
Benzyne is so reactive that no matter how low the concentration was (is) you would get a mess, unless you used Schlenk conditions and trapped it out with a dienophile at low temp.
-
http://orgsyn.org/orgsyn/pdfs/CV5P0632.pdf
It has been shown (A. T. Bottini and J. D. Roberts, J. Am. Chem. Soc., 79, 1458 (1957).) that two mechanisms, elimination-addition (benzyne) and SN2 displacement, are operative in the liquid-phase hydrolysis of halogenated aromatic compounds. The formation of isomeric phenols as a result of the availability of the benzyne route makes the reaction of limited synthetic value. The incorporation of the copper-cuprous oxide system suppresses reaction via the benzyne route, so that the present method has general utility for the preparation of isomer-free phenols. For example, p-cresol is the only cresol formed from p-bromotoluene under the conditions of this preparation.