Sciencemadness Discussion Board - Are these reactions likely to happen?

Are these reactions likely to happen?

Doctor Cat - 10-3-2017 at 14:10

I want to synthesize the following molecule but I couldn't find its synthesis over the internet neither on literature... I could just try to do the experiments but the reagents involved are very valuable to me since I don't have access to them and I had to synthesize them myself... So I just came here looking for a second opinion... It's just a classical Williamson reaction, I don't see why this shouldn't work right?... any suggestion or comments will be greatly received.

Thanks in advance to everyone.

(I don't know why the images look like this... but if you click them they are ok)

[Edited on 10-08-2015 by Doctor Cat]

theAngryLittleBunny - 10-3-2017 at 14:29

With the first one I see no reason why that shouldn't work, if the contentration of KOH is low, that this should work.
But with the second synthesis, I wouldn't be so sure, I mean refluxing it for three hours with NaOH might substitute the Cl-groups to HO-groups, the aldehyde group in the first molecule of the second reaction would make a cannizzaro reaction, where it disproportionates into a carboxylic acid and an alcohol, as aldehydes without alpha hydrogens in hydroxide solutions do, and the ether might hydrolyse, but that seems not so likely to me. You could maybe protect the aldehyde group by forming an acetal with ethylen glycol.

JJay - 10-3-2017 at 14:39

I think the first one would work but would be higher yielding in acidic conditions.

The second one, I'm not so sure about.... I actually kind of doubt that you would get those products. You'd get a Cannizzaro reaction of the substituted benzaldehyde with itself and probably a lot of other undesirable side reactions

Doctor Cat - 10-3-2017 at 14:58

Thanks for your responses...

Is sodium hydroxide really needed for the last reaction? I think that by removing the NaOH lots of problems would be solved... I'm not really sure how the procedure would be in that case, I know that the synthesis of esters by this way has very poor yields but it looks so simple and yet so complex

theAngryLittleBunny - 10-3-2017 at 15:07

And by the way, this is not a williamson reaction, because that what you wanna make is an ester, but the williamson reaction is the synthesis of ethers from alkoxides and alkyl halides. What might work is if you chlorinate the p-chlorobenzoic acid with PCl3 (or if you live in the US and P is not avaliable, SOCl2 works as well) to p-chlorobenzoyl chloride, and instead of thie iodide group, you make an alcohol group on the vaniline by just decomposing the diazonium compound which I presume was an intermediate of the 5-iodovaniline. And then you react the p-chlorobenzoyl chloride with the um.....5-hydroxyvanilin to make the ester, since caboxylic acid chlorides form esters with alcohols. The only problem there is that the p-chlorobenzoyl chloride obviously will also react with the HO-group at the 4-position, which means your yield will most likely be below 50%.

[Edited on 10-3-2017 by theAngryLittleBunny]

JJay - 10-3-2017 at 15:38

You might be able to use sulfuric acid with 5-hydroxyvanillin to form esters with p-chlorobenzoic acid, but I think esterfication will be favored at the 4-position, so the yield of your desired product would likely be very low.

[Edited on 11-3-2017 by JJay]

Doctor Cat - 10-3-2017 at 15:40

I was also thinking on making the silver salt of the 4-chlorobenzoic acid... I happen to have a great amount of silver oxide and the insolubility of the iodide may favor the formation of the ester...

I don't want to work with chemicals like thionyl chloride, it's just too nasty.

AvBaeyer - 10-3-2017 at 19:55

The second reaction will definitely NOT happen as written. Think about the pKa's involved here. The phenolic OH will be deprotonated by the hydroxide which will deactivate the ring to any hope of an Snr reaction. Moreover, the iodide is meta to the aldehyde group and is entirely deactivated for an Snr reaction. Double barrelled chemistry working against you here.

