Sciencemadness Discussion Board

Noob Organic Chemistry question involving Oxamide

Velzee - 19-7-2016 at 20:59

Okay, as you may know, I've acquired a good amount of urea, so I sat down and began to draw a bit, and came up with a few ideas, one of which involve synthesizing oxamide.

As you can see, I came up with the idea of reacting urea and sodium oxalate:


https://i.imgur.com/o57Mntz.jpg

But I'm afraid that the oxalate ion would be too electronegative and won't leave the sodium ion, thus leaving no reaction to occur. I'm just a newbie, and I don't have too much knowledge on electronegativity, so please forgive me, but how, if it is at all possible, could I make this reaction occur? Would just mixing the reactants be sufficient enough?

Thank you in advance!

[Edited on 7/20/2016 by Velzee]

Velzee - 21-7-2016 at 18:38

Okay, I attempted to just mix the reactants(3.00g of Na2C2O4, and 1.35g of (NH2)2CO) together, and no visible reaction occurred. If there was, there was too much water(300mL) present, and it's likely that the product, if any, would be dissolved in the solution, along with Na2CO3.

DraconicAcid - 21-7-2016 at 20:26

I don't know about the reaction of carboxylates with urea, but you don't have to worry about the electronegativity. It's ionic. The sodium is just kinda hanging around the anion, and isn't bonded.

Praxichys - 22-7-2016 at 06:35

Typically this preparation of amides involves refluxing the carboxylic acid with urea for extended periods of time. The sodium ion would definitely mess up the reaction, instead probably forming sodium cyanate with the urea and sodium carbonate as the oxalic acid decomposes.

That said, even if you prepared oxalic acid from the sodium oxalate, the urea-reflux method probably won't work well. Oxalic acid comes as the dihydrate which slows the reaction, and when anhydrous will tend to sublime. This will create lots of problems with clogging the condenser. Further, oxalic acid decomposes into CO2 around the temperatures typically used in this reaction, which will react with the ammonia from the decomposition of the urea and water to form ammonium carbonate and ammonium carbamate which also sublimes and will clog the condenser. (This happens normally with this synthesis in general, but will be much worse with oxalic acid if it can be made to work at all)

My advise would be to use a different method to form the amide. You would be better off running a Fischer esterification to make either the dimethyl or diethyl oxalate, then mixing the ester with a strong ammonia solution and letting it sit for a few days. The challenge with this is the removal of water during the Fischer, especially since the starting acid is more convieniently the dihydrate. This esterification is best done with continuous removal of water with a dean-stark apparatus, but you might get a decent yield with an excess of H2SO4.

Velzee - 22-7-2016 at 18:21

I am in need of methanol, but I began to prepare a solution of ammonia in preparation for the synthesis of oxamide. I used the NaOH + (NH2)2CO method, and it works really well. But, as an amateur chemist does from time-to-time I made a novice error—while the NaOH+(NH2)2CO solution boiled, the funnel trap wasn't attempting to suck in the water as I seen in previous demonstrations, so I decided to sniff the gases that were pouring out of the funnel, but what I didn't do was waft it—I lazily put the funnel near my nose and took a breath. What followed was the worst burning sensation I've ever experienced in my nose.



Listen to your teachers, kids! Waft, and possibly save your life, or at the least, your nose!

[Edited on 7/23/2016 by Velzee]

[Edited on 7/23/2016 by Velzee]

PHILOU Zrealone - 27-7-2016 at 06:05

NH3 is a war gas for a good reason!

Using urea and oxalic acid will probably lead to a little oxa(la)mide (ethandoic acid diamide) but also to parabanic acid (cyclo(-CO-CO-NH-CO-NH-); 2,4,5-imidazolidinetrione; oxalylurea).

Maybe work from oxalic acid and ammonium formate or ammonium acetate and a little urea upon heating water, CO2 and formic acid or acetic acid will boil off.

Or from ammonium oxalate...beware that too strong a heat and dicyanogen may be formed...it is a toxic war gas; explosive superfuel burning very hot with O2 (4525°C).
NH4O2C-CO2NH4 -heat-> H2N-CO-CO-NH2 + 2 H2O -heat-> N#C-C#N + 4 H2O


Loptr - 27-7-2016 at 06:47

The cyanogen won't form below ~300*C though, right?

Praxichys - 27-7-2016 at 07:08

Yes, Ph.Z. brings up a good point. Many ammonium carboxylates readily lose water to form their acid amides, and many of these acid amides can be pyrolyzed into some nasty byproducts.

Although cyanogen is a product of the thermal decomposition of oxalamide, this appears to be a dehydration reaction. Cyanogen readily undergoes hydrolysis back to oxalamide, so it would probably reconvert on cooling unless the water was removed chemically. It would be a mess to try heating ammonium oxalate though. The stuff sublimes almost as badly as ammonium bicarbonate. I don't know that it would be practical to get much of it to its decomposition temperature before it sublimed away.

@Velzee - Any progress?

Velzee - 27-7-2016 at 11:47

Quote: Originally posted by Praxichys  


@Velzee - Any progress?


Unfortunately, not really. I'm still really low on money, so I am unable to purchase any methanol. At my local Walmart, 355mL of 99% MeOH is sold as "HEET" for around $2. It's the best price I've seen locally. I could also buy about a liter of denatured ethanol for about $5. But, unfortunately, I don't have a reflux column to use with the ethanol method. So until all of our bills are paid and if I have some leftover cash available, I'm stalled.

