a123x
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Naphthol anologs of TNP
Well, I've lately been looking into the bromination of aromatic rings for the purpose of then reacting with hydroxide or methoxide to form
hydroxy or methoxy groups. It occured to me earlier today that one might be able to brominate naphthalene and then react with NaOH to form
naphthol...even better would be if the naphthalene could be dibrominated(if possibly I imagine that it'd give mostly the 1,8 or 1,5 dibromo
compound). Then one would could react with NaOH to give dihydroxy naphthalene which I think would be relatively easy to tetranitrate as opposed to
naphthalene which is more difficult to tetranitrate, requiring very concentrated acids for the final nitration. It is sort of a complex procedure just
to make a nitroaromatic compound but I figure that actually carrying out synths is really half the fun of this sort of stuff. Plus, the lead salts of
tetranitronaphthol or tetranitronaphthadiol might prove interesting.
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kclo4
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hey, sorry if there this has already been discussed other places but anyways....
I am planning on making some Napthol after i get a little bit more MNN and turn that into Naphthylamine. I want to nitrate the Napthol like you
would phenol to make picric acid/TNP.
Napthol is C10H7OH. I am guessing it is somewhat similar to phenol and can be produced the same way as phenol, starting from benzene, benzene is
turned into nitrobenzene then that into aminobenzene (aniline) then that is then turned into phenol, but you would replace the benzene with
naphthalene of course, so MNN is turned into C10H7NH2 and then C10H7OH
Also might Napthol like phenol be nitrated in almost the same way picric acid is. however it would make a Trinitro-napthol or maybe a
hexanitro-napthol, but i dont know. Does anybody know how to tell how many nitro- groups will be added, or if it does not nitrate like phenol, then
why not? also if i preformed an expirment of nitrating Napthol, how could i determin what formed?
also would only the aromatic ring with the hydroxily group be nitrated or would the other aromatic ring gain a NO2 group? so you might get
somthing like this
..................................NO2
................................../
...................C...........C
...NO2 -C..........C...........C-OH
............C..........C...........C-NO2
..................C...........C
...................................\
.....................................NO2
Sorry im not good at stuctures and i have no clue where to put the NO2 and OH groups
but i am just trying to show you what i mean. plus the stucture is most likely not needed and it looks like crap, but it might help i dont know.
so in summary:
How and what would form when Napthol was nitrated?
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praseodym
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Wouldnt the nitration of naphthol give you nitronaphthol? It is an electrophilic substitution. Electrophilic substitution of 1-naphthol occurs at the
2- and 4-position while that of 2-naphthol occurs at the 1-position as the hydroxy group in 2-naphthol activates the 1-position.
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kclo4
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Just nitronaphthol? why would it not be dinitro, trinitro and tatranitronaphthols?
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Nick F
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Nitration of napthalene at lower temperatures gives 1-nitronapthalene due to kinetics, since the carbocation intermediate has allylic stabilisation.
At higher temperatures, the nitration becomes reversible and so the equilibrium now favours 2-nitronapthalene as the product because although the
intermediate is less stabilised, since resonance would disrupt the aromaticity of the adjacent ring, the final product is less sterically hindered
since 1-nitronapthalene suffers from peri strain between 1-NO2 and 8-H, which forces the nitro group to be non-planar and this disrupts the
conjugation. 2-nitronapthalene does not suffer from this, and so is the more thermodynamically favoured product.
Conversion of 1-nitronapthalene to 1-napthol followed by full nitration would likely give 2,4,5,7-tetranitronapth-1-ol, full nitration of 2-napthol
would give 1,3,6,8-tetranitronapth-2-ol. Notice that the nitration pattern is actually the same, the only difference is where the hydroxyl group is
(1,3,6,8-tetranitronapth-2-ol = 2,4,5,7-tetranitronapth-6-ol). Both these will suffer from a large amount of peri strain between 4-NO2 and
5-NO2 (or 1-NO2 and 8-NO2), which will reduce stability. The 1,3,5,7-tetranitro pattern, which is less sterically hindered, probably will not form in
large amounts due to the strong directing action of the OH, and nitration at higher temperatures to try to form this thermodynamically favoured
substitution pattern could result in NOx and a black sticky mess due to oxidation (like with uncareful nitration of phenol).
It probably shouldn't be hard to form tetranitronapthols, TeNN isn't too difficult and now you have the very strrongly activating OH.
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