Sciencemadness Discussion Board

100% aqueous NaOH?? - yes it was brought up before

zephler - 10-12-2010 at 16:50

Yes, I used the search engine, unfortunately though, they question was never really answered, instead it was discovered that the 100% NaOH quote was actually a typo! My question is this --> What is 100% aqueous sodium hydroxide solution? Is it simply 1g of sodium hydroxide to 1mL of water? OH GOD IT WAS THAT EASY? There is a vague patent floating around about the N-alkylation of piperidinone (forget about the ketone functionality for now). The authors report that they use this 100% aqueous NaOH solution at 120C with piperidinone (again, forget about the ketone part of the molecule - it is protected) and an alkyl halide to alkylate the secondary amine (in the piperidinone ring). Now the ONLY reason I can assume they used this method, because their paper is in a "green chemistry" context. The method I've always heard of was to use acetonitrile, K2CO3 and a PTC


I think I just answered it above 1gram in 1 mL = 100% - oh man!an!

[Edited on 11-12-2010 by zephler]

[Edited on 11-12-2010 by madscientist]

DJF90 - 10-12-2010 at 17:12

Unfortunately not, that would be a 50%w/v (=w/w) solution.

TheOrbit - 11-12-2010 at 07:13

can u post the reference, there must be any clue

zephler - 11-12-2010 at 12:09

Quote: Originally posted by DJF90  
Unfortunately not, that would be a 50%w/v (=w/w) solution.


I do not think that w/w and w/v are not the same in this case, if they are, can you suggest what 100% aqueous NaOH solution is? I think the answer lies in the denisty of NaOH being 2.13g/mL, so 1 gram of NaOH diluted up to 1 mL with water seems to fit since in (w/v)x100 = % , w = weight of solute (being 1 gram), and v = total volume (being 1 mL) would make a 100% w/v NaOH solution - right?

bbartlog - 11-12-2010 at 12:11

Quote:
instead it was discovered that the 100% NaOH quote was actually a typo!


Well this would be your answer, no? 100% would mean there is no water and it's all NaOH. So it's not an aqueous solution, though I suppose if you melt the NaOH there will be some small amount of H2O present as part of the equilibrium.

spirocycle - 11-12-2010 at 12:15

there are so many measures of concentration, that although 100% mol fraction, and w/w are impossible, it seems that 100% w/v is

Nicodem - 11-12-2010 at 13:22

Quote: Originally posted by zephler  
I think the answer lies in the denisty of NaOH being 2.13g/mL, so 1 gram of NaOH diluted up to 1 mL with water seems to fit since in (w/v)x100 = % , w = weight of solute (being 1 gram), and v = total volume (being 1 mL) would make a 100% w/v NaOH solution - right?

No offence meant, but that is nonsense. 100% NaOH, is just what it says, pure NaOH. It is not a solution in anything.
Quote: Originally posted by spirocycle  
there are so many measures of concentration, that although 100% mol fraction, and w/w are impossible, it seems that 100% w/v is

Possibly, but only if w/v fraction would be possible, but it is not. Dividing weight per volume does not cancel the units, so you don't get a fraction. You are still left with g/L or whatever.

PS: Zephler, opening a thread and asking about something found in an article without providing the reference is considered highly impolite (besides of being naive to expect others will be able to reconstruct the context out of nothing). You were asked to do so and you still did not!

[Edited on 11/12/2010 by Nicodem]

spirocycle - 16-12-2010 at 20:39

Quote:
Possibly, but only if w/v fraction would be possible, but it is not. Dividing weight per volume does not cancel the units, so you don't get a fraction. You are still left with g/L or whatever.



"w/v" is defined as g of solute per 100mL of solution

zephler - 16-12-2010 at 20:54

Quote: Originally posted by zephler  
my apologies for not posting the article - it is several pages long, here is the excerpt from the article 100% NaOH aqueous solution -- so what is your answer??? It is clearly not 100% NaOH - w/v% is the only way I can get close to a good answer - this is very frustrating! I have provided the excerpt below that comes from that patent paper - if this is not sufficient, I will post the entire paper, but this is the meat of the issue. The first section is from the Statement of the Invention section. The second section is from the Examplesection. Very odd connotation - why do they do this? For added protection for the patent - I would assume so - because this is ridiculous!





[Edited on 17-12-2010 by zephler]

[Edited on 17-12-2010 by zephler]

TheOrbit - 16-12-2010 at 23:06

really i see this patent as most stupid , bad written patent i have ever read .

it is an Indian patent , really i never depend on indian papers in my work , we can contact the authors and ask them "pharm_tox@yahoo.co.in"

[Edited on 17-12-2010 by TheOrbit]

smuv - 17-12-2010 at 00:56

Just looking at the reaction, I would say use 1.5eq (0.15 mol) of NaOH in 50 mL of water. I expect even without the NaOH the reaction would go to some extent (just some base will get tied up as the pipiridinium bromide salt).