I do, however, have a relatively strong solution of about 100mL of NH4OH in the fridge at the moment.

I did also notice something from my previous experiment to determine if the NaOH +(NH2)2CO method would be effective. I left the remaining solution(which still contained a good amount of ammonia and reactants and their side products) in a plastic bottle, and after a few days, the bottle leaked and effectively destroyed part of the wooden dresser it was on.

Also, I have a good amount(~30-50 grams) of what I believe to be crude NaOCN, contaminated with Na2CO3 that precipitated out from the main reaction. I have no use for it, so it's just sitting on a small table in my room. If any of you know a qualitative test for the -OCN ion, please let me know, because all I can find is tests for the -CN ions. Also, if there are any interesting reactions involving NaOCN, besides synthesizing NaCN, I'd also like to know.

I'll let you guys know if I make any further progress on this project.

I've still got ~2 kilograms of urea to use, by the way.

CuReUS - 29-7-2016 at 07:06

instead of going via the urea route,why not try to make oxamide using TCT ?
http://chemistry.mdma.ch/hiveboard/novel/000372508.html
read the 6th post by aurelius and the 8th post by rhodium

Velzee - 30-7-2016 at 21:56

Update: So I checked back at Walmart, and they've got the HEET at a $1.56 a pop, so I went to grab three, only to find out that you must be at least eighteen years old in order to purchase them. I sure looked like a fool having to leave empty handed.

@CuReUS, I'll check that out later—its time for me to hit the hay.

Velzee - 3-8-2016 at 09:13

In the process of writing up my progress, but I have a question; is there any reaction between Oxamide and NaOH?

Velzee - 3-8-2016 at 13:48

I completed my first organic chemistry project/experiment! I'll continue where I left off(after making NH3 from urea):

So about a week after I prepared a solution of NH3(I estimate that it was of 10-20% NH3), I obtained ~1 liter of MeOH, which I used to continue on with the synthesis of oxamide:

1. I followed chemplayer's guide to make dimethyl oxalate. I found that I didn't have to leave the solution in the fridge for the 12 hours that they suggested; it only took 3 hours to completely crystallize the ester. Here are some photos I took of the first crystallization:

https://i.imgur.com/gHSl31R.jpg
https://i.imgur.com/OFqZGoB.jpg

I believe the brown color resulted from the H2SO4 I used(it's drain cleaner). The brown color almost completely went away after the second crystallization. Unfortunately, I forgot to take a photo of the second.

2. Drying the dimethyl oxalate was difficult. In the end, I couldn't get it fully dry, probably due to some excess H2SO4. I used a ghetto vacuum filtration apparatus consisting of a Snapple bottle, a plastic funnel, the rubber tubing from my distillation set, some duct tape, and my vacuum cleaner in order to attempt to dry the ester.

I was left with wet crystals that looked like ice: https://i.imgur.com/BKLPv6P.jpg

3. After all of this, all of my research, it came down to this step: would oxamide actually form by simply adding the ester to the ammonia solution? Did I have enough ammonia? I weighed out ~15 grams of the dimethyl oxalate, and stirred it into the ammonia solution. To my surprise, the crystals went from a clear ice-like appearance to a milk-like appearance (even more so than an Mg(OH)2 emulsion): https://i.imgur.com/NJAqw0c.jpg

4. Drying the product was even more tedious than that of the ester. I boiled down the solution, but I did not foresee the slurry of oxamide popping out of the beaker from time-to-time, some eventually landing on the hot plate made me paranoid of the possibility of the evolution of cyanogen gas. I did smell a faint odd odor, but I haven't been harmed, though(I preformed the boiling step outside for the most part). I wound up burning some of the oxamide, but this is the final product: https://i.imgur.com/gH0hwi3.jpg

I've yet to weigh it, but, I must say that this project was fun! I look forward to doing some more organic chemistry projects. The next on my list is the production of malonic acid from 1,3—propanediol.

PHILOU Zrealone - 4-8-2016 at 06:59

Quote: Originally posted by Velzee  
In the process of writing up my progress, but I have a question; is there any reaction between Oxamide and NaOH?

If water is present --> hydrolysis...
H2N-CO-CO-NH2 + 2 NaOH --H2O--> NaO2C-CO2Na +2 NH3(g)

without water...I don't know
maybe some salt forming...
H2N-CO-CO-NH2 + NaOH --> NaHN-CO-CO-NH2 + H2O --> NaHN-CO-CO2NH4
NaHN-CO-CO2NH4 <--==> H2N-CO-CO2Na + NH3(g)

[Edited on 4-8-2016 by PHILOU Zrealone]

PHILOU Zrealone - 4-8-2016 at 07:11

You could have dissolved as much as possible dimethyloxalate ester into methanol and bubble dry NH3 gas into it...

Or alternatively add concentrated NH3 water solution (or methanol solution) into the ester solution.
The methanol doesn't arm because part of the reaction...
CH3-O-CO-CO-O-CH3 + 2 NH3 --> H2N-CO-CO-NH2 + 2 CH3-OH

Moderate heating distillation allows to get rid/recover the CH3OH (with traces of H2O and NH3).