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Author: Subject: Possible synthesis to 1,5,7-trihydroxydecahydronaphthalene-2-carboxylic acid?
kclo4
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biggrin.gif posted on 30-12-2007 at 16:53
Possible synthesis to 1,5,7-trihydroxydecahydronaphthalene-2-carboxylic acid?


Alright so i have been looking all over the internet for this chemical, but i don't know any synonyms for it, i just know its name to be "1,5,7-trihydroxydecahydronaphthalene-2-carboxylic acid" according to ChemSketch, but those types of names are never good for finding the chemical anyways. so if you can find any information on it that would be great!

Anyways, I'm looking to synthesize it, or something similar to it.
I actually don't need that exact chemical. It can have few few hydroxy groups in other places etc.
However, it MUST have the 1, and 7, hydroxyl groups and at least a methyl group in the 2 spot. The methyl group could always be turned into a carboxylic group later.

So, any ideas as to how to make it?

I was originally planning on taking the isomers of dinitronaphthalene. However, i am pretty sure that the most common isomer in the end turns out to be 1,8-Dinitronaphtalene, so that doesn't do much good at all.

Perhaps there are ways to control the type of isomer produced via temperature, pH, catalysts, etc. But i am not aware of them.

well, anyways, I'm just knew to the organic chemistry world, but i have really been trying to study it this last week or two, so any help would be appreciated! :)
thanks

Oh, btw i have attached a PDF document of the chemical and what it looks like - because that will make it a lot easier to visualize making, wont it? :)

Attachment: 1,5,7-trihydroxydecahydronaphthalene-2-carboxylic acid.pdf (20kB)
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kclo4
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[*] posted on 30-12-2007 at 23:41


Alright, here is what i have came up with. ---er damn... the molecule

Methylate Naphthalene into the 2 possible derivatives, 1-methylnaphthalene, and 2 methylnaphthalene.

Separate the two MethylNaphthalenes and take the 2-methylnaphthalene.
Nitrate the 2-methylnapthalene into 2-methyl-1,8-dinitronaphthalene.

All right so now we have the needed methyl group in the right place, and one of the needed nitro groups in the right spot; the 1 position. the nitro group will be turned into hydroxyl later and the methyl group will be oxidized into carboxylic later as well. And the extra nitro group in the 8 position will not be a problem.


Alright! so this is were i am stuck

So, now i need to figure out how to add some sort of functional group to the 5 and 7 positions that can be converted or reacted into a hydroxyl group.
Also i need to find a way to react the Naphthalene like molecule with hydrogen.
Wiki says it can be produced fussed with a catalyst and hydrogen.

So, any creative ideas? :)




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not_important
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[*] posted on 31-12-2007 at 03:19


without digging out the books, here's a possible route.

naphthalene to 1,5-disulphonic naphthalene, may take oleum but possibly just con H2SO4, see old books on dye chemistry/precursors.

Fuse that with KOH-NaOH mixture to get the 1,5-dihydroxy naphthalene (di-naphthol).

Acetylate that and the react with further Ac2O or AcCl to get the mono-acetyl - the ketone. This should mostly for the 2- isomer, the 1,4,5,8 positions are disfavoured I believe.

React that with NaOCl to form the carboxylic acid.

An alternative to those two steps would be to use the Kolbe-Schmitt reaction on the di-naphthol to get the carboxyl on it - same as making salicylic acid from phenol.

Reduce with H2/Ni under high pressure. The first ring is easy, H2/[Rh] at 1 or 2 atmospheres, be the second ring is standard benzene aromatic and takes more extreme conditions or exotic catalysts.
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[*] posted on 31-12-2007 at 06:47


3,5-dimethoxybromobenzene as base chemical to make the corresponding alpha-tetralone.
Then, to get the alpha-keto-CO2H you can use, e.g. NaH and dimethylcarbonate.
Making 2-carbomethoxy-3-tropinone is well documented in the literature, meaning you can use that as a template.
Reduction of the keto-group is "chemoselective" with NaBH4.
Then reduce aromatic ring with an expensive Rhodium catalyst

Then demethylate the 2 ethers.

[Edited on 31-12-2007 by Drunkguy]




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