I guess sterics and lower nucleophilicity keep the 4-anilino group unmolested...

Finally !!! How topethe 100% NaOH solution question answered!!! straight from the patent writer!!!!

zephler - 7-1-2011 at 02:22

it seems that my idea (will get to in a min), that was deemed "nonsense" by a super moderator, turned out to be correct one!

What's worse than being deemed "nonsense", is that no alternative or competing theory was provied by the critic (moderator).

Here is the comments from the moderator (and of course mine):

****************************
Quote: Originally posted by zephler
I think the answer lies in the denisty of NaOH being 2.13g/mL, so 1 gram of NaOH diluted up to 1 mL with water seems to fit since in (w/v)x100 = % , w = weight of solute (being 1 gram), and v = total volume (being 1 mL) would make a 100% w/v NaOH solution - right?
********************************
This is the reply from the "Super Moderator"

No offence meant, but that is nonsense. 100% NaOH, is just what it says, pure NaOH. It is not a solution in anything.

(of course it is qualified with a "No offence"

And here is the answer straight from one of the patent writers
==============================

You rightly pointed out that 100 % aqueous sodium hydroxide meant 1 gram of NaOH topped up to 1 mL with water, i.e., w/v %.

-

Dr Pradeep K Gupta
Scientist
Synthetic Chemistry Division
DRDE, Gwalior (MP)-474002
India

turd - 7-1-2011 at 02:54

Which is of course - to put it mildly - nonsense. 100% is unitless. w/v cannot be unitless unless you work in a non-SI system where weight and/or volume are defined relatively.

Nicodem - 7-1-2011 at 09:38

Quote: Originally posted by zephler  
And here is the answer straight from one of the patent writers
==============================

You rightly pointed out that 100 % aqueous sodium hydroxide meant 1 gram of NaOH topped up to 1 mL with water, i.e., w/v %.

I advise you to come to your senses and stop believing in nonsense. Chemistry is a dangerous hobby. If you are the type of person that relies to fairytales rather than reason and rationality, then you will soon end up in dangerous situations. Always remember that if something is nonsense in a strictly mathematical sense, then you can be sure that is complete and total bullshit in real life situation as well. And if you ever bother to reread the nonsense you and your indian friend claim above, and if you ever manage to evaluate it mathematically, then you might even realize how stupid it appears and why so.

zephler - 7-1-2011 at 15:50

I know where your problem is, you are getting hung up on the units -- no one ever said it was unit less! They write these papers assuming you have some knowledge of chemistry - why would they go to all of the trouble of saying 100% aqueous NaOH if they meant to use just pure NaOH pellets - furthermore how would you ever get 100% pure NaOH? Wouldn't they have described the size of the pellets they used, and how they dealt with the deliquescent problems of NaOH if they used pure 100% NaOH? They clearly state 100% aqueous NaOH more than 5 times in the paper - the only thing one can assume for it to make sense is the w/v way - does this make sense to you now?

zephler - 7-1-2011 at 16:10

whats odd is that you are claiming that you know what the author meant and used, even though the author clearly states that they meant and used Textw/v % . DO you think that they wrote 100% aqueous NaOH more than 5 times instead of just writing "pure NaOH" for some reason - what reason is that? Can you tell me what the resulting concentration is (in w/v%) of 1 gram of NaOH topped up to 1 mL with water? If it's not 100% w/v aqueous NaOH, can you please tell the board how to make a 100w/v% aqueous NaOH

TheOrbit - 7-1-2011 at 16:27

@Nicodem , i think we need no further discussion on this topic, the patent author claimed that so OK when you work with this procedure you must follow his opinion but if you will apply another one , then you are free to apply what you feel is right.

Nicodem - 8-1-2011 at 02:02

Quote: Originally posted by zephler  
I know where your problem is, you are getting hung up on the units -- no one ever said it was unit less!

You did! In you first post in this thread. You asked about what 100% concentration means. Surely I don't have to repeat that % means units cancel? It is a fraction, so it can only be unitless.
Quote:
They write these papers assuming you have some knowledge of chemistry

On the contrary. The peer review system exists so that authors are responsible for the level of knowledge presented.
Besides, what is the paper the inventors from the patent application published their result? Did their mistake about concentration remained unnoticed by the reviewers? Provide the reference for the paper! We had to pull out of you the reference for the patent and you never gave the reference for the paper. We still wait.
Quote:
- why would they go to all of the trouble of saying 100% aqueous NaOH if they meant to use just pure NaOH pellets - furthermore how would you ever get 100% pure NaOH?

They never meant that, as it is also obvious from the reply you got. Since the inventors themselves don't know how to calculate concentrations, you can never know what they meant with that 100% value and even less so about how they come to that nonsense (probably via the same nonsensible thinking as yourself, apparently).
Quote:
They clearly state 100% aqueous NaOH more than 5 times in the paper - the only thing one can assume for it to make sense is the w/v way - does this make sense to you now?

