10-10-02 17:20
No 366926
We really do need a chemist's couch of sorts, for shit like this, haha.
I know this is in no way practical, I am only concerned if it is possible. I am thinking that the following procedure, although dubiously longwinded, would serve to remove one of the hardest bitches of a functional group to get off of an aromatic ring: the hydroxyl group, -OH.
The entire first row is a series of steps that all occur in the first reaction. A diazonium salt of an aromatic ring (ArN2+) attacks a phenol in the ortho position, in a standard electrophillic substitution move, forming the intermediate carbocationic ring.
By far the most stable proton to lose is the hydroxyl proton, being far more acidic than any of the protons on the ring. This gives us our compound in the topright corner.
If one could then condense this intermediate with hydrazine, one would replace the oxygen with a removable group, a nitrogen group, in theis case, a hydrazine. I show in the picture that the formed imine will rearrange to restore aromaticity, forming the phenylhydrazine, but in reality the imine is likely not an intermediate, although it does not matter in the end. The gem amino-alcohol would dehydrate not to the imine first, but likely straight to the aromatic phenylhydrazine (I think, but like I said it really doesn't matter.)
The phenylhydrazine is then oxidized with HgO or another suitable oxidant to rip off the hydrazine. The diarylazo structure is presumably immune to these type oxidants, since they are the products of the oxidation of diaryl hydrazines, ArNH-NHAr with these oxidants.
The diarylazo intermediate is then reduced with Zn/HCl to the aminobenzene/aniline. The aniline is diazotized, and reduced with Na2SO3 to the phenylhydrazine, Ph-NHNH2. This is then subjected to a second oxidation with HgO or similar catalyst to yield the free benzene.
There is no fun in reducing standard phenol to benzene, but there may be plenty fun with substituted benzenes. How's that for "novel".
PrimoPyro
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(Stoni's sexual toy)
10-10-02 17:50
No 366935
> By far the most stable proton to lose is the hydroxyl
> proton, being far more acidic than any of the protons on
> the ring. This gives us our compound in the topright corner.
This will not happen, because this means the phenol would have to give up its aromaticity.
Are you sure the the diazo thingy in the first step attacks ortho?
I'm not fat just horizontally disproportionate.
(Hive Prodigy)
10-11-02 11:21
No 367250
Hasn't the phenol already lost it's aromaticity from the attack? The arenium ions are not planar/aromatic are they? I would think that the phenol proton is much more easily liberated at lower energy than any other proton on the ring?
1.O-H bond is much more polar than C-H.
2.O+ species is immensely electronegative.
3.The ring has been destabilized by the electrophilic attack, and wants to return to a more stable form as easily as possible.
So I don't understand why it can't lose that proton. You're sure that it won't come off? I thought that this was the same mechanism for production of quinones from hydroquinones, except that the driving force is not derived from another phenol group, but from the electrophile destabilizing the ring.
And yes, the diazo salt will attack in the ortho position, most 100% certainly especially if the para position is occupied. It can attack either ortho or para to the phenol, and both of these attacks yield the same result. March's 5th, pg.~677 and around there gave me the idea.
The phenol, as you proved to me earlier, is ortho/para directing, so if any group is para, it will attack ortho. If a meta directing group is meta to the phenol, the phenol direction predominates according to March's 5th, so ortho attack will predominate over meta or para.
The few diagrams depicted all show the diazonium species attacking ortho, but this is likely because they all include a second meta directing group, para to the phenol, which makes ortho more favorable over para.
All this ortho/meta/para business can get confusing.
PrimoPyro
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(Hive Bee)
10-11-02 11:35
No 367251
To synthesize benzene, why not just buy it?
It surely isn't suspicious is it?
If it is, I wonder what's next? water?
I was thinking the same thing too about the aromaticity problem brought up It seems like hydroquinone easily loses aromaticity to an electrophile, but that is a different animal of sorts.
