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Nicodem
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[*] posted on 23-11-2009 at 08:19


Quote: Originally posted by Sedit  
I have reference stating low concentrations of NaDMSO being make from NaOH and DMSO but I feel this is due to it forming equilibrium with the formed H2O. Remove the water and the reaction should be pushed to the right yeilding NaDMSO.

In DMSO the dimsyl ion is approximately 5000 times more basic than hydroxide. Therefore if you have a 1M solution of NaOH it will contain less than 50 ppm water (and the equivalent of dimsyl ions which would be less than 0.003 mol/L). Obviously you can not remove this water by using any type of azeotropic distillation. Perhaps by heating NaOH/DMSO over CaH2 or something like that. Or just buy NaH.




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[*] posted on 23-11-2009 at 08:42


I have a reference though that states above 150C the Dimsyl ion becomes a primary componate and it would seem that at those temperatures the H2O should distill from the mixture correct?

Quote:
Acid/base reactions of DMSO and hydroxide ion, which give ris
to dimsyl anions 26, may be prominent at 150°C. The
basicity of hydroxide increases dramatically with temperature.

http://smartech.gatech.edu/bitstream/1853/2796/1/tps-140.pdf

It would appear that even though the Dimsyl ion is 5000x more basic at room temperature then perhaps thats not so true at higher temperatures. As long as the reaction progressed enough to allow any form of efficient water removal then the equilibrium should Proceed to the right.

I have found a way using electrolysis of a soluble potassium salt to safely generate the dimsyl ion so there is no real need for the hydride anymore.





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[*] posted on 23-11-2009 at 11:08


Quote: Originally posted by Sedit  
I have a reference though that states above 150C the Dimsyl ion becomes a primary componate and it would seem that at those temperatures the H2O should distill from the mixture correct?


It says "prominent", not "primary". It only appears to me that you misunderstood the claims in that paper. Why not checking the original paper they cite? I do not have access to it, but you can always ask for it:
http://www.informaworld.com/smpp/content~db=all~content=a762...

Quote:
It would appear that even though the Dimsyl ion is 5000x more basic at room temperature then perhaps thats not so true at higher temperatures. As long as the reaction progressed enough to allow any form of efficient water removal then the equilibrium should Proceed to the right.

Even if at higher temperatures the hydroxide would be only 1000 times less basic than dimsyl, this would not make much of a difference. You would still not be able to remove the water with toluene/water azeotrope or any such similar thing. Read on the theory of azeotropic distillation to understand why not.


This remained unanswered:
Quote: Originally posted by Sedit  
I have heard of Sulphur being used in place of lithium in birch reductions and it got me woundering about the structure of ammonia/sulphur complex that forms. I know Lithium forms Li(NH3)x structures but how exactly would sulphur bond and do you think it would be possible to form this complex simular to how Lithium bronze complexes are formed in a non polar solvent with anhydrous Ammonia feed into it.

It seems interesting because I can not find any information on this in the slightest only a few veuge references to it being used in S/NH3 systems.

If you would read on the mechanism of the Birch reduction you would understand why sulfur is nonsense. Maybe you misread Sr for S. It would be easier to spot the origin of the misinformation if you would use references. Sulfur is not even a metal and therefore has no free electrons in its crystal stucture. No electrons no ammonia solvated electrons no reduction.




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[*] posted on 23-11-2009 at 11:18


My reference for the Sulfur in ammonia to produce solvated electrons was lost when my computer crashes or else I would not have even needed to ask the question here because the paper stated the structure that formed solvating the electrons. It was a paper documenting the use of solvated electrons in a battery that employed liquid ammonia. I will see what I can do about recovering that file from those lost on my hard drive for you.

I will check in on the original reference stated in the paper as well. I must have over looked that citation thank you. Azeotropic distillation though means the use of a co-solvent to aid in the water removal process. That would not be needed here since the difference in boiling points between DMSO and H2O is so great. You are still more then likely correct in that it would not be pushed all the way to the right but I am still intested in what sort of concentrations can be obtained and the mechanics behind it all.