This only demonstrates the inventors don't know how to calculate concentrations rather then them making a typo or copypasta errors. It does not demonstrate anything else. Certainly it does not demonstrate any of that "w/v" nonsense as there is absolutely no mathematically sound way to come to the value of 100% for diluting 1 g NaOH to 1 mL solution. Such a mistake in patent examples also does not bear any legal consequences because it would not stand up as evidence for the patent invalidation process in a court hearing (assuming the patent was already, or is going to be, granted). It would only be useful in court if the error persisted in the claims section of the patent (which I did not check because I couldn't care less about the legal consequences for some indian company).
Quote:
Can you tell me what the resulting concentration is (in w/v%) of 1 gram of NaOH topped up to 1 mL with water?

Can't you use mathematics? If you would at least try you would soon realize that it is not possible to calculate the mass fraction of the resulting solution (assuming all NaOH even dissolves!) without further data. You need the mass of the solution to calculate the mass fraction! You only have the resulting volume telling you the solution would be 1000 g/L or 25 mol/L, but you can't calculate its percentage without first weighting the solution or at least knowing the amount of water used for the dissolution. If you can not weight the solution, you could find tabular data showing the dependence of density vs. g/L or mol/L and use this data to calculate the percentage.
How about actually trying to make that solution and measuring the concentration in %? I would really like to see how you manage to make a 1000 g/L (25 M) solution of NaOH at room temperature when the concentrated NaOH solution is supposed to be about 50% which is 764 g/L or 19.1 M at the density of 1.529 g/mL.

spirocycle - 9-1-2011 at 11:05

http://en.wikipedia.org/wiki/Concentration#Mass-volume_perce...

I have seen this measurement of concentration used many times in various pharmaceutical products.
But of course, it must be nonsense there too, right?

Nicodem - 10-1-2011 at 10:51

Quote: Originally posted by spirocycle  
http://en.wikipedia.org/wiki/Concentration#Mass-volume_perce...

I have seen this measurement of concentration used many times in various pharmaceutical products.
But of course, it must be nonsense there too, right?

Indeed, it is utter nonsense par excellence. That's why it is explicitly not allowed by IUPAC to use non SI compliant and false % concentrations. The nonsense of w/v % is even explicitly mentioned as not being allowed: http://www.standardbase.hu/tech/IUPAC.pdf

Just imagine how stupid such a system would be if people would actually use it. For example, you have 70% H2SO4(aq). By using such nonsense system, this would be 113% instead. Not only we would have pretend these are percentages (which obviously are not since g/100 mL do not cancel units), but we would actually have concentrations of over 100%. How more absurd can it get, than having a 113% concentration? That's why I said earlier, that if something makes no sense mathematically, it is safely considered as complete bullshit in real life.

TheOrbit - 10-1-2011 at 14:29
Quote:

Just imagine how stupid such a system would be if people would actually use it. For example, you have 70% H2SO4(aq). By using such nonsense system, this would be 113% instead. Not only we would have pretend these are percentages (which obviously are not since g/100 mL do not cancel units), but we would actually have concentrations of over 100%. How more absurd can it get, than having a 113% concentration? That's why I said earlier, that if something makes no sense mathematically, it is safely considered as complete bullshit in real life.


but as he said it is very common to use this expression in pharmaceutical industry for calculating % recovery of active substance .

chemrox - 22-4-2011 at 11:30

I distrust percent concentrations because, as has been mentioned, you never know what is meant when all you're given is "..percent aqueous solution." I looked up that Indian patent a couple of years ago trying to find a way to work with 4-Piperidone HCl. I don't know why they patented the process.. I think it's all dry lab. Maybe they thought if they covered the ground that comes up in search engines under "French speaking chemist.." they could could cash in if someone made it work on a commercial scale. If anyone wants to try the patent out I'd suggest the strongest possible base solution and see if the N-alkylation of 4-piperidone will work that way. I'm curious but not motivated enough to devote time and materials to it.

jon - 22-4-2011 at 14:03

o-protect that bitch with ethylene glycol and be done with it.


Attachment: 31479218-Improved-Prepn-of-1-2-Phenethyl-4-Piperidone-NPP-OPPI-40-3-307-2008.pdf (245kB)
This file has been downloaded 1164 times

Attachment: n-alkyl-4-piperidone.synthesis.pdf (60kB)
This file has been downloaded 671 times

also, this is a goldmine for papers of this sort.

http://www.scribd.com/document_collections/2478670/widget


that is a suckass patent no yeilds stated no analytical data.
it's a wise idea to deal with that carbonyl group first either by protection or otherwise.
i think attempts to directly n-alkylate piperidone are too prone to failure.
[Edited on 22-4-2011 by jon]

[Edited on 23-4-2011 by jon]