The few diagrams depicted all show the diazonium species attacking ortho, but this is likely because they all include a second meta directing group, para to the phenol, which makes ortho more favorable over para.
[blue]
And where is that second meta directing group your'e talking about? I don't see it in your illustrations, your right it is confusing when you have an invisible function on the ring (not included in the diagrams) that does this magic directing of ortho attack.
(Stoni's sexual toy)
10-11-02 11:42
No 367254
> Hasn't the phenol already lost it's aromaticity from the
> attack? The arenium ions are not planar/aromatic are they?
You shouldn't take these mesomeric/resonance structures of reaction intermediates too literally, it's the same approximation that shows benzene to have 3 double bonds (which of course it does NOT have!). You can move the positive charge around the molecule, that doesn't mean that this positive charge really exists at these places. It is delocalized and just a very short lived intermediary anyway.
> I would think that the phenol proton is much more easily
> liberated at lower energy than any other proton on the ring?
If that was the case all the electrophilic substitution reactions on phenols would result in such non-aromatic products having this unsaturated ring and a ketone instead of a phenolic OH. But these reactions all yield a product which is still a phenol.
The energy gained from delocalizing the pi electrons in aromatics is so big that you need very convincig arguments to make a molecule give up its aromatic status voluntarily. Hydroquinone to benzoquinone is an example where it works, most of the time it won't.
I'm not fat just horizontally disproportionate.
(Hive Prodigy)
10-11-02 11:54
No 367257
To synthesize benzene, why not just buy it?
It surely isn't suspicious is it?
If it is, I wonder what's next? water?
You need to go back and read my opening a second time because obviously my very point escaped you the first time. I even had the decency to put "impractical" in the title. It is theory, not practice. MDMA is much "easier" to buy than to synthesize, so why bother making it too? "Just buy it."
And where is that second meta directing group your'e talking about? I don't see it in your illustrations, your right it is confusing when you have an invisible function on the ring (not included in the diagrams) that does this magic directing of ortho attack.
Again, you need to re-read what I said. The attack can go either ortho or para, and I explicitly stated that a)presence of a group at the para position prevents attack there, and b)both para and ortho attacks give the same result, it matters not.
And to Osmium:
You shouldn't take these mesomeric/resonance structures of reaction intermediates too literally, it's the same approximation that shows benzene to have 3 double bonds (which of course it does NOT have!). You can move the positive charge around the molecule, that doesn't mean that this positive charge really exists at these places. It is delocalized and just a very short lived intermediary anyway.
I try not to. I know about the resonance and the 3D reality of these things, but how else am I to depict the goings on if not in this manner? Don't expect me to code a nice little computer generated mpeg of shifts in electron density as the reaction progress.
If that was the case all the electrophilic substitution reactions on phenols would result in such non-aromatic products having this unsaturated ring and a ketone instead of a phenolic OH. But these reactions all yield a product which is still a phenol.
The energy gained from delocalizing the pi electrons in aromatics is so big that you need very convincig arguments to make a molecule give up its aromatic status voluntarily. Hydroquinone to benzoquinone is an example where it works, most of the time it won't.
Hmm, a very convincing point, sadly. Well that just sucks ass then. It was worth a shot though.
But, your statement about the hydroquinone to benzoquinone being a situation in which it works that way, makes me wonder: How do we know that this wouldn't possibly be another situation in which "it works that way"?
That second hydroxyl on hydroquinone/catechol must make the difference. I originally thought that one could reduce dihydroxyl functions with the same process, starting at the introduction of hydrazine stage, but it really isn't likely at all because you would have to get lucky enough to get two molecules of the hydrazine to attack the ring at the same time and react in the same manner. Not likely.
[EDIT]My apologies if this sounded very snappy. I am getting a bad habit of that. My points stand, but the tone can be ignored, there is no malice over here, just frustration.[/EDIT]
PrimoPyro
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