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Topic: Replacing NaOH by Ca(OH)2  (Read 10888 times)

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Offline curiouscat

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Replacing NaOH by Ca(OH)2
« on: January 07, 2013, 07:45:39 AM »
The following reaction works well with NaOH / KOH. I think it ought to work with Ca(OH)2 as well, but wanted to get an opinion. Comments?



Are there other typical reagents for this Halohydrin  :rarrow:  Epoxide step ?

PS. The motivation is Ca(OH)2 is a lot cheaper than NaOH and also avoids the high-solubility NaCl in the effluent.

Offline discodermolide

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Re: Replacing NaOH by Ca(OH)2
« Reply #1 on: January 07, 2013, 07:49:28 AM »
Ammonia may just do it. I don't have the pKb's to hand.
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Offline curiouscat

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Re: Replacing NaOH by Ca(OH)2
« Reply #2 on: January 07, 2013, 08:05:28 AM »
Ammonia may just do it. I don't have the pKb's to hand.

Thanks! That's something I've never considered. Interesting. I've to see the cost of NH3 vs. Ca(OH)2 though.

Cost is a very compelling argument to move from NaOH to Ca(OH)2; it seems about 70% cheaper :

e.g. http://www.lime.org/documents/publications/free_downloads/acid-neut-final-2000.pdf (Conflict of interest caveat!)

Another drawback of NH3 would be the high solublity of NH4Cl again (unless we plan to recover and sell it.  :) ).

Offline Hunter2

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Re: Replacing NaOH by Ca(OH)2
« Reply #3 on: January 07, 2013, 08:29:38 AM »
Ammonia may just do it. I don't have the pKb's to hand.

But with ammonia the risk is to get substitution and get Amine compounds or even Imine.

Offline Borek

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Re: Replacing NaOH by Ca(OH)2
« Reply #4 on: January 07, 2013, 08:50:02 AM »
Ca(OH)2 is a strong base, but its low solubility limits how high you can go with pH. If ammonia is a base strong enough, it shouldn't matter.

Low solubility of Ca(OH)2 can have a negative impact on yield, making the idea not economically viable.

Note that CaCl2 is highly soluble. Perhaps not as soluble (in terms of molarity of the saturated solution) as NaCl, still, not something that you would classify as low solubility.
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Offline discodermolide

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Re: Replacing NaOH by Ca(OH)2
« Reply #5 on: January 07, 2013, 09:23:05 AM »
Ammonia may just do it. I don't have the pKb's to hand.

But with ammonia the risk is to get substitution and get Amine compounds or even Imine.


Granted that is a possibility, but I think it may require some temperature and pressure?
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Offline curiouscat

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Re: Replacing NaOH by Ca(OH)2
« Reply #6 on: January 07, 2013, 09:49:41 AM »
Ca(OH)2 is a strong base, but its low solubility limits how high you can go with pH. If ammonia is a base strong enough, it shouldn't matter.

At 20°C Wikipedia reports a solublity of 0.173 gm/100 ml.

1.73 gm/L =23.37×10-3 gmol/L

[OH-]=46.76×10-3 gmol/L

That yields a pH of ~12.7

Am I doing this right? What sort of pH would one be looking for? With NaOH / KOH we're essentially easily at pH=14, right?

Quote
Low solubility of Ca(OH)2 can have a negative impact on yield, making the idea not economically viable.

True. Good point. I'll have to test this. Though, having slurried Ca(OH)2 would ensure more came into solution as soon as the reaction used it up? Mass transfer would be the critical step.


Offline orgopete

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Re: Replacing NaOH by Ca(OH)2
« Reply #7 on: January 07, 2013, 10:08:28 AM »
I don't think you can use ammonia. It can compete directly in the initial substitution reaction and if it formed the epoxide, it can also open it.

I don't doubt that Ca(OH)2 would work. Would it work well enough to be an economical solution overall, I don't know? I can imagine other problems also being introduced due to the solubility, but since the question is about economics, I could not predict that either.
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Offline discodermolide

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Re: Replacing NaOH by Ca(OH)2
« Reply #8 on: January 07, 2013, 10:12:56 AM »
You can say the same for NaOH/KOH. In aqueous solution they can also substitute or epoxide ring open.
Perhaps the epoxide is removed from the reaction medium that we are not aware of. After all the bottom line has been blacked out!

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Offline curiouscat

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Re: Replacing NaOH by Ca(OH)2
« Reply #9 on: January 07, 2013, 11:32:42 PM »
After all the bottom line has been blacked out!

Nothing mysterious in the bottom line: That said "KOH". We've used mainly NaOH so I'd redacted it out to simplify matters.


Offline discodermolide

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Re: Replacing NaOH by Ca(OH)2
« Reply #10 on: January 08, 2013, 12:10:08 AM »
There I was all night trying to get at the trade secret :P
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Offline curiouscat

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Re: Replacing NaOH by Ca(OH)2
« Reply #11 on: January 08, 2013, 12:14:56 AM »
There I was all night trying to get at the trade secret :P

Ok, I'll show you:


Offline discodermolide

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Re: Replacing NaOH by Ca(OH)2
« Reply #12 on: January 08, 2013, 12:42:42 AM »
I heard about that, it's that wonderful new chiral catalyst that works at 0.001mol% and gives 100% ee. Originated in China, I believe, but based on a EP whose number I've forgotten.
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Offline curiouscat

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Re: Replacing NaOH by Ca(OH)2
« Reply #13 on: January 08, 2013, 08:33:10 AM »
Ok, ran the reaction and it does work! That's the good news.

The bad news is that in addition to the top oily organic layer (mostly epoxide) and bottom aqueous lawyer (mostly CaCl2) I get a middle layer which seems a white turbid emulsion probably containing Ca(OH)2 and other solids (perhaps impurities from the lime CaCO3 / SiO2).

It's proving to be a challenge to break this emulsion. Doesn't seem to settle by itself. Nor on heating it up. An initial attempt to filter it was a disaster.

My material balances are not tying up (i.e. low yields) and I suspect a lot of my epoxide is trapped in this emulsion.

Ideas?

Offline orgopete

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Re: Replacing NaOH by Ca(OH)2
« Reply #14 on: January 08, 2013, 08:45:47 AM »
Easy, replace Ca(OH)2 with NaOH. Your result is about what I was expecting. I think the insolubility of the calcium salts is basically the source of your problems. Add HCl to make CaCl2 and separate.
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