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[*] posted on 23-11-2009 at 21:02


i just had my first piranha accident, it was very disconcerting. That stuff readily burns nice round holes in your skin.
I was having an excellent day, so well that i thought 'i know i'll attend to the bucket of glassware i have accumulating for piranha washing'
I recommend never using this solution at the end of a day.
Anyway i was making up as normal, 500ml of conc sulphuric, 300ml of 30%h202, except it was 50%, well i had a geyser out of the two litre beaker, i was out of the way quick enough except for one drop on my arm, however my beautiful cedar bench top is fucked and i have a shit of a clean up.
Question, i am surprised the reaction was this vigorous, normally i get nothing really when i add the two, is the higher concentration h202 the culprit?
Whats best for neutralising this solution?




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[*] posted on 24-11-2009 at 13:20


I have no idea what the proximate cause of your eruption would be; I could imagine both metal impurities in the H2SO4, or else the possibility that once the amount of H20 in the mix is small enough the H2SO4 just tears the peroxide apart.
But I also have to wonder what's on your glassware that you need a cleaning agent this aggressive. Do you actually have a bucket of stuff that wouldn't get clean with acetone, dilute HCl, 10% ammonia, or plain old detergent and water? Or do you just like to employ maximum force cleaning?
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[*] posted on 24-11-2009 at 16:23


Quote: Originally posted by bbartlog  
I have no idea what the proximate cause of your eruption would be; I could imagine both metal impurities in the H2SO4, or else the possibility that once the amount of H20 in the mix is small enough the H2SO4 just tears the peroxide apart.
But I also have to wonder what's on your glassware that you need a cleaning agent this aggressive. Do you actually have a bucket of stuff that wouldn't get clean with acetone, dilute HCl, 10% ammonia, or plain old detergent and water? Or do you just like to employ maximum force cleaning?


Of course metal impurities, i had just used the sulphuric to rinse through the coil of a condenser brown stained from rusty water, it had been ineffective however obviously had solvated enough metal ions to cause the h2o2 to decompose.

I rarely use pirahna, maybe twice annually, i accumulate bits and pieces that i have otherwise been unable to get clean any other way. Most commonly complex pieces with difficult to get to bits within them. I had probably a dozen small bits and pieces yesterday, and seeing as i'm talking about it i will bore everyone with what happened after the spill. So i spent probably around and hour and a half cleaning everything in the proximity of the ruined bench and floor, then a further 30 minutes cleaning everything i used to clean the area.
Then i made up some fresh pirahna, immersed the several pieces of glass within it and went to move the 2litre beaker across to another bench more out of the way AND THEN DROPPED THE BEAKER moving it across benches.
So the cleaning began again, plus the entire reason for making the solution in the first place was in thousands of pieces across the lab floor.
I laughed, i can't remember ever dropping a beaker, let alone within circumstances such as these.
We should start a 'Science Madness' sitcom with accumulated funny lab stories.




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[*] posted on 25-11-2009 at 19:53


hi,
i have got an easy one for you,

i want to make substantial quantities (well something like 50 grams) of white lead oxide form lead metal, what is the best, (safest) way to transform it?

thx
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[*] posted on 25-11-2009 at 21:46


Quote: Originally posted by Hamilton  
hi,
i have got an easy one for you,

i want to make substantial quantities (well something like 50 grams) of white lead oxide form lead metal, what is the best, (safest) way to transform it?

thx


If you're looking to make lead(ii) acetate, a good way would be displacement of aqueous copper(ii) acetate by lead metal.
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[*] posted on 28-11-2009 at 13:39


Quote:
Anyway i was making up as normal, 500ml of conc sulphuric, 300ml of 30%h202, except it was 50%, well i had a geyser out of the two litre beaker


Quote:
Then i made up some fresh pirahna, immersed the several pieces of glass within it and went to move the 2litre beaker across to another bench more out of the way AND THEN DROPPED THE BEAKER moving it across benches


You spilled pirhana twice in one day? You were making 800 mL? For cleaning glassware? What's the deadline on nominations for the 2009 Darwin Award?
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[*] posted on 28-11-2009 at 14:23


Hehe - butter fingers!

But I have had a (less serious) accident also and would please like some help...

I was going to add MeBr to benzylamine in methanol solution (with potassium carbonate) to make the quaternary salt, but unfortunately I was working on two experiments at once (due to limited time) and added the benzylamine to acetophenone :(

I only experiment on a small (boiling tube) scale and I have no way to separate these by distillation, and I have only a small amount of acetophenone anyway, so I would like to make use of it and not dispose. I have looked (but failed) to see if this reaction can be done in acetophenone; nearest I could find was acetone solvent but many examples do not use a base which seems silly(?!)

