Author Topic: Urea --> NH4Cl?  (Read 2359 times)

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Shane_Warne

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Urea --> NH4Cl?
« on: September 02, 2004, 06:21:00 AM »
How can urea bee quickly hydrolysed to basic ammonium compounds that can be reacted with HCl to form NH4Cl?

I was told that HCl could be used, but won't this just form small amounts of urea.hcl?

thanks
:)

Organikum

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Heating the dry urea expells NH3, collecting...
« Reply #1 on: September 02, 2004, 09:41:00 AM »
Heating the dry urea expells NH3, collecting this in HCl yields NH4Cl, cyanuric acid is left behind.
Of course, neutralisation of household ammonia with HCl seems to be a more comfortable way.


Shane_Warne

  • Guest
HCl and NH4OH=livable
« Reply #2 on: September 02, 2004, 12:44:00 PM »
It does. (make more sense)

this is what the web tells me happens during hydrolysis of amides under acidic conditions:


Hydrolysis under acidic conditions

Taking ethanamide as a typical amide:

If ethanamide is heated with a dilute acid (such as dilute hydrochloric acid), ethanoic acid is formed together with ammonium ions. So, if you were using hydrochloric acid, the final solution would contain ammonium chloride and ethanoic acid.




So urea would yield cyanuric acid and NH4Cl as a byproduct?

anyway that reactions useless and doesn't achieve the goal.

basic hydrolysis will form sodium cyanate.

ning said acid hydrolysis was the way to go, remember?

Maybe he didn't know ammonium carbonate/bicarb/carbamate was wanted.

Theres the hexamine MeAm synthesis, but urea would have been nicer for a NH4Cl source because it would'nt be contaminated with a controlled substance before it's ready for use...

I suppose HCl + household ammonia isn't so bad.  :)




Nicodem

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Hydrolysis of amides
« Reply #3 on: September 02, 2004, 01:11:00 PM »
I have no idea why you don't make NH4Cl from ammonia and HCl, but I guess you have your own reasons for that. I just wanted to correct a detail of what you wrote before someone takes it too seriously.

So urea would yield cyanuric acid and NH4Cl as a byproduct?
anyway that reactions useless and doesn't achieve the goal.
basic hydrolysis will form sodium cyanate.


Urea, when hydrolised in conc. hydrochloric acid yields only aqueous NH4Cl and CO2 gas:

H2N-CO-NH2 + H2O  + 2HCl   ==>    2NH4Cl(aq) + CO2(g)

Basic hydrolysis with NaOH(aq) yields aqueous Na2CO3-NaOH mixture and NH3 gas:

H2N-CO-NH2 + 2NaOH   ={H2O}=>    2NH3(g) + Na2CO3(aq)

No cyanuric acid or cyanate forms in any of the hydrolysis methods.


Shane_Warne

  • Guest
Thanks Nicodem! I couldn't find anything like...
« Reply #4 on: September 03, 2004, 06:26:00 AM »
Thanks Nicodem!

I couldn't find anything like that online, because the biology gangsters have been hogging the phrase 'urea hydrolysis' in part.

Nicodem, I assumed basic hydrolysis would yield cyanate due to the fact that that same page desicribes the following:


Hydrolysis under alkaline conditions
Again, taking ethanamide as a typical amide:
If ethanamide is heated with sodium hydroxide solution, ammonia gas is given off and you are left with a solution containing sodium ethanoate.




and also

Post 442105

(DRIVEN: "NaOCN from urea and sodium carbonate", Methods Discourse) which describes the reaction of solid urea and Na2CO3 forming sodium cyanate. <--which isn't a hydrolysis, but I put two and two together and got atleast umm... 6 or 7?  ;D

Well there's nothing I don't like about Hcl and ammonia water, but I was just looking at different ways.

Plus it sounds like the urea reaction you detailed would involve less water to evaporate, which is good!  :)

Do you know how long the reaction would take, temperature, and also whether a lower concentration of HCl can be used?




Nicodem

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Urea hydrolysis and urea pyrolysis
« Reply #5 on: September 03, 2004, 08:21:00 PM »
Sodium ethanoate is the same as sodium acetate by the IUPAC nomenclature. The reaction of urea and Na2CO3 is, as you have already figured out, something else. Urea tends to pyrolise in HO-CN and NH3 in a reversible fashion. If the NH3 gas escapes you get the HOCN+urea product (biuret) or further on the HOCN+biuret product (isocyanuric acid). Other conditions can lead to melamine (2,4,6-triamino-1,3,5-triazine). But if you have Na2CO3 present at the start you only get NaOCN since the reactive HOCN acid gets neutralized imidiately to its heat stable sodium salt. Hope this clears the confusion.