AvB

CuReUS - 11-3-2017 at 11:32

Quote: Originally posted by theAngryLittleBunny

You could maybe protect the aldehyde group by forming an acetal with ethylen glycol.

but won't the ester also get hydrolysed during the deprotection step ?
Dr cat,as bunny said,the iodo compounds is no use,the diazonium has to be converted to OH instead.But after that IMO,the best bet would be to cut your losses and live with the shitty yield rather than trying any protection strategy.Acetylation of the phenol can't be used because during the deprotection step even the benzoyl ester might get hydrolysed.Benzyl ether can't be used as in the deprotection step,the CHO might get reduced due to H₂,Pd/C.The only protecting group I can think of would be triisopropyl silyl ether(TIPS) but that would be out of reach of the home chemist.So that's why just react the p-chlorobenzoic acid with the 5-hydroxyvanillin and hope for the best.You could try the one pot esterification with TCT -http://chemistry.mdma.ch/hiveboard/novel/000372508.html#Post...

Quote: Originally posted by JJay

You might be able to use sulfuric acid with 5-hydroxyvanillin to form esters with p-chlorobenzoic acid, but I think esterfication will be favored at the 4-position, so the yield of your desired product would likely be very low.

not to mention the huge amount of polymerisation side products formed.

[Edited on 11-3-2017 by CuReUS]

theAngryLittleBunny - 11-3-2017 at 12:05

Quote: Originally posted by CuReUS

Quote: Originally posted by theAngryLittleBunny

You could maybe protect the aldehyde group by forming an acetal with ethylen glycol.

but won't the ester also get hydrolysed during the deprotection step ?
Dr cat,as bunny said,the iodo compounds is no use,the diazonium has to be converted to OH instead.
[Edited on 11-3-2017 by CuReUS]

Uh, he called me bunny O.o, that's so cool *-*
Anyway, dammit, you have to take that thing off afterwards, I didn't think about that .-. well, then this wouldn't work either.
Well, then the only thing that you could do is to just make a fischer esterfication with OH instead of the I on the iodovanilin, which means your yield will most certainly be below 50%.
The life of an organic chemist is just really hard.

clearly_not_atara - 12-3-2017 at 15:11

Oof, that's a strange product. Benzoic acid is not generally nucleophic enough to react with an aryl halide even with a catalyst. However you may be able to form the ester if you use 2-nitrobenzyl chloride to protect the 4-hydroxy, hydrolyze the iodo to a phenol with CuO, and acylate with p-chlorobenzoyl chloride. Deprotection is then achieved by irradiation with loss of 2-nitrosobenzaldehyde.

This is quite difficult and I have to suggest a different precursor. I believe that you can protect vanillin as the dithiane, FC acylate with p-chlorobenzoyl chloride primarily at 5 due to sterics, perform a B-V oxidation to the ester, and deprotect the dithiane with mercuric oxide to obtain the desired product. This is still quite hard, but avoids the difficult hydrolysis and photolysis.

EDIT: you might be able to use tert-butyl chloroformate as the sole protecting reagent:
https://www.researchgate.net/profile/Gregory_Hamilton2/publi...

[Edited on 13-3-2017 by clearly_not_atara]

CuReUS - 13-3-2017 at 06:33

Quote: Originally posted by clearly_not_atara

hydrolyze the iodo to a phenol with CuO

won't a base be used in this step,which might cause a cannizaro ?

Quote:

perform a B-V oxidation to the ester

doing a BV on a naked phenol is asking for trouble,isn't it ?

Quote:

you might be able to use tert-butyl chloroformate as the sole protecting reagent

how ?

clearly_not_atara - 14-3-2017 at 13:23

Quote:

won't a base be used in this step,which might cause a cannizaro ?

Yes, but the Cannizzarro requires strong base. If you use the right ligand copper will react surprisingly fast with aryl iodide and you can minimize losses. It's not perfect but protection is also associated with losses.

Quote:

you might be able to use tert-butyl chloroformate as the sole protecting reagent

The paper mentions that t-butyl esters can be deprotected with ZnBr2 in the absence of water. Water-free deprotection means that the p-chlorobenzoyl ester won't be hydrolysed, which makes things a lot easier. This is really just an easier replacement for o-nitrobenzyl chloride.