So my question is, can I bubble MeBr through benzylamine in solution of actetophenone and expect quaternary salt formation... and what base can I use in acetophenone?
Thank you for any help.
It is a shame - I tried to save time but now have wasted it; but there is a lesson to be learned I suppose!

[Edited on 28-11-2009 by sonogashira]

[Edited on 28-11-2009 by sonogashira]
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[*] posted on 29-11-2009 at 14:07


Well I tried on a small portion in the absence of any suggestions, and I think I have some product - but I will have to try and purify further.

The problem is that I did not use an additional base because I don't know what can work, so I have mono- and di-alkylated contaminants.

Hopefully someone can recommend a base to use because this test result was far from ideal :(

[Edited on 29-11-2009 by sonogashira]
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[*] posted on 5-12-2009 at 17:46
Question on chloroacetyl chloride reactivity with hydroxyls


Chloroacetic acid ClCH2COOH react with primary and secondary hydroxyls in aqueous solution.
I was wondering what would happen with the use of Chloroacetyl chloride, ClCH2-COCl - would the acyl chloride attack the hydroxyls preferably? (obviously then under nonaqueous conditions)?

I.e.what if I wanted to convert ethylene glycol to ClCH2-COO-(CH2)2-COO-CH2-Cl? Would this work without formation of mixed products?




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[*] posted on 6-12-2009 at 09:11


Quote: Originally posted by sonogashira  
I was going to add MeBr to benzylamine in methanol solution (with potassium carbonate) to make the quaternary salt, but unfortunately I was working on two experiments at once (due to limited time) and added the benzylamine to acetophenone :(

Recover your benzylamine by:
- add the acetophenone/benzylamine mixture to water
- add HCl untill acidic reaction
- wash with diethyl ether, dichloromethane or whatever suitable solvent you have (and recover acetophenone from the washes)
- basify with NaOH
- extract the precipitated benzyl amine oil with dichloromethane
- dry over Na2SO4
- evaporate.

Don't waste time and material with an acetophenone/benzylamine mixture as you will never be able to efficiently recover the trimethylbenzylammonium bromide.


Quote: Originally posted by chemoleo  
I.e.what if I wanted to convert ethylene glycol to ClCH2-COO-(CH2)2-COO-CH2-Cl? Would this work without formation of mixed products?

Yes, only the mono and diester of ethylene glycol can form by solvolysis of chloroacetyl chloride in ethylene glycol. Without pyridine the reaction might require a long time and/or heating, but otherwise no O-alkylation can occur, just O-acylations.




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[*] posted on 8-12-2009 at 00:59


I haven't been able to find a good preparative procedure for o-toluenesulfonic acid.
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[*] posted on 8-12-2009 at 04:19


Have you checked references for the preparation of saccharin? I know for a fact that o-toluenesulfonic acid is an intermediate in saccharin synthesis (and the reason the para isomer is widely available is that it's an unavoidable side product).

sparky (~_~)

[Edited on 8-12-2009 by sparkgap]




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[*] posted on 25-12-2009 at 10:09


I've read a lot of peoples opinions on bomex, pyrex, kimax, etc but does anyone have experience with this new-ish pyrex vista?
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[*] posted on 26-12-2009 at 06:42


Is this correct ?
5 mol/L H2SO4 = 81.194 %

Damn my math is so rusty that i am not sure.
I dont think its correct :S


*edit*
Damn i feel stupid.

It should be something like 26%
Still i cant confirm this.



[Edited on 26-12-2009 by User]




What a fine day for chemistry this is.
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[*] posted on 26-12-2009 at 07:50


5 moles/L of H2SO4 = 5 X 98 = 490 gm/L of H2SO4 solution.

So the concentration is 49 % (W/V)

In order to know the concentration in weight / weight (w/w) basis, we need to calculate the specific gravity of the solution.

Here we assume that there is no volume change when H2O and H2SO4 are mixed. And the mixing is done isothermally at 30 Deg C. :)

Perry-Chapter 2-108 gives density of 100 % H2SO4 @30 Deg C as 1.8205

So volume of 490 gm of Acid is 490/1.8205 = 269.16 cc. Therefore volume of balance water in solution is 1000 - 269.16 = 730.84 cc. Also weight of that water is 730.84 gm.