Do you know how long the reaction would take, temperature, and also whether a lower concentration of HCl can be used?

I think most amides get hydrolysed faster with NaOH than HCl, but urea may be an exception, I don't know. You should bee able to notice the end of hydrolysis by the end of CO2 bubbles formation from the boiling stones if you lower the reflux bellow the boiling point for a moment. Usually this is a clear sign in reaction involving CO2 formation (like the Leuckart or decarboxylisations). If I dare making a guess I would say that 6h of reflux would do, but don't forget you need an excess of conc. HCl since it gets neutralised by the forming ammonia (at least a 20% excess). If you will use the 20% HCl put a somewhat larger excess  and reflux for a longer time.


Shane_Warne

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Sodium ethanoate is the same as sodium acetate
« Reply #6 on: September 04, 2004, 02:17:00 PM »
Sodium ethanoate is the same as sodium acetate by the IUPAC nomenclature. The reaction of urea and Na2CO3 is, as you have already figured out, something else. Urea tends to pyrolise in HO-CN and NH3 in a reversible fashion. If the NH3 gas escapes you get the HOCN+urea product (biuret) or further on the HOCN+biuret product (isocyanuric acid). Other conditions can lead to melamine (2,4,6-triamino-1,3,5-triazine). But if you have Na2CO3 present at the start you only get NaOCN since the reactive HOCN acid gets neutralized imidiately to its heat stable sodium salt. Hope this clears the confusion.
It does, thank you!
I think most amides get hydrolysed faster with NaOH than HCl, but urea may be an exception, I don't know. You should bee able to notice the end of hydrolysis by the end of CO2 bubbles formation from the boiling stones if you lower the reflux bellow the boiling point for a moment. Usually this is a clear sign in reaction involving CO2 formation (like the Leuckart or decarboxylisations). If I dare making a guess I would say that 6h of reflux would do, but don't forget you need an excess of conc. HCl since it gets neutralised by the forming ammonia (at least a 20% excess). If you will use the 20% HCl put a somewhat larger excess  and reflux for a longer time.
A non-fuming ~20% concentration would be really nice. The escape of Hcl is the main concern, not only because Hcl is corrosive, but because it'll ruin the ratios.

Time isn't important, it can be left for days

First swim'll try refluxing a 150ml solution containing 60g urea with 323ml 31.45% Hcl (30% hcl weight excess) for ~9hrs, and see how he goes.  :)

Thanks a lot Nicodem. Ill describe how it goes,
bye!

Nicodem

  • Guest
Chemistry calculations
« Reply #7 on: September 04, 2004, 03:42:00 PM »
First swim'll try refluxing a 150ml solution containing 60g urea with 323ml 31.45% Hcl (30% hcl weight excess)

Whenever one usess the word "excess", "ratio" or "equivalent" it is always meant to be MOLAR unless specified diferently. Chemists use chemical units. So did I when I suggested a 20% excess of acid.

323ml of 31.4% HCl is ~3.15mol
60g urea is 1mol
stochyometry is 2HCl : 1urea
therefore your excess is about 57%


Shane_Warne

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you sure? 314g/L (8.6 mol per litre) ...
« Reply #8 on: September 05, 2004, 01:28:00 PM »
you sure?

314g/L (8.6 mol per litre)


314g/1000ml = .314g/ml

323ml * .314g/ml = 101.42g (HCl in 323ml@314g/1000ml)
101.42 / 36.46hclMw = 2.78 mol (39% molar excess)

2.78 * 100 / 2.0  = 139%

3 moles would be a 50% molar excess

Actually I think your calculating from 31.45% w/w and not w/v, but I was calculating from w/v, but 39% excess wasn't what I was aiming for.

Wouldnt you know it, an equimolar has already been mixed, and no excess. <--I must have subconciously known my calculations were dodgy.  ;D

thanks

Nicodem

  • Guest
Yes, the %...
« Reply #9 on: September 05, 2004, 07:07:00 PM »
Yes of course I asumed those were w/w percentage. You cannot measure w/v percentage simply because g/ml can't be shown unitless (g/ml can't be traduced into % without knowing the density - I assumed the density was ~1.12g/ml). You should question the information on the concentration of HCl you have since % can only be w/w or more rarely v/v.

Anyway this does not matter much. If there is to much excess you will have more work boiling the acid off. The important thing is that there is enough, so that you will not get NH4Cl contaminated with urea.