So total weight of 1 lit of solution is 490 + 730.84 = 1220.84 gm.

So concentration of Acid = 100 X (490 / 1220.84) = 40.14 % (w/w)

Hope this is correct :)

gsd

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[*] posted on 26-12-2009 at 14:00


In case anyone wants a very precise calculation, I assumed there could be some volume change upon mixing.

I used the table at the bottom of this page as a source.

I used excel to find the data points for the relationship between P (W/W%) and M (molarity). I assumed the concentration would be linear between the two closest points:
(M,P)=(4.879,37.26) and (5.096,38.58).
P=slope*M+b
slope=(P2-P1)/(M2-M1)=(38.58-37.26)/(5.096-4.879)=6.083
b=P-slope*M=37.26-6.083*4.879=7.581043
We want to know the concentration for M=5.000, so:
P=6.083*5.000+7.581043=38.00

So its 38%, not 40%.



[Edited on 27-12-2009 by per.y.ohlin]

Attachment: sa.xls (115kB)
This file has been downloaded 443 times

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[*] posted on 31-12-2009 at 09:53


Quote: Originally posted by Nicodem  
Quote: Originally posted by sonogashira  
I was going to add MeBr to benzylamine in methanol solution (with potassium carbonate) to make the quaternary salt, but unfortunately I was working on two experiments at once (due to limited time) and added the benzylamine to acetophenone :(

Recover your benzylamine by:
- add the acetophenone/benzylamine mixture to water
- add HCl untill acidic reaction
- wash with diethyl ether, dichloromethane or whatever suitable solvent you have (and recover acetophenone from the washes)
- basify with NaOH
- extract the precipitated benzyl amine oil with dichloromethane
- dry over Na2SO4
- evaporate.

Don't waste time and material with an acetophenone/benzylamine mixture as you will never be able to efficiently recover the trimethylbenzylammonium bromide.


Thank you! I have only just seen the reply! Can I ask you (or someone else!) about similar situation (theoretical this time - no more mistakes!):

If I had acetophenone contaminated with 10% benzaldehyde, could I remove the benzaldehyde by adding DCM and washing DCM-Acetophenone-Benzaldehyde with water? DCM is the only solvent I can get really :(

I can't find information if benzaldehyde forms an azeotrope but I suspect so (though it is a complete guess!). And I believe it is more soluble in water than acetophenone, but maybe DCM dissolves it well also? It is acetophenone I want to recover, just to be clear.

[Edited on 31-12-2009 by sonogashira]
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[*] posted on 31-12-2009 at 10:01


Probably best to oxidise the benzaldehyde to benzoic acid, and then extract that from DCM-acetophenone with aqueous sodium bicarbonate or carbonate, followed by a plain water wash.

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[*] posted on 31-12-2009 at 11:22


Thanks. It is the (expected) product ratio from the oxidation of styrene with H2O2 and palladium chloride, so I don't know if adding oxidizer will be very good in this case (probably should have said before!).

Maybe I could extract both into DCM, wash with water to remove catalyst and excess H2O2, then boil-off DCM... dilute with water and add an aqueous oxidant etc... but it seems rather long procedure.

Preferably I would like to separate the benzaldehyde from the DCM-Acetophenone-Benzaldehyde. I thought maybe bisulfite can work but I think acetophenone will react as well.

I gather you do not think it (benzaldehyde) can be removed just with water washing of the DCM solution? Or by heating above the bp of benzaldehyde?

[I also saw something about 'extractive distillation' which I couldn't really understand (it was a patent!), so maybe there is something I could add (to Acetophenone-Benzaldehyde; after removing DCM) to form a low-boiling azeotrope with benzaldehyde, and leave pure acetophenone behind?... If only!]
[Edited on 31-12-2009 by sonogashira]

[Edited on 31-12-2009 by sonogashira]
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[*] posted on 31-12-2009 at 22:04


How much is 1 mole of 35% formaldehyde?



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[*] posted on 31-12-2009 at 22:44



Formaldehyde - HCHO - M.W. = 30

35 % solution = 35 gm in 100 gm solution = 1.1667 moles in 100 gm solution

So (100/1.1667) = 85.714 gm of 35 % solution of formaldehyde will contain 1 mole.

gsd
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